Just 3 questions/puzzles that seem obvious but aren't
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- Опубликовано: 13 окт 2024
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Did you forget this was your main channel... 😅
Feels like video for your second channel!
you dickhead after "you said if you wanna learn more, look up the boy/girl paradox" I asked my ai chatbot and went back and forth for like 15 mins before I realized the question is ambiguous and doesn't represent probability 😭😭😭
"One of these is tails" does not give you a sample space of 3 possibilities it gives you 2 possibilities. One is tails and they are the same, and One is tails and they are different. The chance of both of them being tails becomes 1/2 when you know 1 is tails.
In the card example, holding an Ace leaves you with a 1/3 chance of holding another ace, because there are 3 cards it could be.
Edit: after thinking about this for a while, my conclusion is that knowing one is tails, but not knowing which, gives you 2 ways they could be different, and 1 way they could be the same. picking one of the 2 options, will result in tails 2/3 of the time. knowing which one is tails, narrows that down to 1 way they could be different and 1 way they could be a pair, so knowing which is tails makes guessing the other 50/50, but knowing just that a tails exists, makes it 2/3 you will choose a tails when selecting one of the options.
Knowing one is an Ace, gives you a 6/10 chance of selecting an Ace when choosing a card.
knowing one is an Ace, gives you a 1/3 chance the other is an ace.
selecting the other reliably requires knowing which is the ace.
You thought about it for hours? Me and my uncle thought about while you were explaining the one with the cards and it’s really easy.
There is one flaw in the card/coin puzzle that I don’t believe I’ve seen anyone point out so far. In both scenarios, you have to count the “doubled” possibility (2 aces or 2 tails) twice. In the coin puzzle, when showing all possibilities, if you mark which tails is the revealed tails it becomes quite obvious that it is still 50/50.
With the scenario in which you say that you are holding A coin as tails, I will mark the guaranteed tails with a capitol T, and other heads/tails with lowercase letters.
The options are:
T h
T t
t T
h T
This is because we do not know which hand the guaranteed tails is in, so we need to count both possibilities.
The same can be said for the card puzzle. We need to account for the ace of spades being in your left hand, and in your right hand (with the ace of hearts in the opposite hand). After doing this, the suit of the ace is irrelevant, and it is back to intuition.
My head hurts less when I watch your other channel.
😂
Have you seen his newest video
@@Randomchannelname2 I got a migraine from his latest comedy skit.
That’s just what happens when you put ratios, averages, and probability in the same video
I forgot you also have an educational channel
Jesus loves you with all His heart ❤️
Haha same
Man that needs to leave some talent for the rest of us. One of the funniest content creators I know and also self-taught himself an upper level college math class and has a degree in electrical engineering
@@L17_8 I support the message and all but how many people do you think have seen this type of comment and decided to accept Jesus into their life? I'm thinking none at all because as a Christian I find these comments annoying as hell and I suspect a non Christian would be even more irked by it.
same
blud stopped making shitposts so he can tell us we're dumb and I'm here for it
He just dropped a shitpost on the other channel, I'll be back
This is his main channel
@@squishypeanut42 Surrrrrre, and the moon isn’t made of swiss cheese. Nice try buddy, we all know the other one is the main channel
@@josiahdsmith5641but it is his main channel... he made this one first but made another channel for the other type of content he liked making.
(also notice the sub difference)
So he has both a brain rot channel and a brain education channel. 😂 love it.
EDIT: I realize that some people who are reading this comment have not seen Zack’s other channel. It is not brain rot comedy, but it is stuff that I would listen to with headphones on. Still though, the guy is an absolute genius and is comedic gold.
Its not brainrot its skits
Perfectly balanced.
Well he's always had the main channel since the days it was called 'Major Prep', the second channel is just the excess sludge that comes out of an engineer's head!
real
I love I'm more excited for this series every time his sick meme shit drops (which is actually intelligent and not just shit posting)
That was an illegal cliffhanger. I need to know more.
He has actually covered this problem in a previous video no need to wait. It’s a video from 5 years ago “the most counterintuitive probability paradox” it’s the same concept in a different situation
Edit: watch the resolved version bc the original video he was actually wrong as the problem stumped him too in the second video he explained the reality
@@smackthebomb I mean, he actually has a very different issue with the card problem VS the coin problem - since he mentions making a follow-up video, I have to assume he did so intentionally.
The coin problem is mathematically fine, it's just a conundrum of when the statistics of two coins are dependent to one another, VS independent. (IE: when you flip two coins, there's a 50% chance to get a heads and a tails, so if one of the coins is heads, the other coin is more likely to be tails. This is fairly intuitive, as you would need to win those 50% chance flips multiple times in a row to get duplicates - so independently they'll always be 50;50, but together, cumulatively, they'll always follow the statistical trend.)
The card problem is just misrepresenting the dataset, when he says "We have an Ace, what are the chances we have another" the problem changes from "What are the ways we can combine 4 different cards" to "What are the possible ways we can combine 3 cards with 2 different Aces)" - it's just misleading because one of the aces, is also one of the 3 cards being combined.
Specifically, the probability set changes to this:
A of S + 2 of Diamonds
A of S + 7 of clubs, or
A of S + A of H
A of H + 2 of Diamonds
A of H + 7 of clubs, or
A of H + A of S)
Simplified to:
A of H/S + 2 of Diamonds
A of H/S + 7 of clubs, or
A of H/S + A of S/H)
you aren’t more likely to get tails because you got heads, the law of large numbers causes the coin flip to converge to 50% as # flips goes to infinity
This is your second channel now xD
God sent His only son Jesus to die for our sins on the cross. This was the ultimate expression of God's love for us. Then God raised Jesus from the dead on the third day. Please repent and turn to Jesus and receive Salvation now before it's too late. The end times written about in the Holy Bible are already happening in the world. Jesus loves you with all His heart ❤️ but time is running out.
@@L17_8jesus got soloed by humans lmao
@@L17_8but jesus is god as well
@@fugcatso god got soloed by humans. Although he does have a pretty high K:D ratio overall.
this was his first channel though
When you flip two coins, the options are HH, HT, TH, or TT. However, when the person holding the coins says he knows that at least one coin is tails, the order of the coins doesn’t matter, therefore the sample space becomes 1. a tail and a heads or 2. two tails.
Another way of thinking about it is when the coin holder guarantees one tails, you can remove that coin from the question altogether since it has a 100% probability of being tails, leaving only a 50% chance for the other coin to be tails.
This, and it applies to the cards as well. In the four card example, you have two possible states: ace of spades or ace of hearts. In each state, you have one ace, and the probability of the second card being an ace is 1/3. Removing one ace from the four will always reduce the remaining group to three, and the probability becomes 1/3, regardless of which ace is removed. The suit of the ace is a red herring (or black herring if it's spades) as it incorrectly suggests that it is the point at which the suit is revealed that collapses the probability space, when the question is simply asking for aces independent of suit, and once the presence of one ace is confirmed it is no longer an uncertainty, and removed from the equation. This IS like the Monty Hall problem but not in the way people think.
Most of what you are saying is correct, but you can't remove a specific coin form the equation(you don't know which one to remove).
While it's true that the chances that the "other coin" is tails is 50%, does not mean that there's a 50% chance for TT when the coin holder guaranteed one tails. This is because we don't know if we have TX or XT.
X could be H or T.
So we could have TT, HT, or TH. So 1/3 chance for TT.
@@MartyThun order doesn't matter, so HT and TH are the same outcome and should not be separated into two different outcomes. Therefore, the outcome of a coin flip is always 50%.
@@templarknight7 yes, put TH and HT together, they appear twice as often as TT.
Same outcome, but twice as likely as TT.
@@MartyThuntwice as likely when you flip two coins. rewatch the video. around 7:23 he starts the example and never says the coins were flipped, simply that he is holding two coins in his hand and he knows one is tails.
The first puzzle can also simply be solved with symmetry. Both glasses contain the same amount of liquid, because we've moved one tablespoon from each glass to the other. So whatever the distribution of coke and juice in the first glass is, the distribution in the other glass must be exactly the same but reversed.
Wow! That's the kind of explanation that makes the solution so obvious that you feel doubly stupid for not seeing it immediately
Ah, right, the soda that was taken out obviously has to be replaced with _something._ Thanks, that makes it even more intuitive for me. 😅
@@DavidSmyth666 No kidding. I stopped the video and thought about it for a few minutes and reached the right conclusion, but now I feel stupid for not realizing what Gordon pointed out.
It's the same reason a shuffled deck of cards split in half has as many red cards in one half as black cards in the other.
Gordon, look, I'm someone who thinks of themselves as smart. But dude, I feel dumb right now.
During the last one, I feel like I spotted the problem with it. By revealing the suit, you're able to know specifically which set it's in. Without revealing the suit, you still know it's going to be in a set of that size, you just don't know which one. In your 4 card example, knowing you have one Ace doesn't make it a 1 in 5 until you know the suit. There's only 3 other cards. The probability was always 1 in 3, but from two possible sets.
I'm having a hard time organizing this thought. If your 4 card deck was shuffled, and you drew two cards from the top of it without looking at them, the odds of you drawing both Aces is 1 in 6. If you draw the first card and look at it, and it's an Ace, your odds of drawing the second one become 1 in 3. You inherently know the suit of the card by having looked at it, so you can narrow down which set you're working in. However, if you draw that card and show it to a friend, and they only tell you it's an Ace, and not which one it is, that doesn't change the fact that you have a 1 in 3 chance of drawing the other Ace as the next card.
Holding two cards, knowing one of them is an Ace, you're holding 1 of 5 hands. You will land in a set that does not contain two of those possible hands, no matter which Ace you're holding. You're really only holding 3 possible hands, (Ace - Ace, Ace - 2, Ace - 7), and caring about which Ace it is in the back half is changing the question subtly.
But knowing which ace it is really does help you probability wise. By knowing it’s the ace of spades or hearts you can remove 2 of the 5 sets from your options however without knowing which ace it is you really can’t remove these and need to consider all of them. So the probability really does change. The problem isn’t that the ace thing isn’t real it very much is, it’s called “the paradox of the second ace”
@@deananderson7714 I understand that part of it, but I'm saying you can always discard 2 of the 5 possible hands. There's no scenario where you're holding an Ace where all 5 hands can be true, because you're holding a part of two possible sets, both of which contain three possible hands. By combining the two sets into one for that part of the problem, the question gets changed from "What is the probability the other card is an Ace" to "What is the probability I'm holding a specific hand." In this context, those questions are incredibly similar, but they are different.
There's a wrong assumption being made somewhere if there are three cards and I'm being told there's a 1 in 5 chance of holding a specific card. I don't, I never will, and there's an underlying issue with the logic of the question if we think there's any way to frame it where the probability is actually 1 in 5 at any point. I think my answer solves why that apparent 1 in 5 is actually the 1 in 3 it should be.
Let's scale it up slightly, and change where the second card is. Let's say you draw the first card, your friend tells you it's an Ace, and the second card of the hand stays on top of the deck and neither of you see it. If you have 6 total cards, now you have 15 possible hands. If only two of those cards are Aces, there's 6 hands that contain no Aces. By knowing we have an Ace of some suit, we'd know we have one of 9 hands. 8 of those hands will be one Ace and a non-Ace card. You could split this in two sets of 4, one for each suit. The suit doesn't matter, you know for certain you can only be in one of those two sets of 4 OR have both Aces, which means your probability of having the other Ace would be at most 1 in 5. That's also what you'd get if you revealed the suit. You can always discard 4 of the 9 hands, you just don't know which 4, and you only have to care about which 4 it is if you care about every possible answer. We only care about the one. I left the second card on top of the deck without drawing it, because it's more clear that the other Ace has a 1 in 5 chance of being drawn this entire time, regardless of how many hands you might be able to make. That top card never had a 1 in 9 chance, so our thinking that it did means there's a flaw in our calculation somewhere.
I'm fully willing to be wrong, here, but I just don't understand why we include all sets it might belong to when only one of them can be true at a time. If we cared about each possible outcome, then it'd be different, but we only care about the odds of it being a second Ace.
@@PressAtoJ Ah- yeah, you're 100% right... xD I think, you're brain works a bit too convoluted for me.
To put it more simply, he changed the problem from "What are the ways we can pair 4 different cards" to "What are the possible ways we can pair 3 cards with 2 different Aces (Where one of the 3 cards is the other Ace)"
So our probability chart becomes this:
A of S + 2 of Diamonds
A of S + 7 of clubs, or
A of S + A of H
A of H + 2 of Diamonds
A of H + 7 of clubs, or
A of H + A of S)
Simplified to this:
A of H/S + 2 of Diamonds
A of H/S + 7 of clubs, or
A of H/S + A of S/H
His problem was very misleading by showing '5' probabilities.
@@amortal3248 I've got the autism, my brain is too convoluted for me too
@@PressAtoJ If you know one card is an ace you also know one of the 3 cards remaining is an ace. regardless of suit.
knowing the suit doesn’t help the probability in any way, the video is wrong.
I have no clue how they came up with 3% from the first example when 3/51 is 5.88%
Bro made this video with the SOLE purpose of making us feel dumb 💀💀💀
Probability changes depending on what information the observer has. Doesn't matter if any new information feels insignificant, the observer must check how the new information changes the sample space.
About your coin problem,
Let's say you have a 100 copies of yourself who all flipped two coins and are holding one in each hand.
So, 25 people with two tails, 25 with two heads, 25 with tails in left and heads in right, and 25 with tails in right and heads in left.
1) If you ask "All those who have at least one Tails, stand up", 75 of them would stand. Out of these 75, 25 would have both coins as tails, hence the probability equals 1/3.
2) If you ask "All those whose left hand's coin is Tails, stand up" only 50 of them would be standing. Out of this 50, 25 of them will have both tails. Hence probability equals 1/2.
3) If you repeat 2 with "right" instead of "left", you will get the same probability.
4) Finally, If you say something like "All those with Tails in THIS hand, stand up", but we (the observers) do not know which hand you are referring to, it does not matter.
We know that no matter if you asked for the left hand or the right, there will be 50 people standing.
And 25 of them will have two tails, making the probability equal to 1/2 again.
Which hand is revealed does not matter, the fact that one of the hands was revealed reduces the sample space to 50.
Also, it is important to note that 1) is fundamentally different from 2) or 3) because no hand is revealed, instead it is only revealed that two heads are not possible.
You're thinking the wrong way. If you were trying to guess which person from a group was the one with two tails this would be correct, but that's not the question that was asked. The question was regarding an individual with tails in one hand, which eliminates any possible combinations that include a heads in that same hand. Doesn't matter if we know which hand, because we only care about the other hand, which is either heads or tails. 50%
Verbally said “f*ck off” several times watching this, positively and endearingly
I dont know which is better, zach star uploading or zach star himself uploading
i'm so proud of myself for getting the first question with the soda and juice right, the rest, i don't play cards, so idk
This just confirms my theory that people never tell me all the info, but still expect a full/perfect answer
Well, in the case of the card problem, he does feed you incorrect information. IE: the number of ways you can combine all of the 4 cards among one of two aces we can guarantee is like so:
A of S + 2 of Diamonds
A of S + 7 of clubs, or
A of S + A of H
A of H + 2 of Diamonds
A of H + 7 of clubs, or
A of H + A of S
(Simplified to
A of H/S + 2 of Diamonds
A of H/S + 7 of clubs, or
A of H/S + A of S/H)
Notice the difference from in the video?
When the question is changed from "What are the possible ways we can combine 4 cards" to "What are the possible ways I can combine 3 cards with 2 different aces", you need to change the dataset we're working with, and he didn't do that - so it's just misleading maths.
@@amortal3248 In the first list of possibilities that you deem correct, the two aces appear twice, once per order. However the other combinations such as A of S + 2 of D only appears once, you have fallen for the trap already. The reason this "paradox" arises is that if he decides, to always whenever there is an ace or more than one ace, regardless of which ace it is, tell us about that ace and its suit, if he says, say, 'Ace of Spades', the Ace-Ace scenario becomes half as likely as the other ones, as he could have said 'Ace of Hearts'. So the chance is 0.5/2.5 = 1/5, same as the other one.
@billnyethescienceguy9918 is very right, it depends on the information. If he instead gave us the information: 'I am going to take two random cards out of these 4. If one of them is the Ace of Spades, I will tell you that I have drawn the Ace of Spades. I will then ask you what the probability of the other card being another ace is.' And then took the two cards, revealed information about the ace of spades, then the chance would be one in three. But proper definitions and clear questions are a rarity on this (and many other math channels) unfortunately. Instead the audience is left both confused and feeling like they're smarter than most for 'knowing' something.
Oh well, perhaps he will clear everything up in the follow-up video, my hopes are low.
Yeah the median one annoyed me because I have a degree in statistics and was waiting for “All of the Above”, then he hit us with the “A or B” like I KNOW you are lying just so All of the Above wouldn’t be an obvious out for people who were guessing
The two coins problem is actually asking two completely different questions depending on what information you give.
When you say _"one of these two coins is tails, what's the chance the other one is too?"_ you're actually asking *"What is the probability that both of these coins are tails given that at least one of them is?"*
When you say _"this coin is tails, what's the chance this other one is?"_ you're actually asking *"What is the probability that this coin is tails given that another (unrelated) coin is tails?"*
Looking at it this way it makes sense that they have different answers. The answer to the latter question is obviously 50% because the two coins are unrelated, but in the first question we have no way to separate the states of the two coins, which complicates the probability.
In the final example when the screen blacks out to prevent us from seeing which hand's coin you say is tails, we are not given the ability to distinguish between the two coins, so it's equivalent to the first question.
Completely agree, this is a good explanation of the distinction.
Nah, it's pretty simple.
If you say "one of them is tail", you need to count both cases separately. The order matters!
50% that the left is tail, then 25% that the right is also tail. Or the other, 50% that the right is the one that we know is tail, and then 25% that the left one is tail too. So 25%+25% = 50% that the other is tail.
And same for the cards, the order matters. The double ace should appear twice in the list, hence the double probability.
@@justfoundit The computation of the probabilities is well understood, the difficulty being presented is what probability needs to be determined.
@@justfounditwhen you say "Nah", which part are you disagreeing with?
The "this coin is tails" is only definitely distinct from the first case if he didn't look at the other one too. If he did look at both, he _may_ still have chosen which to reveal to you based on which one was tails.
I’m glad you’re able and willing to educate us
He screwed up his card example a bit, but the rest was good.
If you had no tails in your hand, then you would say "one of them is heads", so you need to include the cases "one of them is heads". then in 50% of all choices the other one is the same.
For the first one, you can reason from the end state.
Since the volume end up the same, the volume of foreign liquid has to be the same in each glass.
That seems to be the simplest way of looking at it.
thats actually smart...i just did the math but your way is far cooler :)
i also did the math. incorrectly, as it turns out 😅.
To clarify, it's nothing smart. It's a well known tool in mathematics. Now, you know too 😉
As an example: Two trains are on the same railway. On train is on station A, the other train is on station B 100Km away. Both trains depart at the same time, but train B as an issue and instead of going at 100km per hour, it goes at 40km per hour. The two trains, going in the same direction, end up colliding.
What was the distance between the trains 10 minutes prior the collision.
(Try to solve, that may be worth it!)
You can solve for the collision point, then solve for t-10.
Or you can consider that you are seeing the two trains colliding and that they separate at a speed of 60Km per hour. (the catch up speed)
So, the problem becomes: what distance is covered in 10 minutes at 60Km per hour ? (A sixth, so 10km)
@@programathsthanks for sharing! never even considered looking at these problems backwards
I thoroughly enjoyed this. You know why? I got all answers wrong. I love it when I am wrong and learn something new.
Zach, you are amazing and love your videos, no intro, no outro, no music, to the point and the topics I like.
For anyone who doesn't know, the reason why the probability 'changes' is because we're applying the wrong theory to find the probability in both of these scenarios. Let's take the coin scenario:
1. In the presented scenario, we're looking for an ORDERED pair. Let's assume the order is in the format [x, x].
2. No matter what, one coin must be tails.
3. All possible ORDERED pairs for this scenario are [T, T]; [T, H]; and [H, T].
This means that, assuming we don't know the position of the coins, we'll have a 1/3 chance of the ORDERED pair being [T, T]. If we do know the position, let's say the second value is T, we can rule out any pair where the second value is not T. This leaves two pairs and a 1/2 chance of the ordered pair being [T, T].
Now, the reason why I say that we're applying the wrong theory is because the question asked is, "If I have two coins and one is tails, what are the chances that the other is heads?" This means that we are not, in fact, supposed to be looking for an ordered pair, and instead we should be looking for the value of the OTHER coin. In this case, we ignore the known coin. Because there are only two states a coin can be in, both tails and heads will have a 1/2 probability, and so, no matter what the position of the first coin is, the second coin will have a 50:50 chance to be tails. All we have to do to apply this same logic to the card example is twist it a bit.
Sorry if I spoiled the mystery for anyone who was waiting for the second video!
tl;dr: The probability doesn't actually change, we're just applying the wrong theory for the situation.
Also, Zachstar is doing God's work here showing people how statistics can really be extremely skewed
Great reply, you said what I wanted to say more eloquently. 👍
@@IJebronLamesI thanks!
I wish I had a teacher like Zach. His charisma really made thinking about these questions/puzzles fun.
For the last problem:
It's the classic Monty Hall problem; or at least, bears much similarity to it. The trick is that while it doesn't technically matter which one you reveal, it does matter HOW you choose to reveal it.
We are looking at whether the following two scenarios have the same probability:
Scenario 1: I look at both coins. If exactly one of them is tails, I reveal that that one is tails. If both are tails, I randomly choose one to reveal that it's tails. What's the probability that both are tails, given I reveal the left coin is tails, and what's the probability given I reveal the right coin is tails?
Or, Scenario 2: I look at both coins. If at least one of them is tails, I say that at least one of them is tails. What's the probability that both are tails, given that I say at least one of them is tails.
In both Scenarios, there's a 25% chance of double tails, 25% chance of left coin being tails, and 25% chance of right coin being tails.
Zach incorrectly reasons that in Scenario 1, when I reveal the left coin, it's a 50-50 chance of whether that's due to having double tails or only left coin being tails.
In actuality, two-thirds of the time when I reveal the left coin is tails, the right coin is heads - as the 25% chance of double tails is shared evenly between revealing the left and right coins, meaning though there's a 25% chance of double tails, there's only a 12.5% chance of double tails AND I reveal the left coin is tails, as there's another 12.5% chance of double tails AND I reveal the right coin is tails.
Therefore, we're actually comparing the 12.5% chance of double tails AND I reveal the left coin is tails, VS only the left coin is tails.
In Scenario 2, as Zach says, it's one-third because it's a 33-33-33 chance of having double tails, left tails, or right tails.
Edit after reading the comments:
The chance is one-third for both of the above scenarios of having double tails. However, we can also imagine Scenario 3: I look at one coin (randomly choose whether to look at the left coin or right coin). If that one coin is tails, I say that at least one coin is tails.
Scenario 4: I look at one coin same as in 3. If that coin is tails, I reveal that the coin I looked at was tails.
Scenario 3 and 4 would indeed have a probability of 50% that both are tails.
Scenario 4 is easier to explain:
If I reveal it, then there's indeed only 2 options, Tails-Heads or Tails-Tails. I did not even look at the other one so it's impossible for the other coin toss to affect the information I give you. Therefore, it's exactly the same as just flipping one coin and asking what's the probability it's tails.
Scenario 3 is trickier, but it's still 50%. You might think there are "three scenarios" - Tails-Heads, Heads-Tails, Tail-Tail - but if you think of it this way, the probability that it is Tails-Tails is twice as likely. If I flip Tails-Heads or Heads-Tails, I only have a 50% chance of seeing the coin with Tails when looking at a coin at random.
I assumed that Zach was referring to Scenarios 1 and 2, but I don't think he made it clear. So just to cover all scenarios, and to clarify what I mean by it mattering HOW you reveal it, I'm adding this edit
This is the best explanation I've read in the comments so far. So indeed, assuming scenario 1, what Zach says at 7:42 would not be correct and the probability is still 1/3. VERRY confusing, because if feels so obviously true!
But I guess 1/2 = P(TT | left is tails) ≠ P(TT | Zach TELLS you left is tails) = 1/3 (where TT = both tails) since in the latter the probability of TT has been split (between Zach telling you either the left or right one is tails) as you mentioned. So at the end of the video when we cannot see which hand he reveals, it actually doesn't matter because whatever hand he tells you tails is in, the probability is still 1/3.
If we toss out every heads-heads scenario 2 is obviously 33-33-33 like you and Zach says.
For scenario 1: Yeah I think I get your point.
Half the time you are going to say "left is tails", half the time you are going to say "right is tails".
The overall 1/3 chance of tails-tails will be torn between these two, so (1/2)x(1/3)
So in this scenario, if you're saying "left is tails" it's actually 2/3 chance of right coin being heads, exactly like you said!
So what zach is giving us for the 50% chance of tails-tails when you're saying "left is tails" is actually scenario 3: (I think)
You look at one specific coin: If it's tails, you say it straight away, game on. If it's not, we don't do anything and reflip?
For this one you get the monty hall effect, we get information because it's been through a "chance gate" already.
We know it's not heads heads. We know that it's either tails-heads or heads heads. so 50-50, but this one is dodgy, yeah.
To get here, we had to go through chances of (3/4)x(2/3)x(1/2)=1/4 overall chance of it being tail-tails.
(3/4) first to avoid heads-heads. (2/3) to get the specific coin to be tails. (1/2) at the end, the answer Zach want to give us.
I think scenario 2 is the better interpretation of the riddle, and then the answer should be 1/3, right?
To sum up: Yeah, I absolutely agree that it depends on how you got to the point of saying "left is tails". It's super interesting.
@@steventhijs6921 Yeah, I think for it to be 1/2, it has to be a scenario where Zach decides to peek a specific one, then tells us what it is. If it's tails, its game, if it's heads we all go home. We didn't win the monty hall car this time :- )
It's still a really cool riddle, but like @byeguyssry said, it's important how it is revealed. Same thing with monty hall, seen some youtubers give the riddle incorrectly, the wording is super important.
In the Scenario 1, he could have just looked in his right hand just to report to us whether it's heads or tails. It's irrelevant in this case, so the left hand is 50/50. Your scenario 1 is different, so you get a different answer. Neither you nor Zach is wrong, but the selection algorithm is ambiguous.
At the end of the day, it boils down to "Ask different questions, get different answers." Just because they sound the same doesn't mean they are.
So the problem with this is that you assumed he flipped them, we have no idea the probability of anything in this scenario, we don’t know if he flipped them and then checked if they were both heads and turned one to tails, we don’t know if he just set both to tails and then told us that at least one is tails, we don’t know if he set one to and tails flipped the other one, we do not know how the given information is derived and thus we know nothing
Now I want to test the probabilities in a real world application and what the results end up being.
You can’t test it until he tells us how he knows which coin is tails and how he choose to give us that information.
Here would be your results of two card one:
You draw two cards 120 times.
1. 20 times: AĄ
2. 20 times: A3
3. 20 times: A2
4. 20 times: Ą3
5. 20 times: Ą2
6. 20 times: 23
A-ace of spades.
Ą-ace of hearts.
If you get 6. You end your game.
If you get 2. Or 3. You say I have A. Total: 40 times.
If you get 4. Or 5. You say I have Ą. Total 40 times.
If you get 1. You say I have A 50% and Ą 50%. In total: 10 A and 10 Ą.
So no matter what you say A or Ą there is 10/(40+10)=1/5 chance for you to have AĄ as if you just said I have an ace.
@@titaskleinas2682 I mean, you fell for misleading maths - specifically, after he says 'I have an ace, what are the odds I draw another ace" the dataset you're working with isn't
AA, A3, A2, A3, A2, 23
It becomes:
A of S + 2 of Diamonds
A of S + 7 of clubs, or
A of S + A of H
A of H + 2 of Diamonds
A of H + 7 of clubs, or
A of H + A of S)
Simplified to
A of H/S + 2 of Diamonds
A of H/S + 7 of clubs, or
A of H/S + A of S/H
To clarify, the problem changed from "What are the ways we can combine 4 different cards into pairs" to "What are the possible ways we can combine 3 cards with 2 different Aces into pairs" - it's just misleading because one of the aces, is also one of the 3 cards being combined.
More-so, note once you know an Ace is among the cards, the 2 + 3 pair is no longer a possibility, so it wouldn't show up 20 times in your example, but zero (As per the scenario). Yes those 2 + 3 pairs would show up when trying to recreate these statistics with real cards, but we'd be discarding them from our statistics, since they don't meet the scenario standards of 'having at least once Ace'
@@billcook4768 I really doubt that's what he's going to explain - in the first place the issue with the card problem and the coin problem are very different. In the card problem he misrepresented our dataset. IE, he changed our probability pool from "What are the ways we can pair 4 cards" to "What are the ways we can pair 3 cards into 2 aces, while one of the aces is among the 3 to be paired"
The coins problem is mathematically fine, but he's presumedly going to explain why the statistics work differently when coins are flipped independently VS together. IE when you're at this step:
HT
TH
TT
Between the two coins, there's a 4/3rds chance of a tails, and a 2/3rds chance of a head - these odds are skewed from the standard 50:50 coin flip because of the condition "One of these must be tails" - but if we remove this condition by identifying the one with tails, the remaining coin becomes independent and has the 50:50 odds again.
I like to think of it a a probability slider. Imagine you had a slider (Like the scroll bar) if you bring the bar all the way to the left, you have a 100% chance for a tails, and the right has a 50:50. If you bring the bar all the way to the right, you'll have a 50:50, and a 100%. But if the bar is in the middle, it'll be 2/3rds odds for either to be tails and 1/3rd for heads. - at any given time, the coins will add up to 200% odds, though
@@amortal3248 well to begin with he didn't specify where did he got those cards. And maybe one is a joker or both of them are ace of spades. So I made an asumtion that he drew those cards from a deck (in my example deck of 4 cards) and only if he got at least one ace he would start this conversation
2:00 that's how solved it initially too (I did x/(x+1) × (1/(x+1))
but then realized it's obviously the same cause they both have the same volume before and after.
It's like that card trick where knowing the number of red cards in one pile tells you how many black cards in the other
Blud left us with a cliffhanger
For anyone having existential issues with the last puzzle, it always helps to remember that probabilities are not a 'concrete thing' that we are 'measuring.' They are just an approximation that can freely change as more information is added.
Think of it this way: the True chance of drawing an ace is always either 0% or 100% -- the deck is already there and the universe already knows whats on top. But the probability we're calculating is just an estimate based on what info we have, and as info is changed then the estimate changes.
When we could have ANY ace in our two-card hand, then there are tons of ways it could fail (A❤ 5♠️, A♦️J♦️, etc. etc.) But when we hear the suit, even if we dont "care," it still helps us greatly reduce the number of ways our hand can fail (any failure now MUST look like A♠️ and something).
Any information helps our estimate, even if we don't "care" about the info. Whether the dealer is right or left handed might matter, the weather might matter. In the problem we just assume that these factors have absolutsly no effect whatsoever. But its important to remember that the universe already decided the top card of the deck, and ANY info on how it got there can help refine our probability estimation.
This comment is wrong in so many ways....First of all the universe does not decide anything, phrases like this are only said by people with little scientific background. What you are trying to say is that the universe is deterministic, a statement for which not only we don't have much evidence, but we do have evidence for the exact opposite. Given that there are probabilities in the heart of quantum mechanics (and bell's theorem) the evidence at this point in time is showing the universe to be probabilistic.
For the cards (and coins), I figured there's a quite easy way to calculate it: You divide the overall probability of the event (for example both cards being aces) by what you already know (at least one card is an ace). Both cards being an ace would be 4/52 * 3/51 (for a standard deck of 52 cards with 4 aces). The probability that at least one of the cards is any ace can be calculated with 1 - (48/52 * 47/51). The total calculation therefore would be 4/52 * 3/51 / (1 - (48/52 * 47/51)), which as mentioned results in 1/33 or roughly 3%.
For the second example, where we know that one of the cards is the ace of spades, also the overall probability changes - we need the probability of one card being the aces of spades, and the other one being any ace. This probability is 1/52 * 3/51 * 2. (notice that the factor 2 is there because the position of the aces of spade, either first or second, doesn't matter). The given probability is, that one cards is the ace of spades, that's 1 - (51/52 * 50/51). Once again, if you calculate this you get 1/52 * 3/51 * 2 / (1 - (51/52 * 50/51)) which results in 1/17, or roughly 6% and almost double of the first probability.
Please correct me if I'm wrong, I just came up with these calculations on the spot, but as the numbers do seem to be close to the (not precisely mentioned) actual answers from the video I'm fairly confident.
0:44 I got it wrong, but I just noticed there is a simple explanation of why it has to be the same. The total amount of soda and juice is one cup total and each cup has the same amount of liquid. If there is "1- a" soda and "a" juice in one cup, there must be "1-a" juice and "a" soda in the other.
How we mix the liquids is irrelevant, if both cups end with the same amount of liquid, the composition must be mirrored.
Regarding the two coins/two cards issue. The probability changes based on how much information the revealer has, how much they reveal, and what logic they use to reveal it.
For example lets take 2 coin flips, the revealer knows the result of both flips, and will reveal the first flip that is tails.
If the revealer reveals that the second flip is tails, the first flip MUST be heads, so the coins are HT
If the revealer reveals that the first flip is tails, the second flip is a 50% chance, so TH or TT are equally likely.
If the revealer chooses at random when there are 2 tails, we have these possibilities:
TH reveal 1 = ⅓
HT reveal 2 = ⅓
TT reveal 1 = ⅓×½=⅙
TT reveal 2 = ⅓×½=⅙
So, regardless of whether they reveal 1 or 2, the odds are ⅓ that the other coin is T.
If the revealer randomly chooses one coin to look at and declare and will declare whatever side is showing, the odds are 50% the other coin matches.
That is why the Monty Hall problem is hard. The host knows everything and choosing to reveal a goat affects the probabilities post reveal.
If the host randomly revealed a door without knowledge (sometimes revealing the prize), then naive probability works.
I think this is relevant to resolving the problem.
Thank you for uploading on your second channel. This is much easier to understand than the most recent video on your first channel!
this tripped me up so hard, i had to stop the video and just kinda think for a minute
I feel like the last problem has strong under-currents of the monty-hall problem. By revealing which of the hands/doors has it (or doesn't have it in the monty-hall case), you narrow the probability space for the other hands/doors. Though that being said, this problem is still breaking my brain somewhat, so it hasn't solved it for me yet.
I think the answer can be made with assumption. You can simply assume that the ace is spades, and calculate probabilities based off that assumption. So long as the probabilities are the same regardless of which suit you assume it is, it doesn't matter which suit you pick. You know it has to be one of them. It's kinda like thinking two moves ahead in chess? *shrug, was a fun problem.
@@Alec0124 It's just a classic clash of two views on probability. The view that probability depends on the information you have about the situation is Bayesian, and extremely useful in statistics, but it's not always the proper philosophy. You could make a very strong case for the thing you mentioned: that because you know the ace is at least one of the suits, you can arbitrarily assume one of them. It would be foolish not to do it if you were gambling for example. But if you were a pollster, for example, you probably wouldn't want to make an assumption, especially if you don't know the precise relationship between all the possible variables, but at the end of the day it's a tradeoff between type 1 and type 2 error type of thing, and how much you're betting to go with your guts or not, if it's a more complex scenario
Or quantum entanglement stuff. Like how an observer changes hypotheticals for you,by omission of facts? Idk
@@Alec0124The problem with this view is that the empirical probability is 3%, so any method of calculation that gives a different result must be invalid, regardless of how reasonable it seems.
This is different from the Monty Hall problem, as in that the host reveals new information that makes it advantageous to switch.
Here, Zach is doing a very clever slight of hand when he shows the 6 possible hands. If you know that he has an ace, but you don't know which one, he is saying "I have the ace of spades OR the ace of hearts (or both)." When you look at it like that it becomes clear - if he has the ace of spades, he can also have either the 2, the 7, or the other ace; same is true if he has the ace of hearts.
It's the same with the full deck. You can think if it as if you are drawing cards from a deck. The first card you draw is an ace, what are the odds that the next one you draw is an ace? It's going to be 3/51 (~6%) since there are 3 aces left in the deck. Knowing which ace you pull doesn't affect the odds.
5:05 What if they're not from the same pack?
Yes, or from a deck of 51 bought on a Tuesday.
Yeah, it was never said they were from the same deck
The Monty hall problem only works as we assume it does if the host has inner knowledge of the game and we know he undergoes certain rules (He always reveals a goat etc). If those rules aren’t in place (ie the host doesn’t know what’s behind the doors and just picks randomly) then the Monty hall problem as we understand it isn’t the same as in some scenarios the host might just open the one with the car.
In this example I feel like it must be to do with the set up. How do we conduct this game? Do we flip both coins over and over until at least one of them is tails? And what rule does the flipper follow in revealing the tails? Let’s say that he flips both and if there’s at least one tails he rolls a die. If there is only one tail he tells the viewer about it regardless of the die roll. If there are two tails he’ll tell the viewer about the left tail when then die is odd and the right tail when the die is even (the viewer can’t see the die he just knows the assignment is random). Under these strict rules we can eliminate the one in four events when they’re both heads as nothing will come of those. Since our guy reveals tails every time in this scenario nothing has changed. The probability of the other being tails is still 1 in 3. I think the confusion comes from vagueness in the rules of choosing.
Oh and in this new version of the game him saying it in the dark is just the same as him saying it to us directly. Since we know the rules and everything’s consistent it’s 1/3rd regardless. Revealing the tail literally does nothing whether we see it or not.
To get the scenario like the second one with the 1/2 chance we’d have to have a situation where our flipper only announces a tails when he has it in his left hand. The reason this is now 1/2 rather than 1/3rd is now in the heads left tails right case which before would have lead to an announcement nothing happens. So we only get the tails left heads right or the tails left tails right cases. So the flipper’s strategy is what matters and that determines probability.
Yeah it feels like everyone is missing this. The probability for both the cards and the coins, could be literally anything depending the preferences of the information revealer. Hearing one is an Ace/tails, doesn't mean you can include the total probability space for all remaining possibilities, unless that exact information would be certain to be revealed in all cases.
I totally agree. I was about to comment the same thing. Probability essentially depends on the set-up.
My reaction when he presented 2 cards and talked about probability was kind of: "You didn't even tell me those 2 cards were picked from the same deck.".
The problem is ambiguous as it allows for 2 possible scenarios. The gotcha is that the follow-up question tricks people into switching between scenarios, which is completely wrong to do once you're locked in.
I find it's easier to think of 2 rooms each with 100 people, and they all flip exactly 2 coins. In the first room (aka. interpretation), they were all asked to reveal a coin at random. In the second room, they were specifically asked to reveal if they have tails. Then we eliminate everyone who didn't reveal tails. The distribution of flips is not going to be the same between those 2 rooms, and yet everyone can still ask the same question of "Given at least one T, do I have TT?" You can extend the logic further.
It doesn't matter which hand he opens, he doesn't need to open either. This is the Monty hall problem but not in the part of the question you are looking at. The chances of getting one heads and one tails is 50% (H,H H,T T,H T,T), just like the chances of picking the correct door is 33% in the Monty hall problem. When he reveals the door without the car that doesn't change the fact that the chances you were wrong was 66% because he revealed it with inside knowledge. That is why you should switch. Same thing happens here, when he reveals that one coin is tails that doesn't change the fact that one heads and one tails is a 50% chance, the door he opened was the H,H door, nothing else matters at this point. The Monty hall problem happened before the question even started so the revealing of the hands is all for show.
@@reuvencooper8170 the idea of him revealing the so-to-speak HH door is an interesting idea but I’m not exactly sure of your point. If the host only tells us about scenarios where there is at least one tails the chance of HT in any order becomes 66%. The setup and rules absolutely do matter. Like if the rule is that he just skips every HH those aren’t in the probability space.
4:06 If anyone is have problem here just as me, the median is the value that is in the middle when they are sorted from lower to grater for example “2,3,6,9,10” in here 6 is the median and if the number of values is even like “1,3,5,8” you need to get the average of the 2 middle values in this case (3+5)/2=4
But if you add up ((1+3+5+8)/(4))=4.25
Wouldn't that be the medium?
Also if we add up all the salaries of 2024 we get 900k in total thus a median of 225k which would make an increase of 50% compared to 2023.
Which does make sense.
The total female influence over the total in 2023 was ~66.67% (as ((200k+200k)=400k and 400k is ~ 66.67% of 600k)
The male subsequently would be 33.33%.
So our calculation would be this:
0.6667•150% + 0.3333•150%=1.5=150%
Correct me if I am wrong
@@gegecrythat is the difference between the median and the average maybe my first example was a little bit confusing because its average and its median is the same but if there are “1,20,80” the median is 20 just because is in the middle “3,20,100” here 20 is also the median. Is just another way of doing statistics but average is more specific i think. In your first example 4.25 is the average not the median.
Yeah, this bugged me too. Median != Mean/average
Zack can make me laugh so hard I stop breathing, AND teach me useful problem solving skills. Respect.
If the coins are randomly flipped repeatedly and only presented to us when the left coin is tails, then the probability of guessing correctly that both are tails is 50%. If instead the coin flips are presented when and only when at least one coin is tails, then the probability of guessing correctly that both are tails is 33%.
Nah it’s still 50%, there are three “outcomes” but two are each 25% likely
@@skalty9868 Not so, all three outcomes would have been 25% likely out of 100%, but we've tossed the 25% of 2 heads away. Of the remaining 75% of possibilities we consider, each outcome is 1/3.
Aren't both 1/2 and 1/3 chance possible depending on how he decides to call out possibilities? (or any other probability >= 1/3, actually)
Options are TT, TH, HT, HH. Obviously, when it's TT he will say "at least one of the coins is tails". By symmetry if it's HH, he will say "at least one of the coins is heads".
But what about TH/HT?
Case 1: If he says "one of the coins is tails" if there is a tail, then both options result in him saying that at least one is tails. Then there are three scenarios which result in him saying "one of the coins is tails", giving 1/3 chance that the other is a tail.
Case 2: If instead for TH/HT he says "one of the coins is tails" 1/2 of the time, and the other 1/2 he says "one of the coins is heads", then we have a different scenario. 1/4 chance that it's TT and (1/4+1/4)*1/2=1/4 that it's TH/HT AND that he says "one of the coins is tails", giving 1/2 chance that the other coin is a tail.
Case 3: (symmetrical to Case 1) If he always says "one of the coins is heads" if there is at least one head, then the only possibility of him saying "one of the coins is tails" is if it's exactly TT. Giving 1/1 chance that the other coin is a tail.
I think case 2 is most consistent with the setup, looking only at a single coin and saying if it's heads or tails, giving 1/2 chance for the other coin to be tails (when he says the first one is tails). Saying "at least one coin is tails" knowing both coins leaves all 3 cases possible.
I think you're exactly right. It's random reveal vs deterministic reveal.
I was not convinced with question 1 so I did the math myself if you're interested :
Lets assume both solutions of soda an juice have an initial volume of 1000ml.
We'll respectively call these solutions A and B.
Step 1 : You take a tablespoon that contains X ml (where X is a small number, say 10ml or 5ml) from solution A and pour it in solution B.
After step 1,
- solution A contains 1000-X ml of soda and 0 ml of juice
- solution B contains X ml of soda and 1000 ml of juice
Step 2 : You mix perfectly solution B, that now contains 1000+X ml. Now, every sample from solution B has to be in the same proportions of soda and juice as the whole solution B.
After step 2, solution B contains :
- a proportion p_soda of soda, where p_soda = X/(1000+X), with 0
Same but I did it for problem 3
We can find conditional probabilities by dividing the probability of our desired case by the sum of the probabilities of all cases which fit our conditions
Let’s say I randomly pick a card I have a 4/52 chance to get an ace then the next card is 3/51 so our chance of two aces will be 12/2652. We are told we have one ace guaranteed which means we can eliminate any cases with no aces. So we have the described case, ace then no ace and no ace then ace.
The first of those is (4/52)(48/51) =
192/2652
The 2nd is (48/52)(4/51) which will logically be the same
so logically our probability should be
(12/2652)/(12/2652 + 2(192/2652)) which is indeed around 3%(3.0303…%)
If we instead are told for example one of them is the ace of hearts we get
P1 : (1/52)(3/51) + (3/52)(1/51)
= 6/2652
P2: (1/52)(48/51) = 48/2652
P3: (48/52)(1/51) same as P2
So we have
(6/2652)/(6/2652 + 2(48/2652)) ~=
5.8824% about 6%
@@deananderson7714you’re doing too much. I solved it much more simply in my post. Check it out if you’re interested.
@@deananderson7714you’re doing too much. I solved it in a much simpler way. Look at my comment if you’re interested.
Btw, this is my second time saying this. My comment got deleted for some reason?? I don’t think I’m violating any guidelines, right?
@@IJebronLamesI your math is wrong. It assumes he picked up the ace he’s talking about the first time when he could have picked up the ace the second time. That does matter
@@deananderson7714 I mean, you can say my math is wrong, but I’d like to see you actually disprove it. It looks like it holds up to me.
In the last question, it depends on how the situation came about, *not on what information you reveal* about the situation.
Did you choose heads for one and then flip the other one? (50%)
Did you flip them until you got at least 1 heads? (33%)
Did you flip them and then (out of the 1 or 2 sides that you saw), chose between calling heads and tails (or the left or right hand's result)? (50%)
...And if there's both heads and tails, will you reliably prefer saying heads? (33% if you said heads and 100% if you said tails.)
Did you choose both? (Then for you the probability is either 0% or 100%, but for us it depends on how well we know how you'd choose.)
Same with the cards:
Did you draw 2 cards (and discarded both) until you got at least one ace? Then it's 3%.
Did you draw 2 cards (and discarded both) until one is the ace of spades? Then it's 6%.
Did you draw 2 cards, and then discard *one* and draw a replacement until you got at least one ace? Then the chance that you have a second ace is basically zero. (Since it can only happen in the initial draw.)
Did you choose the ace of spades (or any of the others) and _then_ draw a second card;
or did you draw a first card and _then_ choose one of the aces as the second card? The probability of having an ace goes up by 1/3rd.
Did you draw one card each from 2 _separate_ decks of 52 cards (until one is an ace/ the ace of spades)?
Did you draw two cards from a _combined_ pile of two 52-card decks (until one is an ace/ the ace of spades)?
Did you choose both cards?
Yessir the boy/girl paradox drama is back raaaaaa🗣️🗣️
Jesus loves you ❤️
that's the best ad for brilliant I've ever seen. they should give you some extra cash for that one
With the first question, the mixing wouldn't matter right? The amounts would be equivalent regardless of the proportion of juice to soda in the tablespoon
Yep
thats true and made me think a lot about how questions can be structured to make the answer less intuitive. i think the mixing makes the answer a little less trivial.
if he doesnt mix it and says the answer will be the same regardless of the proportion of juice to soda in the tbsp, its a little more obvious that the answer is the final proportion between the glasses will be inverted (juice:soda on left exactly the same as soda:juice on right). this is because we can assume the simplest cases, that the tablespoon is full of ONLY juice or ONLY soda, making the answer much more obvious. it’s the mixing that complicates the question slightly, making it less intuitive.
Well. If the liquids would separate perfectly (like oil on top) then you could theoretically just scoop out the added liquid. So, yes, the problem needs the aspect that those liquids mix. But also yes, the stirring is not necessary.
The answer to your last question is still 1/3 because when the screen blacks out, we receive no new information about which hand has the tails hence the probability doesn’t change for us.
This example reminds me of quantum mechanics where two or more particles can be in an entangled state and have a joint probability distribution (1/3 probability case) but when the state function of one collapses, the probability distribution of the other changes (revealing which hand has tails hence probability changes to 1/2). This is very similar to the famous double-slit experiment where the simple act of observation changes the result.
Card problem was wrong. In both cases of the 4 cards problem shown, it's 1/3rd. You have artificially thrown away one of the two possibilities for the two ace pairing by saying 'the order doesn't matter', but it actually does IF AND ONLY IF you have an ace already but haven't disclosed which ace it is - it can either be Ace of Hearts and then the Ace of Spades, or, if the ace you are holding is the Ace of Spades, it can be the Ace of Spades then the Ace of Hearts. In the Ace of Spades example, the possibility of the Ace of Hearts then Ace of Spades option doesn't exist so you don't get the probability wrong for that one.
The coin problem is based on the same logical error. If you say one coin is tails, yes, the possibilities are limited since you removed HH, but you should have removed either the TH or the HT as well, since one of the coins cannot be heads anymore. Thus you are left with TH and TT, for 50%. Specifying which coin makes no difference, except that it decides whether TH or HT was correct to remove.
How about some simple math instead of reasoning. Imagine 100 people flip 2 coins.
Given 4 possible outcomes to flipping coins, each outcome is an equally likely 25% resulting in the following.
TT - 25
TH - 25
HT - 25
HH - 25
What percentage have at least 1 tails? 75%
What percentage has 2 tails? 25%
So the answer is 1/3.
You cannot remove TH or HT because it is equally likely that you would get either. You would only know that one is impossible if I told you that one hand has a tails.
@@randomlycasual4941 The video was specifically about what happens when he tells you that one coin of the two was tails, so the both-hand percentages for TT, TH, HT and HH are misleading. That's the point... knowing which hand has the tails allows you to remove the TH or the HT specifically, - but if you didn't know which hand you still need to remove one anyway, you just don't know which to remove. So because of a false sense of order dependence you get the wrong outcome. Otherwise, you're telling me that my knowledge or lack thereof can alter the probability of an event that already occurred - absolute nonsense.
Simpler explanation for problem 1 without calculating volumes:
After the switch we end up with 2 solutions of exactly the same volume as before since we moved the same volume back and forth (1tbs in this case).
Since the total volume of liquid in the soda bottle didn't change, whatever the amount of soda missing (aka the amount of soda now in the second bottle) must have been replaced by exactly the same amount of juice and vice versa.
Thus the amounts are equal.
Ok, so i feel like the problem here is in the fact that we use two different probability distributions in the two examples. I.e. if you choose the two coins randomly, then look at them, confirm that one of them is a tails, and then reveal it, it's still 1/3, since you revealed the one that you saw that was a tails, not picked a tails in the first place. I feel like it is pretty much the same problem that in Monty Haul paradox but in reverse
The monty hall problem isn't a paradox, it makes perfect logical sense, it's just unintuitive
@@patrickhector on some level, all paradoxes are
@@emermage tell me the perfect logical sense behind the grandfather paradox
@@emermage
This guy has all the answers!
So what of theseus’s paradox?
@@-ZH @patrickhector ok, i guess we have different definitions of a paradox here. For me grandfathers paradox is a contradiction, that proves impossibility of time travel in single timeline, and theseus's paradox is just a question of how one defines equality
Probability is where information about the world is taken in and a number describing a state of that world is given out. If we were to simulate the world a bunch of times then we can count of the number of times things happened and use them to get probabilities based on the worlds we could be in, and the worlds where an event happened. We restrict the space of worlds to those where someone had two coins in their hands and secretly "revealed" which hand held a tails. These worlds can be split in two: those where that person knows a tail is in their left hand, and those where that person knows a tail is in their right hand, but we have no information to determine which one of those worlds we are in. We just know that we are in a world where there is at least one tails, and it is 1/3. The person holding the coins has different information available, the subset of worlds they could be in is different, and so they would get 1/2.
I found that probability starts making a lot more sense when you think about what state space the situation is occurring in, and what subspace the event of interest occurs in. These spaces are determined entirely by the information, and as different observers can have different information, their state spaces can also be different leading to different probabilities.
You entered the wrong answer in the brilliant ad and it turned green.
When?
@@Tyy-b-un9bl 10:06
How did you get 3%? When I calculated it, I got 5.8% the first time and I have no idea what numbers I'm supposed to plug in the calculator to get 3%.
Here's my thought process: If I have an ace, that's 1 card out of 52. For the other card to be an ace, it has to be any of the other 3 from the total of 51 cards left. So the probability is 3/51 which is about 5.88%.
With the two aces problem, I assumed a standard 52-card deck. If we know the first card is an ace, there are 3 aces remaining out of 51 other cards that can be drawn, which equals a 1 in 17 chance.
I also got that, but that's 5.88%. He claimed it was ~3% before moving to the variation where it's specifically A♠️. Do you have any idea why he says that? I don't.
The 3% comes from conditional probability. P(TwoAces | OneAce) = P(TwoAces ∩ OneAce) / P(OneAce) = (4/52 * 3/51) / (1 - (48/52 * 47/51)) = (0.0045) / (0.15) = 0.03.
You don't know the 1st card is an Ace, just that at least 1 of them is. I believe that is the crux, as shown in the coin question. If you specify which card is an Ace, it narrows the possible outcomes.
@@ccmoose1 I understand this now, thanks.
I once gave the first riddle to a friend of mine who was a chemist. And the first thought she made was that the volume decreases as soon as you pour two different liquids together. Therefore the second tablespoon contains more liquids (when you messure the mass) than the first one. And therefore there is more juice in the soda. Since then I always use mass instead of volume when asking this riddle.
The median was not properly represented in the second example. Median is a powerful tool that more people should be aware of. The value difference between median and average can tell a lot about a data set; loan figures are a perfect example for this indeed.
Thinking about it, I think the card/coin thing is actually way more intuitive if you think about what the inverse, the scenarios that you must reject* for the premise to be true. By saying "one of these is a tails", you're essentailly saying "I flipped both coins until either one (or both) of them was tails". By saying "_This_ one is tails", you're essentially saying "I flipped two coins until the left one was tails". In the first scenario, the only time you'd re-flip both coins is if they were both heads. In the second, you'd re-flip both coins if the left one was heads, no matter what the right coin was. So you rejected more outcomes.
Similarly, when you're drawing two cards, when you say "One of these is an ace", you're saying you rejected all the outcomes where you didn't draw an ace. There's 1326 different two-card hands in a standard deck of 52 cards, but only 112 of them contain one or more aces, so you're rejecting 1214 aceless hands. Of the remaining 112, only 12 of them have 2 aces, or just under 11%. When you say "One of these is the ace of spades", you're saying you rejected all the outcomes where you didn't draw _specifically_ the ace of spades. Out of those 1326, there's only 30 that have the ace of spades, so you're rejecting 1296 of them, and of those remaining 30, 6 have two aces, or exactly 20%.
Since there are more non-aces than aces in the deck, rejecting more outcomes means rejecting more one-ace outcomes than two-ace outcomes.
*When I say "reject", I don't necessarily literally mean reject, so much as "it did not happen, because if it had, the puzzle would not be structured this way"
People tend to forget that probability and randomness is strictly linked to how much you know about the process of "randomization". If you knew how a coin is thrown exactly, you could predict its outcome, if you knew parts of it, you could come up with more accurate probabilities of the outcome, ones that make you right more often.. Same thing in the last exemple of the video, depending on what you know, the random experience changes, and the probabilities associated with each issue evolve too, strictly based on your own knowledge of the situation.
The cards are an inverse Monty Hall. Put yourself in the position of the dealer at the six minute mark. There are five combinations of cards, each with a twenty percent chance of being dealt. You choose a suit for the ace that you will tell the player. What is the chance that each combination will be in the possible remaining combinations? The answer is 50% for each, except the pair of aces, for which it is 100%. There are only three combinations left, but that doesn’t mean that each had an equal 1/3 chance of being left. The chance of the pair being dealt was 20% from the outset, and the dealer’s statement about the suit of one of the cards does not change this.
For the first one, you can take the "table spoon" to the extreme max and use a full cup instead.
So basically you would pour 1 cup of cola into the juice, then pour 1 cup of the solution back into the cup that originally contained the cola.
Now it's very obvious that it's equal.
The reason this works is because there's nothing special about the "table spoon", it's just a arbitrary amount, any amount would work
You solve problems kinda how I do: take stuff to the extreme and just assume that it reflects the original question
I don't understand how the probabilities worked for the cards. assuming he's referring to a 52-card deck, shouldn't both of them be 1/17 (roughly 6%)?
in the first example, he asks for the probability that the other card is an ace, given that the first is also an ace. since there are 4 aces in a standard deck, knowing the first card is an ace pulls one from your deck, leaving you with 3 aces in 51 cards, or 1/17.
in the second example, he adds the info that one of the cards is an ace of spades. this shouldn't change anything however, since knowing what the suites of the cards are doesn't change their outcomes. ((As, Ad), (As, Ah), (As, Ac)) has the same size as ((Ad, As), (Ah, As), (Ac, As)), so the outcome should be the same regardless of if the ace of spades is the known card or not.
edit: if he's asking what's the total probability of pulling both cards as aces, it's 4/52 x 3/51 or 1/221, roughly 0.45%. still lost how he got 3% anywhere?
@@summerlovinxx from what I can tell, he used statistics, not logic. We all know how easily statistics can be skewed. This is just another example of that. Then again, it’s possible I might be the m0r0n. 🤷♂️
Was anyone else constantly waiting for a joke or a punchline? Zach just sounds so snarky all the time even when he’s not even trying to be, it’s really hard to take him seriously.
Isn't the last one because you are combining two separate event into one? Like for the coin, for simplicity's sake (mine mainly) I'll do T for tail, H for head, L for left, R for right, I for info (meaning it's the coin you are talking about) and U for unknown (meaning it's the other coin that could be either tail or head). if one of them is tail, you don't really have 1 in 3, you have 2 in four combination, you have RTI + LTU, RTI+LHU, LTI + RTU and LTI + RHU. This also falls perfectly back to the expected probability. Same with the cards, with the simplified example you have 3 event where your ace is spade, 3 where the ace is hear and two of those happen to look the same. which give you 2 in 6 which is equivalent to the expected 1 in three.
Please stop saying "odds" when you mean "probability." Although they are inextricably linked and often yield the same information, they do have important differences. Prob. = likelihood an event occurs; Odds = P(event happens) / P(event doesn't happen). This error is like mixing up standard deviation with variance. Yes, they typically reveal the same info, but an error is an error.
For the third quiz, the problem is that the statement you make is affected by the coin flips. If you pick a random tails whenever there is one flipped up to say this one is tails. When you do pick a coin to say it is tails, the two possibilities T H and T T are not equally likely since in T T case you had a 50% chance to say the other coin was tails making that half as likely and the probability the second coin is tails returns to 1 in 3. Hopefully Zach explains it better than I did.
Your ace of spades question is dependant on what assumptions we make about what you'd have said (and I'd say the most natural assumption means it isn't correct)
It's only if we assume you'll ask about the ace of spaces in advance, or if some other independent process meant we knew you'd ask about the ace of spaces (eg. say I ask you "Is one of the cards the ace of spades?" and you answered "yes"). That way we can assume your card is sampled from the population of 2-card hands that contain the ace of spades.
But if the fact that you asked about the ace of spades was DEPENDENT on the cards you have (which it must be - you wouldn't tell us you had the ace of spades if neither card was it), the answer changes: it's no longer a sample of 2-card hands containing the ace of spades, but the probability is not the dependent one of P(chose to talk about ace of spades | hand).'
Ie. for all I know, if you'd held the ace of hearts and the ace of spades, you'd have been equally likely to phrase the question as "One of them is the ace of hearts, what's the chance the other is an ace?". So of the double-ace hands, I should only expect the question to be about the ace of spades 50% of the time.
That's why the "this one" part (arguably) makes a difference with the 2-coin case. It makes it clear you're basing the question you asked on a coin you picked (even if you don't show it), so have twice the odds to pick heads in a 2-heads situation than in a heads/tails one. In fact, I'd say even without the "this one" part, it runs into the same issue of assuming you'd talk about "one is heads" rather than "one is tails" in advance: it'd be more natural to assume what you ask about depends on what's in your hand (since again, you wouldn't say "One is heads" if they were both tails, so you must be doing **some** kind of examination of the coins to base your question on.)
I watched this earlier today, left the house and did a bunch of errands, gleefully awaiting my return to see the follow up video and ........ I hope you upload it soon! Great content Zach, I follow you here and on the comedy which is my all time favourite comedy.
Problem 3 reminds me of the principle of restricted choice in bridge. When trying to locate the KQ of a suit in the opponents' hands and you see one, (they should play K or Q at random if they hold both), the options are that that opponent started with KQ, Qx or Kx, all equally likely cases. The chance they have the other as well is closer to 1/3.
I did the math out for the first one and was surprised (though I shouldn't have been) that the proportions were just reversed.
Right before doing the math my vague intuition said "the volume of substance A being added to container B is more unlike substance B than the volume of the mixture being added to container A is unlike substance A" ... and "so I expect container B should end up with more substance A than container A ends up with substance B" - but the latter can't be the case because the volumes end up the same with nothing lost or gained.
I did notice that you specified the volume removed was "a tablespoon" but the total volume of each was not specified, but I just abstracted both of those anyway in my solve. If the volume removed is 1/n of the original volume, then the proportion of foreign substance in each container ends up as 1/(n+1).
I also noticed you just said "solution of" without being more explicit about the concentration. (i.e. is it a 10% solution of coke in water (also a crime)?) I solved assuming you meant each was totally pure and also figured that it would not be an interesting dimension of variation on the problem.
The reason that the last puzzle is different is because the event sizes of 'at least 1 coin is tails' and 'this coin (unknown) is tails' (will be interpreted as coin 'i' is tails with 'i' being 1 or 2 uniformly) are different.
at least 1 coint is tail = {TT,HT,TH}
coin i is tails = {TT, TH} or {TT, HT} with 50% each
Pr[case 1]=Pr[TT|at least 1 point is tail] = Pr[TT and {TT,HT,TH}] / Pr[{TT,HT,TH}] = 1/3
Pr[case 2]=Pr[TT|coin i is tails] = Pr[TT and coin i is tails] / Pr[coin i is tails] = Pr[TT] / (1/2) = 1/2
* case 2 is actually a sort of 'expected value' of probabilities between the cases of different 'i's, a different way to calculate this is defining a different probability with i included with the condition Pr[i=1]=0.5 (i saw a lot of people giving more weight to TT, this is pretty similar).
in total, 'at least 1 coin is tails' =/= 'an unknown coin is tails' = 'one of these coins is tails' (notice that there is no use of the word EXACTLY one)
why? because 'at least 1 coin is tails' tells us something about the correlation between the value of the two coins, while saying that 1 of these coins is tails does not.
Card/coin one: when you reveal a particular card, you’re no longer sampling uniformly.
The third one depends on the means by which the information "one of these is tails" was acquired. Since english is vague, "one of these is tails" could mean either that both were checked, and it wasn't the case that both of them were heads, or that one of them was checked and it was a tail (the one being checked could be random, or one predetermined, as long as the one being checked is independent from the outcome of the flip itself).
In one case you can imagine a game where a flipper sits across from a player. The flipper flips two coins behind a screen, looks at them, and if they are both heads she flips them both again until thats not the case. She then says "at least one was a tails. I will give you $18 if they are both Tails, but you must give me $12 if they are not. You can play this game for me as many times as you would like". If the 1/3 case is right, then the expected value is negative (-$2 per game) and the player will lose money in the long run. If the 1/2 case is right, then they will gradually make money (+$3 per game). You can do this yourself or similate it to verify, but what you can imagine is a big table of results of all the flips, including the HHs cases, and of the ones where the player is given the chance to take the bet, 2/3 of those would be losing. The 1/3 case is correct here.
Now you can imagine a case where the flipper flips the two coins, but only decides to look at one of them (eg. By flipping a third coin labelled "left/right"). She then offers you the bet if the one she checked was a tails, otherwise she flips them both again. In this case, there are 8 possibilities LHH, LHT, LTH, LTT, RHH, RHT, RTH, RTT. The bet is only offered in the LTH, LTT, RHT, RTT cases, of which half are TT and the expected value will be a win: the 1/2 case is correct. It should be clear that this is also the case if the flipper always checks the left coin, or chooses the coin by any procedure as long as long as the outcome doesnt influence the checked coin.
For the example at the end, it's 1/2. My assumptions are that you shuffled the coins and randomised their parity, revealed only one of them, and you repeated the take of the video with that procedure until you happened to get tails. That makes it like the second flipper game.
1) Mixed solutions are gonna represent the same fractional proportion
2) Median salary will only depend on the middle person’s salary. You could have any of these scenarios and that still hold; for example:
Median of 1, 4, 2 is 2.
Median of 1, 2, 10000 is still 2
Median of 1, 3, 10000 is 3
Problem 3: Appears to be an analogue of the sister/brother problem wherein we are told that at least one is a girl or something like that. If you used Bayes formula the specification actually increases the odds.
Will check my work now.
Edit: For statistics majors these problems are bread and butter in any undergraduate intro probability course. For the very last problem my initial thought without writing out the work is that this is an analogue of the monty hall class of problems but Im not sure why. My intuition is still that the specification is useful as compared to the base case
One interesting thing I noted is that the probability of the other card being an Ace given that one of them is the Ace of Spades surprisingly is the same as just the probability of the SECOND card being an Ace given that the FIRST is the Ace of Spades (3/51 is approximately 6 percent as well). I don't really get why that would be the case.
I just did the maths, and the first problem indeed reduces down to 3/51, so the maths checks out. Still confused as to why those probabilities should be the same...
This also works if you change the number of cards in the deck since the probability simplifies to
(number aces - 1)/(number cards - 1) assuming you have more than 1 card
A simple way to explain the soda/juice proportion is to note that both glasses have the same smounts of liquid they started with so equal amounts must have been exchanged.
Regarding the cards and the counselling, I feel you may be planning to touch on this more in your next video. But the answer from my perspective is that there is not enough information to answer the question.
In the simplified example with the cards, then were the question 'Of the pairs which contain an ace, what proportion contain two Aces?' or 'Of the pairs which contain the Ace of spades, what proportion contain 2 Aces?', it would be unambiguous. However, when you are given the information by being told, then you can't calculate the odds without knowing what information they would choose to give you with each holding. For both the questions the answer could be anything from 0% to 100% depending on this.
Supposing that they would always tell you of the Ace of hearts in preference to the Ace of spades, then the fact they told you of the Ace of spades would mean the other Ace was not present. Suppose they would always prefer to tell you of a non Ace were it present, the fact they told you of either Ace would mean both must be Aces. Maybe their preference isn't only to tell you of one card at all, so that when they do that reduces the subset.
*coins* not 'counselling', strange autocorrect 😄
I have a different approach for the last problem. Obviously the way you showed it makes sense, but there is another way that can help understand the final question. Basically we have two independent experiments (throwing two coins and seeing if they are heads or tails). Once you tell me the actual result of one of them, what we are basically doing, is reducing the amount of experiments to one (one coin toss with either heads or tails). And everyone knows that a coin toss is 50:50. However, if you just tell me that at least one landed on tails, you didn't actually give me the result of one specific experiment. You just eliminated a certain set of possible results (in this case two heads).
With this idea in mind, we can understand the final question. Assuming the question is again: "What is the probability that both of them are tails?", then yes, it is going to be 50:50, because you are basically just asking me for the probability of one coin landing on tails. Even if I don't know exactly which coin you are talking about.
I missed this type of content from your channel. I LIVE IT . Before i could understand you at all . But now that i'm older i can . It's fascinating
Revealing the suit of the ace or the hand the coin is in doesn’t change the probabilities.
You know that there is at least one ace. Because we don’t know the suit, the guaranteed ace has a 1/4 chance of being spades, a 1/4 chance of being hearts, a 1/4 chance of being clubs, and a 1/4 chance of being diamonds.
If the ace is spades, the chance the other card is an ace is 3/51
If it’s any of the other suits, the chance is still 3/51
So the probability calculation is
1/4*3/51+1/4*3/51+1/4*3/51+1/4*3/51 = 3/51
You have 4 individual sample spaces depending on what the suit of the guaranteed ace is. Each one has equal probability of the 2nd card being another ace, so the suit of the first doesn’t affect the probability.
The coin example is similar, but a little more tricky to wrap your head around. You know that at least one of the hands has a guaranteed tails. There’s a 1/2 chance the guaranteed tails is in the right hand and a 1/2 chance it’s in the left.
If it is in the right hand, there is a 1/2 chance the left hand coin is tails.
If it’s in the left hand, it’s a 1/2 chance the right hand coin is tails.
So the probability calculation is
1/2*1/2+1/2*1/2 = 1/2
In the game of bridge, each player is dealt 13 cards. If we deal out hands until the first player gets a hand with at least one Ace, the probability they have a second Ace is slightly less than 1/2. If we deal out hands until the first player gets a hand with the Ace of Spades, the probability they have a second Ace is slightly greater than 1/2.
For the soda coke exchange: The easier explanation is, that the amount of liquid is the same after ferrying one spoon back and forth, so all the coke that ended up on the other side has been replaced by apple juice. It doesn't even matter if the coke was properly mixed into the apple juice.
Also the amount would have been different if the spoons would have been exchanged simultaneously (but still as much coke in the apple juice as vice versa).
The coins example: It sounds like after asking "Which one is tails" and getting an answer the probability goes up, but that doesn't take into account that the person holding the coins then makes a choice based on the coins held, so the probability actually doesn't change. It's kind of the reverse goat problem
A different question would be "Is the right hand tails?", depending on the answer the probability for the other hand changes to either 1 for "no" or 1/2 for "yes" from 2/3 before.
So it depends how the coins are prepared, either flipping coins until at least one shows tails, then telling where a coin with tails is vs. determining a hand, flipping coins until that hand holds tails, then announcing that.
So I went back and watched your video on the boy/girl problem (from 5 years ago). Currently scrawling on a paper plate at 4am trying to make sense of it.
I found that if you add another slight tweak to the question, the probability continues to increase.
Say the parent said "I have two kids. One of which is a girl, who is named either Julie OR Natalie." Seeing that it'd be perfectly logical to have two girls, one named Julie and the other Natalie, you add more possible outcomes, and thus more outcomes in which both children are girls.
I drew up a punnett square and Watched the probability slowly climb, as if it were approaching 100%.
I'd love to hear/see you talk about this or at least tell me where I slipped up to get the ever increasing percentage.
Sorry this was so long!!
1:40 There's a way easier way to view this.
As long as the volume of the two is the same, the amount of juice in one *must* equal the amount of soda in the other. It doesn't matter how you stir it or rearrange it.
Great point! The stirring actually didn't matter :o
#3 is just updating our priors, isn't it? It's like Monty Hall kinda
For the probability question, I think that we mistake the quantity of probability as a fixed quantity for a specific event. The whole idea is to be able to predict the future with respect to the number of possibilities we can narrow the future down to. In the cards or coin game, the additional "information" about the card or which hand has the tails helps us narrow down the sample space and thus increase the probability. However, without the information about whether it is the left/right hand we cannot do so. If we would have considered the arrangement to be immaterial i.e. considered HT to be equivalent to TH, then knowing which hand has it, would NOT have made any difference, as it would not have allowed us to reduce our sample space(the denominator in the ratio). However, considering different arrangements as SEPARATE events, as we do here, without knowing which hand has the tails won't help us to reduce the sample space any further. It is because the arrangements being "separately" considered gives vital importance to the knowledge of WHICH hand contains the tails. The reason we find this unintuitive is because we subconsciously think that probability must be definite for a specific situation and/or don't consider the knowledge of whether the ace is a spade or whether the right/left hand has the tails as "additional information". But they ARE additional information in this case, as they help us reduce our sample space and the number of favourable events, in turn affecting the ratio i.e. the probability.
At 5:10 you say that the probability of one card being an ace given that the other one is, is about 3% and it doesn't make sense to me. Given that we remove one ace from the deck, we've got 3 aces left in a deck of 51 cards, so 3/51 is 5.8%, so almosy 6 and definitely not 3%. How did he come up with 3?
I'm glad you were able to take time out of your skit acting career to learn a little engineering stuff!
7:45 The way i make sense of it is, when you specify one of them is tail then you essentially locked in the result for one of them, and only the results of the other one could vary. But if you only say one of them is tail it could be one or the other, none of the two has to be a tail
For the later one when we didn't see which one gets revealed. We could just bruteforce the situation by saying that there are x% chance that the left one is revealed, (1-x)% its the other one. And since in both of these situation its 50% chance that both are tail. It is still 50%
I absolutely LOVE when you do this counter intuitive stuff! Can't wait for the next video!
The boy-girl vids are one of my all-time-fav YT videos.
There's a simpler solution to the first problem. Since you didn't specify the amount of soda in the glasses, it must not matter. Therefore just look at what happens if each contains one tablespoon. Then the first move puts the entire glass of soda into the juice for a 50/50 mixture. Moving half of that back just fills it with the same 50/50 mixture.
I feel smart that I instantly realised that the third one was wrong and why
The first two were fairly obvious with a little thought, although the 3rd has some bad math in it. - specifically in how you look at the odds, since you're not considering how interchangeable certain probabilities are IE, when you say "I have an ace in my hand" - we know it's one of two cards, and that is very important information, and it changes our dataset so that we're not looking at the original 6 possibilities.
First "I'm holding an ace in a 4 card set, what are the chances I'm holding a second ace?"
We can simplify the math here because we have redundant information. We have either Ace of Hearts, or Ace of Spades, it doesn't matter which ace we have, so when we have an ace of spades and a 2 of diamonds, or a ace of hearts and the two of diamonds, it means the exact same thing to us statistically, so we can combine them as one possibility - in Schrodinger's cat fashion, we can call this A of H/S - and likewise, the other ace we don't know about is A of S/H
By simplifying the probabilities in this way, we can look at the possible probabilities/possibilities like this:
A of H/S + 2 of Diamonds
A of H/S + 7 of clubs, or
A of H/S + A of S/H
(This of course could be expanded to:
A of S + 2 of Diamonds
A of S + 7 of clubs, or
A of S + A of H
A of H + 2 of Diamonds
A of H + 7 of clubs, or
A of H + A of S)
^Note how this compares to the set of '5' in the video. A of H + A of S is listed twice - A certain 'break in logic' in the video, there are TWO aces we could draw from, the ace of hearts, or spades, yet only one combination of the two aces is shown in the video.
(To further clarify, the possibilities was essentially changed from "What are the ways we can combine 4 different cards" to "What are the possible ways we can combine 3 cards with 2 different Aces" - it's just misleading because one of the aces, is also one of the 3 cards being combined.
To explain this in an intuitive way, assume you covered up the 'suit' of the aces with stickers or something. After you pick up the first ace, you know you have an ace, but not which one. For your second draw there are 3 cards left, what are the odds you'll draw another ace? 1 in 3, correct! Aren't you smart. See, math is always intuitive when you look at it the correct way.
In other words, the chance of having two spades DOES NOT change by knowing which ace it is. This is misleading math.
The coin example on the other hand doesn't have such a blatant error - when he talks about 'does it matter which one is tails', it's a topic on how independent the flips are.
If you flip just one coin, it's 50:50, if you flip another coin, that coin is also 50:50, however if you flip two coins together, you have a 1/2 chance to get a head and a tails, and a 25% to either get two heads or two tails:
HT
TH
HH
TT
So when you say "I have a tails" you're merely eliminating the two heads possibility.
HT
TH
TT
However, the moment you identify which hand has the tails, you make the other flip independent from other, IE "This coin is a tails; so what is the probability that this other coin was a tails?" 50:50 naturally. (Think of it like a % slider. We know that we have AT LEAST 1 tails, so if my left coin is a tails, that one is 100% tails, and the other is 50:50. However, if the right coin is a tails, then its 50:50 and 100% in the other direction. If we move the slider to the middle, it's 2/3s chance either side is tails: BUT there's also clarity that there is a 0% chance for both to be heads. In other words, among the two coins, there is 4/3rds chance of a tails, and 2/3rds chance of a heads.)
The change in probability in the coins game doesn't come from the reveal, but from the selection:
I flipped the coins until at least one was Heads and in this case it is the "shows side" one: 33%
I flipped the coins until the side i chose beforehand turned up Heads and its the "shows side" one: 50%
In the first case we flipped and had to reflip at 1 of the 4 cases, leaving us with 3 end results (1 of them is HH)
In the second case we flipped and had to reflip at 2 of the 4 cases, leaving us with 2 end results (1 of them is HH)
Same with the cards, if I chose two cards until atleast one of them is "an Ace" it doesn't matter if I tell you the suit.
But if I chose two cards until one of them is the "Ace of Spades", then thats a different probability.
Selection process matters, telling you the suit doesnt. Its all about the difference in cases you need to reject and reflip.
For even more magical effect, you can repeat both experiments, but not tell the other and just write it down or even just think it.
Based on the selection process of your flips the probabilities will reflect which process you used even without you telling the suit/side.
I would assume the last one is because of unequal probability between the possible hands/coins. Just because your possible combinations is now 1/3 instead of 1/6 or whatever it was, doesn't mean that all 3 are equally likely.
This is pretty fascinating. It reminds me of the old question: "If a tree falls in the forest and there's no one to hear it, does it make a sound?" I wonder if the answer changes if a deaf person watches it happen...?
Regarding the ace and coin problem, I feel like it depends on what you are observing. If you are looking for the combination of cards then sure, the possibility changes. But you're tasked with figuring out how likely it is that one of the cards is a particular one, so I feel like it makes sense to look at what card the mistery card could be almost in a vacuum.
It's simpler with coins. Sure, you could have TH, HT or TT, but in the end it doesn't matter because coin #2 only has two possible states it can be and it is either of those states independently of coin #1, so it's always a 50/50
The probability changes when a counterfactual possibility is eliminated. If you have a choice about what to reveal, how that choice is made matters.
In the Aces case, if you were always going to reveal the suit of your Ace, or choose randomly in case you had both, then we eliminate ½ of the cases where you have both Aces (because you could have chosen to reveal the other suit instead). The probability you have both Aces becomes ½ out of 2½ which is still 1 in 5.
This is different than the case where you reveal the suit of one of your Aces, but spades has priority over hearts. I.e. you'll always say Ace of Spades when you have both. Now I can't eliminate any probability from the pair of Aces scenario, and the probability you have both Aces is 1/3.
If instead the Ace of Hearts had priority, then when you say "Ace of Spades" I know for certain you don’t have both Aces.
Another situation: I ask you "Do you have the Ace of Spades"? Now your choice is eliminated, and you might say yes and you might say no. If you say yes, the probability goes up to 1/3, and if you say no, the probability goes down to 0.
In the coin paradox, it again matters how the choice is made. If you were always planning to reveal a tails, picking randomly in case you had both, the probability ends up not changing. If you were always going to reveal your right hand, and it happened to be tails, the probability does change, because you might have said it's heads.
The problem is underspecified. But, since we have no information on how you choose which suit to reveal, or which hand to reveal, we have to assume you chose randomly in the cases where you had a choice.
If we change the original Monty Hall problem by saying after you pick a door, the host lets you choose another door to reveal. If you open the other door and it reveals a goat, now it really doesn't matter whether you switch or not.
Conditional probability can be weird and unintuitive. Another way of thinking about this is the principle that information must be discounted when the choice of what information us revealed is under the control of an adversary.
These paradoxes tend to reduce to a limited space with weird probabilities specifically because you get your information from someone else who already knows everything. It's like the monty hall problem
The card problem is like this because when he has two aces and says "I have an ace", it can either mean he is talking about the spade ace, or the heart ace - he is confessing to having one of two different cases, even though it only sounds like one "he has two aces" case.
So the probability at 6:30 is not 1/5, but 2/6 = 1/3.
With that first one, it doesn’t matter what ratio of juice to soda you pull out of the mixed cup, assuming the volume transferred between containers is exactly the same, the ratio in the final solutions will always be exqctly proportionate.
The last problem comes from asking for an orderless quantity of objects in an ordered set. It simply doesn't make sense to do unless you disregard duplicate quantities with different orders in the set