Monty Hall Problem is what you're thinking about, and yes it's always best to switch door if asked, you have a much greater chance of getting the prize.
You have 2 doors the prize is behind one a 50/50. If you agree to this you agree it's not 2/3. Problem solving is hard for some people. Edit: someone explained it with 100 doors and now makes sense.
Why though? First your chance is 33%. Then after the reveal there is a chance of 50% that your door is right. Just like the other door you could switch to. It's 50/50. Edit: No need to answer, i'm just gonna google it. There probably are thousands of explanations anyways
@@bobdole3926 Imagine the prize is behind the door to the right and you can travel back in time. 1st try: You pick the left door. Host reveals middle door. You stay with your pick: FAIL. 2nd try: You pick the middle door: Host reveals left door. You stay with your pick: FAIL. 3rd try: You pick the right door. Host reveals left or middle door. You stay with your pick: SUCCESS. Staying with your pick only has a 1/3 chance of winning the prize. Changing your mind has a 2/3 chance. I already commented this somewhere else but my brother is teaching maths and said this is a good explanation so...
That thing you were talking about at the start. I watched a video on that explaining it, and statistically your odds are best if you always plan to swap. Because you have a 2/3 chance to pick wrong first guess, when they reveal an option that was incorrect and offer a swap, there is a 2/3 chance you gave up your incorrect choice. Always swaping only results in a loss in that situation when you picked the 1/3 correct choice first
The odds are not static. Odds and percentages are not some thing that describes the choice or the object. It describes how we perceive the likelihood of certain outcomes in the cases of uncertainty. If you find out about the new information, the odds change. If you have 3 identical doors and 1 of 3 doors is "good", while not having other information, you initially have 33% chance to be correct with random guess. But the thing is - unless the result (content behind the door) changes depending on your choice, the outcome behind each door should be predetermined. So, you will always (100%) have "good" result behind one certain door, and always (0%) - behind 2 other doors. Odds - is just our prediction, because we don't know which door is good or bad. Our choice or swapping of our choice does not change the content of the door. So, back to the revealing and swapping. You have 33% chance to be correct. Somebody reveals that other door is incorrect. So, now your odds have risen to 50%, no matter if you change your choice or not. Because now you have one door 100% "bad", and the "good" content is behind one of 2 remaining doors. No matter if you change or not, if you don't have any other information about the remaining doors and your choice is random, you will have 50% odds. Even if you have 1000 doors and thus initially predict that your chances of correct pick are 0,1%. When you make a first pick, and somebody reveals you that 998 of other doors are "bad" - now you have 1/2 chance to be correct, again - no matter if you change your pick or not. EDIT: After learning about this issue, I've found that it is indeed statistically more beneficial to switch the doors... Which at first look (and even second look for most people) does not make sense. I apologize. It is still hard matter for me to understand intuitively.
@@Kyle_VanMan I think A lot of people really over complicate it, and it is a simple as if picked wrong, you'll win and if you picked correct, you'll lose
About the Monty hole problem: it's easier to understand if you refer to the switching door part as flipping your decision: if you started with the right door (33% for that to happen) then you'll switch to an empty door. if you started with a wrong door (66% for that to happen) then you'll switch to the right door. so because there's 66% to pick a wrong door, and because switching doors equals flipping the result then you'll have 66% of winning.
Odds are not intrinsic to situation, and are not static. They change when you get new information. If your initial odds were 33% to be correct, but you find out that one of the other doors was incorrect - now you have 2 doors, one of which is correct. If you don't have any other information about the situation, and your choice is/was random - you now have 50% chance to be correct, no matter if you change or not. Most of the paradoxes - are simply incorrect interpretation of laws of our universe, or using same words in different situations with different meaning. I was explaining this stuff to my sister with the task she was given. There was possibility to shoot the target with some odds (30/50/90%, I think). And you had to calculate probability to hit the target certain amount (or minimum = at least some amount) of times in 10 attempts. And she won't get that if you miss/hit 3 in a row, you still have 30/50/90% chance to miss 4th time. The same chance. It does not change because of your previous choices or their probabilities or their results. Because different attempts are independent of each other. Just like opening the doors. If you look at it like picking one of 2 remaining doors (instead of thinking about previous choice), you will clearly see that it is independent and now 50%. If it is still hard - imagine the doors are identical in look. And somebody covers them behind a wall and shuffles. Now you have no idea what door you was picking before. Is it still 33%/67%? No, it is 50/50. Odds - are simply our way to predict which choice is more likely to give us better outcome in most cases. In this case, it does not matter what exactly % odds are - we know they are equal (if doors are identical and we have no other information, thus making our choice blindly and randomly). So, we know that: - doors 1 and 2 have equal chances. - doors 2 and 3 have equal chances. - doors 1 and 3 have equal chances. When we find out that we no longer have door 3, we now have only doors 1 and 2. And we know that they have equal chances.
@@drakewarrior1013 that's just wrong for the exact reason i explained in my comment. if you have a programmer friend ask him to simulate success rates for always switching and see for youself it's around 66%
@@amitir22 you know, after thinking and reading about this problem again, I'm not so sure about my logic here anymore. Well, I mean, my logic is correct (if you treat couple doors at the end like independent situation and pick randomly, you will have 50% chance of success). Like in the example of picking new door blindly, not knowing which one was picked at the start. But maybe there is actually sound logic in using previous knowledge here too. It is just my brain refuses to treat these situations (starting and after the reveal) as linked. I will have to work on that. Also, I saw a spreadsheet of all possible combinations, results and outcomes - that could be more accurate than just simulation too. And it really has 67% chance of success... I'm baffled, honestly. Never logic seemed so weird to me before. Or, well - almost never.
@@drakewarrior1013 "at the end like independent situation" - lemme stop you right there 😅. it's not an independent situation. please re-read my comment. and treat the switching the doors part as flipping the result. because always: 1. when you pick an empty door and switch - you get the prize door. 2. when you pick the prize door and switch - you get an empty door. now it all comes down to the probability to pick each door when you have 3 doors in front of you. -> 66% for option 1. -> 33% for option 2.
Monty hall problem is an example of inherited odds. One door has a prize and one does not, yes. But it's not 50/50 because the door you picked was not 50/50 when you picked it, and a losing door will always be opened. If it breaks your brain, as it does for a lot of people, imagine 100 doors instead. You pick one. 98 losing doors open. One door is now a loser and one is the winner. Is it still 50/50? Your odds of picking the right door was 1%. You will win 99% of the time by switching in this scenario.
Emm... No matter how many doors was opened, their odds of being correct are now 0%. All of the doors are equal in their chances, right? That means that doors 1 and 2 have equal chances. If you open doors 3-100, that does not change the fact that you pick the door randomly. No matter %, 33% or 1% initially, after the reveal you know that doors 3-100 now have 0% chance of being correct doors. And 100% chance of having good door is now equally spread across doors 1 and 2.
@@drakewarrior1013 I don't understand what you mean. Here is what happens: - You pick one door. - Every door *except* the door you picked and one other door is opened, never revealing a winning door. The only way to win without switching is to guess correctly. Alternatively, your odds are reversed by switching, as the only way to lose is to initially guess the correct door. The initial odds are what matter. Opening doors does not change the odds. Imagine that doors are not opened and the host says "how confident are you? If you'd like, I'll let you win if you lose but lose if you win" because that's basically what this boils down to with some extra showmanship.
With the change the door problem, there is a 2 out of 3 chance that you picked one of the two wrong doors. This means that if they reveal a wrong door after you have picked a door, there is a 2 in 3 chance that they only had one option to choose to reveal a wrong door, as you already picked the other wrong door in 2 out of the 3 potential scenarios. That means in 2 out of the 3 scenarios, the remaining door is the correct door.
Monty Hall problem explained: Out of 3 doors, if there is a 33% chance that your randomly selected door is the correct door, then there must be a 66% chance that the correct door will instead be among the remaining two. Imagine that you are told, "If the correct door is among the remaining doors, you will switch to the correct door, otherwise, you will switch to an incorrect door." TLDR: The key to the puzzle is the condition, "If the correct door is among the remaining doors." We understand that there is a 66% probability that among two unchosen doors, one will be correct. When we are guaranteed that correct doors are *preferred* by the switch, the switch inherits the full 66% probability of the two doors combined. The confusing part of this puzzle is that you are wont to reassess there being 1 correct and 1 incorrect door, among these two doors if chosen randomly you would have a 50% chance of having the correct door I.e. there is no difference in changing. This would be correct if the condition instead stated, "One incorrect door has been eliminated. Which of the remaining doors would you prefer?" In the case that your selected door is revealed to be incorrect, the logic still holds that there was a 33% chance of that door being correct to begin with, and that there was a 66% chance of the remaining doors to contain a correct door. That 66% is then split among the two remaining doors because we have no information discriminating one door from another. If another door is revealed, we still know that our own door, if incorrect, had a 50% chance of being revealed AND that the other unselected door, if incorrect, had a 50% chance of being revealed. See that we do not have information discriminating the selected door from the unselected door in either situation. However, in the Monty Hall case, a revealed incorrect door *must* be an unselected door. Our selecting a door has an impact on which door is revealed. There is no 50% chance that our door was going to be revealed. Or, "Allow me to reveal an unselected door at random, then you may switch." In this case, it would have to be possible to the host to reveal a correct door. This would, however, be silly. He must reveal an incorrect door. We know that there is a 66% chance that one of the unselected doors is the correct door, and we know that the revealed door is NOT the correct door, so we know there is a 66% chance that the *unselected door* that *is not revealed* is the correct door.
"I've seen the video, I'm aware that it's 66%, but between you and me, it's 50%". The best part of the video and it's not even connected to the game itself.
Monty Hall problem: Let's say we switch the order of operations. You pick a door, then you can choose to switch it for BOTH of the other doors, and then the not-it door of those two get revealed. It's really the same thing in effect.
Look, let me explain it really simply: At the start, all 3 doors have 1/3 chance and you pick a door. Then, your choice has 1/3 chance, the other 2 doors have 2/3 chance. Then, the host removes one LOSE door. Obviously your choice still has 1/3 chance. What about the 2 other doors? They had 2/3 together. We know the host removed a LOSE door, so that 2/3 can only apply to the single remaining door. Your door has 1/3, the other door has 2/3.
No. Once you get new information, odds change. Odds are not some static value that gets assigned to your choices. They are simply descriptions of your predictions that some outcomes will happen. You know that they choices have equal chances of success, right? 1 and 2 and 3 (and 4,5,6,etc. - if you have more). No matter how many LOSE doors you reveal, chances of the doors are still equal. 2 "other doors" do not have 66% between them. Once you have one of them revealed, the revealed one is either 100% or 0% (in case it is "bad" door, it is 0%). You recalculate all of the odds after that. All doors together have 100% chance that one of them is "good", right? If they are equal, then you have 1/3, 1/3, 1/3 chances (roughly 33%). When you find out which one of them is "bad", you know it is 0% now, and other doors have 100% combined. In reality, only one of them has 100% chance to be "good", and all others have 0% chance. But if you don't know which is which, you make a prediction and assign the odds. But that does not chance the content behind the doors. Again - it is just description of an agent that tries to make a decision.
More than just looking at the math for the Monty Hall problem, you can find simulators online to find out for yourself if it's actually 50/50. If you can't be convinced by the math, surely seeing it play out as many times as you feel like should help a little. Spoiler: It's not 50/50.
Wtii If you change you have a 66% chance og being right where as you only have a 33% of being right on your first original Pick Therefore you allways change #montyhallproblem
You are right, Richard Changing the door doesn’t make any difference. When the empty door is reviled you get zero additional information about the door you selected. Number of doors does’t matter. Changing the door increase your chances to win only in case the information from host affect your choice directly, and there are only two possible options: 1# you chose the door, and then a host reveal another door and there is a prize. It means that there is no prize behind the door you selected and you obviously need to choose the door that host just opened #2 you choose the door and the host opens the door you just chosen and there is no prize. Obviously you need to change the door in this case. If the host does not open the prize door or the do you chosen, you get no extra information that can affect your chances of opening the prize door. It is not math, it is common sense Revealing the empty door gives you same amount of information as if the host told you that the sky is blue.
Let's say we switch the order of operations. You pick a door, then you can choose to switch it for BOTH of the other doors, and then the not-it door of those two(there must be one) get revealed. It's really the same thing in effect.
@@Fleghm88 it does not work this way. Let’s look at your case with 1000 doors. Imagine there is a 1000 doors, but you are considering only between two of them. You choose the door and then host open all 998 door except two that you was struggling between. How does it affect your choice in the first place? It is the same if he opened this 998 doors before you made your mind what of the rest 2 doors to open. The host does not give you extra information unless he reveal the door that you chosen or the door that has the prize.
The reason the Monty Hall problem is called a problem is specifically because the mathematics behind it defy common sense. It it completely counter-intuitive, yet under the calculations the correct choice is the one that doesn't obviously make sense. Part of the reason behind the confusion is that people keep on forgetting that there are specific conditions on which door the host picks: it cannot be the door which you chose and it cannot be the door that holds the prize. Therefore, the host can only choose a 'wrong' door.
@@weakhunter13 the fact that host can only open the wrong door means that his action is an informational noise and has no affect on your choice. You did not pick his “100% wrong door” in the first place, so it does not affect your choice.
@@Fleghm88 ok let’s imagine the win number from 1 to 1000 is 986 Imagine this two following conversations between us: //DIALOG 1: Me: I pick 442. You: It is not 74. Do you want to change your mind? //DIALOG 2: Me: I pick 442. You: WTii is best you tuber ever. Do you change your mind? Do you get the point? If you tell me what numbers from 1 to 1000 are wrong, you don’t affect my chances to pick the win number by changing my current pick In other words I have to repick 442 only if you tell me that it is a wrong number, or if you tell me that 986 is the right number. All other information from you can not make a repick reasonable
0:00 Round 1
1:58 Round 2
4:48 Round 3
7:12 Round 4
8:36 Round 5
11:30 Round 6
13:26 Round 7
15:15 Round 8
19:04 Round 9
21:09 Round 10
25:16 Scores
This was the epic underdog tail of rAndom. Being last plase whole game and won because of bets!
Monty Hall Problem is what you're thinking about, and yes it's always best to switch door if asked, you have a much greater chance of getting the prize.
You have 2 doors the prize is behind one a 50/50. If you agree to this you agree it's not 2/3. Problem solving is hard for some people. Edit: someone explained it with 100 doors and now makes sense.
Why though?
First your chance is 33%.
Then after the reveal there is a chance of 50% that your door is right.
Just like the other door you could switch to. It's 50/50.
Edit: No need to answer, i'm just gonna google it. There probably are thousands of explanations anyways
@@bobdole3926
Imagine the prize is behind the door to the right and you can travel back in time.
1st try: You pick the left door. Host reveals middle door. You stay with your pick: FAIL.
2nd try: You pick the middle door: Host reveals left door. You stay with your pick: FAIL.
3rd try: You pick the right door. Host reveals left or middle door. You stay with your pick: SUCCESS.
Staying with your pick only has a 1/3 chance of winning the prize.
Changing your mind has a 2/3 chance.
I already commented this somewhere else but my brother is teaching maths and said this is a good explanation so...
@@TechnoSwenk Thank you I was incorrect I used 100 doors to make it clearer.
I love these mini game maps! Great video, that gambling addiction certainly got you, Wtii!
That thing you were talking about at the start. I watched a video on that explaining it, and statistically your odds are best if you always plan to swap. Because you have a 2/3 chance to pick wrong first guess, when they reveal an option that was incorrect and offer a swap, there is a 2/3 chance you gave up your incorrect choice. Always swaping only results in a loss in that situation when you picked the 1/3 correct choice first
The odds are not static. Odds and percentages are not some thing that describes the choice or the object. It describes how we perceive the likelihood of certain outcomes in the cases of uncertainty.
If you find out about the new information, the odds change. If you have 3 identical doors and 1 of 3 doors is "good", while not having other information, you initially have 33% chance to be correct with random guess. But the thing is - unless the result (content behind the door) changes depending on your choice, the outcome behind each door should be predetermined. So, you will always (100%) have "good" result behind one certain door, and always (0%) - behind 2 other doors.
Odds - is just our prediction, because we don't know which door is good or bad. Our choice or swapping of our choice does not change the content of the door.
So, back to the revealing and swapping. You have 33% chance to be correct. Somebody reveals that other door is incorrect. So, now your odds have risen to 50%, no matter if you change your choice or not. Because now you have one door 100% "bad", and the "good" content is behind one of 2 remaining doors. No matter if you change or not, if you don't have any other information about the remaining doors and your choice is random, you will have 50% odds.
Even if you have 1000 doors and thus initially predict that your chances of correct pick are 0,1%. When you make a first pick, and somebody reveals you that 998 of other doors are "bad" - now you have 1/2 chance to be correct, again - no matter if you change your pick or not.
EDIT: After learning about this issue, I've found that it is indeed statistically more beneficial to switch the doors... Which at first look (and even second look for most people) does not make sense.
I apologize. It is still hard matter for me to understand intuitively.
This is the only explanation I’ve ever read that actually makes sense to me. Thx Ashley.
@@Kyle_VanMan I think A lot of people really over complicate it, and it is a simple as if picked wrong, you'll win and if you picked correct, you'll lose
The russian roulette part was simply the best. Total skill and psychological warfare.
About the Monty hole problem:
it's easier to understand if you refer to the switching door part as flipping your decision:
if you started with the right door (33% for that to happen) then you'll switch to an empty door.
if you started with a wrong door (66% for that to happen) then you'll switch to the right door.
so because there's 66% to pick a wrong door, and because switching doors equals flipping the result then you'll have 66% of winning.
Odds are not intrinsic to situation, and are not static. They change when you get new information.
If your initial odds were 33% to be correct, but you find out that one of the other doors was incorrect - now you have 2 doors, one of which is correct.
If you don't have any other information about the situation, and your choice is/was random - you now have 50% chance to be correct, no matter if you change or not.
Most of the paradoxes - are simply incorrect interpretation of laws of our universe, or using same words in different situations with different meaning.
I was explaining this stuff to my sister with the task she was given. There was possibility to shoot the target with some odds (30/50/90%, I think). And you had to calculate probability to hit the target certain amount (or minimum = at least some amount) of times in 10 attempts. And she won't get that if you miss/hit 3 in a row, you still have 30/50/90% chance to miss 4th time. The same chance. It does not change because of your previous choices or their probabilities or their results. Because different attempts are independent of each other.
Just like opening the doors. If you look at it like picking one of 2 remaining doors (instead of thinking about previous choice), you will clearly see that it is independent and now 50%.
If it is still hard - imagine the doors are identical in look. And somebody covers them behind a wall and shuffles. Now you have no idea what door you was picking before. Is it still 33%/67%? No, it is 50/50.
Odds - are simply our way to predict which choice is more likely to give us better outcome in most cases. In this case, it does not matter what exactly % odds are - we know they are equal (if doors are identical and we have no other information, thus making our choice blindly and randomly). So, we know that:
- doors 1 and 2 have equal chances.
- doors 2 and 3 have equal chances.
- doors 1 and 3 have equal chances.
When we find out that we no longer have door 3, we now have only doors 1 and 2. And we know that they have equal chances.
@@drakewarrior1013 wait, you mean the probability in the end is equal between switching and not switching?
@@drakewarrior1013 that's just wrong for the exact reason i explained in my comment.
if you have a programmer friend ask him to simulate success rates for always switching and see for youself it's around 66%
@@amitir22 you know, after thinking and reading about this problem again, I'm not so sure about my logic here anymore. Well, I mean, my logic is correct (if you treat couple doors at the end like independent situation and pick randomly, you will have 50% chance of success). Like in the example of picking new door blindly, not knowing which one was picked at the start.
But maybe there is actually sound logic in using previous knowledge here too. It is just my brain refuses to treat these situations (starting and after the reveal) as linked. I will have to work on that.
Also, I saw a spreadsheet of all possible combinations, results and outcomes - that could be more accurate than just simulation too. And it really has 67% chance of success...
I'm baffled, honestly. Never logic seemed so weird to me before. Or, well - almost never.
@@drakewarrior1013 "at the end like independent situation" - lemme stop you right there 😅.
it's not an independent situation.
please re-read my comment.
and treat the switching the doors part as flipping the result.
because always:
1. when you pick an empty door and switch - you get the prize door.
2. when you pick the prize door and switch - you get an empty door.
now it all comes down to the probability to pick each door when you have 3 doors in front of you.
-> 66% for option 1.
-> 33% for option 2.
Monty hall problem is an example of inherited odds. One door has a prize and one does not, yes. But it's not 50/50 because the door you picked was not 50/50 when you picked it, and a losing door will always be opened.
If it breaks your brain, as it does for a lot of people, imagine 100 doors instead. You pick one. 98 losing doors open. One door is now a loser and one is the winner. Is it still 50/50? Your odds of picking the right door was 1%. You will win 99% of the time by switching in this scenario.
Emm... No matter how many doors was opened, their odds of being correct are now 0%. All of the doors are equal in their chances, right? That means that doors 1 and 2 have equal chances. If you open doors 3-100, that does not change the fact that you pick the door randomly. No matter %, 33% or 1% initially, after the reveal you know that doors 3-100 now have 0% chance of being correct doors. And 100% chance of having good door is now equally spread across doors 1 and 2.
@@drakewarrior1013 I don't understand what you mean.
Here is what happens:
- You pick one door.
- Every door *except* the door you picked and one other door is opened, never revealing a winning door.
The only way to win without switching is to guess correctly.
Alternatively, your odds are reversed by switching, as the only way to lose is to initially guess the correct door.
The initial odds are what matter. Opening doors does not change the odds.
Imagine that doors are not opened and the host says "how confident are you? If you'd like, I'll let you win if you lose but lose if you win" because that's basically what this boils down to with some extra showmanship.
There were no Wtii Snake Impressions this episode.
Current drystreak: 9
-loan
Another amazing video, as expected from Wtii.
With the change the door problem, there is a 2 out of 3 chance that you picked one of the two wrong doors. This means that if they reveal a wrong door after you have picked a door, there is a 2 in 3 chance that they only had one option to choose to reveal a wrong door, as you already picked the other wrong door in 2 out of the 3 potential scenarios. That means in 2 out of the 3 scenarios, the remaining door is the correct door.
Monty Hall problem explained:
Out of 3 doors, if there is a 33% chance that your randomly selected door is the correct door, then there must be a 66% chance that the correct door will instead be among the remaining two.
Imagine that you are told, "If the correct door is among the remaining doors, you will switch to the correct door, otherwise, you will switch to an incorrect door."
TLDR: The key to the puzzle is the condition, "If the correct door is among the remaining doors." We understand that there is a 66% probability that among two unchosen doors, one will be correct. When we are guaranteed that correct doors are *preferred* by the switch, the switch inherits the full 66% probability of the two doors combined.
The confusing part of this puzzle is that you are wont to reassess there being 1 correct and 1 incorrect door, among these two doors if chosen randomly you would have a 50% chance of having the correct door I.e. there is no difference in changing.
This would be correct if the condition instead stated,
"One incorrect door has been eliminated. Which of the remaining doors would you prefer?"
In the case that your selected door is revealed to be incorrect, the logic still holds that there was a 33% chance of that door being correct to begin with, and that there was a 66% chance of the remaining doors to contain a correct door. That 66% is then split among the two remaining doors because we have no information discriminating one door from another. If another door is revealed, we still know that our own door, if incorrect, had a 50% chance of being revealed AND that the other unselected door, if incorrect, had a 50% chance of being revealed. See that we do not have information discriminating the selected door from the unselected door in either situation. However, in the Monty Hall case, a revealed incorrect door *must* be an unselected door. Our selecting a door has an impact on which door is revealed. There is no 50% chance that our door was going to be revealed.
Or,
"Allow me to reveal an unselected door at random, then you may switch."
In this case, it would have to be possible to the host to reveal a correct door. This would, however, be silly. He must reveal an incorrect door. We know that there is a 66% chance that one of the unselected doors is the correct door, and we know that the revealed door is NOT the correct door, so we know there is a 66% chance that the *unselected door* that *is not revealed* is the correct door.
This custom looks awesome. Like an upgraded Uther Party? Nice one Wtii!
"I've seen the video, I'm aware that it's 66%, but between you and me, it's 50%".
The best part of the video and it's not even connected to the game itself.
I'm glad so many people are explaining the Monty Hall Problem! I was about to right my own explanation, but I don't feel the need to do so anymore. :)
This episode is pure gold
What I learned from this: Never bet on OC.
Based on my observation the only speed that really matters in the race, is the avg. speed.
@witi for the racer mini game you need to look at AVG speed and max speed (mainly AVG speed)
Monty Hall problem: Let's say we switch the order of operations. You pick a door, then you can choose to switch it for BOTH of the other doors, and then the not-it door of those two get revealed. It's really the same thing in effect.
OMG THATS HILARIOUS 17:30 😱😭😂
I cant sleep watching "are you lucker" lol🤣
You can simulate Monty Hall, still worth it to switch, you get exactly the same as the math proves
I never expect wtii to win these because he always goes all in on the betting
"2 + 2 = 4"
WTii: "It's bullshit! I disagree!"
7:53 how did you still alive XD
Any Luckers? 👀
0:57 monty hall problem
15:55 this shit made me laugh more than i thought it would :D :D god dammit
Strife TD though?
Look, let me explain it really simply:
At the start, all 3 doors have 1/3 chance and you pick a door.
Then, your choice has 1/3 chance, the other 2 doors have 2/3 chance.
Then, the host removes one LOSE door. Obviously your choice still has 1/3 chance. What about the 2 other doors? They had 2/3 together. We know the host removed a LOSE door, so that 2/3 can only apply to the single remaining door. Your door has 1/3, the other door has 2/3.
No. Once you get new information, odds change. Odds are not some static value that gets assigned to your choices. They are simply descriptions of your predictions that some outcomes will happen.
You know that they choices have equal chances of success, right? 1 and 2 and 3 (and 4,5,6,etc. - if you have more).
No matter how many LOSE doors you reveal, chances of the doors are still equal.
2 "other doors" do not have 66% between them. Once you have one of them revealed, the revealed one is either 100% or 0% (in case it is "bad" door, it is 0%).
You recalculate all of the odds after that. All doors together have 100% chance that one of them is "good", right?
If they are equal, then you have 1/3, 1/3, 1/3 chances (roughly 33%).
When you find out which one of them is "bad", you know it is 0% now, and other doors have 100% combined. In reality, only one of them has 100% chance to be "good", and all others have 0% chance.
But if you don't know which is which, you make a prediction and assign the odds. But that does not chance the content behind the doors.
Again - it is just description of an agent that tries to make a decision.
very good vid
More than just looking at the math for the Monty Hall problem, you can find simulators online to find out for yourself if it's actually 50/50. If you can't be convinced by the math, surely seeing it play out as many times as you feel like should help a little.
Spoiler: It's not 50/50.
Dude, its always 50% chance, it either the prize, or not
Wtii
If you change you have a 66% chance og being right where as you only have a 33% of being right on your first original Pick
Therefore you allways change
#montyhallproblem
Why isn't there canned bread
it was so fucking obvious that turtle was gonna win man, just look at the stats smh.
Gg
You are right, Richard
Changing the door doesn’t make any difference.
When the empty door is reviled you get zero additional information about the door you selected.
Number of doors does’t matter.
Changing the door increase your chances to win only in case the information from host affect your choice directly, and there are only two possible options:
1# you chose the door, and then a host reveal another door and there is a prize. It means that there is no prize behind the door you selected and you obviously need to choose the door that host just opened
#2 you choose the door and the host opens the door you just chosen and there is no prize. Obviously you need to change the door in this case.
If the host does not open the prize door or the do you chosen, you get no extra information that can affect your chances of opening the prize door.
It is not math, it is common sense
Revealing the empty door gives you same amount of information as if the host told you that the sky is blue.
Let's say we switch the order of operations. You pick a door, then you can choose to switch it for BOTH of the other doors, and then the not-it door of those two(there must be one) get revealed. It's really the same thing in effect.
@@Fleghm88 it does not work this way. Let’s look at your case with 1000 doors. Imagine there is a 1000 doors, but you are considering only between two of them. You choose the door and then host open all 998 door except two that you was struggling between. How does it affect your choice in the first place?
It is the same if he opened this 998 doors before you made your mind what of the rest 2 doors to open.
The host does not give you extra information unless he reveal the door that you chosen or the door that has the prize.
The reason the Monty Hall problem is called a problem is specifically because the mathematics behind it defy common sense. It it completely counter-intuitive, yet under the calculations the correct choice is the one that doesn't obviously make sense.
Part of the reason behind the confusion is that people keep on forgetting that there are specific conditions on which door the host picks: it cannot be the door which you chose and it cannot be the door that holds the prize. Therefore, the host can only choose a 'wrong' door.
@@weakhunter13 the fact that host can only open the wrong door means that his action is an informational noise and has no affect on your choice. You did not pick his “100% wrong door” in the first place, so it does not affect your choice.
@@Fleghm88 ok let’s imagine the win number from 1 to 1000 is 986
Imagine this two following conversations between us:
//DIALOG 1:
Me: I pick 442.
You: It is not 74. Do you want to change your mind?
//DIALOG 2:
Me: I pick 442.
You: WTii is best you tuber ever. Do you change your mind?
Do you get the point? If you tell me what numbers from 1 to 1000 are wrong, you don’t affect my chances to pick the win number by changing my current pick
In other words I have to repick 442 only if you tell me that it is a wrong number, or if you tell me that 986 is the right number.
All other information from you can not make a repick reasonable
Hahaha he sucks at betting
Heh
Oc banned cz of gawld
i love this game mode, totally video content