- Видео 6
- Просмотров 51 561
YATAQi
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Добавлен 24 янв 2024
Mathematical puzzles & problems done visually
The Four Dice Puzzle
Four fair six-sided dice are rolled. What is the probability that they can be divided into two pairs which sum to the same value?
Problem Source: Utah Math Olympiad (2016 P2)
Chapters:
0:00 Intro
0:37 Pause
1:01 Solution: What's the Sample Space?
2:38 Solution: Case 1
3:00 Solution: Case 2
3:19 Solution: Case 3
4:15 Solution: Case 4
5:34 Solution: Case 5
7:07 Outro
Difficulty: 5/10
__________________________________________________________________
Animations were done using 3Blue1Brown's animation engine; Manim.
Problem Source: Utah Math Olympiad (2016 P2)
Chapters:
0:00 Intro
0:37 Pause
1:01 Solution: What's the Sample Space?
2:38 Solution: Case 1
3:00 Solution: Case 2
3:19 Solution: Case 3
4:15 Solution: Case 4
5:34 Solution: Case 5
7:07 Outro
Difficulty: 5/10
__________________________________________________________________
Animations were done using 3Blue1Brown's animation engine; Manim.
Просмотров: 6 453
Видео
Does This Polish Puzzle Have a Solution?
Просмотров 22 тыс.14 дней назад
In this quick Polish puzzle, we'll try to prove whether it's possible (or impossible) to cover a 9x9 square grid using only tiles of size 1x5 and 1x6. Problem Source: Letters of Polish Junior Mathematical Olympics, September 2014. Chapters: 0:00 Intro 0:34 Pause 0:58 Equation Attempt 2:31 Solution 5:09 Outro Difficulty: 3/10 Animations were done using 3Blue1Brown's animation engine; Manim.
What's the BEST Full House in Poker?
Просмотров 34 тыс.Месяц назад
Can you figure out what full house in poker maximizes your chances of winning a round? For this problem, we're assuming: • A standard 52-card deck. • Each player draws 5 distinct cards (no shared cards). Chapters: 00:00 Intro 01:22 Pause 01:46 Solution Difficulty: 4/10 Animations were done using 3Blue1Brown's animation engine; Manim.
What's the Value of This Angle?
Просмотров 8462 месяца назад
In this video, we take a look at how we can find the angle inside a decagon that connects two different vertices. Chapters: 0:00 Intro 0:13 Pause 0:38 Solution Difficulty: 2/10 Animations were done using 3Blue1Brown's animation engine; Manim.
How Logic Can be More Useful for Hat Puzzles
Просмотров 8843 месяца назад
Can you use some logic to figure out what color your hat is in this quick puzzle? Difficulty: 3/10 Chapters: 0:00 Intro 1:56 Pause 2:23 Solution Animations were done using 3Blue1Brown's animation engine; Manim.
This Math Puzzle is Easier Than it Looks
Просмотров 7943 месяца назад
In this video, we go through a quick puzzle where we have to fill up the circles with integers given a few restrictions. Can you solve it? Chapters: 0:00 Intro 1:01 Pause 1:26 Solution Difficulty: 1/10 Animations were done using 3Blue1Brown's animation engine; Manim.
Nice animations. Are you using manim?
i’d like to argue that the full house is the 3rd best hand in the game and that a straight flush is the 1st because a royal flush is really just a straight flush
Brute Force
Good explanation and cool animations. Take my sub!
Something went wrong. There is 36 combinations for rolling 2 dice, with 11 possibilities. The ones in question and their probabilities are: 2 and 12 (1/36) 3 and 11 (2/36) 4 and 10 (3/36) 5 and 9 (4/36) 6 and 8 (5/36) 7 (6/36) Because there is only one possibility for the extreme combinations (2 and 12), two for the second extreme ones (3 and 11) and incrementing by one untill we get to the middle combination. (7, the most probable one.) Now multiply the probabilities for themselves to get the chances for two pairs to combine. We get: 2 and 12 (1/1296) 3 and 11 (4/1296) 4 and 10 (9/1296) 5 and 9 (16/1296) 6 and 8 (25/1296) 7 (36/1296) Then we add everything to get (146/1296). Now we multiply this by the possibilities in pairing dice. We have 3: AB and CD, AC and BD, AD and BC. Permutations with the pairs and inside the pairs don't count. This will give (438/1296). This is not the same result. Did I miss something?
There are a few cases that can be overcounted. Like 2, 2, 3, 3 for example - you get a + c = b + d = 5 and a + d = b + c = 5 so it is counted twice. After subtracting these you get the answer as in the video.
I don't understand why you divided by 2! to avoid double-counting the (a, a, b, c) permutations where the "a"s are swapped, but you didn't do that for any of the other partitions or the sample space as a whole.
Short answer is because it was quicker / more obvious through other counting methods. For example, case 3 is (a, a, b, b), and he says "that's 6, which we verify with 4 choose 2" (I'm paraphrasing). You also get 6 by doing (4!/2!)/2!, which is more like what was done for the (a, a, b, c) case.
In general, to count permutations of 4 objects where you have x a's, y b's, z c's, and t d's, you do 4!/(x!y!z!t!),but in many cases those counts will be 1! or 0! in the denominator, and so can be omitted freely.
Shower-thoughts gave me a rough estimate of ~30% - now let's actually watch the video and see how wrong my approach and/or estimation really are ... Okay, the estimation wasn't too far off actually, but my approach was. I assumed stochastical independence where it probably wasn't there. Still fun exercise :)
alot of potential to be a great math youtuber
love the minimalist animation and careful but succinct explanation!
really great video, fun and concise yet complete!
Thanks for making this video I enjoyed it as lot. It was just the right level for me, I could understand the puzzle, I tried to solve it but couldn’t, then I could follow and understand your answer, so I learnt something new :)
Very nice
At the start of the video when you say the question, I paused the video and gave it a few seconds. Then without doing math, I said to myself, “it feels close to 1/4. Imagine my surprise at the end of the video.
I tried to figure it out and got 762/1296, or 127/216, or about 0.588. I'm probably way off.
Yep, I was wrong. I think it was because I did every possible way it can be split into two equal groups, without accounting for the fact that some permutations can be split in multiple ways.
1:57, you say to let a, b, c, & d be distinct integers but that means they must all be different from one another and we clearly don’t have that condition here.
At 1:28 you state that the sample space is the 1,296 ordered 4-tuples of possible dice rolls. But at 1:48 you mention partitioning the set of ordered 4-tuples. That’s not the same thing.
You can define an equivalence relation on the ordered tuples to make them equivalent iff they're the same up to order. This partitions the ordered tuples, and if you then partition the un ordered representatives of those equivalence classes, that is also a partition of the underlying ordered tuples, so it's logically sound. Only thing to remember is to do the combinatorics bit at the end to "rehydrate" those classes and it's all okay. Perhaps more concisely: if you partition the un ordered tuples, you also get a partition on the ordered tuples
This is the most well researched and well produced poker video that I hsve ever seen that is also UTTERLY and funnily pointless 😂😂😅
I’m not sure how to feel about this comment.. but it made me laugh. So thank you I guess haha
Great video
Technically a royal flush does not really exist. It is just a straight flush. Otherwise there would be a royal straight as well. This makes a full house the third best hand.osinh only to 4 of a kind and a straight flush.
Nice video. I would request if you need music, just leave it for bumpers. Not going through the whole video.
POLSKA🔥🔥🔥
GOATed
Any outcome of the form (x,x,x,x) is good: 1st+3rd=2x=2nd+4th. There are 6 such outcomes. Outcomes (x,x,x,y), x≠y and permutations are bad: no matter how you pair the dice, you have a 2x pair and an x+y pair, and 2x≠x+y. (x,x,y,y), x≠y and permutations are good: 1st+3rd=x+y=2nd+4th. There are 6×3×5=90 such outcomes: 6 values for 1st, 3 locations for its twin, 5 values for the other pair. (x,x,y,z), x≠y, y>z, z≠x and permutations are good iff y+z=2x (because x+y≠x+z). There are 6 such outcomes, not including permuations: 2231, 3351, 3342, 4462, 4453, 5564. Including permutations, there are 6×4×3=72 such outcomes: 4 locations for y, 3 locations for z. (x,y,z,w), x>y>z>w and permutations, are good iff x+w=y+z. There are 7 such outcomes, not including permuations: 4321, 5421, 5432, 6521, 6431, 6532, 6543. Including permutations, there are 7×4!=168 outcomes. 6+90+72+168=336 good outcomes, out of total 6⁴=1296. The probability is 336/1296=7/27=0.(259).
I accidentally counted wrong for a,a,b,c test... Need to be careful with heuristics... I thought there were 10 ways but there are only 6 because of course you cannot roll 2.5 or 3.5 as the a... Other than that yeah, fairly standard combinatorics problem
Can someone explain why we are dealing with permutations? Like if we get 5,5,3,3, since each die is the same and there is no rules on how those pairs are chosen, we can rearrange them as 5,3-3,5. Moreover, since we are doing sommes (english not 1st language) 5+3=3+5 so the order in which we arrange each pair doesn't matter either. Like i don't understand how 5533 / 5353 / 5335 / 3535 / ect. makes a difference...
The order of the numbers is important because we are considering the PROBABILITY of a particular event when rolling 4 dice. The order of the numbers correspond to the outcomes for each of the 4 dice. The first number represents the outcome of the first die, the 2nd number represents the outcome of the 2nd die and so on. There are a total of 6 x 6 x 6 x 6 ordered sets of four numbers. This is the number of equally likely outcomes when rolling 4 dice and includes all possible permutations. There is only one way for the outcome 2222 to be rolled. Rolling three 2s and a 6 can be achieved in 4 different ways: 2226, 2262, 2622, 6222. Rolling two 3s and two 5s can be achieved in 6 different ways: 3355, 3535, 3553, 5335, 5353, 5533 Thus, rolling two 3s and two 5s is 6 times more likely than rolling four 2s. If you ignore the order of the numbers, you are not taking into account the number of different ways of achieving that particular outcome.
@Paxed10 makes sense, thanks!
I think he used permutations because each one has the same likelihood of being rolled and that makes the problem much simpler than if you try to do it working with combinations.
Can someone explain why aabb case has 90 cases? I mean, there are 6 ways yo pick a and 5 ways to pick b, then 4!/2!2! ways to shuffle them. Multiplying everything gives 5*6*6=180. Or am i missing smth?
You double-counted the ways to pick a and b
You are basically "mixing" two ways of counting together. If you want to count the permutations of (a,a,b,b), the order does not matter, and you have 6 choose 2 options = 15. If you want to count the available options for each spot (hence the order matters), you have indeed 6*1*5*1 = 30 outcomes. But since you already established that order matters, taking the permutations would count some cases again. Instead, in this scenario (and without loss of generality), consider the different ordered "pairs" and count the different outcomes: (a,a,b,b) gives 30 different outcomes (6*1*5*1) (a,b,a,b) gives 30 different outcomes (6*5*1*1) (a,b,b,a) gives 30 different outcomes (6*5*1*1) hence a total of 90! Please keep in mind that since we took order into consideration, the letters are not as significant (meaning that's why we don't swap a's and b's) as much as the spot whose available options we are counting.
@@Mrjimibanez okay i see, thanks for the explanation
Nice vid. The best full house would not be the one with lower chance of a higher point: if you got a full house and the other does not have anything, its value is a small blind. The best full house is the one granting the opponent the highest chance of a point _just below_ your. What is the full house granting you the highest chance of a Flush _that is not a Straight Flush?_ This could also be a small variation to the present exercise: you have assessed that the best FH is AAAXX, with X being any number between 6 and 9, any rank. You could work on this degree of freedom now, to assess the best ranks to have the opponent get a non-straight flush. Is it better to have both the X of the same rank of the As, or the missing one?
The math test will be easy guys. The test:
3 brown 1 blue
Here’s a related (but mathematically unrelated) question. What are the odds for each player winning the game assuming they only answer with 100% certainty. Additionally, if they can only make 1 guess, what strategy can each player use to maximise their own chances of winning?
Here's my try at the last method: Let W_n be the ways of taking a and b so that a + b = n. W_2 = 1 since 1 + 1 = 2 W_3 = 2 since 1 + 2 = 2 + 1 = 3 W_4 = 3 since 1 + 3 = 2 + 2 = 3 + 1 = 4 It’s not difficult to find that the pattern for W_n is 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1. The number of ways to pick a, b, c, d so that a + b = c + d = n is (W_n)^2 since, you may first use one of W_n ways to pick (a, b) and then independently use one of W_n ways to pick (c, d). Summing over all values of n, the number of ways to pick a, b, c, d so that a + b = c + d = n for any n would be 2*(1^2 + 2^2 + 3^2 + 4^2 + 5^2) + 6^2. Calculating that gives us 146. The final answer consists of 3 ways to divide the sum (I: a + b, II: a + c, III a + d) so 146*3 = 438 would be the answer. However this would overcount all cases in the form of (a, a, b, b) and (a, a, a, a). (a, a, b, b) cases have been counted twice, there are 90 such cases (as in the video), so we must deduct 90 from the sum, (a, a, a, a) cases have been counted 3 times each, there are 6 such cases, so we must deduct 2*6 from the sum. Finally 438 - 90 - 12 = 336, giving us the answer. Personally I find this answer very nice, and it’s somewhat generalizable for more dice faces too. But I still find it tricky to separate out the overcounted cases, so I’d be happy to see if there’s any simpler way to do things.
I did it the same way! Thanks for writing out the formalization so I didn't have to 😅
The board game “Can’t Stop” is literally this, a game where you roll four red dice and split them into two pairs. Maybe that was the question setter’s inspiration.🙂
i'm only a minute in but i'm absolutely charmed by the animation and editing. this is astonishingly high quality!
Super cool! I do this as a pass-time on road trips with license plate numbers instead of dice.
I only watched 2 minutes. I think I know where it's going and if right I'm proud to say I've been saying this for 40 years. Kings over aces because you need two aces preventing other players from having three aces
interesting! good watch, take my like
Nhubusbsygaggvhgs=hgshbhuzbuhsjnisijsuhbhusvghbhushsjishhshgvghdbhushubsnhiuuxhudhuuhsbyyshxh is orangutan
You may argue that having 3 aces is not as good as having 9's over 2's because with the aces still out there it increases your opponents odds of having a good 2 pair or set which would increase the likelihood of you making more money on the hand. If you have 10's over 2's and your opponent has a set of aces you are likely to win a big pot but if you have aces over 6's and your opponent has a set of 2's they are not as likely to put a bunch of there chips in.
Nice video
dumb video. the best full house is AAAKK, saved you 12 minutes of yapping
Nice video. I think that you should've done the calculation on the full house of kings to show that it's worse than aces, and blocking more straight flushes doesn't make up for allowing full houses of aces to beat you.
have you used any external tool other than manim? aniamtions are great. and is the source code on github.
Disappointed you did not cover KKKAA, which is also an unbeatable full house. But otherwise good video.
Just in time.
Definitely need more videos on texas holdem
fun video, i'm witnessing the birth of a channel right here,, i know you'll get more subs in no time
If you have Aces full and the opponent has a str flush then u are done for. The best you can hope for is that they check all the way.
este video es una maravilla
What ruleset are you basing this off? Everywhere I’ve ever played and anywhere i looked online it says that the pair does matter for hand ranking?
Once you have three aces, no other player can, so your full house is already better than any other full house they might have.
The concept that's really being covered here are what we refer to as "blockers" or "removal" cards in a hand. These help to narrow an opponent's range of hands he'd be willing to play based on what's on the board and possible hand combinations.