The answers are so surprising you won't believe them, even after seeing the calculations

[@MindYourDecisions]

I received a nice email from Atanas who has developed code in 3 languages if you want to simulate this result yourself! Check it out on Github: github.com/batinkov/simulations/tree/main/2_puzzles_about_shuffling_cards

[@icarus877]

This is utter gibberish and relies on an understanding of what the term shuffle means. Shuffle means to randomly change the order of cards, in fact the whole idea of shuffling a deck is to not understand how the deck may be shuffled. The question is - if you shuffle a standard deck of cards, what is the expected number of cards that stay in the same spot. The question is false - it should be what number could you expect to stay in the same spot. I can shuffle a deck of cards and 100% guarantee that not one card will be in the same spot, this can be done without intention. I can also appear to shuffle a standard deck of cards and 100% ensure every card is returned to the same spot, this I would have to do with intention and skill.

[@zzmej1987]

The first calculation stops to seem wrong, if you consider that 2 of clubs can be opened before the first ace, and ace of spades can't.

[@kevinhughes8577]

That was my thought as well, but I think actually the solution presented is correct. After ordering the 51 cards randomly, it doesn't actually matter if the ace of spades ends up before the ace that otherwise would be first. Bottom line is, there's 52 places to put it, and only one of them yields the result that "the card following the first ace is the ace of spades". That 1/52 possibility remains true for every arrangement of the other 51 cards.

[@MichaelOnines]

I think that is what balances out the possibility the first ace could be the ace of spades.

[@Zebsy]

Yes I think he has a logic error in the first solution, as when you're considering inserting an ace, that card itself could become the 'first ace' - that is not the case when inserting the two of clubs Edit: Just watched it again and I see what he means - I now think he's right!

[@nickronca1562]

This is how I did it before watching the video: given that both cards are still left the probability of either card appearing next us just as likely. Meaning we only need to determine which card if any is more likely to remain. In order for the A♠️ to remain the first Ace that appears can not be a spades. The probability of that is 3 in 4. In order for the 2♣️ to remain, an Ace must appear before then. In other words, the Two of Clubs can not appear before any of 5 options and the probability that it won't appear first out of 5 options is 4 in 5. So indeed the 2♣️ is more likely to appear next then the A♠️.

[@BeefinOut]

I think it's one of those probability situations where, if you take a complex analysis and multiply all the probabilities through, you end up getting the same answer you would have gotten had you done the simple math. In this case, I think the odds of drawing the two in question before reaching an ace are the same as the odds of the first ace being the ace in question, so, even in that case, the odds are equal.

[@d95mback]

Another way to understand the first problem is that for sure, the ace of spades might be the first ace, but when it isn't, it's _guaranteed_ not to come before the first ace, increasing the probability that it's the card after. The same guarantee is not available for the two of clubs, and these two cancel each other out.

[@matthewedwards9423]

This is an excellent way to explain this apparent paradox in simple language. Well done.

[@Nikioko]

General consideration: The first ace it the nth card. This gives us four possible scenarios: 1. Ace of spades is the nth card and 2 of clubs is among the first n-1 cards: Both ace of spades and 2 of clubs have a chance of 0 to come up next. 2. Ace of spades is the nth card and 2 of clubs in not among the first n-1 cards: Ace of spades has a chance of 0, and 2 of clubs has a chance of 1/(52-n) to come up next. 3. Ace of spades is not the nth card and 2 of clubs is among the first n-1 cards: Ace of spades has a chance of 1/(52-n), and 2 of clubs has a chance of 0 to come up next. 4. Ace of spades is not the nth card and 2 of clubs is not among the first n-1 cards: Both ace of spades and 2 of clubs have a chance of 1/(52-n) to come up next. If you now eliminate case 1 and 4, in which both cards have equal possibilities to show up next, you see that case 2 and 3 have the same odds, but swapped. And thus, the overall chances for both cards coming next after the first ace are equally likely.

[@PaulDean125]

This is the reasoning I like.

[@K1C2]

I like this, but for a given value of n, case 2 and 3 are not equally likely. At the extreme, if n = 49, the probability of case 2 is 0. Somehow, over all possible n's it must average out, but that's less intuitive to me than the original question.

[@Nikioko]

@@K1C2 Yes, for n=49, cases 2 and 4 are impossible. But likewise, for n=1, cases 1 and 3 are impossible. Which averages out again.

[@markphc99]

Flawless logic !

[@Chalisque]

Well said. This is the proper argument.

[@real-webbe]

I struggled to understand the solution to problem 1 - until you demonstrated with a deck of 3 cards, which makes it a lot easier to understand intuitively. I thought problem 2 was easier, but still an interesting result.

[@kaskade7416]

Here's the better way to think about the first problem. Let's replace "next card" with "card at the bottom of the deck". Clearly this problem is equivalent, as once you draw the first ace, the remaining cards are going to be uniformly distributed. So now the problem reads: Wait until you draw the first ace, then look at the card at the bottom of the deck. Is it more likely to be ace of spades or two of clubs? Ok but you are always going to run into an ace at some point before reaching the bottom since there are four of them. So you might as well remove that part as it gives you no information (if you really want to be super precise, conditioning on an event of probability one doesn't change your probability measure). So now the question is just: Look at the card at the bottom of the deck. Is it more likely to be ace of spades or two of clubs? The answer to this is obvious.

[@YCLP]

Best intuition for the first answer I found so far!

[@YOM2_UB]

Another way to prove that the problems are equivalent: Take any two unique shuffles where the ace of spades is at the bottom of the deck. In these two shuffles, it must be the case that the card after the first ace is different from in the other shuffle, or the the other (n-2) cards are in a different order than the other shuffle. Otherwise they would not be unique shuffles. Now swap the ace of spades into the position after the first ace in both shuffles. Now it must be the case that the card at the bottom of the deck is different in the two shuffles, or the (n-2) other cards (which were unchanged) are still in different orders. Therefore these are now two unique shuffles where the ace of spades is the card after the first ace. By the same argument, the reverse is also true. Starting with two unique shuffles where the ace of spades is the card after the first ace and swapping them with the bottom card gives two unique shuffles where the ace of spades is at the bottom of the deck. Since both of these can start with any arbitrary unique shuffles, it must be that there are the same number of unique shuffles where the ace of spades is at the bottom of the deck as there are where the ace of spades is the card after the first ace.

[@kaskade7416]

​@@YOM2_UB if you look at the rest of the sentence, you'll see I do justify it. However if you'd like me to fill in the details for you: Let C be a r.v representing the position of some fixed card like the ace of spades or two of clubs or whatever. Let A be the r.v representing the position of the first ace. We want to show that for any c and a with c > a, the conditional distribution of C is uniform amongst possible c. But this is clearly the case, since we have: P(C=c | C > A, A = a) = P(C=c | C > a) proportional by Bayes to P(C > a | C = c)P(C = c) = P(C = c) which is uniform in c. This is just symbolically saying the same thing as before, that if you start with a uniform distribution then it stays uniform after conditioning.

[@YOM2_UB]

@@kaskade7416 Sorry, didn't mean to imply that you had no justification. Sloppy wording on my part for sure.

[@edsimnett]

Not sure I agree. If I tell you I am going to remove a random ace from the pack, then isn't it obvious the 2C is more likely to be at bottom of the pack than the AS (because there is a 25% chance the AS has been removed)?

[@MichaelPiz]

2:16 - Having learned what factorial is about 50 years ago, I still cannot help reading it as if it's very excited to be here: "A deck of fifty-two cards can be arranged in FIFTY-TWO ways!"

[@NovaDenizen]

I think the argument for the second problem is wrong because it assumes that the probabilities of each card being in its original place are independent, when in fact they are positively covariant with each other. That is, if you know the first is the right place, that makes the rest of them more likely to be right.

[@DuncanBooth]

If you had a deck of two cards then after shuffling you can expect one card to be in the right position but the one thing you can be absolutely certain is that there isn't one card in the right position.

[@hlee4248]

I was thinking about the two-card case too. Isn't this disapprove Q2?

[@karhukivi]

With a random shuffle of two cards, surely it is 50:50 that either card will be in its original position? They are either back in the same order or they are in reverse order.

[@keith6706]

Either both cards are in the right position, or neither are. Which means the average is 1.

[@DuncanBooth]

@@hlee4248 Expectation is 1 card, probability is 50:50 zero or two. So no, it matches the answer but it's a bit surprising that the expected value has P=0

[@karhukivi]

Average of what? The expected probability is a sum of all the expected probabilities.

[@koalaman18]

I also got the first question wrong, but your example with 3 cards was helpful in getting me to visualise the problem and the intuitive reason why the probabilities are equal. Yes, there is a chance that the ace of spades is the first ace, thereby giving it a 0% probability of being the next card after the first ace in that event. However, if the ace of spades is not the first ace, then we know it must come after the first ace, which narrows down the number of cards it could be. For example, if the first ace is the 12th card overall, then any of the remaining 40 cards could be the ace of spades - a probability of 1/40, which is higher than 1/52. So there is a higher probability in these cases. Overall, these probabilities average out to 1/52, exactly the same as the 2 of clubs. I did get the second question right, but do think it is a very interesting result!

[@stevezagieboylo9172]

I was absolutely blown away by the first one. I see that you're right, but my intuition fights it the whole way. For the second one, the answer seemed so obvious to me that I thought it must be wrong.

[@leostvincent8430]

Looks to me like the important line missing from the "Intuitively" list for problem 1 is "If the 2 of clubs is drawn before the first Ace, then only the ace of spades can appear (after the first ace)"

[@YouTubeallowedmynametobestolen]

So there is a higher chance of the next card being the ace of spades, right?

[@jacemandt]

The intiution-building example of a 3-card deck is helpful, but also tricky because it actually does depend on the contents of the deck: in a 3-card deck that contains *no other aces*, the probability of the 2 being right after the 1st ace is still 1/3, but the probability of the lone ace being right after the 1st ace is 0. So I'm not sure this example is as convincing (to everyone) as you said, since it *does* depend on the makeup of the deck.

[@keith6706]

But that goes against the implied assumption that both options have a possibility of happening with the deck. If the ace of spades is the only ace, obviously one of the two options is not possible, so the question is meaningless. Hell, just say neither of the cards are in the deck, and it's obvious there's no possibility of either option.

[@DqwertyC]

While the math still works with 3 cards, I think a more useful miniature 'deck' would be 4-cards: The Ace and Two of Spades and Clubs.

[@kcolloran]

@@DqwertyC The problem is listing all the possiblities becomes much harder when you have to do 4!. 24 options is so many more possibilities to write out than 6.

[@Chris-hf2sl]

The problem with only using 3 cards is that there is an unrealistic number of aces in the 'pack'. He had 2 aces, but had there only been 1, the result would have been very different. Neither of these represent the situation with a full deck. Using 4 cards with two aces would be more convincing.

[@dustinbachstein3729]

Puzzle 2 was easy, but puzzle 1 was very fun and challenging to me because Ididn't have your elegant solution in my head. Very nice, thanks! :) 1:44 Here's a concrete counter-argument why the "intuitive" solution is wrong: The error is in case 1. "If the first ace is not the ace of spades, then either card is equally likely." This is wrong, because the 2 of clubs might already be out, while the ace of spades is certain to be still in the remaining deck. So case 1 favors the ace of spades, while case 2 obviously favors the 2 of clubs, and now it's no longer counter-intuitive that both are overall equally likely :)

[@edsimnett]

but why are the two cases the same? Note also that there is a 3 in 52 chance that you draw a non-spade ace as the first card, so you have a third branch to the tree (I think).

[@pchartmann]

I'm not clear why it is not 49! instead of 51! (2:50) since there is no way to have the first ace after the 49th card since there are 3 aces other than the ace of spades in the deck. If the 49th card is the 1st ace, then the 50th and 51st card must also be aces.

[@Schadock_Magpie]

In the 51! way to arrange all cards but the ace of spade, there are 6*48! in witch the last 4 card will be the aces, and the ace of spade will be in 49th position. we still need to count them, and I think you got stuff mixed up.

[@steluhututui5502]

Something if off on problem 2. Let n=2 you can't expect to get the same position after each shuffle. Maybe something like 1-(1-(1/n) )^n if you're to assume that each event is independent of each other. It's not really independent event. One outcome will influence the others.

[@chap0120]

Theres a big flaw here in the use of the word "shuffle." @MindYourDecisions uses that to mean "randomly order" but in common parlance, it means "riffle" (i.e., divide the deck in two and zip the two halves together). A bunch of sequential riffle shuffles will eventually randomize the deck, but the way this is phrased allows for considerable ambiguity. It's clear what's intendee once you watch the solution, but the problem alone allows for the significant probability of the first two cards remaining unmoved because of how most people use the word "shuffle."

[@michaelhallock1428]

I had the same thought. In this case, "shuffle" means shuffle sufficiently to achieve complete randomization.

[@powerofk]

Which can only be done via computer RNG calculations. Even then, it depends on the quality of the randomized @@michaelhallock1428.

[@SmileyEmoji42]

This is about mathematical problems not real world. You can assume a single riffle shuffle if you like but, in the spirit of the channel, it must be a perfect, mathematical riffle, i.e. alternate cards. In this case exactly one card will have the same position and it must be either the top or the bottom card.

[@PsychoSavager289]

I'm poor at statistics, so I wrote both problems as Python programs and ran them both 1 million times. For the first, I got chance of Two of Clubs next as 1.924 % (about 1 in 52) and the Ace of Spades as 1.933% (also about 1 in 52). For the second problem, I got the chance of a card being in the same position as 1.925% (also, as you guessed, about 1 in 52). I don't know if the simulated result differences are significant, but they seem to match the mathematical calculations.

[@scaninteractive7259]

I think one of the reasons that people have trouble with this is because they think about the likelihood of picking AN ace versus A two. In those cases because one ace has been removed from the deck the likelihood is lower. Picking any other ace is 3 in 51 where as picking any two is 4 in 51. Making is more likely you would get a two rather than an ace. Right?

[@wsandell]

For question 2 you can expect 1 card to be in the same position *on average*, right? It could be none, or more than one, but after multiple trials it will average out to one. As presented it made it sound like you should expect it to be exactly one each time.

[@3GOD]

Correct.

[@jamessanchez3032]

If there were 4 cards, there would be 24 possible combinations. Only 8 of those 24 combinations would have exactly 1 card where it was before. But, indeed, sometimes there would be more than 1 card in the "right" place (and sometimes fewer), so it averages to 1.

[@darbyl3872]

Zero is the most common result, which averages to 1 when you occasionally get 2, 3, 4....

[@danmerget]

Presh is using the word "expected" in the mathematical sense. In mathematics, the "expected value" of a random variable is the sum of each possible value times the probability that the value will occur. For example, if you have a box with 3 red balls and 2 white balls, and you take two balls from the box at random, then you have the following odds of drawing each combination of colors: 30% - two red balls 60% - one red and one white 10% - two white balls Given the above probabilities, the "expected number" of red balls you drew is (2 * 30%) + (1 * 60%) + (0 * 10%), which adds up to 1.2. So you "expect" to draw 1.2 red balls. It's obviously impossible to draw exactly 1.2 red balls, but the intuitive idea is what you stated: if you do the experiment many times, it'll tend to average to 1.2 red balls.

[@geoffstrickler]

Using a “perfect shuffle”, you can expect 0 or 2 to be in their original positions, depending on whether the original “bottom” card ends up on the bottom (in which the original top card will be on top, total of 2. If the original bottom card isn’t on the bottom, then you can expect 0. 1 is the average, you can only actually have exactly 1 if you’re not using the perfect shuffle. Most people can’t reliably perform a perfect shuffle, but many magicians and “card mechanics” have mastered it.

[@snowfloofcathug]

Edit: I was confused about it averaging out to 1/52 when the latest position it could be would be 50th for the ace and 49th for the two of clubs, it could never be 1/52 on its own. But I wrote a Python program and after running it a few times with 52000 shuffled decks it does indeed hover around 1000 for both each time

[@MichaelOnines]

Try the n=5 arrangement of a deck comprised of 4 aces and a 2 of clubs. There are 24 arrangements out of 120 where the 2 of clubs is the card immediately after the first ace, and 24 arrangements where the ace of spades is the card immediately after the first ace. Thinking on it further you might also want to look at the n=5 deck with 3 aces, a 2 of clubs and another card that is not an ace.

[@mse326]

The key thing is where the first Ace is doesn't matter. There are 52! ways to order cards. If you remove a card (Ace of spades or 2 of clubs) the remaining 51 cards have 51! was of being ordered. Everyone of them has a 1st Ace regardless of where it is, and the removed card can be inserted immediately after that. The card will have restrictions in where in the order that can be (it can never be the 1st because there needs to be an A before it and as you said there are some at the backend) but that is just accounting for the other ways the cards in a full deck can be ordered that don't have the card following the 1st A

[@jimmyh2137]

The card can be in 52 positions. Only one of those will satisfy the requirement (card being after the first ace) Lastest position to satisfy the answer is indeed the 50th because "the first ace" can't be in position 50-51-52 but that doesn't matter because there will always be a "first ace" in some position (1-49) and you always have only 1 out of 52 possibile positions (52 cards = 52 positions, only 1 of those is good) to make the thing true. For example position 1 or position 52 will always be wrong. But you still have 1 position out of 52 possibilities to make it right. You can think of it like this: 51 "aces" and a "2". The only correct solution is obviously "position 2", but you still have 52 places to put the card in (51 always wrong, 1 always right). Still 1/52

[@keith6706]

As I mention below, it's a bit of a trick question, because the question can be rephrased more simply as "What are the odds that card A is in position X in a shuffled deck of N cards?" Clearly position X can be specified *almost* any way you want: first card, tenth card after you turn over a 3, fifth from the bottom of the deck, it doesn't matter. The odds of card A being in that position are always 1/N, which means the odds of any other specific card being in that position is also 1/N. The one thing that changes that is if the card you are looking for is in a position relative to that same card: obviously, a card can't be in a position different than the one it is in, just as "the card after the last card in the deck" is a meaningless statement. So "position X" is restricted to "positions in a deck that can actually exist".

[@iamthefrankleton]

Monte Carlo simulation is a beautiful tool.

[@taflo1981]

If P1 were to ask about the probabilities of AoS and 2oC appearing *anywhere* after the first ace (not necessarily directly after), the intuition would be correct, because the probability would be 3/4 for AoS and 4/5 for 2oC. But the probability for AoS is independent from the position of the first ace, while the probability for 2oC decreases the later the position of the first ace is. Simultaneously, the probability that some card that we know to appear after the first ace actually appears immediately after increases with the position of the first ace. This creates a bias in favour of AoS, and "magically" that bias exactly nullifies the discrepancy between the 3/4 and 4/5 from the beginning.

[@andydaniels6363]

The second result is nifty, but I don’t know that I’d call it surprising. Having an expected value of 1 doesn’t quite mean that one can “expect one card to stay in place,” at least not with the common understanding of that phrase. The expected value is a weighted average that might not even be one of the possible outcomes. For example, in a two-card deck either both cards remain in place or neither does, with equal probability; exactly one is impossible. Similarly, the expected value of rolling a standard die is 7/2, which obviously isn’t one of the possible rolls. In this case we can say that it’s likely that at least one card will remain in place, but that conclusion depends on our knowing something about the distribution as well. For instance, if we had a weird way of shuffling that left every card in place with a probability of 1/52 and left none in place otherwise, the expected value would still be 1, but the chances of that happening are slim.

[@Ninja20704]

This is slightly related to the second problem, but you could ask what is the probability that no card ends up in its original position, or at least 1 card does. Suprisingly, the answer is approximately 1/e and 1-1/e respectively. The approximation gets better the more cards in the deck you have. Otherwise, great video.

[@edsimnett]

In general the chance of you doing something n times with a 1/n probability, and that thing NOT happening, is 1/e (quite a useful thing to remember)

[@yurenchu]

​​@@edsimnett Be careful there! You're talking about n _independent_ (sub-)events (for example, n tosses of an n-sided die, and the event that in none of those tosses the die lands "1"; so probability is [(n-1)/n]^n = (1-1/n)^n ), whereas the original commenter is referring to an event that consists of n _dependent_ sub-events (in this case, probability is ∑ [(-1)^k]/k! , from k=0 to k=n ). But yes, surprisingly both probabilities approach 1/e for increasing n .

[@sharky98]

The way you explain the problem 1 with text got me even more confused at that part: "Every ordering also can be obtained by arranging the 51 other cards, and then inserting the *card* (52 positions)." My brain locked in there confused about that "(52 positions)". And I stopped the video, thinking that no, you cannot insert it in any of the 52 positions, you can only insert it at the single position it fit the requirements in each of the 51! orderings! Then I decided to continue watching what I missed and you said exactly that in the next sentence! 😅

[@ijustwannaleaveacommentony6511]

mind boggling - more arrangements in a deck than there have been seconds since the universe began, that is so hard to conceptualize

[@philiptaylor7902]

Where there are 2 cards in the deck there cannot be just one in the same position, either both are the same or neither is the same, but I suppose the average is one is in the same position……

[@rogerkearns8094]

Yes, it's imprecisely worded. He must mean, _on average, 1._ (That does hold for two cards, of course.)

[@MichaelOnines]

That's what expected value means.@@rogerkearns8094

[@yurenchu]

​@@rogerkearns8094 It is precisely worded. Look up "expected value" (also called "expectation value"), it's a mathematically defined concept.

[@MichaelGowland]

You are absolutely right, even after seeing the calculation, I don't believe you, at least not in the case of the first problem. The issue is this, even after explicitly mentioning the possibility that the first ace is the ace of spades, you do not go on to factor that into your calculations. Here is what I get when I do: The question is rather unclear as we do not know if the two of clubs was drawn before the ace at the starting point of the question. However, I am going to assume for now that the two of clubs has not yet been dawn. Given these assumptions, there are two options: 1) The first ace is the ace of spades, with probability ¼ =0.25 2) The first ace is not the ace of spades, with probability ¾ = 0.75 In the case that the first ace is he ace of spades there are two ways of looking at the question. One view is that the two of clubs will necessarily be drawn before the ace of spades as there is no ace of spades left in the pack to be drawn. The other way of looking at it is that the ace of spades has been drawn first. The actual wording of the question marginally favours the first reading, so I will go with that. In this case there is a probability of 0.25 option one applies, with a probability of 1 of drawing the two of clubs first and a probability of 0 drawing the ace of spades first. The probability that case two applies is 0.75. In this case, given the assumptions I have made, until one of the specified cards is dawn, the probability of drawing either of them is the same (as the fact that we have already dawn a different ace makes no difference to the probability of drawing the ace of spades). Thus, in the case both cards have a probability of 0.5 of being drawn first. So, we have four possibilities: Case 1, with the Ace of Spades dawn first, with probability 0.25*0 = 0 Case 1, with the two of clubs drawn first, with probability 0.25*1 = 0.25 Case 2, with the ace of spades drawn first, with probability 0.75*0.5 = 0.375 Case 2 with the two of clubs drawn first, with probability 0.75*0.5 = 0.375 Thus the probability of the ace of spades being drawn first is 0.375 + 0 = 0.375 and the probability of the two of spades being drawn first is 0.375 + 0.25 =0.625. If you prefer the interpretation that in case one the ace of spades has been drawn first, the probabilities switch. If we do not assume that the two of clubs has not already been drawn, that makes a crucial difference as no we have to take into account conditions in which it is no longer in the pack to be drawn. In particular, if we take my reading of what it means to draw the two of spades first we would then have to take into account the possibility that neither card would be drawn first. We can’t, however, assess the probability of that as we would have to know how many cards had already been drawn prior to the starting ace.

[@swampertdeck]

Intuïtion told me 1/52 and 1 immediately, but it’s really cool to see the math behind it.

[@xpusostomos]

Too lazy to watch the video... Did we really need math to observe that the chance of any card being in a place is 1/52, there's 52 cards, so chances are 1 is in the same spot?

[@viljami.haakana.laulut]

The most important thing there is to notice that the expected value of a sum is the sum of expected values even for non-independent random variables. Thus if you have an exam where the taskbis to match two equally sized lists (say, people to their years of birth), the expected number of correct answers will always be 1 if you're purely guessing, no matter if you make different guesses, same guesses or a mixture.

[@danquaylesitsspeltpotatoe8307]

@@xpusostomos Only the people who cant work it those cant do the maths, so tis only maths people who cant work it out in their mind with common sense!

[@canniballissimo]

In simple way we have 2 cards. After a shaking we have the same positions or other. Not 100%

[@michaelwisniewski6047]

I think P1 is a great example of the importance of framing the question. To get the result quite easily, it's important to know that you're asking about the probabilities BEFORE any more information is known, ie before any cards have been uncovered. And of course, that's simply the same as asking what is the probability of a given card appearing in a given location in the deck. BUT the answer will be different if you're asking about these probabilities the moment someone has already reached the first ace. At that point they have more information, crucially they know if they have already uncovered the 2♣️ and if they have, then the probability it will follow the first ace is 0. Now, people unaware of the distinction, may try somehow include this possibility in their a priori probability intuition. But if you made it very clear to them what the distinction is, they may get the answer right quickly.

[@hamishblair1370]

Great problems, I took a bit of convincing on that first problem before it clicked! I got the second one right, but it is interesting to think of a deck of cards where n=2. There are only two orders they can end up, one order has both cards in the same position as the original order (2 cards in the same position), the other has neither (zero cards in the same position) So on average you would expect there to be one card ((2+0)/2)=1) in the same position as the original order, but in reality you could never actually expect there to be exactly one card in the same position, because with a deck of two cards it can never happen!

[@birchgarden2238]

1st problem is just trying to confuse you with the 3 other aces. You could take any card, and then ask whether A of spades or 2 of clubs comes next, that's why it's the same probabilities. In the scenario that A of spades comes before any of those 3 cards you selected, the question is just whether 2 of clubs appears before or after A of spades, which is also the same probability. Hence, the whole 1st problem = same probability.

[@markskid9942]

Fo P2 you could derive the probability function P[X=k]~(1/n!)(n choose k)[(n-k)!/e]. The expected value drops out very simply. Although the solution given is correct, this probability function is different to the that implied in the video - which is a binomial distribution (being the sum of independent Bernoulli RVs). I may be wrong, but I don’t think they are independent with constant probability 1/52.

[@SirFreddyPL]

Thought the same, if these events were independent the probability of all cards at the same place would be (1/52)^52, much lower than the true probability (1/52!).

[@yurenchu]

​@@SirFreddyPL Thank you! I too had my doubts, but I couldn't put my finger on it. However, your argument about all cards in the same place illustrates the problem nicely and clearly. Again, thanks to the both of you! :-)

[@yurenchu]

Actually, the exact probability function is Pr(X=k) = (1/n!)*(n choose k)*round((n-k)!/e) for 0≤k

[@hippophile]

1) I guessed this right. In order to understand the problem, I considered the cases where the first ace was the ace of spades and the case where it was a different ace (and the relative probabilities of those two events). If you consider this early on, your intuition should be that it is not obvious - because either the ace is a spade (so only the 2 club is possible, lowering the chance of the spade ace by around 25%) or it is a different ace ( in which case no card before was the spade ace so the spade ace probability is enhanced by about 25% assuming the first ace is about 1/4 of the way along). From that intuition you can estimate the probabilities as about equal. 2) The second problem: why would it not by (1/52 )* 52? And why would that be unexpected? You fooled me into thinking something surprising would arise, so maybe the answer IS surprisingly not surprising!!!

[@jimv9210]

In the video he mentions that 52! is so large that every shuffled deck has an order that likely has never been encountered before. It's also true that if you were to add seven cards to the deck, the number of possible arrangements would be roughly equal to the number of elementary particles which make up the visible universe!

[@orisphera]

Notes on the first one: There's only one ace of spades. So, if it's the first ace, it's not after the first other ace of spades. Therefore, it's after the first ace iff it's after the first other ace If it's the only ace, this doesn't work. In this case, it can't be after the first one, so two of hearts is more likely. The OP's solution doesn't work because there's no space for it in the ace of spades case

[@WarmWeatherGuy]

2:22 "There is exactly 1 spot to place the ace of spades to follow the first ace" but there are many places to place the ace of spades so it is the first ace. You have ruled out the ace of spades being the first ace in your calculation.

[@MushookieMan]

Yes, I wish he had given a proper explanation of the result (which is probably true)

[@NicD]

The quoted sentence does cover those cases: a spot where the ace of spades becomes the first ace makes it not follow the first ace.

[@underdoneelm7721]

Another way of thinking about the 1st problem is to ask whether its more likely that the two of clubs or the ace of spades is more likely to occur after the first ace other than the ace of spades. In this case, its very obvious that they're equally likely since the ace of spades can't be the first ace other than the ace of spades. However, these are also the exact same question due to the fact that if the ace of spades is the first ace, then it definitely won't be after the first ace other than itself.

[@troyj3292]

After getting my head around the full set of rules for problem 1 (I think most of the confusion is how it's worded), it becomes a bit more obvious why the probability is the same. What I'm still struggling with is why the probability works out to 1/52 or 1/n. The statement "For each of the 51! orderings there is exactly one spot to place the to follow the first ace," why doesn't it matter that there are positions in the resulting ordering that are not possible (thinking mainly of bunching the 2C and all the aces at the end).

[@sahanaandshamita6161]

Even if you consider the case in which you take out ace and the card following first ace is 2, of you keep the card in between the 2 and first a, it still makes a probability. And if the last 4 cards with the ace are like A,2,A,A, it doesn't make a difference, think again, about this taking out and putting the card in again, I was confused too in the first, if you still have confusion, feel free to reply the question you are posing. Hope this helps!

[@davidjoch]

2nd problem: if n = 2, there are 2 ways the cards (1;2)) could end after the shuffle: 1;2 or 2;1 => so the expected chance that one card is in the same spot agaon is 50 % or not?

[@YCLP]

For P1: I think the 2 of clubs being 1/52 is intuitive to everyone. However the spade A also being 1/52 is less easily intuitive. My initial (incorrect) reasoning was that the space A is the first A in 1/4 of cases, so the probability is smaller. This is incorrect because in the other 3/4 of cases the spade A is guaranteed to be in the remaining cards, making the probability in those cases >1/52. Why it's the probability exactly 1/52 then? I think presh gave an intuitive answer. I saw another great comment which highlighted that drawing 'the next' card is the same as drawing 'the bottom' card. Clearly the spade A has a 1/52 chance to start on the bottom.

[@JohnDoe-ti2np]

If you find it intuitive that the 2 of clubs (or any non-ace) is 1/52, then the aces also have to be 1/52, because the probabilities have to add up to 1.

[@YCLP]

@@JohnDoe-ti2np Conditions (such as draw until the first A) seem to be influencing A-odds because you've now seen one already. The club 2 on the other hand is much more intuitive because it is clearly not influenced by the condition.

[@JustSchalke04]

@@YCLP "This is incorrect because in the other 3/4 of cases the spade A is guaranteed to be in the remaining cards, making the probability in those cases >1/52." this was the hint i needed to understand it finally. Thank you

[@JohnDoe-ti2np]

@@YCLP I don't disagree with your latest remark, but I'm just pointing out that *if* the probability of every non-ace is 1/52, then it logically follows that the probability of every ace is 1/52. This is because the probability of *some* ace must be one minus the probability of a non-ace, which is 1 - 48/52 = 4/52. And of course the four aces are equally likely as each other, so this total probability of 4/52 must be equally split among them.

[@YCLP]

@@JohnDoe-ti2np Ah yes, I didn't fully understand it the first time. You are right.

[@nebraskajoel7838]

I started by examining a special case where the first ace appeared on the 42nd card, leaving 10 cards left. Both probabilities have 2 cases. Case 1 is if the desired card has appeared during the first 42 cards, and Case 2 is if the desired card hasn't appeared during the first 42 cards. In Case 1 the probability of hitting the desired card next is 0 (because it's already appeared earlier), and in Case 2 the probability of hitting the desired card is 1/10. So the probability of hitting either desired card is 1/10. After that I figured that it didn't matter which card in the deck you were on, so the probabilities are equal.

[@edsimnett]

doesn't that ignore the fact that we know the AS has not appeared in cards 1-41, but we do not know if the 2C has?

[@johnschmidt1262]

I got it wrong, but my first gut reaction was the ace was more likely because it is impossible to pull the Ace of Spades before the first ace! Cool it balanced out like that!

[@dead2thewind]

As to the 2nd problem,i used to do a bar bet. If they could get through a deck of cards, turning over 1 card at a time, and simultaneously counting up, without naming the card turned over they win a drink. So they count A, 2, 3, 4... J, Q, K, A, 2..... If yhey match the card they buy a drink

[@SmileyEmoji42]

So the trick is that "after the first ace" doesn't convey as much information as it feels that it should - We could replace it with "after the 10th card" or "after the first heart" and the result would still be 51!/52! as we are always just inserting 1 card into a single, well defined, position in each permutation of the 51 remaining cards.

[@SpinyBadger]

Problem 1 was tricky, but once you remember that 2C might have come out before the first ace (which an ace obviously can't), it looks a lot simpler. Problem 2 just seems obvious to me as a sum of n*(1/n), so I'm not sure what I'm missing.

[@WarGiver]

When I thought of the first question I thought about it more and realized that most of the question was unneeded detail and framing. If forget its asking "After the first ace" because its not actually important. All it does is START the deck AT the first ace, and ask which of the two cards is more likely to be the top card. It would mean the exact same thing as if the deck is shuffled which of the two cards is more likely.

[@Rondon0905]

Alternative approach for the 2nd problem: For n=1 it's obviously E_1 = 1. Now we take an additional card A. The new deck has n+1 cards. With probability 1/(n+1) the additional card A is shuffled to its original spot (1 correct) and we get additionally the expected value for the n cards deck (E_n). On the other hand, with probability n/(n+1) card A lands on the position of another card, let's call it B. Because B can't land correctly anymore, we will treat B landing on the original position of A as correct because then we again have the expected value for an n cards deck. But we will subtract 1 in the 1/n cases in which it actually does land on A's spot because it isn't really correct (E_n - 1/n). So E_{n+1} = 1/(n+1) * (1 + E_n) + n/(n+1) * ( E_n - 1/n) = E_n. By induction, this is always 1.

[@2dark4noir]

As is I'd call this one the better solution. Summing over expected values as done in the video is justified only if the individual observables are independent. That wasn't shown to be true. This solution doesn't have this pitfall. Nice 👍

[@Rondon0905]

@@2dark4noir Thank you =) But I am afraid I have to correct you: E[X+Y] = E[X] + E[Y] holds even if X and Y are independent or dependent. But we can still both like this solution better because it helps with the variance! Var[X + Y] = Var[X] + Var[Y] only holds if X and Y are linearly independent. So we can't do the trick in the video. But we can use the recursion to get the variance of the number of cards that stay in the same position in a deck of n cards, X_n. For a deck of n+1 cards, we get that the variance is the same as for a deck of n cards, Var[X_n+1] = 1/(n+1) * Var[1 + X_n] + n/(n+1) Var[X_n - 1/n] = 1/(n+1) Var[X_n] + n/(n+1) Var[X_n] = Var[X_n]. And because Var[X_2] = 1, this variance is always Var=1 :)

[@2dark4noir]

@@Rondon0905 don't be afraid, you're indeed correct :D thanks for pointing that out~ really interesting insight

[@mipplemep]

For P2, let's do the same experiment of trying it out on 3 cards. I have a three card deck with card values 1,2,3. Let's check if a card is in the same place after shuffling for all 6 permutations: 1,2,3 yes; 1,3,2 yes; 2,1,3 yes; 2,3,1 no; 3,1,2 no; 3,2,1 yes. By experiment, for a 3 card deck, there is only a 2/3 chance that a card is in the same position for a random shuffle.

[@globecyclist]

But on average there is exactly 1 card in the same position: 3 cards same, 1 card same, 1 card, 0 card, 0 card, 1 card = 6 cards in 6 permutations.

[@Ashebrethafe]

There are three permutations with one card in the same place, and one with all three in the same place. So the expected number of cards in the same place is one, since there are a total of six across the six permutations. For four cards, there’s one permutation with 4 matches, six with 2 matches (1243, 1324, 1432, 2134, 3214, 4231), eight with 1 match (1342, 1423, 2314, 2431, 3124, 3241, 4213, 4132), and nine with no matches (2143, 2341, 2413, 3142, 3412, 3421, 4123, 4312, 4321), for a total of 24 matches in 24 permutations.

[@yurenchu]

​​@@Ashebrethafe Fun fact: if the number of cards in the deck is even, then the number of permutations with 0 matches is one more than the number of permutations with exactly 1 match. However, if the number of cards in the deck is odd, then the number of permutations with 0 matches is _one less_ than the number of permutations with exactly 1 match.

[@Bowshewicz]

A funny little "paradox" with the expected number result: In a deck of two cards, the expected number of cards you'd find in the same spot after a shuffle is still 1 -- as explained in the video. And yet it would be quite unexpected to find exactly one such card after shuffling a deck of size 2! The math and why it works that way is obvious, but I like the apparent contradiction between the mathematical construction of "expected value" and our expectations of the physical result.

[@peterthomas5792]

Another issue of incompletely specified initial conditions. Shuffling a deck of cards is VERY different for 'randomising the order of a deck of cards'. *Never* does a shuffle properly randomise the order, so the puzzle is insolveable.

[@kerriann04]

For the first question, I figured they were equally likely, even without the math, just because the number and suit of one card doesn't impact the number and suit of another card. Any card is equally likely to be in any position. Probably too simplistic, but that's how I worked it out.

[@bizarnikreaturanikolivkrea3829]

Yes one of the few times i got everything right! 🎉🎉🎉 And when i realised it will be everytime one Card it was for second mindblowing for me! Great math problems!

[@mikefochtman7164]

I don't like the first problem solution because it ignores that there are 3 other aces. "Every ordering can be obtained by arranging the 51 other cards and then inserting the ace of spades (52 positions)" is flawed. Take the example that the three remaining aces are at the very last positions (49, 50, and 51). Now the ace of spades can NOT take on any of the 52 positions, since the 'first ace' cannot be in the 51 position. Nor can the 'first ace' be in the 50th position. The 'first ace' cannot be any later than the 49th position, meaning the ace of spades cannot be in 52 possible positions, but rather only 49. So possible locations where the ace of spades will be just after the 'first ace' is 51! * 49. The possibility of the 2 of clubs is a tiny bit different, since the 'first ace' could include the ace of spades and the 'first ace' then cannot be any further down the deck than position 48. So for the 2 of clubs to be after the 'first ace', we have 51! * 48.

[@MichaelOnines]

It can take on any of those positions. The 51 card deck is arranged, then you choose a random position for the ace of spades to be placed - anywhere from before the first card and after the last card for a total of 52 positions. Now you assess whether the ace of spades appears immediately after the first of the other 3 aces in the pack. That assessment is no different if you place the 2 of clubs there or the ace of spades there. If you put the ace of spades in a spot that happens to be BEFORE the first of the remaining aces, it becomes the first ace and can't itself be the next card.

[@mikefochtman7164]

@@MichaelOninesWell it took me a while, but I've come around. What bothered me is that for the ace of spades scenario, of the 52 positions, 3 of them could never be 'winning positions' (first card, last card, or next-to-last). Yet for the two of clubs, of the 52 positions 4 of them could never be 'winning positions' (first, last, next-to-last and next-next-to-last). But working out all possible combinations with 'n' up to 5, I gained some insight as to why this doesn't change the overall odds. Although there are fewer positions in which the 2 of clubs can be a 'winner', there is more permutations that it 'wins' in those positions (because there is slightly better odds of an ace being in front of it). Or put another way, while the ace of spades scenario has more possible positions in which it still 'wins', fewer of those positions are 'winning' permutations.

[@SuspenduAuGaffa]

8:46: "One lakh of cards". Yes! I've started using this term after hearing our Indian colleagues use it. It's very handy for that particular order of magnitude.

[@jaybingham3711]

P1 fails linguistically. If the very first card turned up/over (the card on top of the deck) is an ace...and in fact...the As...there is precisely ZERO probability/likelihood that "the next card" that gets flipped over will be...another As (assuming a correctly packaged deck...free of duplicates.) Hence, you necessarily must choose the 2c as the correct answer (as presented 0:13). The solution disconnects from this presentation. It instead evaluates probabilities associated with a scenario in which the As can’t/isn’t deemed to be the “first” ace. In reality, even after thoroughly shuffling a deck, the As can very much find itself right on top of the deck.

[@madeconomist]

For the second problem: The presumption that there are 52 places that the first card in a deck being shuffled are equally likely to go is not true in most real cases. If a human shuffles competently using the standard two-hand method, there is no possible way for the top card to be on the bottom after one shuffle. And it is highly unlikely that the original top card would be on the bottom after two or even three shuffles. To see what I mean, try placing the four aces on the bottom of a deck. Shuffle once and lay out the deck and see where the aces are. They won't have moved far from their original spots. In my experience, it takes about 7 shuffles to achieve something approaching true randomness. I am not trying to bash you. You make great videos and I love your channel. I just wanted to add something to contemplate. Cheers!

[@RoyalPurple667]

The solution for problem 2 is especially funny when you set n=2, where the statistical expectation for the number of cards that show up at their old position is still 1, but that is impossible to happen.

[@empathogen75]

Not impossible, it’s just 50/50. Which makes me think the argument must be wrong.

[@empathogen75]

Actually I just figured out why it’s not wrong. Half the time there will be zero cards in the same position, and half the time there will be two, so the expected value is 1.

[@jeffreyestahl]

I'm glad you brought up this type of problem. Back in 1980 while in college, I went to my math professor with the following question - In probability problems we always talk about a fair die roll or a fair coin; we also say when talking about cards that it's 'perfectly shuffled'. I told him the die and coin fairness was easily described through physics (i.e. the closer that the geometric center of the die or coin was to the center of gravity of the die or coin, the more likely it was to be fair), but I mentioned that no such external determination could be made for a deck of cards, so what did a perfectly shuffled deck of cards look like? He mentioned that it wasn't really known, but perhaps we could discover it for ourselves and began introducing me to the concept of chaos mathematics. I realized that a perfectly shuffled deck (regardless of size but requires the inclusion of complete sets of card types) is actually quite ordered, not random. Thank you for this.

[@jimlocke9320]

Some slot machines, such as video poker machines, electronically shuffle the cards, They use something called a random number generator (RNG) to select the cards. The random number generator uses some kind of random input in conjunction with a pseudorandom algorithm. The RNG might generate a large integer, which gets divided into 52 equal length ranges, with something left over. Depending on which range the number falls in, a card is chosen. If the number falls in the "something left over", the computer discards the RNG output and tries again. For the second card, the RNG output might be divided into 51 ranges and something left over. In this way, the computer can "shuffle" a deck of cards and create a "perfectly shuffled" deck. The algorithms are design reviewed and tested extensively. If there were any pattern, the players might pick up on it and improve their odds. There are people who obtain slot machines through legitimate channels and do extensive reverse engineering and testing to try to locate bugs, which they then can exploit on the machines located in casinos.

[@jeffreyestahl]

@@jimlocke9320 The thing to remember is that not even RNGs are perfectly random. If you start them with the same seed, they will yield the same sequence of numbers. This is done to test various gaming AI systems. Also, RNG algorithms are usually based on advancing an index through a typical transcendental number like Pi or Euler's Constant. (multiple iterations are run which advances the index along the number, the resulting digit that's plugged into the output stream is based on a sequence of 8 to 16 digits depending on the level of security built in) So again, not truely random. Industrial algorithm typically use the Unix time to determine the RNG seed, but that doesn't always happen. BTW, I've designed quite a few RNG algorithms in my lifetime. (these are frequently used for encryption algorithms which use RNGs and a determinable seed built from keys) They're really not that complicated. And, yes, some very enterprising mathematicians HAVE gathered sufficient data in casinos to bank the slot machines. Once identified, those casinos refuse access to the individuals involved. However, that takes time, observation, and demonstrates that in the end Math rules all.

[@KevinK-s3g]

Something I find interesting with P1 is that it only holds true if there is at least one other ace in the deck. Otherwise the probability for the ace of spades following the first ace goes to zero. Not sure how to understand that mathematically, but i still find it interesting!

[@SmileyEmoji42]

Not interesting if you posed the question succinctly knowing that there is only one ace: "Which is the most likely card to follow the ace of spades: The ace of spades or the two of clubs?" - See - it would just seem silly.

[@adb012]

@@SmileyEmoji42 ... No. By phrasing the question that way you are assuming that the only ace is the ace of spade ;-)

[@SmileyEmoji42]

@@adb012 I know that. My point was that the OP should not find it interesting that it is only true for more than one ace. This is because, if P is interesting NOT(P) should logically also be interesting since they divide the problem space into the same two sets and yet here I show that NOT(P) is not interesting and hence P should not be seen as interesting 🙂

[@KevinK-s3g]

My comment was regarding the example at the end with a deck size of n=3. It was pointed out that in a deck of size n the probability of a certain card following the first ace would always be 1/n, whether the card was an ace or not. However, that rule only holds true if the deck contains more than one ace. I found it to be an interesting constraint and it wasn't mentioned in the video.

[@Krydax8]

My favorite way to think about problem one: What are the odds the 2 of clubs comes BEFORE the first ace (and thus its probability of being the winner goes to zero)? 1/4! cause the deck is 4 aces. What are the odds that the first ace you get is the ace of clubs? 1/4! so there is a 1/4 chance that each option is negated, thus equaling out!

[@yurenchu]

First of all, it's called probability (or chance), not odds. Odds are something different. Secondly, the probability that the Two of Clubs appears before the first ace, is 1/5 , not 1/4 . However, the scenario that {the Two of Clubs appears before the first ace while the Ace of Spades is the first card after the first ace} is equally likely as the scenario that {the Ace of Spades is the first ace while the Two of Clubs is the first card after the first ace}.

[@Keldor314]

For the first question, the next card is more likely to be the 2 of clubs than the ace of spades since there's always the chance that the first ace we drew was the ace of spades, and we can't draw the same card twice. If the first ace was any other ace (maybe we know because it was revealed as the first ace), then the probability is equal. On the other hand, you have discarded some number of cards up until you reach the first ace. The 2 of clubs could be among these, but the ace of spades cannot. This means that if the first ace is not the ace of spades, then the probability of it being revealed next is higher because there are fewer possible positions for it to inhabit (precisely one of which is the next card drawn) than the 2. We might make the guess that the number of cards before the first ace is about 1/5 of the whole deck, since aces partition it into 5 chunks (note that the real number is slightly less, since there are only 48 non-ace cards in the deck). Thus, the 1/4 chance of the first ace being the ace of spades seems to outweigh the 1/5 chance of the 2 of clubs being on the wrong side of the first ace.

[@CardCaptor32584]

Sorry if I am misunderstanding problem 2. When they say a "proper" shuffle, does the problem mean a perfect Faro shuffle? If that's the case it's guaranteed to be 2 if you are doing an out shuffle once or 0 if you are doing an in shuffle once. If you want all the cards to come back to their original positions, you can do 8 perfect out shuffles. Easier to do with a brand new deck of cards.

[@noobatredstone3001]

I believe the problem means a probabilistically proper shuffle, where the position of each card is completely random.

[@r.s.hildebrant2012]

The confusing intuition in the first problem is from whether we're solving given the "First card drawn is an Ace" or the "First drawn Ace"...

[@mathmannix]

I don't understand the first problem. An unshuffled deck (after you remove the jokers and "ad cards") will always be (from the bottom, or face-side) Spades (A, 2, ... Q, K), then Diamonds (A, 2, ... Q, K), then Clubs (K, Q, ..., 2, A), then Hearts (K, Q, ..., 2, A). If you cut the cards evenly into two stacks, and then do exactly one "proper" shuffle, then the most likely outcome is that either the first four cards on top will be 1. Ace of Hearts 2. King of Diamonds 3. 2 of Hearts 4. Queen of Diamonds, or 1. King of Diamonds 2. Ace of Hearts 3. Queen of Diamonds 4. 2 of Hearts. If you cut the right side larger (say 24-28), then the top three cards will be the (unshuffled because they didn't get matched in the shuffle) 1. Ace of Hearts 2. 2 of Hearts 3. 3 of Hearts And if you cut the left side larger (say 28-24), then the top four cards will be the (unshuffled) 1. Queen of Clubs 2. King of Clubs 3. King of Diamonds 4. Queen of Diamonds followed by either 5. Ace of Hearts 6. Jack of Diamonds 7. 2 of Hearts 8. 10 of Diamonds or 5. Jack of Diamonds 6. Ace of Hearts 7. 10 of Diamonds 8. 2 of Hearts So there is almost zero chance of it being anything but a high Diamond (10-K) after the first Ace (which will be the Ace of Hearts).

[@coreyburton8]

thank goodness they are equal- that's the only answer that made sense when you read the problem out. I didn't see the issue

[@stuartdavidson162]

MIND. BLOWN. Cheers Presh :)

[@BleuSquid]

The first question as posed is a bit ambiguous. It does not specify starting deck status, so I assumed it was a new unused deck. It does not specify shuffling method or degree, so I took it to mean a single standard riffle shuffle. An unshuffled deck usually has ordering of As, 2s, 3s, ... Ad, 2d, 3d, ... 3c, 2c, Ac, ... 3h, 2h, Ah, ... A riffle shuffle would result in either: As, 3c, 2s, 2c, 3s, Ac, ... Ad, 3h, 2d, 2h, 3d, Ah or 3c, As, 2c, 2s, Ac, 3s, ... 3h, Ad, 2h, 2d, Ah, 3d When placed face down, and turning over the top card in order: the card that follows the first ace will be either 3 diamonds, or 2 diamonds, if it were a perfect shuffle. Even if it weren't a perfect riffle shuffle, neither a spade nor a club are even possible in the first half of the deck. They'd be all the way at the bottom! tldr; If you mean to refer to a properly randomised ordering, please state that in the givens!

[@BleuSquid]

If anyone saw my original comment, I had to delete it because I had the new deck order completely backwards.. I was thinking face down instead of face up!

[@letao12]

I don't know anyone who interprets "shuffle" as being only one standard riffle. In math of course it always means perfectly randomly. But even in a tabletop game setting, completely separate from any mathematical context, if the instructions say "shuffle the deck" they would mean to shuffle it as randomly as you can. Certainly you would never play a game where the deck was only riffled once from an ordered arrangement?

[@mihailghinea]

Another really easy way to understand P.1 is to think that the problem is identical to replacing the "after the 1st Ace" with "after the 1st King"(or any other card, except the 2 of clubs and the Ace of Spades). Or comparing the probs for any other 2 cards. As a pro poker player, I love this problems because they make great hustle side-bets opportunities at live tables 😅

[@corvididaecorax2991]

I'm not sure your analysis for the second problem is correct, since the card positions aren't completely independent. If the first card isn’t in the first position then it is also taking up a spot that another card has as its original position, meaning both are out of place. You would be correct for dice, since they roll independently and multiple can show the same number, but cards are a bit more complicated as each position can only have one card. Or at least that is my perspective. I could be convinced that I am wrong.

[@wightboy12345]

I was also wondering this. I’m not sure if it’s at least partially cancelled out by the idea that if the first card is in the correct position, the remaining cards have a higher chance of being correct as well.

[@corvididaecorax2991]

@@wightboy12345 That seems like a good point. The chance of multiple is not completely independent either.

[@exeaxe]

to the first, when inserting 2 of clubs (or whatever) there couldn't be 52 positions considering a deck includes 4 aces so the first ace must appear at the 49th position or earlier

[@ThomasGutierrez]

Great puzzles as always. There are a couple of card-deck nuances to perhaps consider for future puzzles. The term "shuffle" colloquially (in the context of these probability puzzles means) "selecting one random configuration of the deck from the 52! possible ones" However, for real decks, a shuffle is a very mechanical, often imperfect, interstitial "riffle" (interlacing between two nearly equally separated halves) and the number of times you shuffle, the number of times you cut, and initial configuration of the deck can GREATLY change the probabilities. For example, try Problem 1 with a new deck after shuffling only once and you will see that the ace of spades will never follow the first ace (it will always be near the bottom while the two of clubs will always be near the top). Some magic tricks and cheating methods rely on manipulating these initial conditions and riffling methods. Perhaps you can do a future puzzle on the famous Faro shuffle (perfectly interweaved shuffle). It can be shown that only eight so-called "out" Faro shuffles can return a deck to its initial configuration!

[@rickdesper]

For the first problem: rather than caring about all 52 cards, we can reduce the problem to a 5-card deck with the four aces and the 2 of clubs. For the purpose of this problem, it only matters what order these five cards appear. It's fairly clear that each of the five will appear 1st 1/5 of the time. If the two appears first, it cannot be the first card after the first ace. In this case, the ace of spades will be the 2nd ace 1/4th of the time. If the ace of spades appears first, then the next card will be the two of clubs 1/4 of the time. So, summing the probabilities over these two special cases, we see that two of clubs and the ace of spades will be equally likely. The other three cases are when one of the other aces appears first. It's fairly obvious that, should this happen, it's equally likely for the next card to be the two of clubs as the ace of spades. A bit more rigor would be required to justify this reduction in the size of the deck. Think of that as "an exercise for the reader." The second problem is very basic probability when "Expected Value" is introduced. Let X_i be the event "card i remains in the same place after the shuffle,"" then X = sum_i X_i is the number we're looking for.

[@grezamisoit]

For the second solution, I don"t understand why you consider all the Xk as being independent. If 51 cards are at the good position, the 52nd can't be at the wrong position. So all the Xk are not independent. I do think that this avoid using your solution.

[@gabrielrockman]

The problem is that when we're drawing cards and not getting an ace, we're not putting these cards back into the deck. So the portion of the deck that are aces starts at 1/13th, but its constantly increasing every time that we draw cards without getting an ace. If we start with a full deck, and immediately draw an ace, then there's a 1/17th chance that the next card is an ace. But if we're not starting with a full deck when we draw our first ace, then the chance that the next card is an ace is always higher than 1/17th, and quite often is higher than 1/13th. We're not reshuffling after each card that we draw. These aren't independent events. That's what makes this seem counterintuitive.

[@averagejuveenjoyer1994]

In second task. Consider a random permutation and its N shifts. Each of N cards will be on its place exactly once. N same positions for every group of N permutations. Groups are disjoint. So average number is 1.

[@JhettJones]

Good thing we're talking EV in Q2, otherwise that deck of exactly 2 cards is pretty awkward, being impossible to ever have 1 card in the same position. >.>

[@johannesvanderhorst9778]

8:54 imagine shuffling a deck of Grahams number of cards, and then checking whether one of those cards are in their original position. This will take a while. 😝🤣

[@brianrichter8255]

Problem 1 intuitive answer: Consider a variation of problem 1: is it more likely for the ace of spades or the two of clubs to appear immediately after the first *non-spade* ace (ace of clubs, diamonds, or hearts). Of course, the ace of spades or two of clubs is equally likely to appear after the first non-spade ace. Now compare this variation to the original problem 1 as stated. Every case in the original problem where the ace of spades was the first ace corresponds to exactly one case in the variation where the ace of spades did not appear after the first non-spade ace. Therefore, the problems are equivalent, and the probability of both cards is equal.

[@SimonClarkstone]

That is a great explanation! Thanks for making it intuitive for me.

[@edsimnett]

100% not true if the first ace appears as the 49th card. In fact unless you watched for the 2C as you were turning cards you *know* the AS has not gone, but you you do not know the 2C has not gone, so it is much more likely that the next card will be the AS.

[@ovoremek]

Both results are surprising and counterintuitive.

[@alanjones4358]

Another way to understand it: The first ace will be a non-spade 3/4 of the time and there will be an average of 39 cards left. So 3/4 times 1/39 = 1/52.

[@George4943]

Random facts. The simplest shuffle is to cut the deck. None of the cards is in its original position. There are 51 ways to cut. In a perfect riffle shuffle of 26-card stacks either the top card or the bottom card stays in place. Ace cases: Ace of spades on top is 1 chance in 52. 1/51 club 2 next. Spade A (again) zero. Other A on top. Spade A 1/51, club 2 next is 1/51. Ace of spades first ace is 1 chance in 4. 1/51 club 2 chance. Spade A again zero. Other A first. Spade A 1/51, club 2 next is1/51.

[@bryn.1tbs]

*disclosure: I've opened a few hundred new decks at a bridge club (I tried to do it first and by myself when possible so I could spread them out and get them out of position) and studied and simulated shuffling, but that was over a decade ago and my math/logic is not what it was when I was younger so it evens out? I'm confused a little by the "intuition" on the first. Even if they're shuffled once standardly, if the cards were perfectly cut, the top card is equally likely to be the A♤ or the 2♧. After a random shuffle I would have said the rank of the cards is independent of which specific card it is since the 2♧ could have been before the first Ace. For the second, I would have guessed that it would have been less likely as the deck grew.

[@3GOD]

For problem 1, I dove into the conditional probabilities given the 1st ace appears in position n. The results were interesting: Pr(2 clubs at n+1 | first ace at n) = (49 - n)/(48*(52-n)) for n = 1 to 49 Pr(Ace of spades at n+1 | first ace at n) = 3/(4*(52-n)) for n = 1 to 49 For the first 12 values of n, the conditional probability for the 2 of clubs is slightly larger. For n = 13, the conditional probabilities are equal. For n greater than 13, the conditional probability for the ace of spades is greater (and grows much much greater for n close to 49). We can then compute Pr(2 clubs immediately follows the first ace) and Pr(Ace of spades immediately follows the first ace) using the law of total probability -- we take a "weighted sum" of the conditional probabilities above with "weights" of Pr(first ace at n), where Pr(first ace at n) = (4*(52-n)!*48!)/(52!*(49-n)!) for n = 1 to 49 It is interesting how these "weights" shrink as n increases and perfectly balance out the differences in the conditional probabilities for the distinct values of n. In the end, by the law of total probability, both probabilities end up being 1/52 as Presh correctly stated (via a much simpler argument). These conditional probability computations also help illuminate some reasons for the wrong intuition that the 2 of clubs is more likely to follow the first ace. There is about a 66.2% chance that the first ace will appear within the first 12 cards, and for each of those values of n, the conditional probability for the 2 of clubs is slightly larger than that for the ace of spades. On the other hand, there is a relatively low probability of the first ace appearing toward the end of the deck (far less than 1% chance after n = 40 for example), but at that point, the conditional probability for the ace of spades is much greater than that for the 2 of clubs. P. S. I wish I could show the plot of these conditional probability values along with the weights that I created in Excel. A picture is worth a thousand words in this case.

[@michaellautermilch9185]

I fully explored the first problem, believing his solution to be wrong. Turns out it's right after all. There are 2 special effects going on. The ace of spades may be the first ace (25% probability) in which case it can not be followed by itself. This is stated as intuitive. On the other hand, when the first Ace is not the Ace of Spades, we know that the Ace of Spaces has not yet been "used up", but the 2 of Clubs may have come before the first Ace. Specifically, there's a 20% chance that the 2 of Clubs is used up before the first Ace appears. Lots of other cards (20% of the deck) will also have been used up before the first Ace, on average, which greatly inflates the chances (+25%) that the next card is the Ace of Spades. The +25% comes from the inverse of the average number of used up cards when the first Ace is seen, 1/(1-20%).

[@Qwentar]

With Faro shuffles, two cards will be in the same position: the top and bottom cards.

[@davidklein8608]

At first I thought this was addressing questions of card order assuming the deck started out in factory order of ace through king alternating red and black suits and that shuffle meant to divide the deck by half (approximately) and fold the cards together. A standard shuffle isn't truly random and would likely result in a two coming right after an ace and the top and bottom cards not changing position.

[@Raj-xt4fk]

Another sol to p2 Let we are finding the expectency of k cards to be in same position So when areanging, we keep k cards and arrange the other (52-k) cards in (52-k)! Ways. So the probability of its expectency is (52-k)!/52! According to question, we want to maximise this expectency. So we have to maximize the numerator. As k cant be 0, k=1 So we expect 1 card

[@szlatnik]

I think for problem 2 expecting that a shuffle will completely randomize the deck is probably not accurate. You probably have to shuffle a certain number of times to truly randomize the deck, and even then things like "the top card going to the bottom of the deck" are still unlikely unless you just spread the cards out on the table and mix them up that way

[@mbmillermo]

There are really two problems to consider -- one in which the first Ace is never the Ace of Spaces, and a second one in which the first Ace might be the Ace of Spades, thereby ruling it out as the subsequent card. Apparently, that second problem was the one we were working on because the Ace of Spades and the Deuce of Clubs are equally likely as the subsequent card in that scenario. See below. If the first Ace is the 49th card, then the Ace of Spades can be the 50th card, but the Deuce of Clubs cannot be 50th because the 50th card then has to be an Ace. The farther you go through the deck, the more likely you will have passed the Deuce of Clubs, but you will never have passed the Ace of Spades. If the first Ace is the kth card, and it is not the Ace of Spades, then the probability that the k+1st card is the Ace if Spades is 1/(52-k), because it always is one of the remaining cards. In the same circumstance, the probability that the k+1st card is the the Deuce of Clubs would be 1/(52-k), but only if it is one of the remaining cards. So, to get the right answer for the Deuce, we have to multiply times the probability that it is one of the remaining cards, which is one minus the probability that it was one of the first k-1 cards. We have effectively selected, at random, k-1 cards from the 48 non-Ace cards, so the probability is (k-1)/48 that the Deuce was already passed over, and the probability is (48-k+1)/48 that it remains in the deck. So our final probability for the Deuce of Clubs being the k+1st card is [(48-k+1)/48]*[1/(52-k)] = (48-k+1)/[48(52-k)]. Thus, the two probabilities are only equal when k=1, but when k=49 the Deuce has zero probability and the Ace has probability 1/3. That analysis was based on the assumption that the Ace of Spades cannot be the first Ace. If the Ace of spades *can* be the first Ace, then 1/4 of the time it will be the first Ace and, the Ace of Spades cannot then be the card that follows immediately after the first Ace. The probability that the Deuce is the card immediately following the first Ace is unchanged, but the probability that the Ace of Spades immediately follows the first Ace is now reduced greatly to 3/4 of what it was: 3/[4(52-k)]. That means the Deuce is more likely for small values of k ( if the first Ace comes early), the Deuce of Clubs and Ace of Spades are equally likely to be the subsequent card when k=13, and the Ace of Spades is more likely than the Deuce of Clubs for k > 13. But what happens when we average across the 48 possible positions for the first Ace? Well, it's tricky because we have to refer to the Negative Hypergeometric distribution and do weighted averages. Let's do it in R: > ace=rep(0,49) > deuce=rep(0,49) > for (k in 1:49) {pak

[@mbmillermo]

I should add that I think the analysis in the video is correct, and that's a much simpler way to think about it and get the answer. We get the same answer: 1/52. I just like to mess around with probabilities and R, etc., so I did it anyway.

[@edsimnett]

​@@mbmillermo Aren't you solving the a priori average case? The question asks you to make a decision at the point the first ace appears. If we say that the dealer merely says "first ace" then the count of the deck at that point is critical (e.g. obviously 2C if first ace is the first card and obviously AS if it is the 49th card)?

[@mbmillermo]

@@edsimnett -- Hi Ed. I want to answer your question but I might not understand it correctly. I solved two different problems. I'll try, though. The count in the deck is important, but we aren't asked to make the call based on how many cards have been dealt. I think of it as making the call in advance -- predicting which card will follow the first ace whenever that ace appears. So we do have to average across all the possible positions in the deck where that first ace might be found. Of course, as shown in the video, all that weighted-averaging math can be avoided by thinking intelligently. I did it the less intelligent way -- brute force weighted averaging -- and got the same answer.

[@edsimnett]

@@mbmillermo If you solve a priori without taking account of the count (sorry) then yes 100% agree a coin toss is the right strategy. My point is that you can improve on that very significantly, a priori, by saying 2C if the first ace appears before card ~9 and AS if afterwards. Importantly, the rules are silent on a number of critical things (do you watch the cards discarded, do you know if the ace is the AS and what is the count [you say "we aren't asked" I say "the question did not tell us one way or the other" #reasonable people can disagree]), despite also implying you are making the call (the way the problem is described) at the point of the first ace being drawn, not before the game starts.

[@mbmillermo]

@@edsimnett -- Yes, Ed. Good points. Apparently, the odds reverse at the 13th card, so AS is more likely after 13 cards have been drawn, but DC is more likely before that. And that is assuming we haven't seen the discarded cards. We could work that out, of course, but the only card of interest would be the DC. I guess not seeing the DC in the discard pile would mean it is a lot more likely to be drawn immediately after the first ace. In fact, DC will remain more likely than AS as long as it isn't in the discard pile! That's because AS can be the first ace but DC (obviously) cannot play that role.

[@tobyfitzpatrick3914]

I can't say that I was surprised here. The chances of the 2 of Clubs appearing before the first Ace negates its advantage. The answer to the 2nd problem is intuitively obvious.

[@MrConverse]

The expected number of cards in their original position is one. But what’s the probability of getting exactly one card in its original position? Of getting zero? Of getting more than one?

[@yurenchu]

Suppose a deck of n cards: - Total number of permutations: P(n) = n! - Number of permutations with no card in its original place: W(n) = (n-1)*(W(n-2) + W(n-1)) , with W(1) = 0 and W(2) = 1 ==> W(n) = P(n) * { ∑ [(-1)^k]/k! , from k=0 to k=n } - Number of permutations with exactly one card in its original place: V(n) = n*W(n-1) For example: n = 4 : P(4) = 4! = 24 , W(4) = 9 , V(4) = 8 n = 5 : P(5) = 5! = 120 , W(5) = 44 , V(5) = 45 So the probabilities are: Pr(0 cards in original position) = = W(n)/P(n) = { ∑ [(-1)^k]/k! , from k=0 to k=n } which approaches 1/e ≈ 36.8% as n increases. Pr(exactly 1 card in original position) = = V(n)/P(n) = n*W(n-1)/n! = W(n-1)/(n-1)! = W(n-1)/P(n-1) = { ∑ [(-1)^k]/k! , from k=0 to k=(n-1) } which too approaches 1/e ≈ 36.8% as n increases. Pr(more than 1 card in original position) = = 1 - W(n)/P(n) - V(n)/P(n) = 1 - W(n)/P(n) - W(n-1)/P(n-1) = 1 - 2*{ ∑ [(-1)^k]/k! , from k=0 to k=n } + [(-1)^n]/n! which approaches (1 - 2/e) ≈ 26.4% as n increases. I hope that helps.

[@РусланГорельчик]

2. P = 1 - (51/52)^52 ~ 0.63. In my opinion, this is how probability is considered. What about a deck of 2-3-4 cards?

[@TosiakiS]

Wrong. This is not "the probability at least one card is in the same position," this the "average number of cards in the same position." In other words, if there are 2 cards in the same position, that has a weight of 2, not 1.

[@aliasmask]

This is an old card trick where "at least" 1 card is in the correct position and the answer to that is 0.63.

[@AwestrikeFearofGods]

If you have a deck of 2 cards the expected value after shuffling is 1, but you can expect to never have a result of 1.

[@Nikioko]

Both are equally likely. The chances to be drawn are each the same for the remaining card. If the first ace drawn is the ace of spades, the 2 of club is more likely, since the ace of spades is already drawn. But if the first ace isn't the ace of spades, the 2 of clubs might already have been drawn before the ace showed up, and then the ace of spades is more likely, as the 2 of clubs can't show up again.

[@fredrik3685]

Thanks

[@nickronca1562]

Can you please read my comment and tell me where I went wrong in the logic of my comment please?

[@Nikioko]

@@nickronca1562 OK, let's see. Just say that the first ace is the 8th card. Now, if it's ace of spades, the chance of 2 of clubs coming up next is 1/44 if it wasn't among the first seven cards and 0 if it was. Ace of spades coming up next has a chance of 0 in any case. If the first ace is NOT ace of spades, both ace of spades have a chance of 1/44 to come up next if 2 of clubs wasn't among the first seven cards. If it was, then its chance is 0, while the chance for ace of spades remains 1/44. If you compare these two scenarios and take out the cases where the chances are equal (0 for ace of spades being the first ace and 2 of clubs among the first 7 cards, and 1/44 for ace of spades not the first ace and 2 of clubs not among the first 7 cards), the remaining unequal chances are the same: If the first ace is ace of spades and 2 of clubs is not among the first 7 cards, the chance for ace of spades is 0, and the chance of 2 of clubs is 1/44. If the first ace is not ace of spades and 2 of clubs is among the first 7 cards, the chance for ace of spades is 1/44, and the chance of 2 of clubs is 0. Of course, the first ace being the 8th card is an arbitrary decision, but this consideration works for any other position as well.

[@CreamusG]

It's amazing how the answer is always 1 considering that in a deck of 2 cards you can never get just 1 card in its starting position.

[@SmileyEmoji42]

It's an average, an expectation. With 2 cards you have either 0 or 2 in the same position for an average of 1

[@Jeff-gd8ev]

I believe the 2nd solution ignores the mechanics of shuffling. In a typical shuffle, you take 1/2 the deck in your left hand and half in your right, then merge them with an approximate alternating procedure. Of course the alternating is not perfect, but the probability that the top card remains the top card is approximately 50%, and is certainly not 1/52. If you shuffle the deck again, the probability that the original top card moves down the deck increases, however even after an infinite number of shuffles, the original top card will not be in a truly random location, but have a higher than average position due to its starting position.

[@RemyWillard]

8:25 What about a deck of exactly 2 cards? Isn't it then impossible for exactly 1 card to be in the same spot?

[@3GOD]

Yes, but the average over many trials would be 1. Half the time, no cards remain in the same spot, and half the time both cards remain in the same spot.

[@RemyWillard]

@@3GOD While the average would indeed be 1, expecting the impossible makes zero sense. This just strikes me as a rare oversight from Presh, who usually covers every possibility I can think of and several other possibilities I never even considered.

[@3GOD]

@@RemyWillard It is really just an unfortunate overlap of the word "expectation" from probability theory with our English understanding of the word. It is not intuitive to "expect" something that cannot happen, but the average result (or "expected value") of a random variable need not be any of the possible values of the random variable. As an example of sorts, maybe the average income in the US is $52,356.7685, but no one can actually earn that exact amount (they cannot earn a fraction of a cent).

[@quangpham4372]

@@RemyWillardyou miss the point of “expected value”. If you play a coin toss, the expected number of head is 0.5, even though for each toss you can’t have half a head

[@ssifr3331]

My first thought for the first question was it was supposed to be equal but then I reread the statement and thought if the first ace was the ace of spades then the probability is zero so 2 of clubs is more likely LOL

[@ClarenceJudd]

Think about a game with 5 outcomes. Any of these outcomes will end the game. The Ace of spades will be one outcome, the two of clubs is another outcome, and the other three aces are the other 3 outcomes. All 5 outcomes have an equal probability. If any ace other than the ace of spades is the games outcome, than the game ends with both the Ace of spaces and the two of clubs still available to be drawn from the deck with equal probability. The remaining two outcomes are also for the ace of spades or the 2 of clubs, and both of theses have equal probability.