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I'm attempting the following question from MIT Discrete Math Course and I want to know if my Proof for the given question is correct or not.

Question:

I have a deck of 52 regular playing cards, 26 red, 26 black, randomly shuffled. They all lie face down in the deck so that you can’t see them. I will draw a card off the top of the deck and turn it face up so that you can see it and then put it aside. I will continue to turn up cards like this but at some point while there are still cards left in the deck, you have to declare that you want the next card in the deck to be turned up. If that next card turns up black you win and otherwise you lose. Either way, the game is then over.

Either,

  1. come up with a strategy for this game that gives you a probability of winning strictly greater than 1/2 and prove that the strategy works, or,
  2. come up with a proof that no such strategy can exist.

Answer:

No such Strategy exists.

Proof by Contraction:

  1. Assume there exists a strategy S that produces a probability of winning (drawing a black card) strictly greater than 1/2.
  2. This implies that strategy S wins for more than half of the possible 52! deck arrangements.
  3. For every deck arrangement where S wins with a black card, there is a corresponding arrangement where the positions of the black and red cards are swapped.
  4. In this swapped arrangement, S would win if we redefined winning to be drawing a red card.
  5. Therefore, if S wins for more than half of the deck arrangements when winning is defined as drawing a black card, then for every one of those arrangements, there is an alternative arrangement where the strategy would have produced a win if winning were defined as drawing a red card.
  6. This means we have more than 52!/2 arrangements where the strategy wins and similarly more than 52!/2 arrangements where the strategy loses.
  7. But this is a contradiction since it implies that the total number of winning and losing arrangements exceeds 52!, the total number of possible combinations.
J. W. Tanner
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AzharKhan
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  • Not sure I follow. Yes, for any winning strategy for black the same strategy would produce a win for red if all colors were swapped. Why does that suffice? In any case, this question is a duplicate. – lulu Aug 05 '24 at 15:02
  • There seems a lot of intuitive fudging in that answer. First, the strategy for red in black look the same, but they obviously are not, but are inverses. It is not impossible that the strategies for red and black won't work given the same deck. For example, if your strategy for black is "choose the next after the first red card," and you play similar for red, then a deck that starts red/black/red will win for both colors. It is worth writing it out more formally, rather than intuitively. – Thomas Andrews Aug 05 '24 at 15:11
  • Specifically, consider the deck as a set $S\subset{1,2,\dots,52}$ of size $26.$ Then the reversed deck, with red and black swapped, is just the complement of $S.$ – Thomas Andrews Aug 05 '24 at 15:13
  • @Lulu The reason why that suffices is explained in step 5. We have total of 52! possible deck arrangements. Suppose wining is defined as black. S wins for more than half of the 52! deck arrangements. For every one of those arrangements for which S won, there is an alternative arrangement in total 52! arrangments where S losses but S would have produced a win if winning were defined as drawing a red card. – AzharKhan Aug 05 '24 at 15:40
  • @Lulu We have more than 52!/2 arrangements for which S wins and since we agreed that there are alternative arrangements for every one of those arrangements where S losses (but would have won if winning was defined as red) therefore we have more than 52!/2 arrangements where the strategy wins and similarly more than 52!/2 arrangements where the strategy loses which is a contradiction. – AzharKhan Aug 05 '24 at 15:40
  • I think you'll find that the solutions posted on the duplicate are a lot easier to follow. – lulu Aug 05 '24 at 15:45
  • @ThomasAndrews You're right. The example you gave works for both colors. I somehow did not considered that. That invalidates my proof right? – AzharKhan Aug 05 '24 at 15:51
  • @lulu I actually wanted to have my own unique solution to the problem, so that's why I posted the question. I wanted to verify my proof. – AzharKhan Aug 05 '24 at 15:58
  • As I say, I can't follow your proof. – lulu Aug 05 '24 at 15:59
  • I couldn't follow your proof entirely, because you were using vague terminology, only allowing me to guess successively what you meant. But it seemed like the direction you were going implied it was likely in error due to that kind of example. @AzharKhan – Thomas Andrews Aug 05 '24 at 19:55
  • This question is similar to: guess the color of the next card, what is the payoff of this game?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Lucenaposition Aug 06 '24 at 00:55
  • Related: https://math.stackexchange.com/questions/4827620/winning-strategy-for-game-guessing-if-next-number-is-prime – Henry Sep 11 '24 at 22:28
  • Hi @ThomasAndrews , I'm working on improving my ability to do proofs. So I wrote another proof for this question. I would really be thankful to you if you could have a look at it. No one seems to be responding. Here is the Link: https://math.stackexchange.com/questions/4969735/verification-of-proof-optimal-strategy-for-a-card-drawing-game?noredirect=1#comment10641121_4969735 – AzharKhan Sep 12 '24 at 17:11

1 Answers1

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I'm not sure I agree with the redefining winning part of your argument. Consider it for 2 or 4 cards and you'll see why I'm not too sure about it.

Solution:

Consider the similar game where instead of guessing the next card, you guess the last card in the deck. Then indeed this game is identical to before, since any strategy that would work in the former would work in the latter, since it must treat the cards remaining in the deck identically. But the probability of this card being black or red is $\frac{1}{2}$, irrespective of what cards you see, since the card is fixed once the deck is shuffled. Hence, you cannot beat $\frac{1}{2}$.

Sharky Kesa
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    This is a nice solution (+1) Perhaps it would be better to post it on the duplicate, as I expect this version will soon be closed. – lulu Aug 05 '24 at 15:13
  • @lulu I deleted my previous comment, which was too hasty. I now believe that Sharky Kesa's answer is both valid and elegant, and I have upvoted my answer. That is, they are right that you can equivalently assume that you need the last card to be black. – user2661923 Aug 05 '24 at 15:13
  • @lulu Fair enough, I've posted it there now – Sharky Kesa Aug 05 '24 at 15:16
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    I'll upvote it there as well. – lulu Aug 05 '24 at 15:19
  • @SharkyKesa Would this proof be correct: The probability of the next card being black at any point in the game is b/(r + b). This is greater than 1/2 only when b > r. Due to random shuffling, every possible arrangement of the deck is equally likely. So there will be half of the arrangements where starting from the top we will always have b >= r (thus making Pr(next card black) >= 1/2) and similarly half of the arrangments where r >= b thus making Pr(next card black) <= 1/2. Thus probability can never exceed 1/2. Would greatly appreciate your response – AzharKhan Aug 17 '24 at 12:11