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A friend shared this article, which includes a description of this solitaire card game:

On snowy afternoons, you like to play a solitaire “game” with a standard, randomly shuffled deck of 52 cards. You start dealing cards face up, one at a time, into a pile. As you deal each card, you also speak aloud, in order, the 13 card faces in a standard deck: ace, two, three, etc. (When you get to king, you start over at ace.) You keep doing this until the rank of the card you deal matches the rank you speak aloud, in which case you lose. You win if you reach the end of the deck without any matches.

What is the probability that you win?

My friend observed "Surely it can't be as easy as just $(\frac{12}{13})^{52}$, right?" I agreed and tried to perform a rigorous analysis, but found that the problem quickly became intractable--or at least more effort than seemed worthwhile--and decided to just wait for the solution to be posted.

When it arrived, it said that $(\frac{12}{13})^{52}$ ≈ 1.5573% was indeed the solution. I have my doubts, though.

First of all, I ran simulations of a million games, and found a winning percentage of about 1.6361%, which differs from the given solution to a degree that appears significant.

Secondly, my intuition tells me that, whatever card is drawn first, if it's not an ace (ending the game immediately), then there are only three cards of that rank remaining, improving the chances of avoiding a loss each time the player speaks that rank aloud later in the game. Similar logic holds for every draw but the last, leading to a slightly elevated winning chance, which agrees with my simulations.

So: Am I correct? If so, is there a straightforward formula for the winning probability?


marked as duplicate by user940 Jan 8 '17 at 22:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It's not a duplicate. There is no choice in the game of this question. In the other question the player chooses a rank of the card. $\endgroup$ – Paul Jan 8 '17 at 22:43
  • $\begingroup$ It looks like a dupe to me. Unfortunately there don't really seem to be any distinctive search terms that could have helped me find it. $\endgroup$ – Sean Jan 8 '17 at 22:50
  • $\begingroup$ The other answer has no choice, either, so it is a duplicate. It also quotes a calculated value that agrees well with your simulation. $\endgroup$ – Ross Millikan Jan 8 '17 at 23:14
  • $\begingroup$ Yes. Like I said: a dupe. $\endgroup$ – Sean Jan 9 '17 at 17:22

The important part is "in order." You don't get to choose what card to call. You have to go in order. There is no improvement made by your choices because there are no choices to be made. Even if more cards of a rank have come up, it is also possible that fewer of that rank will come up. It doesn't matter, because the outcome of the game is fixed before the first card is turned.


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