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It is known that the possible shuffles of a deck of cards is $52!$, or ~$80658175170943878571660636856403766975289505440883277824000000000000$ different combinations.

I have become aware of a game known as Idiot's Delight. The game is played in this manner:

  • A deck is shuffled, and held in your left hand, face down.
  • You begin pulling cards from the back of the deck, flipping them face up on top of the deck. You continue this process.
  • When you have at least 4 cards, begin looking at the last 4 cards.
  • If the first and last cards of the last four cards are the same suit, you may remove the two cards between them.
  • If the first and last cards of the last four cards are the same value, you may remove all four cards.
  • If the last four cards do not satisfy these conditions, continue drawing from the back of the deck, until you run out of face-down cards.

Look at this example

[Face Down Cards][$A$ ♠][$2$ ♣][$3$ ♦][$4$ ♠]

Here, the $2$♣ and $3$♦ can be removed, and then play continues.

You can win a game by correctly "matching" all the cards, or removing all cards from the face down section and containing no cards face up. This forces the last play to have the first and fourth card match in number.

Correct me if I'm wrong, but I believe that this game is possible to be won, albeit if difficult. Assuming it is possible to be won, how would I calculate how many potential shuffles out of $52!$ would be possible to be won, if the game is played perfectly, and how could I design a proof for this?

Perfect play, for the sake of this question, encompasses not knowing the order of the deck, but making the decision to remove cards whenever there is a possibility of such.

EDIT: Assume for the sake of this question that a shuffle is a completely random selection of one of $52!$ permutations.

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  • $\begingroup$ Be careful - the word "shuffle" is a confusing term, there are $52!$ permutations of a deck. A "shuffle," on the other hand, is the result of an operation, and how many shuffles there are depends on the method of shuffling. $\endgroup$ – Thomas Andrews May 29 '15 at 21:42
  • $\begingroup$ Excellent point! The question has been edited for clarification. $\endgroup$ – Sam Weaver May 29 '15 at 21:44
  • $\begingroup$ By the way, you haven't said what defines winning. $\endgroup$ – Thomas Andrews May 29 '15 at 21:44
  • $\begingroup$ I definitely intended to include that, I must have missed it when I wrote the question. Edited. $\endgroup$ – Sam Weaver May 29 '15 at 21:46
  • $\begingroup$ It is possible that the percentage of shuffles that can be won assuming truly "perfect" play (i.e. you know all the cards and their ordering, including future ones) is much greater than the EXPECTED chances of winning assuming the shuffle is random and you don't know what the ordering of the future cards will be, just the cards you have seen. Which one do you mean? Note that assuming omniscent "perfect" play (knowing future cards in order) is perhaps unreasonable if you are trying to assume some sort of "randomness" in the original set up. $\endgroup$ – user2566092 May 29 '15 at 22:06

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