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.