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I'm only a high school senior, but I love math a lot, and my teachers say I am pretty good at it.

I have a solitaire game that I like which involves no skill whatsoever. You take a standard deck of 52 cards and flip up cards one at a time saying, "Ace, two, three, four, etc., Jack, Queen, King, ace,..." as you flip the cards.

If you say the same rank as the card you flip up, you lose. So an Ace can't end up in the 1st, 14th, 27th, or 40th position. I know how combinatorics work, but since each card's probability of being flipped up depends on the rank of every single previous card, I can't figure out how to work this problem.

What I tried to do was find the probability that a four card combination had none of a specific rank in it (because there are four places where no rank can be for each rank if you want to win) and then take that to the thirteenth power to accommodate each rank.

The resulting probability was about 1 percent, which seems to fit as I've played about 300 games of it and won three times. However, I want to check this answer and see if I got it right and if not, how to get the right answer.

user285953
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  • This is a case where counting the number of winning decks (then dividing by the total number of decks) is going to be a better way of determining the exact probability than trying to do a string of 52 conditional probabilities (as you've noticed). If there were 52 distinct ranks, then this problem be exactly about "derangements"; so I recommend you read about those to get an idea of how to approach this problem. https://en.wikipedia.org/wiki/Inclusion–exclusion_principle#Counting_derangements – Greg Martin Nov 01 '15 at 19:44
  • There's also a great exposition of the inclusion-exclusion method, as applied to derangements and more generally, in the book Concrete Mathematics by Graham, Knuth, and Patashnik. (And I guess I should say explicitly that your method of taking a probability to the 13th power isn't going to be exactly right, even though it's pretty close, because the various events are not independent of one another.) – Greg Martin Nov 01 '15 at 19:45
  • These generalized derangement problems can be solved using orthogonal polynomials; see for example @joriki's answer to http://math.stackexchange.com/questions/73341/whats-the-general-expression-for-probability-of-a-failed-gift-exchange-draw – Tad Nov 01 '15 at 20:13

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I looked at the link at wikipedia and used the derangement formula with thirteen cards. Then I took this result to the fourth power to account for all four suits. I got approximately 1.8%, a reasonable statistic. Did I do this right, or did I fail to take another variable into account?

Edit: I used a rook polynomial to calculate that the exact probability of winning the game is 1309302175551177162931045000259922525308763433362019257020678406144/52! or approximately 1.6 percent.

user285953
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