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I have a very simple problem.

Lets assume that I have a well shuffled deck of 52 cards. I start drawing the top card always and when the card matches its rank I lose. J=11 Q=12 K=13.

If there were only 13 cards I could easily use the $\ \frac{!n}{n!}$ for derangements in order to solve this. The problem is that there are 52 cards so when I pass 13 I start from 1 again so I don't know what is the probability to win. Example of the game

1st card: 4 - Continue
2nd Card: A - continue
3rd Card: K - Continue
4th Card: K - Continue
5th Card: 6 - Continue
6th Card: 9 - Continue
7th Card: 10 - Continue
8th Card: A - Continue
9th Card: J - Continue
10th Card: 3 - Continue
11th Card: 2 - Continue
12th Card: 8 - Continue
13th Card: A - Continue
1st card: 5 - Continue
2nd Card 2 LOSE

So actually I have to count from 1 to 13 4 times and if I draw all 52 cards then I win. What's the probability?

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    See also the answers at [this duplicate question](https://math.stackexchange.com/questions/495991).2018-07-31

1 Answers 1

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The answer is given in this Wikipedia section on generalized derangements. The number of permutations of $52$ cards with $4$ copies each of $13$ ranks that leave none of the ranks in place is

$ \begin{align} &\int_0^\infty (4!\cdot L_4(x))^{13}\mathrm e^{-x}\,\mathrm dx =\int_0^\infty\left(x^4-16x^3+72x^2-96x+24\right)^{13}\mathrm e^{-x}\,\mathrm dx \\\\\\ =&1309302175551177162931045000259922525308763433362019257020678406144 \end{align} $

(computation), where $L_4(x)$ is the fourth Laguerre polynomial. Since there are

$ 52!=80658175170943878571660636856403766975289505440883277824000000000000 $

permutations of the cards in all, your chances of winning are

$ \frac{1309302175551177162931045000259922525308763433362019257020678406144}{80658175170943878571660636856403766975289505440883277824000000000000} $

$ =\frac{4610507544750288132457667562311567997623087869}{284025438982318025793544200005777916187500000000} $

$ \approx0.0162\;. $

See also What's the General Expression For Probability of a Failed Gift Exchange Draw, Laguerre polynomials and inclusion-exclusion and Counting some special derangements.

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    I love that you have written out the numerator and denominator entirely as integers and not expressions. It is hauntingly beautiful in a way.2012-12-01