I want to calculate the probability to get a set of cards two times in a card game, respectively how many two times I would have to play to get it with a specific probability. The card game is "Schafkopf" which is very popular in Germany: It's played with $32$ cards and $4$ players, so every player gets 8 cards. I assume that there are $\pmatrix{32\\8}$ permutations, so the probability is $\frac{1}{\pmatrix{32\\8}}$ which is about $10 000 000$. For the formula I use a Bernoulli-chain (don't know if that's the English name; It's this one: $$P^n_p(X=k)=\pmatrix{n\\k}\cdot p^k \cdot (1-p)^{n-k}$$ I hope you know what I mean). Assuming that $p = \frac{1}{\pmatrix{32\\8}}$ my formula would be: $$P^n_p(X=2)=\pmatrix{n\\2} \cdot p^2 \cdot (1-p)^{n-2}$$
Is this formula correct? Or have I made a mistake?
First we give a precise version of the problem. We play the game twice. The probability that we get the same cards in both games is indeed $\frac{1}{\binom{32}{8}}$, which is about $9.5\times 10^{-8}$.
For simplicity call this number $p$.
The next day we again play the game twice, and then we play the game again twice the next day, and so on for $N$ days. We want to find the number $N$ of days we must play so that with probability $a$ we get the same cards twice on at least $1$ day.
The probability we get the same cards at least twice is $1$ minus the probability we get the same cards $0$ times on the $N$ days. And the probability we get the same cards $0$ times is $(1-p)^N$. So we want $$1-(1-p)^N=a,$$ or equivalently $$(1-p)^N=1-a.$$ Take the logarithm (say the natural log, but it doesn't matter). We get $$N\log(1-p)=\log(1-a),$$ and now we know $N$.