Ok, game is as follow, with spanish cards (you can do it with poker cards using the As as a 1)
You shuffle, put the deck face bottom, and start turning the cards one by one, saying a number each time you turn a card around —> 1, 2, 3; 1, 2, 3; etc. If when you say 1 a 1 comes out, you lose, same with 2 and 3. If you finish the deck without losing, you win.
I know some basics of probabilities, but is there a way to calculate the probability of winning the game, given a random shuffled deck?
Best Answer
For $i,j\in\{1,2,3\}$, let $a_{i,j}$ denote the number of $i$ cards being dealt with number $j$ spoken. We have $\sum_j a_{i,j}=4$ and for a winning game $a_{i,i}=0$. The number of winning positions for a given $(a_{i,j})$ is $$\frac{18!}{a_{2,1}!a_{3,1}!(18-a_{2,1}-a_{3,1})!}\cdot\frac{17!}{a_{1,2}!a_{3,2}!(17-a_{1,2}-a_{3,2})!}\cdot\frac{17!}{a_{1,3}!a_{2,3}!(17-a_{1,3}-a_{2,3})!}. $$ We need to sum this over all $(a_{i,j})$ and divide by the total count $$ \frac{52!}{4!4!4!40!}.$$ (Actually, we need just let $a_{1,2}, a_{2,3}, a_{3,1}$ run from $0$ to $4$ and this determines $a_{1,3}=4-a_{1,2}$ etc.) The final result is $$p=\frac{58388462678560}{7151046448045500}=\frac{24532967512}{3004641364725}\approx 0.008165 $$ (I just noted that Harold has performed a Monte Carlo simulation with matching result)