I have just started a game with my wife. We rolled $2$ six sided dice twice in this order:
- She rolled $\{5,6\}$.
- Then i rolled $\{5,6\}$ as well.
- Then she rolled $\{2,5\}$.
- Then i rolled $\{2,5\}$ as well.
What were the odds of me getting the same numbers twice?
The key point is to determine the probability of a single match. Since "match" disregards order, that takes a computation. Rolls $(a,b)$ with $a\neq b$ have two matches, but rolls $(a,a)$ have only one. Thus there are two ways to get a match:
Either the first player throws a double and then the second throws the same double, $\frac 16\times \frac 1{36}$. Or the first player throws a non-double and then the second throws a match up to order, probability $\frac 56\times \frac 2{36}$
Thus the probability of a single match is $$\frac 16\times \frac 1{36}+\frac 56\times \frac 2{36}=\frac {11}{216}\approx .051$$
To do it twice we must square, so the final answer is $$\left(\frac {11}{216}\right)^2=\frac {121}{46656}\approx .0026$$