What is the probability of a person winning a dice game on the first roll, given that he/she won the game? Let Jill and Jack be two people. They play a game with a dice. The first person to roll $6$ wins the game. Jill goes first.
I want to compute the probability that Jack won on his first roll given that he won the game.
It can be shown that the probability of Jill winning is $6/11$ and the probability of Jack winning is then $5/11$. Obviously the probability of Jill winning on her first roll is $1/6$, but since Jack rolls second, the probability is different, although I am not sure what. Suppose it is $x$, then this would be a conditional probability question: i.e., computing the probability of Jack winning first intersected with the probability of him winning the game, so $x+5/11$, divided by $5/11$.
 A: The probability of Jack winning on his first roll includes the probability that Jill doesn't win on her first roll ($\frac56$). Hence $\frac56×\frac16=\frac5{36}$. The conditional probability is then
$$\frac{5/36}{5/11}=\frac{11}{36}$$
A: According to Bayes' formula,
\begin{align*}
\mathbb{P}(\text{Jack wins on first roll} \, | \, \text{Jack wins})
&=
\mathbb{P}(\text{Jack wins} \, | \, \text{Jack wins on first roll})\frac{\mathbb{P}(\text{Jack wins on first roll})}{\mathbb{P}(\text{Jack wins})} \\
&=
\frac{\mathbb{P}(\text{Jack wins on first roll})}{5/11}.
\end{align*}
The probability that Jack wins on first roll is
\begin{align*}
\mathbb{P}(\text{Jack wins on first roll})
&=
\mathbb{P}(\text{Jack wins on first roll} \, | \, \text{Jill doesn't win on first roll})\mathbb{P}(\text{Jill doesn't win on first roll}) \\
&=1/6 \cdot 5/6 = 5/36,
\end{align*}
since the event that Jill doesn't win on first roll is contained in the event that Jack wins on first roll, and since the dice are independent.
All in all,
\begin{align*}
\mathbb{P}(\text{Jack wins on first roll} \, | \, \text{Jack wins})
=
\frac{5/36}{5/11} = \frac{11}{36}.
\end{align*}
