Alice and Bob are playing dices, Alice begins.
If the current player gets a 6, he wins. If he gets 4 ou 5, he plays again. Else, the other player plays.
Let $p_n$ (resp. $q_n$) be the probability of the event $P_n$: "Alice (resp. Bob) plays at the turn $n$", and $r_n$ be the probability of the event "The game is over at the turn $n$".
- Determine a recurrence relation between $p_{n+1}$, $p_n$ and $r_n$.
I think I am ok for this question: $$p_{n+1} = \frac 2 6 p_n + \frac 1 2 q_n = \frac 13 p_n + \frac 1 2 (1 - r_n-p_n) = \frac 1 2 \overline{r_n}-\frac 1 6 p_n,$$ and $$q_{n+1} = \dots = \frac 1 3 \overline{r_n} + \frac 1 6 p_n.$$ we easily have $\overline{r_n}= {\left(\dfrac{5}{6}\right)}^{n-1}$, which can be insert in the above formulae.
- Deduce the probability that Alice wins in $n$ turns, and then in less than $n$ turn.
I'm stuck here : we want to compute : $P(P_n \cap R_{n+1} ) = P(R_{n+1} | P_n ) P(P_n) = \frac 1 6 p_n$... But I don't see how to compute $p_n$.