Alice and Bob dice game Alice and Bob are playing a dice game. They take turns rolling a fair $6$-sided die, with Alice going first. The first person whose roll repeats one that has already been seen loses. Before play starts, what’s the probability that Alice will win the game?
The probability to roll repeats one on a single round is $\dfrac{1}{6}\times\dfrac{1}{6}=\dfrac{1}{36}$. In order for Alice to win the game, these repeats one must happen on the single rounds. The probability of rolling the first repeats one to on the $k-th$ round is $\dfrac{35}{36}^{k-1}\dfrac{1}{36}$. So the probability of Alice wins the game is $\sum_{n=1}\dfrac{35}{36}^{2(k-1)}\dfrac{1}{36}=\dfrac{1}{36}(1+\dfrac{35^2}{36^2}+(\dfrac{35^2}{36^2})^2+(\dfrac{35^2}{36^2})^3+...=\dfrac{1}{36}\dfrac{1}{1-\dfrac{35^2}{36^2}}=\dfrac{36}{71}$
Can someone check my solution for me please? Thank you.

An infinite sum seems inappropriate, since the game for sure ends on the $7$-th toss or earlier. Alicia wins if (i) Bob loses in the first round, or (ii) the second, or (iii) the third. 
(i) The probability that Bob loses in the first round is $\frac{1}{6}$. For whatever Alicia tossed, the probability Bob matches it is $\frac{1}{6}$.
(ii) For Bob to lose in the second round, the first three tosses have to be different, and there must be a match on the fourth toss. The probability the first $3$ tosses are different is $\frac{5}{6}\cdot \frac{4}{6}$. Given that they are different, the probability of a match on the fourth toss is $\frac{3}{6}$. So the probability Bob loses in the second round is $\frac{5}{6}\cdot \frac{4}{6}\cdot \frac{3}{6}$.
(iii) We leave it to you to find the probability Bob loses in the third round. Then add up.
