Dicey Game
For certain game, David has to roll two dice. If the sum of them is 7, he wins. If the two dice have the same number (a pair), he gets another chance. Otherwise, he loses. He can roll the dice at most 3 times, after which he loses even if he gets a pair.
What is the probability that he wins, given that he obtained a pair the first time? Please answer with a fraction of the form x/y.
Let \(W\) be the event that David wins by getting a 7, and let \(P\) the event of getting a pair, and
let \(L\) be the event of losing. The sample space is \(S = \{W, L, PW, PL, PPW, PPL\}\). Let \(A\) be the event of winning
and let \(B\) be the event in which David starts with a pair. We see that \(A = \{W, PW, PPW\}\) and \(B= \{PW, PL, PPW, PPL\}\).
We now compute the probability of each possible outcome: \[\Pr[W]=\frac{1}{6},\; Pr[L]=\frac{2}{3},\; Pr[PW]=\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36},\]
\[ \Pr[PL]=\frac{1}{6}\cdot\frac{2}{3},\;\Pr[PPW]=\frac{1}{36}\cdot\frac{1}{6},\; \Pr[PPL]=\frac{1}{36}\cdot\frac{5}{6}.\]
Then \(\Pr[ A \cap B ] = \frac{1}{36} + \frac{1}{216} = \frac{7}{216}.\)\(\Pr[B] = \frac{1}{36} + \frac{1}{9} + \frac{1}{216} + \frac{5}{216} = \frac{1}{6}.\)\(\Pr[ A\;|\; B] = \frac{7}{36}.\) Alternatively, we can observe that the condition is equivalent to removing one of David's tries.