The game of craps and dice The game of craps involves the repeated tossing of a pair of dice. In the game of craps, Nancy throws a five on the first toss of a pair of dice. ["Five" means that the sum of the number of dots on both dice is equal to five.] The rules call for her to keep on tossing the dice until she gets a five again, in which case she wins, or until she gets a seven, in which case she loses. What is the probability that she will win. My attempt: $P$ $($Sum of five occurs$)=$ $(2/6)+ (1/6)(2/6)+(1/6)^2(2/6)+...$ $P$ $($Sum of five occurs$)$ $=$ $\sum_{x=1}^\infty (2/6)(1/6)^{x-1}$ $P$ $($Sum of five occurs$)=$ $(2/6)/1-(1/6)= 2/5$ I know the answer is correct. However I am not sure if this is the way how you would show the working. Is there any other way we can show how to get this probability with a formula?

On each roll, the probability of total 5 is $4/36$, and the probability of total 7 is $6/36$.

Notice that if Nancy rolls any total other than 5 or 7, her situation has not changed at all: she still needs to keep rolling until the dice show 5 or 7, and will get the same results. So her probability of a win after that roll is the same as it was before.

$P(\mbox{win}) = P(\mbox{roll 5}) P(\mbox{win} \mid \mbox{roll 5}) +P(\mbox{roll 7}) P(\mbox{win} \mid \mbox{roll 7}) + P(\mbox{roll other}) P(\mbox{win} \mid \mbox{roll other})$

$P(\mbox{win}) = \frac{4}{36} \cdot 1 + \frac{6}{36} \cdot 0 + \frac{26}{36} P(\mbox{win})$

$\frac{10}{36} P(\mbox{win}) = \frac{4}{36}$

$P(\mbox{win}) = \frac{2}{5}$.