In a dice game craps, Alex rolls a pair of fair dice.
- If he gets $7$ on the first roll, he wins immediately
- If the result is any number $\neq 7$, he keeps rolling the dice until he gets that number again (he wins) or he get a $7$ (he loses).
Example: if Alex gets $8$ on the first roll, he keeps rolling the dice. After a number of rolls, if he gets $8$, he wins. If he gets $7$, he loses. I think of a probability but it seems infinite cases. I'm thinking of a conditional probability.
- What is the probability that Alex will win this game?
- What is the probability that Alex's final roll will be $7$?
The probability for a $7$ is $\frac 16$, the probability for $x\ne 7$ varies from $\frac1{36}$ to $\frac5{36}$. For example, when trying to get $8$ (probability $\frac{5}{36}$), the probability of succeeding before getting a $7$ is $\frac{5}{5+6}$ as the $36-5-6$ rolls giving neither $7$ not $8$ can be ignored. The probability of first rolling $8$ and then winning is therefore $\frac5{36}\cdot \frac{5}{5+6}$. By adding, the total probability of winning is $$\frac1{36}\cdot \frac{1}{1+6}+\frac2{36}\cdot \frac{2}{2+6}+\frac3{36}\cdot \frac{3}{3+6}+\frac4{36}\cdot \frac{4}{4+6}+\frac5{36}\cdot \frac{5}{5+6}+\frac6{36}+\frac5{36}\cdot \frac{5}{5+6}+\frac4{36}\cdot \frac{4}{4+6}+\frac3{36}\cdot \frac{3}{3+6}+\frac2{36}\cdot \frac{2}{2+6}+\frac1{36}\cdot \frac{1}{1+6} = \frac{6557}{13860}.$$ The probability that the game end by rolling a $7$, is $$1- \frac{6557}{13860}+\frac16=\frac{9613}{13860}. $$