Suppose you roll two dice. Find the probability that you roll an 8, provided you roll a 7 or 8 first?

Question

In the gambling game "craps," a pair of dice is rolled and the outcome of the experiment is the sum of the points on the up sides of the six-sided dice. The bettor wins on the first roll if the sum is 7 or 11. The bettor loses on the first roll if the sum is 2, 3 or 12. If the sum is 4, 5, 6, 8, 9, or 10, that number is called the bettor's "point." Once the point is established, the rule is as follows: If the bettor rolls a 7 before the point, the bettor loses; but if the point is rolled before a 7, the bettor wins.

Given that 8 is the outcome on the first roll, find the probability that the bettor now rolls the point 8 before rolling a 7 and thus wins. Note that at this stage in the game the only outcomes of interest are 7 and 8. Thus, find P(8|7 or 8)

Answer 1

I suppose we are rolling two dices with $6$ faces, and we know we have rolled a $7$ or an $8$ (summing the two scores of the dices), I'm not sure this is the case. Given that:

Every combo has the same probability to be rolled, ie $\frac{1}{36}$.

You can roll $7$ in 6 different ways, and $8$ in 5 different ways. They sum up to 11. Only $5$ of them are good so the probability is $\frac{5}{11}$.

If you don't understand that you can blindly apply the Bayes Formula:

$A$ is the event: you get a $8$

$B$ is the event: you get a $7$ or a $8$.

Of course $P(A)=\frac{5}{36}$, $P(B)=\frac{11}{36}$ and $P(B|A)=1$.

The Bayes Formula says:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

Thus $P(A|B)=\frac{5}{11}$.

Answer 2

In view of this answer, I think the question should be worded differently. It should be "Assuming that you have rolled 7 or 8, what is the probably that you have rolled 8?) "