Winning at craps
Two fair dice are rolled. If the first total is 7 or 11 you win; if it is 2, 3 or 12 you lose. Any other first total (4, 5, 6, 8, 9 or 10) becomes your “point”, and the dice are then rolled again and again until either the point comes up again (you win) or a 7 comes up (you lose).
Find the probability of winning.
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
Once the point is set, rolls that are neither the point nor 7 change nothing. What is the probability that the point comes before a 7?
The number of ways to roll each total: 4 and 10 have 3, 5 and 9 have 4, 6 and 8 have 5, and 7 has 6.
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. Condition on the first roll. With a point set, only rolls of or 7 matter, so the point wins with probability , where is the number of ways to roll .
Step 1: the first roll. A natural (7 or 11) has ways out of 36: probability .
Step 2: making the point. For a point with ways, the probability of rolling it first is , and then of hitting it before a 7 is . The points come in pairs with the same :
Step 3: add up.
So the player wins about 49.3% of the time: the house edge on the pass line is about 1.4%.