Math prob

All of these new crap games being posted on here reminded me of a math question for craps.

What are the odds three 6-4's (easy 10) will roll before a natural number (2, 3, 7, 11 or 12) is rolled?

What are the odds one 6-4 (easy 10) and one 6-3 will roll before a natural number?

Thanks...

Quote: allinriverking

All of these new crap games being posted on here reminded me of a math question for craps.

What are the odds three 6-4's (easy 10) will roll before a natural number (2, 3, 7, 11 or 12) is rolled?

What are the odds one 6-4 (easy 10) and one 6-3 will roll before a natural number?

Note that there is 1 way to roll a 2, 2 to roll a 3, 6 to roll a 7, 2 to roll an 11, and 1 to roll a 12, so there are a total of 12 ways to roll a natural number.

For the three 6-4 problem, first, calculate the probability that one 6-4 will be rolled before a natural number.
In this case, we can ignore all other rolls; there are two ways to roll a 6-4 (namely, 6-4 and 4-6) and 12 to roll a natural, so the probability of rolling the 6-4 first is 2/14, or 1/7.
The probability of rolling two 6-4s first = the odds of doing it once x the odds of doing it a second time = 1/7 x 1/7 = 1/49
The probability of rolling three 6-4s first = the odds of rolling 2 6-4s first x the odds of doing it again = 1/49 x 1/7 = 1/343.
Thus the probability = 1/343, and the odds are 342-1 against.

For one 6-4 and one 6-3, it is the sum of:
(a) Rolling a 6-4 before a 6-3 or a natural, then rolling a 6-3 before a natural = (4 / (4 + 12)) x (2 / (2 + 12)) = 1/28
(b) Rolling a 6-3 before a 6-4 or a natural, then rolling a 6-4 before a natural = (4 / (4 + 12)) x (2 / (2 + 12)) = 1/28
The sum = 1/28 + 1/28 = 1/14, so the odds are 13-1 against.
(The reason you check for 6-4 before 6-3 in (a) is, if you roll 6-3 before 6-4, it becomes part of (b); similarly, in (b), if you roll 6-4 before 6-3, it becomes part of (a)).