- #1

Adeimantus

- 113

- 1

You make a sequence of rolls with a pair of dice. What is the probability that all six place numbers, 4-5-6-8-9-10, come up in any order before you roll a 7? In his book New Complete Guide to Gambling, John Scarne gave the calculation

P = (24/30)(20/26)(16/22)(12/18)(8/14)(4/10) (roughly 13.66 to 1 odds)

without comment as to why it was right. I have my doubts that it is exactly right, but I think it is pretty close. It looks to me like he decided that since each number comes up an

**average**of 4 ways, you can just calculate the probability as if they all came up in exactly 4 ways. I don't think this is right, but it's awfully close to the answer I got using a different approximation: I calculated the probability of getting 4-5-6-8-9-10 in that order before 7

P(4,5,6,8,9,10 in order before 7) = (3/30)(4/27)(5/23)(5/18)(4/13)(3/9)

and assumed that all 6! = 720 such sequences have roughly the same probability. This gave me a total probability, expressed as odds, of about 14 to 1.

Is there an easy way to get the exact answer that doesn't take pages and pages of arithmetic? Thanks.