Kili:  The probability is greater than zero.  Get over it.

alanback: Wow, that's rather uncharacteristically snappish! Chill out Karma Daddy-O!

playBunny: not like his more laid back easy going style on DG eh !!

paully:   Actually, "Get over it" pretty much summarizes my approach to most problems :-)

Kili: The probability of getting exactly ten doubles in 45 throws is (1/6)^10*(5/6)^35 = 2.79990879 × 10^-11.

The probability of getting exactly five double-6s in exactly ten doubles in 45 throws is (1/36)^6*(5/36)^4*(5/6)^35 = 2.89408577 × 10^-16.

Someone correct me if I'm wrong. ;-)

Thad:  I think the relevant quantities in context would be ten or more doubles and five or more boxcars.  As far as your calcs go, I think the 10 doubles figure has to be multiplied by the combinations of 10 items chosen from 45, which I think is 45!/(10!*35!).  A comparable adjustment needs to be made to the second calc, but I don't have time to work it out :-)

alanback: Right, I forgot the combination factors. Oops! Hurrying = Bad Math!! ;-)

Thad: You are wrong.

The chance of rolling exactly 10 doubles in 45 rolls is (1/6)^10 * (5/6)^35 * C (10, 45), where C (x, y) gives you the number of ways to pick x elements from a set of y elements. Rolling exactly 5 double sixes, and exactly 5 other doubles out of 45 rolls happens with chance (1/36)^5 * (5/36)^5 * C (5, 45) * C (5, 40).

The former is slightly less than 9%, the latter is slightly more than 0.1%.