Re: Probability of the dealer having the same up card in BJ
The way you did it is for exactly 7/10 times with an exact required order. Do you care about the order, or do you simply want 7/10, in any order?
If you don't care about the order, you multiply (1/52)^7*(51/52)^3 by C(10,7). The C(10,7) is how many ways you can take 7 things out of 10 (120). Note that this equation is the same as Bruce's, with K = 7.
-- Homer
|