[fnord_too]

This is similar to a random walk type problem. You are basically looking for the probability you reach 21 before your opponent. I am tired and don't seem to have mental access to my bag of math tricks, so I'll give you a brute force sort of approach for a race to 3 points.

In a race to three points, the game can last 3, 4, or 5 rounds.

For three: You win .62^3 times, he wins .38^3 times
For four: You win .62^3*.38*c(4,3) times he wins .38^3*.62*c(4,3) times
For five: You win .62^3*.38^2*c(5,3) times and he wins .38^2*.62^2*c(5,3) times

Take the weighted average of your win percentage over the weighted average of his. There are 2 ways end in 3 rounds, 8 in 4, and 20 in 5.

I think there is an easy generalized solution to this, but I may be wrong. There is at least an algorithmic one that you can probably extrapolate from the above example if you follow the notation. (if the notation is foreign, I can explain it better, but as I said, I am tired, and as I didn't say, I am lazy).