[iversonian]

This problem got my interest. I spent a few minutes working on it and got an answer of the probability of you winning this prop bet at 14%.

Just to clarify the problem, you said that it was 101 games, first to 51. Except with ties, nobody might get to 51 before you hit 101 games. I'm guessing you meant first to 51 regardless. For my program, I assumed every game had a win/lose outcome, with a win% as given by the EV % on twodimes. I'm sure it doesn't change the outcome too much.

The probability of, in 101 games, of winning 0 games, losing 101 (with the nines, p = 0.553 per game) is:

(1-p)^101

The probability of winning one and losing 100 is:
(n choose r) * p^r * (1-p)^(n-r)
where n = games = 101, r = games won = 1
---- (n choose r) = n! / ( r!*(n-r)! )

Probability of winning 51 to 101 games is:
SUM r=51 to 101: (n choose r) * p^r * (1-p)^(n-r) = 0.8579

Probability of AK winning 0 to 50 games should be:
SUM r=0 to 50: (n choose r) * p^r * (1-p)^(n-r) = 0.8579
where p = .447
which checks out.

Public Function PropBet()
Dim n, r As Integer
Dim prob, sum As Double

n = 101
prob = 0.553
sum = 0#

For r = 51 To 101
sum = sum + Factorial((n)) / Factorial((r)) / Factorial((n - r)) * prob ^ r * (1 - prob) ^ (n - r)
Next

MsgBox sum

End Function