Non-Poker Related Probability Question

[aloiz]

He has a 62% chance of winning any given point, not the match itself.

Not completly sure if this is right but I get:
.62^21 + Sum(.62^21*.38^i*C(20+i,i),i,1,20) = 94.1%


aloiz

[Lost Wages]

Ah, knew there had to be something. I was missing the "each point" part.

Lost Wages

[TomCollins]

This is wrong because some of the ways it is not possible to win 21-20 have your opponent losing the last point. For example in a 5 point match, it is impossible to have this sequence:

WWWWWLL

[BruceZ]

[ QUOTE ]
This is wrong because some of the ways it is not possible to win 21-20 have your opponent losing the last point. For example in a 5 point match, it is impossible to have this sequence:

WWWWWLL

[/ QUOTE ]

Aloiz has it right. I got the same thing. He is only counting sequences that end in a win for the 62% guy. That is accomplished by the term C(20+i,i). For example, if the final score was 21-20, the wins for the opponent can occur in any of the first 20+20 = 40 points, but not on the final 41st point.

The probability of winning 21-0 is (0.62)^21. The probability of winning 21-1 is 21*(0.62)^21*(0.38)^1. We multiply by 21 because the opponent can win any of the first 21 points but not the last point. The probability of winning 21-2 is C(22,2)*(0.62)^21*(0.38)^2. And so on up to the probabiilty of winning 21-20 is C(40,20)*(0.68)^21*(0.32)^20. All together we have:

(0.62)^21*[ 1 + 21*(0.38)^1 + C(22,2)*(0.38)^2 + C(23,3)*(0.38)^3 + ... + 40*(0.38)^20 ]

= (0.62)^21*sum{n = 0 to 20}C(20+n,n)*(0.38)^n

=~ 94.1697%.