Non-Poker Related Probability Question

[The Dude]

Here's a simple probability question. And, if you don't mind, can you explain to me how to do this type of problem. Ones like it seem to come up fairly often for me, and I'd like to not have to keep bothering other people for answers. Thanks.

Me and my opponent are playing table tennis. Let's assume that I have exactly a 62% chance of winning any given point. If that is the only information to go on, what is the probability that I will reach 21 points before my opponent? (Let's ignore things like, advantage depending on who's serve it is, win-by-two rule, and such.)

[Lost Wages]

Uh...maybe I'm misunderstanding your question but since you win by getting to 21 first and since you have a 62% chance of winning it's seems obvious that you have a 62% chance of getting to 21 first /images/graemlins/confused.gif.

Lost Wages

[TomCollins]

I'll give you a hint. His chances of winning are much higher than 62%. If he has a 99% chance of winning each point, his chances of losing 21 points is much less than 1%.

[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

[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).

[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

[pzhon]

Here are two possible approaches.

First, sum the probability of reaching each winning score over the possible scores. (Aloiz used this method.) To win 21-10, you need to win 20 out of the first 30 points, then win the last point, so the probability is (30 choose 20) p^21 q^10, where p=.62 and q=1-p=.38.

i=20
Sum (20+i choose i) p^21 q^i
i=0

= .941697.

Second, imagine playing 41 points, even if the match is decided before then. You win if you have at least 21 points. You reach a score of 31-10 with probability (41 choose 10) p^31 q^10.

i=20
Sum (41 choose i) p^(41-i) q^i
i=0

= .941697.

[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%.

[TomCollins]

41 Points is the easiest way by far, great idea! I now see my error, he DOES accurately account for this possibility. He has many different ways to get 20 of the first n (1 to 40), and then MUST win the last point.

[BruceZ]

[ QUOTE ]
41 Points is the easiest way by far, great idea!

[/ QUOTE ]

Both involve summing over a combinatoric times an exponential for 21 terms, so they are about the same difficulty computation-wise.

[TomCollins]

Anyone else have more of these interesting problems? I particularly liked the horse problem. Post a new thread if you do.