probability question help

Sky Kalkman · science forum beginner Joined: 27 Jun 2006 Posts: 1

I'm working on a sports-related probability question and need some help
from whoever's interested:

Team A has probability a of flipping heads each round.
Team B has probability b of flipping heads each round.

The teams alternate flipping one coin each round until one team scores
ten points (ten heads).

Given any score in the middle of the game, how can I compute the
probability of Team A winning?

For example, if Team A is winning 6-3 and it's Team A's turn, I want
the probability that Team A flips 4 heads before Team B flips 7 heads.

I seem to be running into infinite series that aren't easy to simplify.

Thanks for your help

-Sky

danheyman@yahoo.com · science forum beginner Joined: 18 Jul 2005 Posts: 33

This is a type of random walk problem. It's not the classical RW
problem because each players score is nondecreasing. I suggest writing
recursions as one does for the RW. Let p(i,.j) be the probability that
A wins when he has i heads and B has j heads and it's A gets to flip,
i,j<or=10. Let q(i,j) be the same when B gets to flip. Start with
p(10,j)=1, j<10 and p(i,10)=0,i<10 [and the same for q]. Then

p(i,j)=aq(i+1,j)+(1-a)q(i,j)
q(i,j)=(1-b)p(i,j)+bp(i,j+1)

and you start the recursion from i=9 and j=9 and keep decreasing iand j
until they reach 0.

There's a chance that the recursions can be solved analytically (at
least for the probability generating functions of p and q), but at
worst, this is a system of linear equations in about 200 unknowns. [I
didn't calculate the number of unknowns exactly, but there are about
10X10=100 p's and the same number of q's.]

Dan Heyman

Sky Kalkman wrote:

Ray Koopman · science forum Guru Wannabe Joined: 25 Mar 2005 Posts: 216

If A has m points, B has n points, and it's A's turn to flip, then
P(A wins) = P(X + 10-m <= Y + 10-n) = P(X - Y <= m-n), where X and Y
are independent Negative Binomial variables with parameters (10-m,a)
and (10-n,b). But I too have not found a way to express P(A wins)
without involving infinite series.

Sky Kalkman wrote:

iandjmsmith@aol.com · science forum beginner Joined: 13 Sep 2005 Posts: 15

Sky Kalkman wrote:

Scott Collier · science forum beginner Joined: 30 Jun 2006 Posts: 1

I don't know if this is right but my logic is as follows.

At the beginning of the game
p(A) = p (B)
hence p(A) = 0.5 , p(B) = 0.5

If at any stage in the "match" - when the scores are equal (ie Score A = 8,
Score B =

p(A) = p(B), hence p(A) = 0.5 , p(B) = 0.5
----
If lets say A gets a "head-start" and starts with a score of 5 (B starts at
0), then
p(A) = 2 x p(B)
p(A) = 0.66666666
p(B) = 0.33333333
-----

So the way I would work it out would be as follows. If ..
TS = target score,
AS = A team's score,
BS = B team's score.

p(A wins) = (TS-BS) / [(TS-AS)+(TS-BS)]
p(B wins) = (TS-AS) / [(TS-AS)+(TS-BS)]

ie. When score is 5-0 (in favour of team A)

p(A wins) = (10-0) / [(10-5)+(10-0)] = 10/15 = 2/3 = 0.666666
p(B wins) = (10-5) / [(10-5)+(10-0)] = 5/15 = 1/3 = 0.333333

When score is even at 4-4
p(A wins) = (10-4) / [(10-4)+(10-4)] = 6/12 = 1/2 = 0.5
p(B wins) = (10-4) / [(10-4)+(10-4)] = 6/12 = 1/2 = 0.5

But as I said before - I might be wrong

Scott

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