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probability question help
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Scott Collier
science forum beginner


Joined: 30 Jun 2006
Posts: 1

PostPosted: Fri Jun 30, 2006 12:34 pm    Post subject: Re: probability question help Reply with quote

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 =Cool
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



"Sky Kalkman" <skyking162@gmail.com> wrote in message
news:1151433178.987226.37640@b68g2000cwa.googlegroups.com...
Quote:
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
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iandjmsmith@aol.com
science forum beginner


Joined: 13 Sep 2005
Posts: 15

PostPosted: Fri Jun 30, 2006 11:00 am    Post subject: Re: probability question help Reply with quote

Sky Kalkman wrote:
Quote:
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

Consider the simpler mini-game where the mini-game terminates when the
first head is flipped.

When A starts, the probability that A wins is a/(a+b-a.b) and B starts
the next mini-game if there is one.
When B starts, the probability that A wins is 1-b/(a+b-a.b) or
a.(1-b)/(a+b-a.b) and A starts the next mini-game if there is one.

Now let P(m,n,True) be the probability that A wins starting from A has
m heads and B has n and A starts and let P(m,n,False) be the
probability that A wins starting from A has m heads and B has n and B
starts.

P(m,n.True) = a/(a+b-a.b) . P(m+1,n,False) + b.(1-a)/(a+b-a.b) .
P(m,n+1.True)
P(m,n.False) = a.(1-b)/(a+b-a.b) . P(m+1,n,False) + b/(a+b-a.b) .
P(m,n+1.True)

We also have P(10,n,b) = 1 for n <= 9 and P(m,10,b) = 0 for m <= 9

So we can calculate P(x,y,True) and P(x,y,False), in the order (x,y) =
(9,9),(9,Cool...(9,n),(8,9),...(8,n)...(m,9)...(m,n)


This is the same as Dan's solution (or is intended to be, at least) but
allows you to solve the 200 equations in 200 unknowns in a simple
fashion.


Ian Smith
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Ray Koopman
science forum Guru Wannabe


Joined: 25 Mar 2005
Posts: 216

PostPosted: Thu Jun 29, 2006 10:31 pm    Post subject: Re: probability question help Reply with quote

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:
Quote:
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
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danheyman@yahoo.com
science forum beginner


Joined: 18 Jul 2005
Posts: 33

PostPosted: Thu Jun 29, 2006 7:09 pm    Post subject: Re: probability question help Reply with quote

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:
Quote:
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
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Sky Kalkman
science forum beginner


Joined: 27 Jun 2006
Posts: 1

PostPosted: Tue Jun 27, 2006 6:32 pm    Post subject: probability question help Reply with quote

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
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