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

Posted: Tue Jun 27, 2006 6:32 pm Post subject:
probability question help



I'm working on a sportsrelated 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 63 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

Posted: Thu Jun 29, 2006 7:09 pm Post subject:
Re: probability question help



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)+(1a)q(i,j)
q(i,j)=(1b)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 sportsrelated 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 63 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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Ray Koopman science forum Guru Wannabe
Joined: 25 Mar 2005
Posts: 216

Posted: Thu Jun 29, 2006 10:31 pm Post subject:
Re: probability question help



If A has m points, B has n points, and it's A's turn to flip, then
P(A wins) = P(X + 10m <= Y + 10n) = P(X  Y <= mn), where X and Y
are independent Negative Binomial variables with parameters (10m,a)
and (10n,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 sportsrelated 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 63 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

Posted: Fri Jun 30, 2006 11:00 am Post subject:
Re: probability question help



Sky Kalkman wrote:
Quote:  I'm working on a sportsrelated 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 63 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 minigame where the minigame terminates when the
first head is flipped.
When A starts, the probability that A wins is a/(a+ba.b) and B starts
the next minigame if there is one.
When B starts, the probability that A wins is 1b/(a+ba.b) or
a.(1b)/(a+ba.b) and A starts the next minigame 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+ba.b) . P(m+1,n,False) + b.(1a)/(a+ba.b) .
P(m,n+1.True)
P(m,n.False) = a.(1b)/(a+ba.b) . P(m+1,n,False) + b/(a+ba.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,...(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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Scott Collier science forum beginner
Joined: 30 Jun 2006
Posts: 1

Posted: Fri Jun 30, 2006 12:34 pm Post subject:
Re: probability question help



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 "headstart" 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) = (TSBS) / [(TSAS)+(TSBS)]
p(B wins) = (TSAS) / [(TSAS)+(TSBS)]
ie. When score is 50 (in favour of team A)
p(A wins) = (100) / [(105)+(100)] = 10/15 = 2/3 = 0.666666
p(B wins) = (105) / [(105)+(100)] = 5/15 = 1/3 = 0.333333
When score is even at 44
p(A wins) = (104) / [(104)+(104)] = 6/12 = 1/2 = 0.5
p(B wins) = (104) / [(104)+(104)] = 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 sportsrelated 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 63 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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