23 Aug 06
Originally posted by ThudanBlunderThe probability of Bill getting "m" heads with his (n+1) coins is:
Bill has n+1 coins and Ben has n coins. They both throw all of their coins simultaneously and observe the number that come up heads. Assuming all the coins are fair, what is the probability that Bill gets more heads than Ben?
(n+1)choose(m)/2^(n+1)
The probability of Ben getting (m-1) or less heads with his "n" coins is:
SUM{(n)choose(i)/2^n}, where i=0...(m-1)
The probability of both these events happening simultaneously is simply their product. Now, to count all the possible outcomes we use the double sum:
P = SUM{(n+1)choose(j)/2^(n+1) * SUM{(n)choose(i)/2^n}}, where j=1...n+1, and i=0...j-1
For example, for n=10 at some point we would calculate the probability that Bill got 3 heads (j=3), and Bill got 0, 1, and 2 heads (i=0,1,2), at some other point we would calculate the probability that Bill got 8 heads (j=8), and Ben got 0, 1, 2, 3, 4, 5, 6, and 7 heads (i=0,1,2,3,4,5,6,7), etc...
I don't know if there's a good way to simplify the above expression, or if there's an easier "aha!" solution, but this one will give you the correct answer if you grind it out.
23 Aug 06
3 edits
Originally posted by PBE6I'm not sure if this is worth an "aha!":
I don't know if there's a good way to simplify the above expression, or if there's an easier "aha!" solution, but this one will give you the correct answer if you grind it out.
Probability of Bill getting more heads = probability of Bill getting more tails = 50% (as they can't both happen).
23 Aug 06
2 edits
Originally posted by ThudanBlunderYes I had:
I'm not sure if this is worth an "aha!":
Probability of Bill getting more heads = probability of Bill getting more tails = 50% (as they can't both happen).
If A has k heads and n+1-k tails and B has j heads and n-j tails then
P( k > j ) = P( n+1-k > n-j ) (*)
by symmetry of heads/tails. Finally, for integers k and j
n+1-k > n-j if and only if ( k < j or k=j )
so the two events in (*) partition the sample space so each have probability 1/2.
Yours is definitely "aha" though!