# Coin game probability - simple set up, tricky to calculate

• Apr 18th 2011, 03:58 PM
DannyOcean
Coin game probability - simple set up, tricky to calculate
Suppose Bob flips a coin N times, and Jeff flips a coin N+1 times (N is an arbitrary positive integer). What are the odds that Jeff gets strictly more heads than Bob?

I'm stuck here.
• Apr 18th 2011, 05:20 PM
DannyOcean
Actually, I think I may have found a solution, but would love for someone to verify.

Assume Bob and Jeff both flip the coin N times. Either Bob gets more heads (Event B), Jeff gets more heads(Event J), or they tie (Event T). Obviously P(B) = P(J), and P(B) + P(J) + P(T) = 1. We can also see that P(J) + (1/2)*P(T) = 1/2 with just a little algebra.

Now, if Jeff flips one more coin after this (making a total of N+1 flips for Jeff), what is the probability he has strictly more heads than Bob after this extra flip? If he was already behind Bob, then he can't have strictly more heads than Bob now no matter what the outcome of the extra flip. If he was already ahead of Bob, then he'll have strictly more heads no matter what the outcome of the extra flip. And if they were tied, then there's a 1/2 chance that he will now have strictly more heads than Bob and a 1/2 chance he won't.

So via the statement P(J) + (1/2)*P(T) =1/2, we can say that the probability Jeff gets more heads in N+1 throws than Bob gets in N throws is 1/2.

Does this look sound?