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This is an extension of the $n+1$ vs n problem here: Probability of $5$ fair coin flips having strictly more heads than $4$ fair coin flips

So a common way to think about this is to say that after $n$ tries, both players have the same expected number of heads. So player $A$ has a $50\%$ chance of winning because he gets a head with $50\%$ chance on his last try. But if he gets $x$ more tries than player $B$, does that still stand? He wins with probability $1 - \frac{1}{2^x}$? I did the calculations for a few $n$'s and $x$'s and it doesn't seem to be the case.

What's the answer here? Thanks!

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narcissa
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    It isn't the same because A could be losing at the n flip mark and then win after the final x flips. This doesn't happen with only one extra flip. – Michael Biro Jun 15 '13 at 16:01
  • That's what I thought. Is there a way to get a simple equation for all $n$ and $x$ though? How can you solve this for general $n$ and $x$? – narcissa Jun 15 '13 at 16:11

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A is flipping $n+x$ coins and will get $u$ heads with probability $C(n+x,u)/2^{n+x}.$ Also B is flipping $n$ coins and will get $v$ heads with probability $C(n,v)/2^n.$ The range on $v$ is from $0$ to $n$, and in favorable cases for A to win, the range on $u$ is from $v+1$ to $n+x$. This gives the probability for A to win as $$\frac{1}{2^{2n+x}} \sum_{v=0}^n \sum_{u=v+1}^{n+x} C(n,v)C(n+x,u).$$ Maple was unable to get a closed form for the double sum of binomials here.

In the case where $x=2$, wherein A has two more coins than B, the calculation gives the following numerators for $n=1,2,3,4,5,6$: $$11,\ \ 42,\ \ 163,\ \ 638,\ \ 2510,\ \ 9908.$$ In each case for the probabilities one has to divide by $2^{2n+2}=4^{n+1}.$

coffeemath
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