Calculate Probability of Winning Head and Tails Game

[haiha]

I ve got this simple problem and I can can not find the solution.
Two men play head and tails. Each man flips the coin n times, every time he has head, he should add one point to his score. At the end, they compare the scores, whose score has more points , the man is the winner.
A is the probability of a man lose or win, B is the probability when the two have the same points. Calculate A and B?
Thanks

 

[daveb]

The first thing you want to do is show us where you're stuck. As a hint, find the probability that person 1 rolls m heads, then find the probability that the other rolls more than m heads.

 

[haiha]

Thanks for your answering me
My way is to find the B value meaning the two are equal. So i tried to calculate the probabiltiy of 1st man to have m heads, then the 2nd also must have m heads, which is called Pm. Then I will sum up Pm with m running from 0 to n. But the formula is very long and i think there must be mistakes somewhere..
P0 = 1/2^(2n)
P1=2.n/2^2n
P2=..

 

[chroot]

Your sum is correct. Are you sure you have to actually evaluate the sum, or can you just leave your answer in terms of the sum?

This is not a trivial sum to compute.

- Warren

 

[haiha]

Can anyone find the general formula for Pm?, at least one that can be computed using a program.

 

[chroot]

"Pm" is simple -- it's a Binomial distribution:

<br /> p \left( m \right) = \left( {\begin{array}{*{20}c}<br /> n \\<br /> m \\<br /> <br /> \end{array} } \right)p^m \left( {1 - p} \right)^{n - m} = \left( {\begin{array}{*{20}c}<br /> n \\<br /> m \\<br /> <br /> \end{array} } \right)2^{ - n} <br />

where p is the probability of heads (1/2), n is the number of flips, and m is the number of heads.

Remember that the probability of both men getting m heads is actually [p(m)]^2.

- Warren

 

[haiha]

So I can not calculate P0, when the two men both have no heads ?

Oh I m sorry, I thought (n/m) is a division!.

 

[AlephZero]

This is not a trivial sum to compute.

You can get an approximate answer by replacing the binomial distribution by a normal distribution. The pdf of the "two scores are equal" distribution is the square of the pdf of the normal distribution.

Since (e^{-x^2})^2 = e^{-2x^2} this is similar to a normal distribution so you can integrate it.

The approximate probability of equal scores after n tosses is 1/\sqrt{n\pi}.

 

[listle]

please explain me how do this