Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say we have Tom and John, each tosses a fair coin $n$ times. What is the probability that they get same number of heads?

I tried to do it this way: individually, the probability of getting $k$ heads for each is equal to $$\binom{n}{k} \Big(\frac12\Big)^n.$$ So, we can do $$\sum^{n}_{k=0} \left( \binom{n}{k} \Big(\frac12\Big)^n \cdot \binom{n}{k}\Big(\frac12\Big)^n \right)$$ which results into something very ugly. This ugly thing is equal to the 'simple' answer in the back of the book: $\binom{2n}{n}\left(\frac12\right)^{2n},$ but the equality was verified by WolframAlpha -- it's not obvious when you look at it. So I think there's a much easier way to solve this, can someone point it out? Thanks.

share|cite|improve this question
up vote 14 down vote accepted

The probability John gets $k$ heads is the same as the probability John gets $n-k$ heads since the coin is fair.

So the answer to the original question is equal to the probability that the sum of Tom's and John's heads is $n$.

That is the probability of $n$ heads from $2n$ tosses which is indeed $\frac{1}{2^{2n}}{2n \choose n}$.

share|cite|improve this answer

As you have noted, the probability is $$ p_n = \frac{1}{4^n} \sum_{k=0}^n \binom{n}{k} \binom{n}{k} = \frac{1}{4^n} \sum_{k=0}^n \binom{n}{k} \binom{n}{n-k} = \frac{1}{4^n} \binom{2n}{n} $$ The middle equality uses symmetry of binomials, and last used Vandermonde's convolution identity.

share|cite|improve this answer

The problem with this answer is that its not correct to begin with. Assume the problem pertains to two individuals throwing 2 coins then

the total outcomes is indeed $ \frac{1}{2^m} , m= 2n$ however the amount of possible outcomes is not $\binom{2n}{n}$ so in the case of each having 2 coins the resulting outcomes is:

1 head and two tails, this can be done 2 ways $2*2 = 4$

2 heads , this can be done only 1 way $1*1=1$

This is a total of $5$ and $\binom{2*2}{2} = 6$ this is of course not equal and difference of 1 holds as you continue.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.