Probability of that the number of heads in tossing a coin n times and n+1 times be equal

Question

If we toss a fair coin ${n}$ times and another fair coin ${n+1}$ times; what is the probability of that the number of heads be equal?

Answer 1

It's possible without the clever trick you were offered in the comments. If the number of heads for the first coin is $m$, then the probability that the second coin will achieve the same number is ${n+1 \choose {m}} (\frac{1}{2})^{n+1}$.

Therefore the chance that the first coin gives us $m$ heads and the second coin gives the same result is ${n \choose m} (\frac{1}{2})^{n} {n+1 \choose m} (\frac{1}{2})^{n+1} = (\frac{1}{2})^{2n+1} {n \choose m}{n+1 \choose m}$

You need to sum this over all integer $m$ between $0$ and $n$. $(\frac{1}{2})^{2n+1}$ is independent of $m$ so is not a problem. To sum the remaining part you will need to use some identities of binomial coefficients.

The trick that was offered in the comments makes this much simpler: if the first coin landed $m$ heads, then the chance that the second coin will land $m$ heads is the same chance as that it will land $m$ tails, by symmetry. So the chance of the second coin landing the same number of heads as the first coin, is the same chance that the second coin will land as many tails as the first coin landed heads. But if the second coin lands $m$ tails then it lands $n+1-m$ heads, so the total number of heads both coins land is $m+n+1-m=n+1$. So the whole problem is equivalent to asking what is the chance that out of $2n+1$ tosses we will get exactly $n+1$ heads, and this is immediate without any sums.