Let $X$ and $Y$ be two independent random variables with respective distributions $B(n+1,\frac{1}{2})$ and $B(n,\frac{1}{2})$. I am trying to determine $\mathbb{P}(X>Y)$. So far, I have written that:
$$\mathbb{P}(X>Y)=\sum_{k=0}^n \mathbb{P}(X>k)\mathbb{P}(Y=k)=\frac{1}{2^{2n+1}}\sum_{k=0}^n {n \choose k}\sum_{j=k+1}^{n+1}{n+1 \choose j}.$$
However, at this point, I am not sure how to compute the double sum involving the binomial coefficients. Is there any way we can re-arrange the terms to make use of the identity $\sum_{k=0}^n {n \choose k}=2^n$?
Best Answer
Alicia and Beti each toss a fair coin repeatedly. Alicia does it $n+1$ times and Beti does it $n$ times. Alicia wins if her number $X$ of heads is greater than Beti's number $Y$ of heads.
We find the probability that $X\gt Y$, that is, the probability that Alicia wins.
After $n$ tosses by Alicia, and $n$ by Beti, there are three possibilities:
(i) Alicia is leading. Then, whatever happens on her $(n+1)$-th toss, Alicia will win
(ii) Beti is leading. Then whatever happens on Alicia's $(n+1)$-th toss, Beti will win.
(iii) They are tied. Then, depending on the result of her $(n+1)$-th toss, Alicia will win or lose.
By symmetry (i) and (ii) are equally likely. And in Case (iii), Alicia has probability $1/2$ of winning.
Thus the probability that Alicia wins the game is $1/2$. So $\Pr(X\gt Y)=1/2$.