Let $X$ be the number of heads in $4$ tosses of a fair coin by Person 1and $Y$ the number of heads in $4$ tosses of a fair coin by Person 2. Assuming independence, the value of $\mathbb{P}(X=Y)$ is? My attemp:
$\mathbb{P}(X=Y)= \mathbb{P}(X=Y=0)+\mathbb{P}(X=Y=1)+\mathbb{P}(X=Y=2)+......+$
$=\mathbb{P}(X=Y=0) +\mathbb{P}(X=1)\mathbb{P}(Y=1)+\mathbb{P}(X=2)\mathbb{P}(Y=2)+......+$
$=\mathbb{P}(X=0)^2+ \mathbb{P}(X=1)^2+\mathbb{P}(X=2)^2+......+$
$=\dbinom{4}{0}^2\left(\dfrac{1}{2}\right)^8+ \dbinom{4}{1}^2\left(\dfrac{1}{2}\right)^8+\dbinom{4}{2}^2\left(\dfrac{1}{2}\right)^8+\dbinom{4}{3}^2\left(\dfrac{1}{2}\right)^8+\dbinom{4}{4}^2\left(\dfrac{1}{2}\right)^8$
$=\dfrac{1}{16}\dbinom{8}{4}=0.273$
Is it correct?
Yes, indeed it is so. In series form:
$$\begin{align}\mathsf P(X=Y)&=\sum_k \mathsf P(X=k)\mathsf P(Y=k)\\&=\sum_{k=0}^4 \binom 4k^2 2^{-8} \\&= 2^{-8}\sum_{k=0}^4\binom 4k\binom 4{4-k} \\&= 2^{-8}\binom 84&{\text{via Vandermonde's convolution}\\\sum_{k=0}^r \binom mk\binom n{r-k}=\binom{m+n}r}\end{align}$$