expectation of number of hands shaken at round table There are $x$ people from country A and $y$ people from country B. They sit around a table and shake hands with people on their left and right, but they only shake hands if they're from the same country. We want to know the expected number of handshake made. my thought: since A and B countries only shake hands with their own, so the total expectation is the sum of individual country. Given we have x people from A, there are x+y-1 ways to place two people together, and ${x \choose 2}$ ways to select people from country A, 2 ways to sit them in a 2-people slot, so expectation of handshaken for country A people is $\frac{2{x \choose 2}}{x+y-1}$. However I am not very sure about my approach. Any suggestions will be greatly appreciated!

Your approach seems fine.

Alternatively,...

In each seat, the probability the person sitting there shakes the hand to the left is $$\frac{x}{x+y}\frac{x-1}{x+y-1} + \frac{y}{x+y}\frac{y-1}{x+y-1}=\frac{x(x-1)+y(y-1)}{(x+y)(x+y-1)}$$

So if $X_i$ is the number of handshakes to the left from seat $i,$ we have that $X_i=0\text{ or }1.$ And $X=X_1+\cdots+X_{x+y+1}$ is the total number of handshakes to the left from any seat, which is the total number of seats.

But $E(X)=\sum E(X_i),$ and $$E(X_i)=P(X_i=1)=\frac{x(x-1)+y(y-1)}{(x+y)(x+y-1)}$$

So: $$E(X)=(x+y)E(X_1)=\frac{x(x-1)+y(y-1)}{x+y-1}$$