Question

Please read all three pictures to get full understanding, no need to complete Monte Carlo simulation. Just numerically price expectation:

Interview question (JP Morgan) You have 10 blue balls and 8 red balls, lined up in a queue. What is the expected number of balls that have a neighbor of a different color?

Answer 1

Let m be no. of red balls
Let n be no. of blue balls

Now we have the following$P(X_i = 1) = P( i.is.red). P( i + 1.is.blue | i.is.red)+ P( i.is.blue). P( i + 1.is.red | i. is. blue)\\ = \frac{m}{m+n}\frac{n}{m+n-1}+\frac{n}{m+n}\frac{m}{m+n-1}=\frac{2mn}{(m+n)(m+n-1)}\\\\ E[X]=\sum_{i=1}^{m+n-1} \frac{2mn}{(m+n)(m+n-1)} = \frac{2mn}{m+n}\\\\ E[X]= \frac{2\times 10 \times 8}{10+8}= \frac{80}{9}\\$