You select $n$ balls from an urn that has infinitely many blue and red balls. The probability that you choose a blue ball is $1/2$ and the probability that you choose a red ball is also $1/2$. Let $X$ denote the random variable that determines the number of color changes, i.e. for $n = 2$ if you first choose a red and afterwards a blue ball then $X = 1$ since there is one color change. Moreover, you place the balls in a cyclic order, i.e. $b_1, b_2, \ldots, b_n, b_1$ so there is one additionally possible colour change from $b_n$ to $b_1$. Is my argumentation correct, that $X$ is not binomially distributed because the probability for a change in the last pull is higher than for the other pulls since you have this cyclic ordering?
Edit: How could I compute the expected value of colour changes?
Best Answer
It's not binomial because while there can be from $0$ to $n$ changes, we always get back to the color we started with, so there must be an even number of changes.