Law of Iterated Expectations example I have a question that I hope can be shown using the LIE. There is a urn which contains 3 marbles (2 white and 1 black). The person who gets to pick the black marble gets to win $100 whereas the individual who picks the white marble walks away with nothing. The marbles that are taken out are not replaced. Assuming that there are only 2 individuals, how to show, using LIE that it does not matter whether you are the first or second to pick the marble?
Thanks
 A: Denote:
Y = the second guy's earnings
X = the first guy's earnings
Now, let's prove that E(X) = E(Y), using LIE (law of iterated expectations)
E(X) = 2/3 * 0 + 1/3 * 100 = 100/3
E(Y) = E(E(Y|X)) = prob(X=100) * E(Y|X=100) + prob(X=0) * E(Y|X=0)
prob(X=100) = 1/3
E(Y|X=100) = 1/2 * 0 + 1/2 * 0 = 0
prob(X=0) = 2/3
E(Y|X=0) = 1/2 * 0 + 1/2 * 100 = 50
Substitute these numbers into the former equation, we find E(Y) = 100/3
So E(X) = E(Y)
A: $\newcommand{\E}{\mathbb E}$Let $X$ be the number of black marbles appearing on the first draw, so $X=\text{either 0 or 1}$, and let $Y$ be the number of black marbles on the second draw.  The problem is to show that $\E(X)=\E(Y)$.
The law of iterated expectations, sometimes called the law of total expectation, tells us that
$$
\E(Y) = \E(\E(Y\mid X)).
$$
So we look at $\E(Y\mid X)$:
$$
\E(Y\mid X) = \left.\begin{cases} 1/2 & \text{if }X=0 \\ 0 & \text{if }X=1 \end{cases}\right\} = \begin{cases} 1/2 & \text{with probability }2/3, \\ 0 & \text{with probability }1/3. \end{cases}
$$
Given that, you should be able to find the expected value of this random variable $\E(X\mid Y)$.  Then the question is whether that's equal to $\E(X)$.
