A carnival game consists of three dice in a cage. A player can be a dollar on any of the numbers 1 through 6. The cage is shaken, and the payoff is as follows. If the player's number doesn't appear on any of the dice, he loses his dollar. Otherwise, if his number appears on exactly $k$ of the three dice, for $k = 1, 2, 3$, he keeps his dollar and wins $k$ more dollars. What is the expected gain from playing the carnival game once?

Here are the probabilities:

$$ \Pr\{X = 3\} = 1/216 \\ \Pr\{X = 2\} = 3(6/216) - 3(1/126) = 5/216 \\ \Pr\{X = 1\} = 3(36/216) - 3(11/216) = 75/216 \\ \Pr\{X = 1\} = 125/216 $$

(There are 36 ways that a specific die is $k$, but in 10 of them one of the other die is also $k$ and in one of them both are).

Here's the calculation:

$$ \begin{align} \E[X] &= -1 \cdot \Pr\{X = 0\} + 1 \cdot \Pr\{X = 1\} + 2 \cdot \Pr\{X = 2\} + 3 \cdot \Pr\{X = 3\} \\ &= - 1 \cdot \frac{125}{216} + 1 \cdot \frac{75}{216} + 2 \cdot \frac{15}{216} + 3 \cdot \frac{1}{216} \\ &= - \frac{17}{216} \\ &= -0.07\ldots \end{align} $$

Turns out you loose ever so slightly. This did not match my intuition.