Question: Suppose you enter a contest where your probability of coming in first place is $0.21$ and your probability of coming in second place is $0.35$. If you come in first place, you win $62$. If you come in second place, you win $12$. If you don't come in first or second place, you have to pay $64$. What is the expected value of this game?
1. **State the problem:** We want to find the expected value of the game given the probabilities and payoffs.
2. **Recall the formula for expected value:**
$$E = \sum (\text{probability of outcome} \times \text{value of outcome})$$
3. **Identify the outcomes and their probabilities:**
- First place: probability $p_1 = 0.21$, payoff $x_1 = 62$
- Second place: probability $p_2 = 0.35$, payoff $x_2 = 12$
- Neither first nor second place: probability $p_3 = 1 - p_1 - p_2 = 1 - 0.21 - 0.35 = 0.44$, payoff $x_3 = -64$ (since you pay 64)
4. **Calculate the expected value:**
$$E = p_1 x_1 + p_2 x_2 + p_3 x_3$$
$$E = 0.21 \times 62 + 0.35 \times 12 + 0.44 \times (-64)$$
5. **Calculate each term:**
$$0.21 \times 62 = 13.02$$
$$0.35 \times 12 = 4.20$$
$$0.44 \times (-64) = -28.16$$
6. **Sum the terms:**
$$E = 13.02 + 4.20 - 28.16 = 17.22 - 28.16$$
$$E = -10.94$$
7. **Interpretation:** The expected value of the game is $-10.94$, meaning on average you lose 10.94 per game.
**Final answer:**
$$\boxed{-10.94}$$