Solution: 2014 Fall Final - 21

1. Possible Outcomes and Corresponding Values:
   There are four possible outcomes when flipping two fair coins:

• First coin is heads (H), second coin is heads (H): $(H, H)$
• First coin is heads (H), second coin is tails (T): $(H, T)$
• First coin is tails (T), second coin is heads (H): $(T, H)$
• First coin is tails (T), second coin is tails (T): $(T, T)$

2. Winning Amount for Each Outcome:
   For each outcome, we determine the amount $ X $ that you win:

• $(H, H)$: The first coin is heads, so we don't lose.
  The second coin is heads, so we win 1 dollar.
  Thus, $ X = 1 $.
• $(H, T)$: The first coin is heads, so we don't lose.
  The second coin is tails, so we don't win.
  Thus, $ X = 0 $.
• $(T, H)$: The first coin is tails, so we lose (i.e., we win $-1$ dollars).
  The second coin is heads, so we win 1 dollar, but since the first coin being tails means we lose, this overrides the second coin's result.
  Thus, $ X = -1 $.
• $(T, T)$: The first coin is tails, so we lose (i.e., we win $-1$ dollars).
  The second coin is tails, so we don't win. Thus, $ X = -1 $.

3. Calculate the Expected Value $ E(X) $:
   The expected value $ E(X) $ is calculated using the formula:
   $E(X) = sum_{i} P(X = x_i) \cdot x_i $
   Each of the four outcomes has a probability of $ \frac{1}{4} $, as the coins are fair and independent.
   Therefore, we have:

• $ P(X = 1) = \frac{1}{4} $ (from the $(H, H)$ outcome)
• $ P(X = 0) = \frac{1}{4} $ (from the $(H, T)$ outcome)
• $ P(X = -1) = \frac{1}{2} $ (from the $(T, H)$ and $(T, T)$ outcomes combined)

4. Substitute the Values into the Expected Value Formula:
   $E(X) = 1 \cdot \frac{1}{4} + 0 \cdot \frac{1}{4} + (-1) \cdot \frac{1}{2}$
   $E(X) = \frac{1}{4} + 0 - \frac{1}{2}$
   $E(X) = \frac{1}{4} - \frac{2}{4}$
   $E(X) = -\frac{1}{4}$

Therefore, the expected value of $ X $ is $ \boxed{-\frac{1}{4}} $.