Question

Toss three fair coins. Player A wins if exactly 2 heads land up. Otherwise Player B wins. Is the game fair?

Answer 1

To determine if the game is fair, we can analyze the outcomes of tossing three fair coins and calculate the probabilities for each player's winning conditions.

When tossing three coins, there are a total of \(2^3 = 8\) possible outcomes. The outcomes when three coins are tossed can be listed as follows:

1. HHH
2. HHT
3. HTH
4. THH
5. HTT
6. THT
7. TTH
8. TTT

Next, we'll determine how many of these outcomes correspond to Player A winning (exactly 2 heads) and how many correspond to Player B winning (not exactly 2 heads).

**Player A wins if there are exactly 2 heads:**
- The outcomes with exactly 2 heads are:
- HHT
- HTH
- THH

This gives us 3 outcomes where Player A wins.

**Player B wins if there are not exactly 2 heads:**
- The outcomes that do not have exactly 2 heads are:
- HHH (3 heads)
- HTT (1 head)
- THT (1 head)
- TTH (1 head)
- TTT (0 heads)

This gives us 5 outcomes where Player B wins.

Now, we'll calculate the probabilities for each player:
- The probability that Player A wins (exactly 2 heads) is:
\[
P(A) = \frac{\text{Number of winning outcomes for A}}{\text{Total outcomes}} = \frac{3}{8}
\]

- The probability that Player B wins (not exactly 2 heads) is:
\[
P(B) = \frac{\text{Number of winning outcomes for B}}{\text{Total outcomes}} = \frac{5}{8}
\]

Since \(P(A) \neq P(B)\) (specifically, \(P(A) = \frac{3}{8}\) and \(P(B) = \frac{5}{8}\)), we can conclude that the game is **not fair**.

Player B has a higher probability of winning. Thus, the game favors Player B.