Suppose the probability for A to win a game against B is 0.4. If A has an option of playing either a “best of 3 games'' or a “best of 5 games match against B, which option should be chosen so that the probability of his winning the match is higher? (No game ends in a draw.)

Explanation

To determine whether A should choose to play a "best of 3 games" or a "best of 5 games" match against B, we need to calculate the probabilities of A winning in both scenarios.

### Step 1: Define the probabilities
Let \( P(A) = 0.4 \) (the probability that A wins a game) and \( P(A') = 1 - P(A) = 0.6 \) (the probability that A loses a game).

### Step 2: Calculate the probability of winning a "best of 3 games"
In a "best of 3 games," A needs to win at least 2 games out of 3. The possible winning scenarios are:
1. A wins the first 2 games and loses the 3rd.
2. A wins the 1st and 3rd games and loses the 2nd.
3. A loses the 1st game and wins the next 2.
4. A wins all 3 games.

We can calculate the probabilities for each scenario:

1. **Win first 2, lose 3rd**:
\[ P(A) \times P(A) \times P(A') = 0.4 \times 0.4 \times 0.6 = 0.096 \]

2. **Win 1st and 3rd, lose 2nd**:
\[ P(A) \times P(A') \times P(A) = 0.4 \times 0.6 \times 0.4 = 0.096 \]

3. **Lose 1st, win 2nd and 3rd**:
\[ P(A') \times P(A) \times P(A) = 0.6 \times 0.4 \times 0.4 = 0.096 \]

4. **Win all 3 games**:
\[ P(A) \times P(A) \times P(A) = 0.4 \times 0.4 \times 0.4 = 0.064 \]

Now, we sum these probabilities:
\[ P(\text{Win best of 3}) = 0.096 + 0.096 + 0.096 + 0.064 = 0.352 \]

### Step 3: Calculate the probability of winning a "best of 5 games"
In a "best of 5 games," A needs to win at least 3 games out of 5. The possible winning scenarios are:
1. A wins 3 games and loses 2.
2. A wins 4 games and loses 1.
3. A wins all 5 games.

We can calculate the probabilities for each scenario:

1. **Win 3, lose 2**: The number of ways to choose 3 wins from 5 games is \( \binom{5}{3} \):
\[ \binom{5}{3} \times P(A)^3 \times P(A')^2 = 10 \times (0.4)^3 \times (0.6)^2 = 10 \times 0.064 \times 0.36 = 0.2304 \]

2. **Win 4, lose 1**: The number of ways to choose 4 wins from 5 games is \( \binom{5}{4} \):
\[ \binom{5}{4} \times P(A)^4 \times P(A')^1 = 5 \times (0.4)^4 \times (0.6) = 5 \times 0.0256 \times 0.6 = 0.0768 \]

3. **Win all 5 games**:
\[ P(A)^5 = (0.4)^5 = 0.01024 \]

Now, we sum these probabilities:
\[ P(\text{Win best of 5}) = 0.2304 + 0.0768 + 0.01024 = 0.31744 \]

### Step 4: Compare the probabilities
- Probability of winning "best of 3 games": \( P(\text{Win best of 3}) = 0.352 \)
- Probability of winning "best of 5 games": \( P(\text{Win best of 5}) = 0.31744 \)

Since \( 0.352 > 0.31744 \), A should choose to play the "best of 3 games" to maximize his chances of winning.

### Final Answer
A should choose the "best of 3 games" option.

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