A, B, C in order, cut a pack of cards, replacing them after each cut, on condition that the first who cuts a spade shall win a prize. Then A’s chance of winning is

Options

A. `16/37`
B. `12/37`
C. `9/37`
D. `14/37`

Answer

To solve the problem of finding A's chance of winning when A, B, and C take turns cutting a pack of cards, we can break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- There are 52 cards in total, with 13 of them being spades.
- The game continues until one player cuts a spade, and the players take turns in the order A, B, C.

2. **Calculating the Probability of Cutting a Spade**:
- The probability of cutting a spade (winning) in one turn is:
\[ P(\text{cutting a spade}) = \frac{13}{52} = \frac{1}{4} \]
- Therefore, the probability of not cutting a spade (losing) is:
\[ P(\text{not cutting a spade}) = 1 - P(\text{cutting a spade}) = 1 - \frac{1}{4} = \frac{3}{4} \]

3. **Finding A's Winning Probability**:
- A can win in several scenarios:
- A wins on the first turn.
- A does not win, B does not win, C does not win, and then A wins on the second turn.
- This pattern continues indefinitely.
- The probability that A wins on the first turn is \( \frac{1}{4} \).
- The probability that A wins on the second turn (after B and C do not win) is:
\[ P(A \text{ wins on 2nd turn}) = P(\text{not A}) \times P(\text{not B}) \times P(\text{not C}) \times P(A \text{ wins}) = \left(\frac{3}{4}\right)^3 \times \frac{1}{4} \]
- This can be generalized for any turn \( n \):
\[ P(A \text{ wins on } n\text{th turn}) = \left(\frac{3}{4}\right)^{3(n-1)} \times \frac{1}{4} \]

4. **Summing the Infinite Series**:
- The total probability \( P(A \text{ wins}) \) can be expressed as an infinite series:
\[ P(A \text{ wins}) = \frac{1}{4} + \left(\frac{3}{4}\right)^3 \cdot \frac{1}{4} + \left(\frac{3}{4}\right)^6 \cdot \frac{1}{4} + \ldots \]
- This is a geometric series where:
- First term \( a = \frac{1}{4} \)
- Common ratio \( r = \left(\frac{3}{4}\right)^3 = \frac{27}{64} \)

5. **Using the Formula for the Sum of an Infinite Geometric Series**:
- The sum \( S \) of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
- Plugging in the values:
\[ S = \frac{\frac{1}{4}}{1 - \frac{27}{64}} = \frac{\frac{1}{4}}{\frac{37}{64}} = \frac{64}{148} = \frac{16}{37} \]

6. **Final Probability**:
- Thus, A's chance of winning is:
\[ P(A \text{ wins}) = \frac{16}{37} \]