A pack of playing cards was found to contain only 51 cards. If the first 13 cards, which are examined, are all red, what is the probability that the missing card is black?

Answer

To solve the problem of finding the probability that the missing card from a pack of 51 cards is black, given that the first 13 cards drawn are all red, we can follow these steps:

### Step 1: Understand the Problem
We have a standard deck of 52 playing cards, consisting of 26 red cards (hearts and diamonds) and 26 black cards (clubs and spades). One card is missing, leaving us with 51 cards. We need to find the probability that the missing card is black, given that the first 13 cards drawn are all red.

### Step 2: Define Events
Let:
- Event A: The missing card is red.
- Event B: The missing card is black.
- Event E: The first 13 cards drawn are all red.

### Step 3: Use Bayes' Theorem
We want to find \( P(B | E) \), the probability that the missing card is black given that the first 13 cards are red. By Bayes' theorem:
\[ P(B | E) = \frac{P(E | B) \cdot P(B)}{P(E)} \]

### Step 4: Calculate \( P(B) \) and \( P(A) \)
Since there are 26 red cards and 26 black cards, the probabilities are:
\[ P(A) = P(B) = \frac{1}{2} \]

### Step 5: Calculate \( P(E | A) \) and \( P(E | B) \)
1. **If the missing card is red (Event A)**:
- There are 25 red cards left. The probability of drawing 13 red cards from these 25 is:
\[ P(E | A) = \frac{\binom{25}{13}}{\binom{51}{13}} \]

2. **If the missing card is black (Event B)**:
- All 26 red cards are available. The probability of drawing 13 red cards from these 26 is:
\[ P(E | B) = \frac{\binom{26}{13}}{\binom{51}{13}} \]

### Step 6: Calculate \( P(E) \)
Using the law of total probability:
\[ P(E) = P(E | A) \cdot P(A) + P(E | B) \cdot P(B) \]
Substituting the values:
\[ P(E) = \frac{\binom{25}{13}}{\binom{51}{13}} \cdot \frac{1}{2} + \frac{\binom{26}{13}}{\binom{51}{13}} \cdot \frac{1}{2} \]

### Step 7: Substitute into Bayes' Theorem
Now we can substitute back into Bayes' theorem:
\[ P(B | E) = \frac{P(E | B) \cdot P(B)}{P(E)} \]
\[ P(B | E) = \frac{\frac{\binom{26}{13}}{\binom{51}{13}} \cdot \frac{1}{2}}{\frac{\binom{25}{13}}{\binom{51}{13}} \cdot \frac{1}{2} + \frac{\binom{26}{13}}{\binom{51}{13}} \cdot \frac{1}{2}} \]

### Step 8: Simplify the Expression
The \( \frac{1}{2} \) and \( \binom{51}{13} \) terms cancel out:
\[ P(B | E) = \frac{\binom{26}{13}}{\binom{25}{13} + \binom{26}{13}} \]

### Step 9: Calculate the Final Probability
Using the identity \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), we can simplify:
\[ P(B | E) = \frac{26}{25 + 26} = \frac{26}{51} \]

Thus, the probability that the missing card is black is:
\[ \boxed{\frac{2}{3}} \]