Cards are dealt one by one from a well shuffled pack until an ace appears. Find the chance that exactly n cards are dealt before the first ace.

Answer

To find the probability that exactly \( n \) cards are dealt before the first ace appears, we can follow these steps:

### Step 1: Understanding the Problem
We are dealing cards one by one from a standard deck of 52 cards until we encounter the first ace. We want to calculate the probability that the first ace appears after exactly \( n \) cards.

### Step 2: Probability of Not Drawing an Ace
For the first \( n \) cards dealt, we need to ensure that none of these cards is an ace. There are 48 non-ace cards in a deck of 52 cards. The probability of drawing \( n \) non-ace cards in a row is given by the combination of choosing \( n \) cards from the 48 non-aces.

The total ways to choose \( n \) cards from 52 cards is \( \binom{52}{n} \).

Thus, the probability of drawing \( n \) non-ace cards is:
\[ P(\text{no ace in } n \text{ cards}) = \frac{\binom{48}{n}}{\binom{52}{n}} \]

### Step 3: Probability of Drawing an Ace
After dealing \( n \) non-ace cards, the next card (the \( n+1 \)-th card) must be an ace. There are 4 aces in the deck, and after dealing \( n \) cards, there are \( 52 - n \) cards left.

The probability of drawing an ace next is:
\[ P(\text{ace on } (n+1)\text{-th card}) = \frac{4}{52 - n} \]

### Step 4: Combining the Probabilities
Now, we combine the probabilities from Step 2 and Step 3 to find the total probability that exactly \( n \) cards are dealt before the first ace appears:
\[ P(n) = P(\text{no ace in } n \text{ cards}) \times P(\text{ace on } (n+1)\text{-th card}) \]
\[ P(n) = \frac{\binom{48}{n}}{\binom{52}{n}} \times \frac{4}{52 - n} \]

### Step 5: Simplifying the Expression
Using the formula for combinations, we can express \( P(n) \):
\[ P(n) = \frac{\frac{48!}{n!(48-n)!}}{\frac{52!}{n!(52-n)!}} \times \frac{4}{52 - n} \]
This simplifies to:
\[ P(n) = \frac{48! \cdot (52-n)! \cdot 4}{52! \cdot (48-n)! \cdot (52-n)} \]

### Final Expression
After further simplification, we can express the final probability as:
\[ P(n) = \frac{48! \cdot 4 \cdot (52-n)!}{52! \cdot (48-n)! \cdot (52-n)} = \frac{4 \cdot (49-n)(50-n)(51-n)}{49 \cdot 50 \cdot 51 \cdot 13} \]

### Conclusion
Thus, the probability that exactly \( n \) cards are dealt before the first ace appears is:
\[ P(n) = \frac{4 \cdot (49-n)(50-n)(51-n)}{49 \cdot 50 \cdot 51 \cdot 13} \]