Cards are drawn one by one at random from a well-shuffled full pack of 52 playing cards until 2 aces

Q: Why do we use combinations instead of permutations?

Great question! In the first phase, we only care about the composition of the set of $n-1$ cards (i.e., one ace and $n-2$ non-aces), not the specific order in which they were drawn. Combinations are the perfect tool for selecting groups where order doesn't matter.

Q: Why is the denominator in the second phase $53-n$?

Think about the state of the deck. We started with 52 cards. We have already removed $n-1$ cards. Therefore, the number of cards remaining is $52 - (n-1) = 53-n$. This is the size of the sample space for the $n$-th draw.

Q: Why is the upper limit of $n$ equal to 50?

Consider the worst-case scenario. You could draw all 48 non-aces first. That takes 48 draws. You still haven't drawn any aces. The 49th card is an ace, and the 50th card is the second ace. You cannot go beyond 50 because there are only 48 non-aces in the deck.

Question

Cards are drawn one by one at random from a well-shuffled full pack of 52 playing cards until 2 aces are obtained for the first time. If $N$ is the number of cards required to be drawn, then show that $P_r \{N = n\} = \frac{(n-1)(52-n)(51-n)}{50 	imes 49 	imes 17 	imes 13}$ where $2 \le n \le 50$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

For the second ace to appear on the $n$-th draw, the universe must have conspired in a very specific way during the first $n-1$ draws. If you had drawn zero aces, you would not be ready for the second one. If you had drawn two, the game would have ended prematurely.

Therefore, in the first $n-1$ draws, you must have drawn exactly one ace and $n-2$ non-aces. This is our 'setup' phase.

We have 4 aces and 48 non-aces in the deck. The number of ways to choose 1 ace from 4 is $\binom{4}{1}$, and the number of ways to choose $n-2$ non-aces from 48 is $\binom{48}{n-2}$. The total number of ways to choose any $n-1$ cards from 52 is $\binom{52}{n-1}$.

Thus, the probability of this first phase, $P_1$, is:

$$P_1 = \frac{\binom{4}{1} 	imes \binom{48}{n-2}}{\binom{52}{n-1}}$$

### The Climax

Now, the stage is set. You have drawn $n-1$ cards, meaning the deck is smaller. The number of cards remaining is exactly $52 - (n-1) = 53-n$.

How many aces remain? Since you successfully drew exactly one ace in the first phase, there are $4 - 1 = 3$ aces left in the deck. For the experiment to end on the $n$-th draw, this specific card must be one of those 3 remaining aces.

The probability of this, $P_2$, is simply:

$$P_2 = \frac{3}{53-n}$$

### The Grand Unification

Since these two phases are sequential, the total probability $P(N=n)$ is the product of $P_1$ and $P_2$:

$$P(N=n) = \left( \frac{\binom{4}{1} 	imes \binom{48}{n-2}}{\binom{52}{n-1}} ight) 	imes \left( \frac{3}{53-n} ight)$$

Now, let us perform the algebraic magic. We expand the combinations into factorials using the identity $\binom{n}{r} = \frac{n!}{r!(n-r)!}$:

$$P(N=n) = \frac{4 	imes \frac{48!}{(n-2)!(50-n)!}}{\frac{52!}{(n-1)!(53-n)!}} 	imes \frac{3}{53-n}$$

Watch closely as the terms begin to dance. We rearrange the fraction to group the factorials:

$$P(N=n) = \frac{12 	imes 48! 	imes (n-1)! 	imes (53-n)!}{(n-2)! 	imes (50-n)! 	imes 52! 	imes (53-n)}$$

### Final Calculation

Look at the beauty of the cancellations. The term $\frac{(n-1)!}{(n-2)!}$ simplifies to $(n-1)$. The term $\frac{(53-n)!}{(50-n)!(53-n)}$ simplifies to $(52-n)(51-n)$.

Finally, the ratio $\frac{48!}{52!}$ becomes $\frac{1}{52 	imes 51 	imes 50 	imes 49}$. When we combine these, we arrive at the elegant result:

$$P(N=n) = \frac{12 	imes (n-1) 	imes (52-n)(51-n)}{52 	imes 51 	imes 50 	imes 49}$$

Simplifying the constants $\frac{12}{52 	imes 51}$ gives us $\frac{1}{13 	imes 17}$. Thus, we have proven that the final probability is:

**$$P(N=n) = \frac{(n-1)(52-n)(51-n)}{50 	imes 49 	imes 17 	imes 13}$$**

Visualized Solution

Visualizing the Experiment & Defining $N = n$

Analyzing the First Phase: First $n - 1$ Draws

Calculating Probability $P_{1}$ for the First Phase

Analyzing the Second Phase: The $n$ -th Draw

Calculating Probability $P_{2}$ for the $n$ -th Draw

Combining the Probabilities: $P (N = n)$

Expanding the Combinations using Factorials

Simplifying the Factorial Terms

Final Algebraic Reduction & Conclusion

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Climax

The Grand Unification

Final Calculation

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Visualized Solution

Visualizing the Experiment & Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N=n

Analyzing the First Phase: First .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n−1 Draws

Calculating Probability .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​ for the First Phase

Analyzing the Second Phase: The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n-th Draw

Calculating Probability .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​ for the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n-th Draw

Combining the Probabilities: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(N=n)