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In the book "An Undergraduate Introduction to Financial Mathematics" there is a following simple problem: "Suppose cards will be drawn without replacement from a standard 52-card deck. What is the probability that the fourth card drawn will be an ace given that the first three cards drawn were all aces?"

The correct answer in the book is given as 4/52 * 3/51 * 2/50 * 1/49 - What strikes me a bit odd is that we already know that the first three draws are aces ("given that"). Why is the correct answer not simply 1/49? I mean first three draws came and went and we are only concerned about the probability of drawing the ace on the fourth draw. To me the answer would be for question "What is the probability that first four cards drawn from the card deck without replacements are all aces?"

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2 Answers 2

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$1/49$ is the correct answer.

It is the probability for obtaining the first four positions for four the aces, given that we obtained the first three positions when shuffling selects any four from the fifty-two positions without bias.

Let $A$ be the set of positions for the four aces, then we have:

$$\begin{align}\mathsf P(A\supseteq\{4\}\mid A\supseteq\{1,2,3\}) &= \dfrac{\mathsf P(A\supseteq\{1,2,3,4\})}{\mathsf P(A\supseteq\{1,2,3\})} \\[1ex] &=\dfrac{\left.\binom44\middle/\binom{52}4\right.}{\left.\binom 33\binom{49}{1}\middle/\binom{52}4\right.} \\[1ex] &=\dfrac1{49} \end{align}$$

Matching your intuition.

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$\color{blue}{\text{Here is the solution for problem 5 in the same book:}}$

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$\color{blue}{\text{Here is the solution for problem 7 in the same book:}}$

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$\color{blue}{\text{Conclusion: Obviously, the solution of problem 7 must be corrected as it missed the conditional probability:}} $ $\color{blue}{\text{1) Since 3 Aces have already been drawn without replacement, then 49 cards with one Ace left. }}$ $\color{blue}{\ \ \ \ \ \text{Hence, the probability of getting the Ace card is: } \frac{1}{49}.}$ $\color{blue}{\text{2) The probability of getting an Ace card given that the first 3 cards were Aces: }}$ $$\color{blue}{P(\text{4th is Ace|first 3 are Aces}) = \\ \frac{\text{P(4th is Ace }\cap \text{first 3 are Aces)}}{\text{P(first 3 are Aces)}}=\frac{\text{P(first 4 are Aces) }}{\text{P(first 3 are Aces)}}=\\ \frac{\frac1{52}\cdot \frac1{51}\cdot \frac1{50}\cdot \frac1{49}}{\frac1{52}\cdot \frac1{51}\cdot \frac1{50}}=\frac1{49}}$$

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