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Suppose two cards are drawn from a standard 52 card deck without replacement. Assuming all cards are equally likely to be selected, what is the probability that both cards are not aces?

My Solution

A = Event that first card is an ace

B = Event that second card is an ace given that first is an ace

C = Event that both cards are aces

D = Event that both cards are not aces $$P(A) = \frac{4}{52}$$ $$P(B) = \frac{3}{51}$$ $$P(C) = \frac{4}{52}*\frac{3}{51} = \frac{1}{221}$$ $$P(D) = 1 - P(C) = 1 - \frac{1}{221} = \frac{220}{221}$$

Actual Solution

A = Event that first card is not an ace

B = Event that second card is not an ace given that first is not an ace

C = Event that both cards are not aces $$P(A) = \frac{48}{52}$$ $$P(B) = \frac{47}{51}$$ $$P(C) = \frac{48}{52}*\frac{47}{51} = \frac{188}{221}$$

Why my solution that P(D) = 1 - P(C) is wrong?

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    $\begingroup$ Somewhat ambiguous question. Suppose the two drawn cards are called A and B. Do you mean A is not an ace and B is also not an ace? It could be interpreted as A and B are both not aces (as a pair) but one of them could be an ace. As with all good math word problems, it is important to understand the question clearly first before attempting to solve it. $\endgroup$
    – David
    Commented Jul 20, 2016 at 11:18
  • $\begingroup$ @David I am now confused with my question after seeing your comment. But the actual answer states its 188/221 so I assume A is not an ace and B is also not an ace. $\endgroup$
    – jblixr
    Commented Jul 20, 2016 at 14:50
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    $\begingroup$ Yes that seems to be the correct interpretation that neither of the $2$ drawn cards are aces so you drew $0$ aces. An "easy" way to compute this probability is imagine we remove all $4$ aces from the deck of $52$, leaving us with $48$ cards only. We have $48$ ways to draw the first card and $47$ ways to draw the 2nd card. Now imagine we put the deck back to $52$ cards and draw $2$ cards. We have $52$ ways to draw the 1st card and $51$ ways to draw the 2nd card. Therefore, the probability of drawing no aces for the first $2$ cards is $(48*47)/(52*51)$. $\endgroup$
    – David
    Commented Jul 21, 2016 at 3:20
  • $\begingroup$ In the above comment, we can ignore the "missing" divide by $2$ for both the numerator and the denominator since they will cancel out. Conceptually, this is about the simplest way I can explain the answer. That is, remove the aces to guarantee no aces drawn the draw $2$ non ace cards which is exactly what we want. Then put all the aces back and draw $2$ cards . Then compute the ratio and that is the probability (about $85$%). $\endgroup$
    – David
    Commented Jul 21, 2016 at 3:45

5 Answers 5

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This may be a language issue. You calculated the probability that not both cards are aces, whereas the problem asks for the probability that both cards are not aces.

Since you overloaded your event variables, I'll define new ones: Let $E$ be the event that the first card is an ace and $F$ the event that the second card is an ace; then the event that not both cards are aces is $\overline{E\cap F}=\overline E\cup\overline F$, and the event that both cards are not aces is $\overline E\cap\overline F=\overline{E\cup F}$.

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If you want neither of the first $2$ drawn cards to be aces then it is simply $ 48 \choose 2$ / $52 \choose 2$ which is $188/221$. Here we are simply choosing $2$ cards from the $48$ non aces and are dividing by the total number of possible $2$ card pairs from the full deck of $52$.

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The probability that both cards are not aces is the complement of the event that at least one of the cards selected is an ace. You overlooked the possibility that exactly one of the two cards selected is an ace.

The probability that both cards are aces is $$\frac{4}{52} \cdot \frac{3}{51} = \frac{1}{221}$$
The probability that the first card is an ace and the second card is not an ace is $$\frac{4}{52} \cdot \frac{48}{51} = \frac{16}{221}$$ The probability that the first cards is not an ace and the second card is an ace is $$\frac{48}{52} \cdot \frac{4}{51} = \frac{16}{221}$$ Therefore, the probability that at least one of the two cards is an ace is $$\frac{1}{221} + \frac{16}{221} + \frac{16}{221} = \frac{33}{221}$$ Hence, the probability that both cards are not aces is $$1 - \frac{33}{221} = \frac{188}{221}$$ which can be found more simply by using the method you included in your post.

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If $C$ is the event that both cards are aces, then $1 - P(C)$ is the probability that at least one of the two cards is not an ace.

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What you need to be calculating (using your A,B) is:

$$(1-P(A))(1-P(B))$$

which is the probability that the first card is not an ace, and the second is an ace given that the first is not.

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