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I need to find the probability of NOT getting an ace card on two draws from a deck of 52 cards.

My first thought (which I really think is correct) was to get the probability from taking $\frac{48}{52}\frac{47}{51}=\frac{\binom{48}{2}}{\binom{52}{2}}\approx0.85$. Isn't this correct?

Then I though about it in another way. There are $\binom{52}{2}$ ways too choose two cards from 52. To not get an ace, you can to choose two out of 12 values (where the ace is excluded) in $\binom{12}{2}$ ways, and then you can choose $\binom{4}{1}=4$ different cards from each of the chosen values. Then I'm thinking you could calculate the probability by taking $$\frac{\binom{12}{2}\binom{4}{1}^{2}}{\binom{52}{2}}\approx0.80$$ This doesn't give the same as the first method I used. I'm obviously missing something, but can't really figure out what is wrong. I would appreciate some guidance on how I should think differently doing the later method! (I suppose the probability is approx. 0.85 as I got in the first place.)

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    $\begingroup$ You haven’t counted the case when you get $2$ cards of the same denomination - say, two kings. $\endgroup$ Commented Mar 28, 2022 at 20:22

2 Answers 2

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Your first thought was correct. Your second method finds the probability of finding two cards of different ranks, neither of which is an ace. To correct your count of favorable cases, you need to add the cases in which both cards are of the same rank and are not aces. There are $12$ ways to choose the rank of such cards and $\binom{4}{2}$ ways to choose two cards of the selected rank. Hence, the desired probability is $$\Pr(\text{two non-aces}) = \frac{\dbinom{12}{2}\dbinom{4}{1}^2 + \dbinom{12}{1}\dbinom{4}{2}}{\dbinom{52}{2}} = \frac{\dbinom{48}{2}}{\dbinom{52}{2}}$$

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In the second method you don't take the case that both cards have the same value into account. That would be $12 \cdot \binom42$ many possibilities. When you add those, you get the same result.

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  • $\begingroup$ Your technique of finding two different ways of solving a problem is great by the way! A great teacher at a summer school I went to said "A theorem worth proving once is worth proven twice." (and he meant in two different ways, not repeating the same idea. $\endgroup$
    – flukx
    Commented Mar 28, 2022 at 20:24

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