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Consider a scenario where Ace is considered as 1 (a number card) and the face cards (King, Queen, Jack) are not number cards. We have to find the probability of 3 cards drawn consecutively (without replacement) being the same number. My friend solved it using the following: $$P = \frac{^4C_1\cdot ^{10}C_1\cdot ^3C_1\cdot ^2C_1 }{^{52}C_3} = \frac{12}{1105}$$ where we choose any one out of the 4 types, and then we choose a number card, and then we choose the same number card in the other types twice.

I did the same problem like the following: $$P = \frac{10}{13}\cdot \frac{3}{51}\cdot \frac{2}{50} = \frac{2}{1105}$$ where $\frac{10}{13}$ is the probability of choosing a number card first, $\frac{3}{51}$ is the probability of choosing the same number card out of the rest of 51 cards and $\frac{2}{50}$ is the probability of choosing the same number card out of the rest of 50 cards.

Could you please explain as to why we are getting different answers and which one is correct (I guess it has to do something with $6$ multiplication error in either one of the answers arising from problems in permutations (as $6$ = $3!$) of the types of cards)?

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  • $\begingroup$ How are you interpreting face cards? $\endgroup$
    – lulu
    Commented Dec 23, 2017 at 14:45
  • $\begingroup$ There are not considered as number cards $\endgroup$
    – Siddharth Venu
    Commented Dec 23, 2017 at 14:46
  • $\begingroup$ You should add that to your question, it is certainly not obvious. $\endgroup$
    – lulu
    Commented Dec 23, 2017 at 14:46
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    $\begingroup$ There are $10$ number cards, and $\binom 43$ ways to choose three cards after the number type has been chosen. As there are $\binom {52}3$ ways to choose three card hands, the answer is $\frac {10\times \binom 43}{\binom {52}3}=\frac 2{1105}$. $\endgroup$
    – lulu
    Commented Dec 23, 2017 at 14:49
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    $\begingroup$ I think it's unclear what your friend meant by "then we choose the same number card in the other types twice" - he ended up mixing up your method and his. You can pick a number (10 possibilities) and then pick a type not to be chosen, that is, $\frac{10 \times 4}{52 \choose 3}$. $\endgroup$
    – Pedro M.
    Commented Dec 23, 2017 at 14:52

2 Answers 2

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The first method is incorrect as the numerator distinguishes between hands with the same cards but different order, but the denominator does not. Thus the first method would say that the probability of drawing $\{2\heartsuit, 2\spadesuit, 2\clubsuit\}$ was $\frac 6{\binom {52}3}$ where it is clearly $\frac 1{\binom {52}3}$ (as all specified hands are equally probable). In other words, you were correct in your intuition that the factor $6=3!$ arises from the possible permutations of a three card hand.

The second method is correct. Another way to see it is to note that there are $10$ possible number values and $\binom 43$ ways to choose a three card hand with those number values, so the probability is $$\frac {10\times \binom 43}{\binom {52}3}=\frac 2{1105}$$

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Consider 4 objects namely A,B,C,D
Now, what your friend has done is first choose one out of 4, then one out of 3 and then one out of 2. And therefore Cases like $ABC$ and $BAC$ are counted separately while they shouldn't be. Your Answer to this question is correct

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