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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)?

Siddharth Venu
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  • How are you interpreting face cards? – lulu Dec 23 '17 at 14:45
  • There are not considered as number cards – Siddharth Venu Dec 23 '17 at 14:46
  • You should add that to your question, it is certainly not obvious. – lulu Dec 23 '17 at 14:46
  • Sure! I have added it now (sorry for the confusion) – Siddharth Venu Dec 23 '17 at 14:47
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    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 $\frac {10\times \binom 43}{\binom {52}3}=\frac 2{1105}$. – lulu Dec 23 '17 at 14:49
  • I don't understand the first calculation. What are the four types? If you mean you choose the suit of the first card then the first calculation is wrong as you are imposing an order on the "good" hands but not on the general hands. – lulu Dec 23 '17 at 14:51
  • I guess the 2nd method is right then... Could you please elaborate on why the first method fails to give the correct answer? (Add it as an answer so I can accept it) – Siddharth Venu Dec 23 '17 at 14:52
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    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, $\frac{10 \times 4}{52 \choose 3}$. – Pedro M. Dec 23 '17 at 14:52
  • Please add your explanations in answers so I can accept it – Siddharth Venu Dec 23 '17 at 14:54

2 Answers2

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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 $$\frac {10\times \binom 43}{\binom {52}3}=\frac 2{1105}$$

lulu
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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

Anvit
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