1
$\begingroup$

I have a question about finding a probability of 3 of a kind using combinations. This is a solution I originally came up with: $$\frac{\binom{13}{1}*\binom{4}{3}*\binom{12}{1}*\binom{4}{1}*\binom{11}{1}*\binom{4}{1}}{\binom{52}{5}}$$ That is, choose a face value for 3 cards, choose a value from the rest of 12 cards for another card, and choose another value for the remaining card.

When I compared my answer with the provided one it was wrong, so I looked for a solution. This is what I came across: $$\frac{\binom{13}{1}*\binom{4}{3}*\binom{12}{2}*\binom{4}{1}*\binom{4}{1}}{\binom{52}{5}}$$ The logic is the same, but we choose 2 face values out of the remaining 12 rather than choosing 1 value from 12 and another value from 11.

This is the part that I don't quite understand. Why instead of doing $\binom{12}{1}*\binom{11}{1}$ we do $\binom{12}{2}$? Based on that logic, I could then say that $$\frac{\binom{13}{3}*\binom{4}{3}*\binom{4}{1}*\binom{4}{1}}{\binom{52}{5}}$$ should have been the solution as we would just pick our corresponding 3 values straight away and then choose the cards themselves(out of 4). I feel really confused about that. Can someone tell me where my logic is flawed and why we choose 1 out of 13 and then 2 out of 12 instead of choosing 1 out of 13,1 out of 12 and 1 out of 11 or choosing 3 out of 13?

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

When you do $\dbinom {12}1 \dbinom {11}1$, the order matters, i.e. $JK$ is considered to be different from $KJ$, so there would be double-counting.

To ignore the order, we either use $\dbinom {12}2$, or we divide the above by $2!$, and the number we obtain from both methods would be equal.

For the final equation, after you chose the $3$ numbers, you must also choose which of the three is the one in the 3-of-a-kind. There are $\dbinom31=3$ choices for that.

Incidentally:

$$3\binom {13}3 =\frac {3\times13!}{10!3!} = 13 \times \frac{12!}{10!2!}= \binom{13}1\binom {12}2$$

so using that method will also arrive at the solution after considering the choice for the special card.

$\endgroup$
3
  • $\begingroup$ Thanks for the clarification. I would like to ask you 2 more questions for clarification and then I will accept your answer: 1) For the final equation, you formally get 3 by $\binom{3}{1}$ (choose 1 value out of the given 3)? 2) When we choose the first face value isn't it the order going to matter too(simillar to $\binom{12}{1}\binom{11}{1}$)? $\endgroup$
    – Alex.Kh
    Commented Oct 4, 2020 at 16:17
  • $\begingroup$ 1) Yes indeed. I will add that. 2) Yes, because the repeated card is different (in status, if you will) from the other two cards: Consider $AAA23$, $222A3$ and $2223A$. The first two are different but the last two are the same, combinationwise. $\endgroup$
    – player3236
    Commented Oct 4, 2020 at 16:25
  • $\begingroup$ Got that, thanks for the answer $\endgroup$
    – Alex.Kh
    Commented Oct 4, 2020 at 16:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .