How Many Hands Having 3 Cards of a Kind are There?

Question

The conditions are as follows:

1. Standard 52-card deck, 4 suits.
2. 5-card hand.
3. There must be three of a kind, but only this (i.e., no 4 of a kind or full house).

I know this has been asked already, but I am not convinced by their methods (and I am not convinced by my method for the same reason). My method gets the right answer, but there is one thing I don't understand about it. Here:

52 cards to pick for the first card of the three of a kind, 3 cards to pick for the second, and 2 for the last. Then we divide by 3! to eliminate the identical hands created by ordering.

Then, for the last 2 cards, we have 48 choices for the first one (for we cannot choose the fourth card of the 3 of a kind rank) and 44 for the second (because the other suits of the same rank as the first filler card cannot be used). Then we divide by 2! to eliminate the permutations of the last two cards. In total, we get 54912 hands.

The issue I have with this is what the reason is why we don't again divide again. Isn't it true that we have found all the arrangements for a hand that goes AAABC, where we have eliminated ordering for AAA and B and C, but not for the two groups? In other words, why is it that it is alright like this, when we could put instead BCAAA, or BAAAC, etc.?

Thanks!

Answer 1

Although it doesn't explicitly say so in the question, it is customary with a hand of cards to assume that order does not matter. So BCAAA is the same hand as AAABC, and the procedure you have proposed counts this hand once only: first choose the triple, then the fourth card, then the fifth.

You could devise an alternative procedure for choosing the hand: choose a card of one value, a card of another, then three of a third value. However this would be something you would do instead of your procedure, not as well as your procedure. What you have done counts every admissible hand, and does not count any hand twice, so it is correct.

Hope this helps.

Answer 2

First step: choose a number to have three of. Second step choose three of this number. Third step, choose two other numbered different to this number. Choose one card from each of these. Multiply together and divide by the total number of possible choices.

So the answer is $\frac{13 \times {4 \choose 3} \times {12 \choose 2} \times 4 \times 4}{52 \choose 5} = \frac{54192}{2598960} = \frac{4756}{21655}$

Answer 3

You've eliminate the ordering when selecting the groups, so you then do not need to eliminate the ordering when combining these selected groups.

• Select a triple of cards (same face, any suit). Eliminate the order of selection to place them in an unordered pile. $$\frac{\overbrace{(13\times 4)}^{52} \times 3 \times 2}{3!}$$

• Select two unpaired cards (any suit, both different faces from each other and the triple). Eliminate the order of selection to place them in a second unordered pile. $$\frac{\overbrace{(12\times 4)}^{48}\times \overbrace{(11\times 4)}^{44}}{2!}$$

• All you then need to do is combine the unsorted piles in an arbitrary way and you have a single unsorted pile; a hand. Just push them together. $$1$$ Thus the count of distinct ways to draw exactly a three of a kind is: $$\tfrac{52\times 48\times 44}{2} = 54,\!912$$