What is the number of 7 card poker hands which contain 3 of a kind and the other 4 are not the same?

Question

Im solving this problem in two ways but i don't know which is correct, because they give different results.

First Way

• I calculate the number of hands for the 4 possible forms of accepting hands:

1. (aaa - bbb - c) : $\binom{13}{2} \binom{4}{3}^2~\binom{11}{1} \binom{4}{1} $
2. (aaa - bb - cc) : $\binom{13}{1} \binom{4}{3}~\binom{12}{2} \binom{4}{2}^2 $
3. (aaa - bb - c - d): $\binom{13}{1} \binom{4}{3} ~\binom{12}{1} \binom{4}{2} ~\binom{11}{2} \binom{4}{1}^2$
4. (aaa - b - c - d - e): $\binom{13}{1} \binom{4}{3} ~\binom{12}{4} \binom{4}{1}^4$

• Add all of them and I get: $\left| A \right| = 10062624$

Second Way

• I calculate all the unordered seven card hands of the form {aaa - ****}:

  $\left| A_1 \right| = \binom{13}{1} \binom{4}{3} \binom{48}{4}$

  Choose 1 from the 13 ranks, choose 3 from 4 of the cards in that rank and then choose 4 of the 48 remaining cards.

• I calculate all the unordered seven card hands of the form {aaa - bbbb}:

  $\left| A_2 \right| = \binom{13}{1} \binom{4}{3} \binom{12}{1} \binom{4}{4}$

  Choose 1 from the 13 ranks, choose 3 from 4 of the cards in that rank, choose 1 from the 12 remaining ranks and then choose 4 of the 4 cards in that rank.

• And then I subtract $\left| A_2 \right|$ from $\left| A_1 \right|$ and I get: $\left| A_2 \right| - \left| A_1 \right| = 10117536$

It seems that when using the second method, I might be counting the (aaa - bbb - c) sequence in 7-card hands twice, because when I subtract it from the total count, I end up with the same result as the first method. Can someone explain why this is happening?

Answer 1

As you noticed : $10117536 - \underbrace{\binom{13}2\binom 4 3^2\binom {11}1\binom 41}_{\large 54912} =10062624$

So indeed your second method was overcounted the pattern aaa-bbb-c .

Why? After selecting a triple of one rank, $\big(\binom{12*4}{4}-\binom{12}1\binom 44\big)$ tries to count selection of 4 cards from the remaining ranks that are not all the same rank, but this includes the possibility of selecting a second triple. Thus overcounting like so:

• Select 3 aces, then 4 cards( including 3 duces and singletons, etc ), or
• Select 3 duces, then 4 cards( including 3 aces and singletons, etc ), or
• $\quad\vdots$

$\therefore$ You do need to discount having that second triple.