5 Card Poker probability with specific cards already dealt

Question

I'm trying to calculate probability of a 5-Card Poker hand where two cards (a single player's hand) have already been dealt.

Calculating probability of receiving Two-Pair is a good example that should answer all of my questions within itself.

Here is my calculation of 5 Card Poker odds of achieving Two Pair without any cards yet dealt:

$$\binom{13}{2} \cdot \binom{4}{2}^2 \cdot \binom{11}{1} \cdot \binom{4}{1} = 123,552 / \binom{52}{5} = 4.7539\% $$

However, considering the probability of achieving Two Pair with a 7,2 Offsuit in the player's hand, here is my calculation:

$$ \binom{2}{2} \cdot \binom{3}{1}^2 \cdot \binom{11}{1} \cdot \binom{4}{1} +\left[\binom{11}{1} \cdot \binom{4}{2} \cdot \binom{2}{1} \cdot \binom{3}{1}\right] = 792/\binom{50}{3} = 4.0408\%$$

• The first 4 binomials are for the occurance of repeats of 7 and 2 followed by any other random card of any suit, hence 2 Choose 2 for two of two specific cards and the 3 Choose 1 squared because each of the two cards would be one of the 3 remaining suits. 11 Choose 1 for any rank of the 11 remaining, and 4 Choose 1 for any suit, as they don't matter.
• The 4 binomials in brackets are for the occurrence of a new pair followed by one of the previous cards (7 or 2). 11 Choose 1 for any card not 7 or 2, 4 Choose 2 because it must be two suits of that same rank. 2 Choose 1 for one of the two existing ranks (7 or 2) and any of the 3 remaining suits of this rank therefore 3 Choose 1. I add both binomial sets together to get 792 out of the 3 card flop with 50 remaining cards (52 minus 7 and 2) to get 4.0408%.

I am not sure if this is correct, or if I need to subtract the Full House occurrences from this. Any feedback would be greatly appreciated. I tried testing this out by comparing the probabilities to a Mississippi Stud game (5 card poker) and getting the EV of this scenario, but the EV did not match up.

Answer 1

[edit: with apologies] Reinterpreting the question as: "What is the probability of forming two-pair if you already have 2 and 7 in your hand, and must draw three more cards?"

• You draw another set of 2 and 7, and one other card: $\binom 3 1^2\binom{11} 1\binom 4 1$ ways
• You draw either another 2 or 7, and another pair: $\binom 2 1\binom 3 1\binom{11}1\binom 4 2$ ways
  • Out of $\binom{50} 3 $ ways to draw any 3 cards from 50.

$$\dfrac{\binom 3 1^2\binom{11} 1\binom 4 1+\binom 2 1\binom 3 1\binom{11}1\binom 4 2}{\binom {50}3}$$

Which is as you had.