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.

  • $\begingroup$ You're question is unclear IMO. For example: "Pre-deal ($52$ cards) here is my calculation" - Calculation of what? And what does "Pre-deal" even mean? Another example: "two cards (a single player's hand)" - since when are two cards consist a single player's hand? $\endgroup$ – barak manos Oct 6 '16 at 4:03
  • $\begingroup$ That is the calculation of getting two-pair in 5 card poker with ALL 52 cards still remaining, none removed. $\endgroup$ – chriskgregory Oct 6 '16 at 4:05
  • $\begingroup$ Right... well, the preceding statements says something about $2$ cards which have already been dealt. Why is that no longer the case then? $\endgroup$ – barak manos Oct 6 '16 at 4:06
  • $\begingroup$ for your second example it seems pretty straightforward - In 5 card poker you are dealt TWO CARDS. You must make a hand with the upcoming 3 cards (2 + 3 = 5 Card Poker) $\endgroup$ – chriskgregory Oct 6 '16 at 4:07
  • $\begingroup$ I'm trying to display my method of calculation with 52 cards - then my second calculation where there are 2 cards removed, and you are using those to achieve two pair. I do not believe my second calculation is correct, therefore by displaying my method of arriving to my conclusion I figured the reader would easier be able to correct me. $\endgroup$ – chriskgregory Oct 6 '16 at 4:09

[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.

  • $\begingroup$ I'm trying to understand but I'm having a hard time following. I am not educated and didn't even attend high school to my great regret, if you can try to elaborate a bit more I would greatly appreciate it. $\endgroup$ – chriskgregory Oct 6 '16 at 4:12
  • $\begingroup$ Well, you could choose a king pair and a queen pair, or a pair of aces and pair of threes, or.... $\endgroup$ – Graham Kemp Oct 6 '16 at 4:12
  • $\begingroup$ Ah I'm following. I thought that was what was in the []s $\endgroup$ – chriskgregory Oct 6 '16 at 4:14
  • $\begingroup$ Or am I miss interpreting what 2 and 7 removed means $\endgroup$ – Graham Kemp Oct 6 '16 at 4:14
  • $\begingroup$ So I have 7clubs and 2 diamonds. There are still 50 cards remaining in the deck. For my calculation of Two Pair using the 50 remaining cards, I took two scenarios and added them together - 1: 7272x 2: 7266x (66 being replaceable with any other pair). I wrote it as 11 Choose 1 by 4 Choose 1. $\endgroup$ – chriskgregory Oct 6 '16 at 4:15

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