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Poker - 52 cards
4 suits of 13 (rank) cards each

Poker probability

A flush is 5 cards of the same suit
The chance of a flush in 5 random cards is:
$$\frac { \binom{13}{5} \binom{4}{1}} {\binom{52}{5}} = 0.001981 $$
I know the above is correct as I got it from link above

In (this) poker you start with 2 cards and then bet and then receive 5 additional
You make the best 5 card hand from the 7 cards
My question is if you are dealt 2 suited cards - what is the chance of make 5 suited cards with 5 additional random cards to come?
To make it easier get at least 3 of the suit in the next 5 cards?

  • 50 cards

  • 11 of the suit you need 3 of

  • 39 other cards

This is for exactly 3:

$$\frac { \binom{11}{3} \binom{39}{2}} {\binom{50}{5}} = 0.057706 = 16.33:1 $$

But the real question is make a flush so 4 or 5 also count.
Is this correct?

$$\frac { \binom{11}{3} \binom{39}{2} + \binom{11}{4} \binom{39}{1} + \binom{11}{5} } { \binom{50}{5} } = 0.064 = 14.6 :1 $$

This article says 6.5%
I have also read a flush is worth +4% but that number may include an adjustment for getting beat by a higher flush which is not part of my question.

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  • $\begingroup$ How many opponents do you have? Is it one, or are we considering n opponents, r of whom get to the river, because, I think (especially) in the second case, unless you make a bunch of unrealistic assumptions, things will get much more complicated than the types of situations you're considering above with binomials - is that what you're after, or are you okay with making a bunch of idealised, and possibly unrealistic assumptions? $\endgroup$ Feb 6, 2016 at 10:54
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    $\begingroup$ @hodgenovice That comment is pushing rude and I am not going to argue the point with you. Number of opponents is immaterial. A down card in the deck or in a hand is an unknown card. The possible boards is not effected by number of opponents. That link to wiki "poker probability" in my question does all the probabilities with zero dependence on number of players in the hand. Good day to you. $\endgroup$
    – paparazzo
    Feb 6, 2016 at 14:30
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    $\begingroup$ The article actually says "around 6.5% of the time", so your answer agrees and your calculation seems reasonable to me. $\endgroup$
    – user1566
    Feb 6, 2016 at 16:28
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    $\begingroup$ @Frisbee Sorry mate, wasn't trying to be rude. I just wanted to know what situation you were wondering about since I typed out an answer for the case of 1 opponent but then realised you might only be interested in games of 6 opponents. With more opponents, it changes the assumptions you'd want to make, so I was just checking what you were thinking behind the question before I posted an answer. Anyway, seems you didn't appreciate what I said, so I won't pursue this any further. Just want to point out I wasn't trying to be rude or offensive. $\endgroup$ Feb 6, 2016 at 18:16
  • $\begingroup$ @hodgenovice Cool then we disagree on if number of opponents is material to me getting a flush. If you want to present an answer that shows it is material for sure I will not vote you down. $\endgroup$
    – paparazzo
    Feb 6, 2016 at 18:29

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