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Assume you're given 2 cards of the same suit, and the next three cards (flop) are also the same suit as your hole cards. What is the probability of that happening?

So I figured out that 2 cards have been dealt to you meaning there are only 50 cards remaining and 50C3 represents the # of different combinations that could be the flop. Since I already have 2 cards of the same suit, there remains only 11 cards of that particular suit making 11C3 the # of combinations to get 3 of those cards from the pile of 11.

Hence, the probability is: 11C3/50C3

That's what I'm thinking, please feel free to correct me. Thanks!

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  • I know that flopping Straight Flush is about 0.001%, although i don't know why i still remember it ;) Nevertheless, not a great odd to remember.
    – user1165
    Jan 11, 2015 at 22:39
  • Um okay, mine def doesn't equal that..but does my approach make sense or am I missing something?
    – user2582622
    Jan 11, 2015 at 22:45
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    i'm not good at computing these kind of probabilities. Actually there's exactly a similar question as this here which gives an odd of 0.005% instead of 0.001%.
    – user1165
    Jan 11, 2015 at 23:04
  • Are you presuming that you have zero-gapped suited connectors? It's easier to flop a straight flush with JTs than it is with J7s.
    – Chris Farmer
    Jan 12, 2015 at 1:18
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    This question has been asked and answered on the internet about a thousand times. It is, as vlzvl points out, really not an interested odds calculation, generally. Any odds less than 1% is generally not worth thinking about IMHO. But, it's good that you're thinking about odds. Just think more about your odds of hitting a double-belly buster when you put your opponent on a range that includes your out cards.
    – jmsimpson68
    Jan 12, 2015 at 19:12

3 Answers 3

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The worst case for a possible straight flush is holding something like A2s, AKs, A5s, etc., where there's only one possible way to flop the straight flush. In that case, the probability is one in 50C3, or 19600.

The best case is 45s..TJs, which is 4 in 19600, because there are 4 ways to flop the straight flush. Hands like 58s are 2-ways, hands like 79s are 3-ways.

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There are 40 straight flushes in a deck: 10 per suit (A-5, 2-6, ... , 10-A) times 4 suits.

For a random draw of 5 cards (e.g. your 2 hole cards plus the 3 flopped cards) there are 2,598,960 possible combinations: (52! / 47!) / 5!

Since 40 of those are straight flushes, the odds of flopping a straight flush are: 1 in (((52! / 47!) / 5!) / 40) = 1 in 64,974

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  • This appears to be assuming a random set of hands, when the OP states suited cards.
    – BowlOfRed
    Apr 25, 2015 at 0:03
  • Yep, I completely ignored that. I don't know why. Presuming that we start with 2 cards that can possibly make a straight flush on the flop, there will be exactly 1, 2, 3, or 4 flops that will make a straight flush. There are 50C3 (19600) possible flops. So the odds are 1, 2, 3, or 4 in 19600, depending on the starting cards. (Which has already been stated in prior answers).
    – DonS
    Apr 27, 2015 at 13:00
  • This is the correct answer to "flop a flush". If the OP flops a straight flush then they are suited.
    – paparazzo
    Mar 27, 2018 at 13:39
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First a comment about the way you used nCk; 11C3 is ALL the possible ways you could draw 3 cards from the 11 remaining cards of your suit, but it isn't the case that any three cards of your suit will give you a straight flush (e.g. Suppose you hold the 2h4h, well, one of the combinations that gets counted by 11C3 is 9hAhTh, but that flop doesn't give you a straight flush). What you've calculated is the probability of flopping a flush given you've been dealt two suited cards.

Now, as @TacticalCoder pointed out, the probability of flopping a straight flush once you've been dealt two cards depends on which two cards you've received; that being said, there is only ever 1,2, or 3 ways for you to flop a straight flush so the probability is either 1/(50C3), 2/(50C3), or 3/(50C3).

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  • ...or 4 in 50C3, with hands 45s to TJs. The former, for example, can flop A23, 236, 367, or 678.
    – Lee Daniel Crocker
    Apr 27, 2015 at 19:50

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