I once flopped a straight flush. Only it was the idiot end of the straight flush, and I lost all chips (of course). The other player had also flopped a straight flush - the higher end straight flush.

What are the odds of this?

I know the odds of one hand flopping a straight flush (1:20,000 or so). What about two hands?

related: Probability of flopping straight flush


In order for two people flopping a straight flush we need:

  1. a connected three straight flop, all cards having the same suit. No cards can be lower than a three or higher than a queen.
  2. player 1 holding two connected cards of that same suit making the higher or the lower straight.
  3. player 2 holding two connected cards of that same suit making the last possible straight.

Odds of this happening can be calculated by calculating each indiviual requirement and multiply them together.

  1. There are a total of C(52,3) = 22100 possible flop combinations. 345, 456, 567, 678, 789, 89T, 9TJ, TJQ are all the possible straights, which are 32 unique combinations (8 straight 4 suits). Odds are therefore 32 / 22100.
  2. Player 1 now has to get either the two lower cards, or two upper cards, which are only two combinations, since the suit is now known. There are a total of C(49,2) = 1176 different holding player 1 can have. Odds are therefore 2 / 1176.
  3. Player 2 now has to get the remaining combination. There are a total of C(47,2) = 1081 different holdings for player 2. Odds are therefore 1 / 1081.

Multiplied together we get 2.278 * 10^-9 or 1 : 438,980,588.

Odds of you specifically getting screwed is 1 : 877,961,175, which is twice as low.


Start with flopping a straight flush. The odds are 1 / 64974.

The only way to flop the idiot end is:

Then all 4 suits. So there are 32 possible idiot end of a straight flush.
Chance of you have the idiot end is:
32/combin(52;2) = 32/1326 = 2.4133% = 1 / 41.4375

There is only 1 flop that gives you the idiot straight flush
1/combin(50;3) = 1/19600 = 0.0051%

The combined probability is:
1.23126173546342E-006 = 1 / 812,175

Now what is the chance your opponent has the exact two cards they need:
1/combin(47;2) = 1/1081 = 0.0925%
Do your opponent last as you know your hole cards and flop. Given you have the idiot end you are only beat 1/1081. You have no redraws - you cannot win the hand. You have to go broke here as they could be jamming the nut (or lower) flush.

The combined probability is:
1.13900253049345E-009 = 1 / 877,961,175

  • It is indeed C(47,2) and not C(48,2). My bad. – Raymond Timmermans Mar 27 '18 at 12:55
  • @RaymondTimmermans I ran the numbers without looking at yours to see if I got the same. Even if I plug in 48 I do not get the same number as you. – paparazzo Mar 27 '18 at 12:56
  • I think my numbers are correct now. Your number is the same as mine, and the other number is simply half of that. – Raymond Timmermans Mar 27 '18 at 14:24
  • @RaymondTimmermans And u get the check mark. – paparazzo Mar 28 '18 at 9:44
  • I answered first and got more upvotes, I only made a minor calculating error. Plus who cares really? – Raymond Timmermans Mar 28 '18 at 10:03

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