# Got beaten three times by the same 4 of a kind, then won with a straight flush. In the span of 141 hands. What are the odds?

I hope anyone can help me, the math goes way over my head. I just had the most unlikely game. At a 9 person hold'em table, I first got beaten by three separate people, all with 5555. Then I won a hand with a straight flush. All this happened within the space of 141 hands. I'm just really curious what those odds are, but not sure how to calculate them. Thanks in advance.

• This is not trivial to calculate as far as I know - there are 148,476,719,998,694,000,000,000,000,000,000 different ways to deal a holdem hand to 9 players and then have a board come out and that's even before the combinations of ways the betting can go and probabilities of people's actions resulting in the outcome you describe. If you are willing to make some assumptions to simplify the math (like if you just pretended every hand was dealt to a river with no betting involved) then you might be able to approximate an answer. – 3N1GM4 Dec 15 '16 at 12:47
• This is actually impossible to calculate because the players per flop / per turn / per river is unknown. Also hand ranges for players are unknown. I'm probably forgetting a LOT of other things. I imagine the odds of this are so low that (if on-line) the odds of players disconnecting and other freak incidents would also have to be considered, and how do you expect anyone to do that? – user4555 May 22 '17 at 19:49
• You would solve this with a binomial distribution. – paparazzo Sep 21 '17 at 17:23
• Could you write an answer to show us how @Paparazzi? – 3N1GM4 Jan 18 '18 at 13:06
• @3N1GM4 See this poker.stackexchange.com/questions/4087/…. Forget the straight flush and just put in probability for 5555. Treat it as 141 * 8 individual hands. Multiple players cannot get 5555 unless it is on the board. I saw an answer where timmerands went at it different but I never compared if it was the same answer. – paparazzo Jan 27 '18 at 16:46