View Full Version : 3-Flush Flops

06-19-2003, 01:20 PM
I recently starting playing on a new online poker site. While I'm a firm believer in random number generators and laugh at people who cry about 'juiced cards' at these sites - I recently saw a string of cards that just didn't seem right to me. Obviously it could have just been one of those odd strings of cards that will always come in the long run - and that simply stood out due to the fast nature of online play - but I wanted to throw it out here for some statistical ball play.

Game: Holdem
Players: 10-handed

I noticed that there seemed to be a lot 3-flush (all same suit) flops hitting during my play at this table. I therefore requested a hand history and found the following.

Hands Played: 113

3-Flush Flops: 12

My question is this...what *should* be the statiscal odds of seeing a 3-flush flop? I believe its around 1.2%. Therefore in 113 hands, on *average* I would expect to see one or two 3-flush flops.

Soo...what are the statistical odds of seeing what I saw? Is it so far out of the realm of possibility as to indicate a potential flaw in the site's logic?

If so...what do you figure this info is worth? ;-)


06-19-2003, 06:14 PM
tcdiscenza are you the player (or were you) that listed your home as The Vatican on UB? If so we used to play together quite often.

06-19-2003, 11:44 PM
LOL - yup - that's me. Don't play a lot on UB anymore - but occasionally. UB isn't the site in question here - it's another one that I've started to play on. What's your screenname on UB? I don't think I recognize 'Jimbo'.

Also - you do know the story behind 'Vatican' right? On UB it says 'representing' not 'from.' Since I'm an atheist - the chance to 'represent' the Vatican was just too good to pass up! ;-)

- Todd

06-20-2003, 07:21 AM
The chance of twelve or more three-flushes occurring in 113 hands is exactly 1.43%.


I think you made an error in your original calculation... It's a lot higher that 1.2%. The chance of seing a 3-flush flop is actually almost exactly 5.2%, calculated thusly;

(52/52) x (12/51) x (11/50) = 0.0517647 ~ 5.2%
i.e. the first card can be anything, the second card must be one of the remaining twelve of that suit in a fifty-one card deck and the third must be one of the remaining eleven of that suit in a fifty card deck.

06-20-2003, 10:59 AM
Hi tc,

I played as Jimbo Moran at UB. We normally played 3/6 together if I remember correctly. Great story about representing The Vatican. /forums/images/icons/smile.gif After all you did not look Polish to me! /forums/images/icons/smile.gif

06-20-2003, 04:10 PM
I knew I was missing something obvious...