Odds of four of a kind

[Jamey Saunders]

This hand just happened in a SNG where the flop came 444. It made me curious what the odds are that someone was holding the last 4. There were 4 people in the pot at a 9-person table.

My math ability sucks, so I'd appreciate any help I can get.

[kiddj]

Assuming you don't have it, the probability that one of the other 3 have it is 0.128 (or 7.8:1 against)

[Jamey Saunders]

Thanks Kiddj. Just what I wanted to know.

[fnord_too]

I think the first responder gave an answer assuming random hands, but that is probably not going to reflect reality (unless these players play EVERY hand). You need to consider the pf action, and what the range of hands someone would come in with. For instance, if there was an early raise and 4 people saw a flop of 444, and none of the four people would play A4s or 45s in that situation (and by extension 4xs, 34s, 45o, etc.) there is zero chance that one of your opponents has quads yet. If they would all play any suited ace or suited connector, in addition to some other hands, you need to count all the hands they would play that do NOT have a 4 and divide that by all the hands they would play in total for each opponent. Then multiply these numbers together and subtract that number from 1.

In practice, lower trips on board reduce the chances of someone having quads, and pre flop raises reduce the chance of someone having quads. High trips, especially aces, have a far greater chance of hitting someone since there are more ace hands that people will play.

[Jamey Saunders]

Good points, and ones that I would definately take into consideration. This being a freeroll, though, the quality of players wasn't all that good, and it's a freaking miracle that there were only four of us to the flop. But Kidd was right on -- barring all the very good points you brought up (meaning just on a random hand) what were the odds.

BTW, nobody had the other 4. I had A-T and caught a ten on the turn for the boat.