odds of quads getting beat by quads

[ICantRaed]

Last night had quad 5s. Showdown came, beat by quad 7s. Each of us were dealt a pair and then two showed up on the table.

I realize that if you are considering this sort of thing, quad 5s are at the low end of the quad hands and can be beat by many - but just for fun, what are the odds of two people getting quads at the same time.

Then for fun on top of that, what are the odds of getting quads which beat your own?

I am less interested in the exact figures than I am in the equations.

[ICantRaed]

The equation for getting beat by higher quads is pretty easy I think.

There are 13 different values for the four of a kind (2,3,4...A).

I held the 5, so that left 3 (2,3,4) smaller than me, and 9 (6,7,8,9,T,J,Q,K,A) larger than me. So 3/12 times I would beat someone if I knew they also held quads and I had quad 5s, so 25% of the time I would win it?

As for the larger "likelihood of two people getting quads at the same time" I keep running into headaches.

My first inclination is to say that even though there are 9 cards we are looking at (the 5 on the board and the two in each player's hand at the showdown) there are 8 cards we care about (quads * 2), so an equation like 8/52 * 7/51 * 6/50 * 5/49 * 4/48 * 3/47 * 2/46 * 1/45
or 1 in 752,538,150

Is that correct?

[ICantRaed]

I don't think that can be correct because using the same logic you don't get the correct likelihood of just one person getting quads.

[RiverDood]

I'd recommend letting go of that approach. You're treating it as a two-handed game, which it's not. Any of the other eight (or seven) players could have quads. That needs to be factored in. You'll do backflips trying to clean up the math.

I'm trying to do this in a 5-minute break, so excuse the hastiness, but start by postulating that you need a board with two pairs. What are the odds of that? About 1/21.

Next, you need two of the 10 players' hands to match those board pairs with their own pocket pairs. Let's start with one pre-identified hand (you.) You can match either board pair, so the odds of you matching is 4/47 x 1/46, or 1/540. So the odds of you hitting quads is 1/11350.

You've now got 9 slots where the other "magical" pair could occur. The odds of the other magical pair in any particular hand are (2/45*1/44), or 1/990. The odds of it being somewhere in the 9 hands is approximately 9/990, or 1/110. So your odds of the whole wonderful thing are 1/11350*110, or something close to
1/1,248,500

Still a rare event. If you're in a whimsical mood, email the site and ask for a bad beat jackpot to be paid to you. Tell them you won't play anymore until you get it.

[ICantRaed]

Thank you very much for taking the time to do that and for explaining the logic on it - I appreciate that.

As for trying to get my money back - it was at PokerRoom and as far as I know they don't have one of those money back "bad beat" systems. To be honest I just sort of laughed at it - I clearly saw that it was possible as it was heading towards the end, but I surely wasn't counting on it.