poker math question

Can someone tell me approximately how many hands it would take for someone to get a bad beat in Texas holdem. At least Aces full of of Queens beat by 4 of a kind or better. Assuming every one (all 9 players) played every hand that had a possibility, all the way to the river. 2 cards must play. Thanks in advanced.

Quote: AxelWolf

Can someone tell me approximately how many hands it would take for someone to get a bad beat in Texas holdem. At least Aces full of of Queens beat by 4 of a kind or better. Assuming every one (all 9 players) played every hand that had a possibility, all the way to the river. 2 cards must play. Thanks in advanced.

According to the Wiz, it's about 58,000

http://www.wizardofodds.com/games/texas-hold-em/bad-beat-jackpots/

You really need to know what constitutes a bad beat as it will vary among casinos.

Hollywood Park Casino near LA used to have any aces full beat by quads or better, then they changed to aces full of 9s beaten. Some casinos use Aces full of tens. And at a few casinos quads have to lose to quads or better.

Check this site: http://www.pokerlistings.com/strategy/general-poker/the-truth-about-bad-beat-jackpots

^^^ It's often quads beaten (both hole cards) east of Nevada.

And yeah, it's roughly 1 in 58k for aces full of queens beaten by quads in the "no-foldem" scenario. But unfortunately preflop folding can alter probabilities quite a bit, by factors of 2 or more.

Quote: AlanMendelson

You really need to know what constitutes a bad beat as it will vary among casinos.

Hollywood Park Casino near LA used to have any aces full beat by quads or better, then they changed to aces full of 9s beaten. Some casinos use Aces full of tens. And at a few casinos quads have to lose to quads or better.

Check this site: http://www.pokerlistings.com/strategy/general-poker/the-truth-about-bad-beat-jackpots

I thought I made it clear. Aces full of queens beat by a four of a kind or better. Lets assume 2 of the Aces must be in your hand.

Quote: tringlomane

^^^ It's often quads beaten (both hole cards) east of Nevada.

And yeah, it's roughly 1 in 58k for aces full of queens beaten by quads in the "no-foldem" scenario. But unfortunately preflop folding can alter probabilities quite a bit, by factors of 2 or more.

Thank you

Also just to be clear on this, the one in 58k probability that mickeycrimm pointed out and I agreed with is the probability for a full table of 10 players actually. So an individual hitting the loser's share would be 1/10th of this. Or 1 in 580k.

I also just noticed you wanted 9-handed, blah. In that case, modifying the Wiz's data, the probability of the table triggering a bad beat 9-handed in a "no-foldem" scenario should approximately be: C(9,2)/C(10,2) x 0.00001721 = 36/45 x 0.00001721 = 0.000013768 = 1 in 72,632. Unfortunately, once again folding complicates matters a bit. The practical odds of triggering the bad beat may be over 1 in 200,000. You'd have to run a simulation with estimated opening hand ranges for that, which is a lot messier.

The biggest hand I ever got beat with was four 8's. Unfortunately, it was at the Horseshoe, no bad beat. The toughest beats I ever took was flopping top set and getting beat by fourth pair. It happened to me twice. Fourth pair is an underpair to the board. You have to catch runner runner and make quads to beat top set in that spot.