Bad Beat Jackpot Probability

What is the probability of a 10-handed Texas Hold 'em hand ending with the loser holding four eights or better? The rules stipulate that both hole cards must play in the winning and losing hands, and that a losing four of a kind must also contain a pocket pair.

The Wizard has posted this information on the Wizard of Odds site: here. For the situation you describe, it is .00000525, or 1 in 190,476

Quote: cas202

The rules stipulate that both hole cards must play in the winning and losing hands, and that a losing four of a kind must also contain a pocket pair.

Actually, its usually even tighter than that.

Hands that have quads, must have a pocket pair. Hands with a straight flush must use both hole cards for the best 5 card hand.

With those rules, the following two scenarios would NOT qualify:

AK QQ AAAQQ - The winner is using both hole cards, but doesn't have a pocket pair.

KcQc 7c6c JcTc9c8cXx - The loser has a Jack high SF and isn't using his 6c.

Quote: DJTeddyBear

Actually, its usually even tighter than that.

Hands that have quads, must have a pocket pair. Hands with a straight flush must use both hole cards for the best 5 card hand.

With those rules, the following two scenarios would NOT qualify:

AK QQ AAAQQ - The winner is using both hole cards, but doesn't have a pocket pair.

KcQc 7c6c JcTc9c8cXx - The loser has a Jack high SF and isn't using his 6c.

Actually, these hands, AK QQ, with this board AAAQQ would qualify. The quad aces are the winning hand, and do not require a pocket pair, only the losing quad.

You are correct that the second scenario fails to qualify, since the "best" straight flush for the small side only uses one hole card. However, I believe the Wizard's table took that into account.

Quote: Ayecarumba

Actually, these hands, AK QQ, with this board AAAQQ would qualify.

Maybe it depens on the casino.

I was in a hand where I had KT vs QQ with a board of QTTT7. I asked if it would have qualified if the board had QTTTQ and was told no. This was Showboat, AC. Other casinos have also confirmed that winning OR losing quads requires a pocket pair.

Now that Harrah's AC combined the Bad Beat, it's impossible to find the old rules. Heck, it's impossible to find the current rules too!

Quote: DJTeddyBear

Other casinos have also confirmed that winning OR losing quads requires a pocket pair.

This is usually the case. I'm making an assumption that OP is asking because Foxwoods' BBJ is nearing $500k. The only mistake OP made regarding the qualifications at Foxwoods is that ALL quads must contain a PP, not just as the loser.

Quote: DJTeddyBear

Maybe it depens on the casino.

I was in a hand where I had KT vs QQ with a board of QTTT7. I asked if it would have qualified if the board had QTTTQ and was told no. This was Showboat, AC. Other casinos have also confirmed that winning OR losing quads requires a pocket pair.

Now that Harrah's AC combined the Bad Beat, it's impossible to find the old rules. Heck, it's impossible to find the current rules too!

You are correct. I have come across both types of "rules", sometimes in the same place. Some card rooms ran promotions where they would pay a regular bad beat (Aces full of Jacks or better, beaten by any quad, both pocket cards must be used), and a bigger jackpot for certain quads over quads both with pocket pairs. These were special promotions, and they come and go.

Given the OP's description, the Wizard's table fit. Note that the Wizard's figure is based on any of the ten hands getting beat, assuming all ten hands always go to the showdown. This is far from the actual condition, so the "real world" probability of encountering a bad beat is lower, but I don't know how much lower.

On a somewhat related note, Mohegan Sun runs a high hand jackpot every 4 hours.

You only need to use one card, and there does NOT need to be a showdown.

Yes this is about Foxwoods, and yes, the winning four of a kind must contain a pocket pair too. Just trying to calculate what the jackpot size is where it becomes worth it to play there since they take an extra dollar out of the pot for the jackpot.

In my experience, the jackpot drop is always taken from the pot if there are the minimum number of players to qualify for a jackpot (typically, 5). Players do not have the option to to opt out, so if you are going to post a blind, you are going to contribute.

I think the break even on the jackpot would have to be north of $1,904,762, as the probability of a single player at a ten player table having a pocket pair of 8s or greater and getting beaten by four of kind or better (with the other player also using both pocket cards, and having a pocket pair if quads) is .000000525 or 1 in 1,904,762 (per the Wizard's table.) However, I am not sure if that is right since individual players do not contribute every hand. I'll start a new thread on calculating break even points.