I am wondering how many hands played on avg would it take for a bad beat jackpot of losing with quad 8s or better happen(must use both hole cards). I do not know how to do the math. Thank you so much for anyone willing to answer and help. Cheers!
"Bad beat" is a term that can mean having an outstanding chance of winning a bet, only to still lose. The term can be used in any form of gambling but is most commonly applied to poker. Many poker rooms offer a progressive jackpot for very unlikely bad beats. Various other rules are added to ensure that only surprising bad beats win. Below I present tables of bad beat probabilities, starting with the most liberal rules, and ending with the most stringent. The most stringent rules, the "Bad Beat Type 3", are the most common, in my experience.
Following are the rules for a type 1 bad beat.
- Both the bad beat and winning hand must be the best possible combination of five cards. In cases where the same hand can be created multiple ways (for example player has AK and the board shows AAKKQ) the player's hole cards will take priority.
- Both the bad beat and winning hand must make use of both hole cards.
- A full house must be beaten by a four of a kind or higher.
The rules for a type 2 bad beat are the same as type 1, plus any four of a kind, whether the bad beat hand or winning hand, must contain a pocket pair.
The rules for a type 3 bad beat are the same as type 2, plus a full house may not make use of a three of a kind entirely on the board.
For your requirements:
Type 1 Type 2 Type 3 Four 8's or higher 0.0000064 0.00000525 0.00000519
Numbers were generated from a simulation of 2.5 billion rounds, so that should give you an idea of the amount of hands.
And if my math is correct (this is debatable), this works out to be a bad beat jackpot every 12,975 hands.
This seems fairly likely. But this is because in this situation you are basically taking every hand, with every player, to showdown, so it seems far more likely than in the real world, where the majority of us play.