View Full Version : EV of PartyPoker bad beat jackpot?

12-22-2004, 11:21 PM
Seeing the Party Poker bad beat jackpot over $210,000 made me wonder what the crossover point is for it to be +EV from a jackpot standpoint to play in these games.

Quad 8's have to be beaten, and I assume they drop $1/hand for JP. I'm also not sure how they break down the payout.

Seems like this would involve estimation of how much you pay for the drop ($1/round maybe) and the odds of having an insane beat like this occur. Anybody have a clue?



12-23-2004, 06:04 AM
I think if you approach this you need to simplify to get to an approximate result as quickly as possible.

Assumption1: full ring, ten hands, nobody sitting out.
Assumption2: you are winning 1/10 of the jack pot, i.e. you do not care if your are the winner, 2nd best hand (the badly beaten) or any other participant of the hand.
Assumption3: You are paying 1/10th of the drop per hand.
Assumption4: All hands are raked since you are only paying the drop in reked hands and can win it only in raked hands.

You have to factor in that you need to use both hole cards.
Then, I would break it up in three occurences: RF beat RF (this should be easy because you exactly 3 connected suits on the board in the range of Q,J,T to 5,4,3) and the respective hole cards are a given. The other two occurences, RF beats quads and quads beat quads might be more tricky to pin down exactly.

12-23-2004, 01:27 PM
I believe there is no easy way to calculate frequency for hitting the JP. The best way is to observe over time how often it hits.

Playing the JP is EV+ whenever:

JP > 0.5 x (ave number of hands between JPs) / 0.7

In the above equation 0.5 represents the JP drop and 0.7 represents the fraction of the JP actually awarded (20% to reseed and 10% adminstrative fee).

So if the JP hits every 200,000 hands, the JP needs to be greater than $142,000 to show a profit.