Calculating Progressive Breakeven Point for Poker
September 2nd, 2010 at 5:33:46 PM
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Are the calculation of breakeven points different when a player does not need to contribute to the jackpot pool each decision in order to qualify?
Specifically, for a bad beat poker jackpot, players do not contribute evenly to the jackpot, since the only required put ins are two blinds. With 10 players at the table, each gets to look at 8 starting hands for free each round. Does this lower the amount of the bad beat jackpot break even point, since participants can choose to play if they have two qualifying cards, or opt out for nothing, 80% of the hands?
Some thoughts:
If players only stayed in if they had qualifying pocket cards, the odds of a jackpot situation would be slimmer than if everyone stayed in with any two cards, but would the minimum cost per player be optimized?
What would the jackpot have to be if the minimum qualifying bad beat hand is quad 8's losing to a higher hand, with both hands using both pocket cards, and if higher quads, also having a pocket pair. From the Wizard's Bad Beat Jackpots table, the probability of any one of the ten hands getting beat in this scenario is .00000525 or 1 in 190,476. So the odds of any single player getting beat is this divded by 10, or 1 in 1,904,762. Does this mean the "breakeven" is $1,904,762? (ignoring the fact that the jackpot is divided between the table, so the winner collects only 50%-60%.)
Specifically, for a bad beat poker jackpot, players do not contribute evenly to the jackpot, since the only required put ins are two blinds. With 10 players at the table, each gets to look at 8 starting hands for free each round. Does this lower the amount of the bad beat jackpot break even point, since participants can choose to play if they have two qualifying cards, or opt out for nothing, 80% of the hands?
Some thoughts:
If players only stayed in if they had qualifying pocket cards, the odds of a jackpot situation would be slimmer than if everyone stayed in with any two cards, but would the minimum cost per player be optimized?
What would the jackpot have to be if the minimum qualifying bad beat hand is quad 8's losing to a higher hand, with both hands using both pocket cards, and if higher quads, also having a pocket pair. From the Wizard's Bad Beat Jackpots table, the probability of any one of the ten hands getting beat in this scenario is .00000525 or 1 in 190,476. So the odds of any single player getting beat is this divded by 10, or 1 in 1,904,762. Does this mean the "breakeven" is $1,904,762? (ignoring the fact that the jackpot is divided between the table, so the winner collects only 50%-60%.)
Simplicity is the ultimate sophistication - Leonardo da Vinci
September 3rd, 2010 at 2:46:22 PM
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Here are key rules from Foxwood's website:
•Quad 8s or better must be beaten in order to qualify for the Bad Beat Jackpot. Both the qualifying losing hand and the winning hand must use their two hole cards. For quads to qualify, the player or players must have a pair in the hole.
•A maximum of $1 will be removed from all pots that reach $20 ($100 for $10/$20 Hold’em) to fund the Bad Beat Jackpot. There must be a flop for the $1 to be removed from the pot.
•In order for a hand to be eligible to qualify for a Bad Beat, the pot must meet a minimum of $20 ($100 for $10/$20 Hold’em) and a minimum of four players must be seated and dealt in at the start of the hand.
•The primary Bad Beat Jackpot will be paid out as follows: 50% will be awarded to the losing hand, 25% to the winning hand and the remaining 25% divided equally among the remaining players seated at the game who received cards at the start of the hand.
As of today, the jackpot is $485,289
•Quad 8s or better must be beaten in order to qualify for the Bad Beat Jackpot. Both the qualifying losing hand and the winning hand must use their two hole cards. For quads to qualify, the player or players must have a pair in the hole.
•A maximum of $1 will be removed from all pots that reach $20 ($100 for $10/$20 Hold’em) to fund the Bad Beat Jackpot. There must be a flop for the $1 to be removed from the pot.
•In order for a hand to be eligible to qualify for a Bad Beat, the pot must meet a minimum of $20 ($100 for $10/$20 Hold’em) and a minimum of four players must be seated and dealt in at the start of the hand.
•The primary Bad Beat Jackpot will be paid out as follows: 50% will be awarded to the losing hand, 25% to the winning hand and the remaining 25% divided equally among the remaining players seated at the game who received cards at the start of the hand.
As of today, the jackpot is $485,289
Simplicity is the ultimate sophistication - Leonardo da Vinci
September 5th, 2010 at 4:24:04 PM
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It doesn't get multiplied by ten. The expected value of the jackpot is as follows, J is the size of the jackpot and P is the 190k to 1:
EV=1/10*J/2*P+1/10*J/4*P+8*(1/10*J/(8*4)*P
To find the break even point, set EV>=1 because the table loses a dollar basically every hand.
EV=1/10*J/2*P+1/10*J/4*P+8*(1/10*J/(8*4)*P
To find the break even point, set EV>=1 because the table loses a dollar basically every hand.
