bad beat jackpot size for 0 ev
 

i am trying to figure out the break even point of the party bbj. here is my thinking, let me know if i am right?

chance of bad beat = 1/155,000 hands. taken from http://www.math.sfu.ca/~alspach/comp46.pdf.

bbj payouts
Loser = pot * .7 * .5
Winner = pot * .7 * .25
Other players = (pot * .7 * .25) / 8

Now, here are my 5/10 stats from PT, this months 10k sample size.
Hands won = .06 (6%)
Hands raked = .7 (70%)
bbj drop = .50 (50 cents)

so... before a jackpot is hit you will lose

(155000 * .06 * .70 * .5) = 3,255$

so you will need to win 3,255$ from the bbj to break even. i use the worst case, the players splitting the remaining pot, and get this.

3,255 * 8 = 26040 amount of money to be split.
26040 / .7 / .25 = bbj size before the split. 25% of 70% of the pot = 26040. mean the bbj size must be 148800.

So about 150k, does this make sense?

Re: bad beat jackpot size for 0 ev
 

It depends on how you play. If you play at a table where everyone limps preflop, checks flop and turn and river unless they have the required hands or better (then they cap river).

Then you get no rake and no contribution to the jackpot. The chance of getting the jackpot increases because of more hands played/hour and that people at a table like that can play all hands that make for a potential bad beat (64s etc). Of course you have to calculate the chance that the pokerroom voids the winnings because of the play pattern at the table.

So the EV would be your share of the potential jackpot.

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If you assume normal play at a normal table (ie at what level the share of the jackpot is equal to the extra rake) bonus whores have some sums for Party.

$2-$4 $193,205.51
$3-$6 $222,151.66
$5-$10 $247,629.10
$10-$20 $275,848.94
$15-$30 $262,267.69

Re: bad beat jackpot size for 0 ev
 

why does this number go down from 10/20 to 15/30???

Re: bad beat jackpot size for 0 ev
 

do you have a way to prove those jackpot sizes?

% of hands raked, % of hands won, % of rake when u win the hand?