bad beat jackpot size for 0 ev

[rt1]

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?