http://wizardofodds.com/games/pai-gow-poker/
From the website:
Possible Outcomes in Pai Gow Poker
Outcome Probability
Player wins both 28.61%
Tie 41.48%
Banker wins both 29.91%
House Edge in One-on-One Pai Gow Poker
Status House Edge
Player 2.73%
Banker 0.20%
Combined 1.46%
House Edge as Banker
by Number of other Players
Players House Edge
1 +0.20%
2 -0.02%
3 -0.10%
4 -0.15%
5 -0.19%
6 -0.21%
My question is how to figure out the house edge based on number of other players.
My math does not match the Wizard's.
My way:
If you are banking, the odds of beating one player is .2991.
The odds of beating both players is .2991 * .2991 = .08946, correct?
And the odds of both players beating you is .2861 * .2861 = .08185, correct?
The rest of the time it is a push = 0
Assume all players are betting 100.
So .08946 times you will win 190 (10 commission) = 16.9974
And .08185 times you will lose 200 = -16.37
For a total avg win of .6274
Which is .3137 per 100, or .3137% EV.
I figure my error is because I'm ignoring the .82869 times I beat one player but lose to the other but how do I solve for it?
Thanks
My reason for asking is that I've seen several places say that if you can book 13 times your player bet as the banker every other hand then the game is neutral EV, but that isn't quite accurate right? That is only true if the 13x is spread evenly among 6 other players. If one player is betting 15x and the rest are betting 1x, then in still won't be profitable, correct?