Two Plus Two Publishing LLC Two Plus Two Publishing LLC
 

Go Back   Two Plus Two Poker Forums > General Gambling > Other Gambling Games

Notices

Other Gambling Games Discussion about other gambling games.

Reply
 
Thread Tools Display Modes
Old 06-05-2012, 01:11 AM   #1
Pooh-Bah
 
Join Date: Aug 2003
Posts: 4,335
Pai Gow math question.

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?
Eihli is online now   Reply With Quote
Old 06-05-2012, 03:18 AM   #2
veteran
 
Join Date: Jun 2011
Posts: 2,263
Re: Pai Gow math question.

Your biggest error is that you are ignoring commission when you earn 1 win and 1 tie:

ResultMathProbabilityPayout for 2 $100 betsReturn per $200 bet
2 wins0.2991^20.08946081+19017.00
1 win and 1 tie2*0.2991*0.41480.24813336+9523.57
2 ties0.4148^20.1720590400
1 win and 1 loss2*0.2991*0.28610.1711450200
1 loss and 1 tie2*0.2861*0.41480.23734856-100-23.73
2 losses0.2861^20.08185321-200-16.37
TotalAdd above figures1.0000varies0.465

Your advantage as banker from the Wizard's individual win/loss/tie numbers is $0.465/$200 = 0.2325%.

This is still off by the Wizard's actual numbers (0.02% advantage when banking). This is probably due to the hands that are dealt are not independent from one another, which will also yield different probabilities for each category in the table. The proper way to calculate it is to count all the possibilities for a 3-handed game. Random simulation over a sufficient number of deals may be used as well. It's also possible I screwed up the math.
tringlomane is offline   Reply With Quote
Old 06-05-2012, 10:12 AM   #3
Pooh-Bah
 
Join Date: Aug 2003
Posts: 4,335
Re: Pai Gow math question.

You don't pay commission on ties.

But duh... I was ignoring the 1 win/loss 1 push.
Eihli is online now   Reply With Quote

Reply
      

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off



All times are GMT -4. The time now is 04:35 PM.


Powered by vBulletin®
Copyright ©2000 - 2013, Jelsoft Enterprises Ltd.
Content Relevant URLs by vBSEO 3.6.0 ©2011, Crawlability, Inc.
Copyright © 2008-2010, Two Plus Two Interactive