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View Full Version : Table game HA: Harrah's "Texas Hold 'Em"


Tapin
08-16-2005, 12:10 PM
So I saw this on a blog (Poker Grub (http://www.pokergrub.com/#112413495047321463)), and I'm curious if anyone here is willing to do the math.

The game is a table game called "Texas Hold 'Em" (how imaginative). Five marked betting areas: "Ante", "Flop", "Turn", "River", "Bonus". The table in question had stakes of $5-100 for the ante and $5-25 for the bonus. You place your bets on the ante & bonus (presumably both are required), and get dealt a pair of hole cards. If you like your cards, you put twice the ante in the "Flop" betting area (if not, you fold and lose your ante). The turn and river then come out, and apparently you can either bet the ante in each betting area after the card, or you can check your way through to a showdown against the dealer.

If you win with less than a straight, you get paid 1:1 on the Flop, Turn, and River bets. If you win with a straight or better, you get paid 1:1 on the Ante, Flop, Turn, and River bets. There is no dealer qualifier.

The bonus betting area is based on your pocket cards:

1000 to 1 for AA and AA for dealer
30 to 1 for AA
25 to 1 for AK suited
20 to 1 for AK unsuited
15 to 1 for AQ, AJ suited
10 to 1 for AQ, AJ unsuited
5 to 1 for pocket high pair JJ, QQ, KK
3 to 1 for pocket low pair

I assume, although it's unstated, that the house gets the bonus bet if you don't hit any of the bonus payouts. I also assume that a bonus bet is not required, but that could easily be a mistaken assumption.

The speculation made in the blog entry is that the "real money" is made in the bonuses, and that the game might actually be unprofitable for the house. I'm a little skeptical (paying 30:1 for a 220:1 occurrence, and the house is losing money?), but figured I'd post here in case anyone was interested in either doing the HA math, or walking me through how it should be done so I can crunch the numbers myself.

Any takers?

LetYouDown
08-16-2005, 12:24 PM
Well, just for the bonuses:

A-A vs. A-A = 1/C(52,4) = 270724 to 1
A-A = 220 to 1
A-K suited = 4/C(52,2) = 330.5 to 1
A-K offsuit = 12/C(52,2) = 109.5 to 1
A-Q & A-J suited/offsuit numbers are the same
High pair (J, Q, K) = 18/C(52,2) = ~72.666 to 1
Low pair (2,3...10) = 54/C(52,2) = ~23.555 to 1

You should only hit the bonus 9.5% of the time.

Tapin
08-16-2005, 12:32 PM
[ QUOTE ]
You place your bets on the ante & bonus (presumably both are required)

[/ QUOTE ]

[ QUOTE ]
I also assume that a bonus bet is not required, but that could easily be a mistaken assumption.

[/ QUOTE ]

I proofread g00t.

I have no idea whether it's required or not.

LetYouDown
08-16-2005, 12:38 PM
What happens when the pot is split? Does tie go to the house? This is not an insignificant factor.

Also, The reason the payouts are disproportional to the odds of actually having the hand can be realized in this scenario:

Say I'll pay you 2-to-1 on any specific hold 'em hand. I have all 1326 listed as paying 2-to-1. The probability of having any individual hand is still 1/1326, but the EV of all of them must be considered. In that case, you're guaranteed to win 2-to-1 for hitting your 1326-to-1 shot. If no one has worked out the EV on the bonus, I'll do it when I get back from lunch.

TomCollins
08-16-2005, 01:19 PM
EV = 1000* 1/ 270275
+ ((6 * 30) + (4 * 25) + (12 * 20) + (8 * 15) + (24 * 10) + (18 * 5) + (54 * 3)) / 1326 = .00370+ .85369 = .8579 = HUGE HOUSE EDGE!!!!