Holdem Challenge

A cardroom I know is offering a new game to its players: Holdem Challenge. Here's how it works:

Once you place your bet, 3 sets of hole cards are dealt face up. Out of these, you choose one hand as your holdem starting hand. A flop, turn, and river are dealt out. If the hand you chose wins, you get paid even money on your bet. If the hand you chose doesn't win, then the house collects your bet. EDIT: A split pot is a push.

How big is the house advantage in this game if you play perfectly (always choose the starting hand with the greatest edge)?


The following variations are possible. Which ones increase/decrease the house edge, and by how much?
1. Option to double on any pair.
2. Option to double on any ace.
3. A final hand that is a pair of aces gets paid 1.5:1 (at least one ace must be in the hole).
4. If you make 3 of a kind (or 4 of a kind) with a pair in the hole, the house pays 2:1.

[Guernica4000]

Is this game one on one vs. the dealer/casino or are there other players with hold cards visible?

The game is played player vs. house only. Multiple players can play, but only the 3 hands are shown (each player individually chooses one of the hands).

Any suggestions on how to go about figuring this out would also be much appreciated.

[AaronBrown]

It's an interesting question. I'll play with it a bit to see if I can't get an approximate answer. My gut is there's a huge house edge, at least before the special payoffs.

To get an exact answer, I'd email feedback@PokerStove.com. Andrew Prock is a brilliant and friendly guy with the software to answer your question.

I agree with the house having an edge before doubling becomes an option. I'm not sure how often one hand will come up that is better than 50% against two other hands, but it seems like it would not be often. Obviously you would want to double on these hands whenever possible. Will these hands more often contain an A or a pair?

[AaronBrown]

For what it's worth, I ran a 100,000 hand simulation. It was pretty simple-minded. It picked the hand from among the three that has the best chance of winning against two other randomly-picked hands, rather than against the two that were actually there. For example, it always picked J6 unsuited over a pair of 2's, but the relative value of those two hands depends on what the third hand is. That made the code much easier to write, but it gave away some value.

Anyway, the best hand going in won 40% of the time.

Aaron, thanks a lot for your help here. I also considered running a simulation, but then I couldn't think of how to test the cases of the additional rules in terms of beating the game and devising a strategy about when to double if you are permitted. I guess I could try two possible options: doubling whenever the hand is a favourite and doubling only when the hand is better than 50%.

Would you be able to post/send me your code so I can make some modifications as necessary? Thanks!

Naively, there's only about 305,000,000 possible combinations of hands.
With some reduction for suit-equivalence, that might well be brute-forceable.

[ QUOTE ]
Naively, there's only about 305,000,000 possible combinations of hands.
With some reduction for suit-equivalence, that might well be brute-forceable.

[/ QUOTE ]

Do you know how to do this?