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?

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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).


[/ QUOTE ]

I'm trying to figure out the optimal betting strategy if these two rules are combined. I figure you would double any time you have an edge on the house. Without the 1.5:1 on the A, you would double anytime your A is better than 50% to win (correct?). But with the extra 1.5:1 on pairing your aces, how much lower can you go? I first thought 40%, because then you would technically breakeven any time you paired your A (1.5:1 is worth 40%). But then I thought that it wouldn't be this low because you won't always pair your A when you double. So how should I figure this out?

[ QUOTE ]
1. Option to double on any pair.
4. If you make 3 of a kind (or 4 of a kind) with a pair in the hole, the house pays 2:1.

[/ QUOTE ]

Also, how often will you finish with a set when you hold a pair (given all your outs are live)? Remember, there are only 46 unknown cards instead of the usual 50. Using that number, how much lower than 50% can you go with your pair in terms of doubling?

[ QUOTE ]
[ 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?

[/ QUOTE ]

Dunno, I need to think about it a bit more. My (notoriously unreliable) back-of-the-napkin calculation suggests that it will take on the order of 50 days to brute force.

[BOTW]

David claims it has a small 0.5% house edge (if it is his same game and hasn't been tweaked in the interim):

Sklansky Essay on playing pairs (http://www.twoplustwo.com/dsessay8.html)

I don't recall ever seeing a follow-up to this essay, though I think the name may also be "Poker Challenge".

[SumZero]

[ 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 ]

That is the number of combination of hands, when you take into account the number of boards that you are evaluating too you see 418,597,840,861,200 possibilities to enumerate. You are right that you may be able to suit simplify but that will not buy you that many orders of magnitude and will also be adding complexity to the enumeration/calculation. Even if you could consider 1 million hands (here I include the board as a hand) a second you are still looking at over 13 years of calculation time.

The bad about that is it seems hard. The good about that is it seems like people likely haven't been able to do it and of all the casino games that might be beatable, obscure new games like this that are hard to calculate are the most likely.

The other fun game that someone described (was it Sklansky again?) was the WSOP one where you anted 1 and then looked at your hands and either got to surrender (and lose the one) or bet 5 or bet 10. If you bet 10 then only about 1/3 of the dealer's hands would call you (using simple blackjack scoring for unpaired hands) and if you bet 5 then about 2/3 of the dealer's hands would call you. And if they call you then it is an all-in hold'em battle with the board being dealt.

That is also computationally challenging, but simpler than the game you ask about.

I think that you can reasonably expect that the EV for the perfect strategy is already proven to be negative for the player. Otherwise it would not have been offered by the casino. I think Sklansky wrote somewhere that such a proof is necessary for any new games to be picked up by the casino.

This is a varation on Sklansky's game. In the actual game, the player is only allowed to double on any pair and this cuts the HA to <0.5%, assuming perfect strategy. The cardroom that is offering this game has definately not done a detailed analysis of its advantage. I know that the above rules reduce the HA and maybe even swing the game into the player's favour. I was wondering by how much. This room is experimenting with the rules using small sample sizes so I'd like to maximize my advantage while I still have a chance.

[Borno]

I too have read that article and posted a question not as detailed or as smart as TKO's...

David Sklansky if you could please clarify the actual rules of your game that would really help. I think its kind of vague in the essay in terms of whether you offer odds on high hands and if they require a side bet to recieve odds. Or if a sidebet is just required for the jackpot. Also, what are the odds for each hand, i.e. str. flush. 3oak, etc.

Also do you have any idea about the variation of that TKO mentiones in his post - how will that effect the HA?

Thanks!

Any math / comp sci types that can figure TKO's questions out would be a huge help!

[AaronBrown]

My algorithm was very simple. I dealt the three hands, then looked up in a table available at Wizard of Odds the chance of each one beating two randomly picked hands in hold'em. I bet on the one with the highest chance, dealt the board, and checked which hand won.

The problem, as I mentioned, is that the best hand against two random hands is not necessarily the best of the three for betting. An obvious example is when the best hand shares a card with another hand, then you're often better off going with the third hand, even if it's weaker on average.

[ QUOTE ]
An obvious example is when the best hand shares a card with another hand, then you're often better off going with the third hand, even if it's weaker on average.

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Can you give an example of when this might occur?

Because the rules described originally are likely the rules that I will see when I show up next, I'd be interested to know what my strategy should be doubling pairs/aces (see my post a few above this one for my "theory" on how to play these). I know that you should for sure double when you are better than 50%, but I'm specifically curious as to how to alter this strategy with the extra payout system.

I suppose these are my questions:
a) If you hold a pair that is less than 50% to win, what are the odds that you will finish with a set and win (46 unknown cards)?
b) If you hold an A that is less than 50% to win, what are the odds that you will finish with a pair of aces and win (46 unknown cards)?