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#11
10-24-2005, 05:32 PM
 Guest Posts: n/a
Re: Holdem Challenge

[ QUOTE ]
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).

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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?
#12
10-24-2005, 06:43 PM
 Guest Posts: n/a
Re: Holdem Challenge

[ 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.

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Do you know how to do this?

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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.
#13
10-24-2005, 08:56 PM
 BOTW Junior Member Join Date: Sep 2002 Posts: 6
Re: Holdem Challenge

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

I don't recall ever seeing a follow-up to this essay, though I think the name may also be "Poker Challenge".
#14
10-25-2005, 02:35 AM
 SumZero Member Join Date: Jul 2004 Posts: 73
Re: Holdem Challenge

[ 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.
#15
10-25-2005, 05:04 AM
 Guest Posts: n/a
Re: Holdem Challenge

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.
#16
10-25-2005, 11:21 AM
 Guest Posts: n/a
Re: Holdem Challenge

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 &lt;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.
#17
10-25-2005, 12:24 PM
 Borno Senior Member Join Date: Nov 2004 Location: Poker Land Posts: 214
Re: Holdem Challenge

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!
#18
10-25-2005, 01:29 PM
 AaronBrown Senior Member Join Date: May 2005 Location: New York Posts: 505
Re: Holdem Challenge

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.
#19
10-25-2005, 02:36 PM
 Guest Posts: n/a
Re: Holdem Challenge

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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 ]

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)?

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