Let's Guess AT This One

[George Rice]

That method won't provide you with the exact answer. Weighting, for example XYo as a 12, is inaccurate. You would need to consider whether the suits are contained in the opponent's hands.

Let's look at AA vs 76o. Your method sees 6*12 occurances of this, all at the same edge for the AA. But that isn't the case. You have 6*2 cases of AA over 76o where no suits are duplicated (the best odds for the 76o hand), 6*8 cases where one of the suits are duplicated (worse for 76o), and 6*2 cases where both suits are duplicated (worst of all for 76o).

[karlson]

As I noted, I did the second sum explicitly for all 1225 hands. Some savings can be done, but they would be more complicated than the 4,6,12 weighting, as you correctly note. However, the first grouping is perfectly legit, as AhAd will have the exact same results as AsAc, in total (e.g. for the case when AhAd is against 7s6c, there is a corresponding case when AsAc is against 7h6d).

[George Rice]

The way I'm looking at the problem now is that if the average win is 60%, to pick a number, I would solve for X as follows:

A(100(.60-.40)-X) = (B+C)*X

So X = 20A/(A+B+C)

Where:
A is the number of hands you have the advantage
B is the number of hands your opponent has the advantage
C is the number of hands neither has the advantage

Is this how you figured for X?

[karlson]

Sure, looks fine. You may want to simplify to X = (A/(A+B+C))*(.6-.4)*100.

The part that you need the computer for is getting that average win figure.

[George Rice]

If this game was played against a casino where the casino showed its cards, X was the non-returnable "ante", and you had option of betting $100 or folding, then a game like this (or some variation) might actually be spread some day. Of course, X would be high enough to gurantee positve ev for the casino. You could even allow the player to win back the ante by just raising it's value (probably doubling "X"). Maybe pay a premium for quads and straight flushes, similar to Caribbean Stud. And you can bet up to "Y" times the ante.

Hmmmm. "Hawaiian Hold'em"

[George Rice]

[ QUOTE ]
The part that you need the computer for is getting that average win figure.

[/ QUOTE ]

I thought you already did that.

[karlson]

I did....and posted my answer.

[karlson]

Look, if hold'em challenge didn't make it, then this is doomed, as it's about 100 times more boring for the player.

BTW, David, what was the end result with hold'em challenge? It did get spread at some point, no?

[BugsBunny]

Without looking at any possible answers I would guess 10-15 dollars. 50% of the time you have an edge and play. 50% of the time you fold. The max edge you can have is probably about 60% - but that's the extreme. Even assuming that that would bring x to about 30. If we assume the average edge is half of that that would be a 30% edge so x=15. I suspect the average edge is smaller than that so I'll say that x should be 10$.

[Redsox]

Ok. So you call obviously 50 % of the time. I estimate when you call you win about 60% of the time. That means that in 100 chances, calling 50, you win 30, and lose 20. That makes for a profit of $1000 ($30*100-$20*100). If you pay on EVERY hand, regardless, you would have to pay $10 a hand to make it an even money bet. Correct?