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Old 10-24-2005, 05:32 PM
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Default 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).


[/ 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?
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