[Josh W]

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even if you are beat with a probability of 95% if hijack and button have an ace each and they have aces with a chance of 80% and you are beat with 75% if they both have aces it is a call!

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Thank you for using math, really.

However, I don't quite understand what you are saying.

I think you are saying (and I fully realize these are somewhat hypothetical numbers, but I'm just trying to understand your example)....

I'm beat 95% of the time both have aces.
Both have aces 80% of the time.

What's the 75% number representative of?

Thanks,

Josh

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i don't believe my hypothetical numbers. i posted them just to show how sure you can be that you are beat and have to call anyway.

the 75% represented the assumption that you lose 75% of the time if hijack or button doesn't have an ace. to make it perfectly clear: i don't think that your chances to win are quite as bad as my numbers suggest. it was just to show how far you can go with the assumptions and still have to call.

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Thanks for clearing that up. I assumed that you didnt' have carte blanche faith in the numbers you presented, which is why I threw in the disclaimer in the first place.

However, if you do the math with 'your' numbers (in quotes because they aren't necessarily what you believe), I only lose 91% of the time. Since I'm getting 13:1, this would make it a call.

However, it doesn't take into account any probability to chop (most likely with the button). If I'm chopping at all, this means it's neutral EV or better to fold.

Again, this assumes the numbers presented, which are fictitious. However, I'm throwing this analysis out there to show just how close it is.

Whether you (by "you", I don't necessarily mean just jjacky, but anybody here) think it's an obvious call...you should rethink it.

At the very worst, it's immensely close.

[note that in my opinion, the 95% is about 95%, the 80% is more like 97%, and the 75% is probably about 80%]

Josh