Re: Winrate Theory
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Why are you assuming that the other players are breaking even?
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I'm not. I'm merely thinking/stating if the worst "sustainable" winrate is X, shouldnt the highest be the converse of that, like -X + X = 0.
I dont know, to me it seems easy to find the worst possible lossrate that could be sustained, which is just someone not ever playing a hand. As I assume someone playing a hand will somehow bungle themselves into pos EV situations; obviously, someone lets say moving 2k in blind every hand will have a lower loss rate, but then in effect they're blinding 2k/hand now, arent they?
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Shouldn't the max theoretical winrate be (7.5bb/100)*9, i.e 67.5bb/100, i.e. the max amount that all the other players at the table put together could be losing?
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I think the numbers would fall if you won every pot would be .75ptbb/hand, or 75ptbb/100. Although your numbers look good as you need to take away the fact 1 out of 10 you're putting the blind money in there, too.
But thats not what Im saying, I think. I'm wondering (and really no clue about this!) that if it's possible to determine the worst winrate - which I have a hunch is not playing a hand - (over the long haul) then couldn't/shouldn't the positive be projected best. As in, because all things end up equal, (minus rake) you cant win more then the biggest loser.
The ONE problem with what I'm saying is this is dealing with just 2 people on a 10 person table, so maybe that alone disproves the worst loss rate isn't the same as a positive best win rate!
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