Re: Normal or anormal variance
Assuming your percentages are correct, the probability you lose all three is (.1)^3 = 1/1000.
The probability you lose 2 of three is 3*.9*(.1)^2 = 27/1000
The probability you win 2 of three is 3*(.9)^2*.1 = 243/1000
The probability you win all three is (.9)^3 = 729/1000
This may seem like an absurd longshot, but every time you have a string of 1000 all-ins as a 9-1 favorite, you can expect to lose three in a row. So, to be honest, nothing is 'normal' or 'abnormal' in retrospect. If this was the first time you've ever played poker, then it would surely be considered improbable, but if you've got time under your belt, you can expect something like this to happen every once in a while. Another way of thinking of this is that if you go 1000 all ins favored 9-1 and never lose three in a row, you got 'lucky' (but would you call someone lucky if their Win/Loss string looks like LLWLLWLLWLLWLLWLLWLLWLL...?).
Just rest assured that after 1000 of these 9-1 favored AIs, it's very very likely you'll be between 88% and 92% overall.
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02-13-2007, 07:00 PM
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