Re: Approximating Multiple Opponents
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If I am facing two opponents, and my hand will beat opponent #1's likely hands 50% of the time, and opponent #2's likely hands 70% of the time, what is the equation for approximating how often my hand will beat both opponent #1 and #2?
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.5*.7 = 35%
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If I assume two opponents at 50%, I can deduce that I will win 33% of the time. I *guess* I arrive at that approximation by: (.5 + .5)/3 = 33%.
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Whoa! Try: .5*.5 = 25%.
HTH,
gm
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