If you can compute the equity in the two pots as if they were independent events, and come to the correct conclusion, then for mathematical purposes they are independent events. You don't have to get into the complicated math of "If x then y." You can if you want to, but it's not necessary.
Statistics: Posted by k3nt — Wed Feb 20, 2008 4:36 pm
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Summing the equities would still be valid though.
Statistics: Posted by redhouse — Wed Feb 20, 2008 3:41 pm
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What I was hoping to do was to explain why you CAN do it that way, since the same thing threw me off when I started thinking about it.
I also like excession's example of running it twice for the separate pots, as it clearly illustrates the problem I was trying to address.
I'm still not getting the claim, though, that the two are independent events. They're connected by an "if-then" statement: namely, if you win main pot, you must win side-pot. Clearly not the case if you run it twice. Then you can win main pot and lose side-pot, the one impossibility in point of fact.
Statistics: Posted by Aisthesis — Wed Feb 20, 2008 3:37 pm
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In fact, there's logically no connection between the pots at all. It would be possible to do a variation of "running it twice" here: deal 5 cards to decide who wins pot 1 and then (shuffle up and) deal 5 different cards to decide who wins pot 2. That wouldn't change the EV calculation, but it gets rid of the illusion that the two pots are inherently interdependent.
You did it right the first time, and Aisthesis is too smart for his own good.
Statistics: Posted by k3nt — Wed Feb 20, 2008 11:00 am
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