Re: * * New Theory of Poker Study Group Session 1 * *
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1) This one seems simple. If the pot is 10 bets and there's only one card to come. I have top pair and I know the other player is on nothing more than a flush draw. I bet for value, opponent calls, because he has odds. To simplify, let's say that the odds of the flush hitting are 4-1. So I bet, because my EV is (12*0.8) - 1 = 8.6bets?
The other player calls because his EV is 12*0.2 - 1 = 1.4 bets. I already discounted the bets he had to put in.
But in terms of the fundamental theorem of poker, neither of us has gained ANYTHING, since we both played it as if the hand were turned up. So the gain in this situation is 0, for both players? Kinda like taking odds in craps?
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Lets look at 2 possible scenarios to see how this relates to the fundamental theorem of poker. Lets make one additional assumption to simplify things that no matter what the river is you will both check and showdown the hand.
Scenario A: 10bb in pot + 1bb turn bet + 1bb turn call = 12bb
12bb*.8-1bb = 8.6bb
12bb*.2-1bb = 1.4bb
Scenario B: 10bb in pot + 0bb turn bet + 0bb turn call = 10bb
10bb*.8 = 8bb
10bb*.2 = 2bb
Therefore, if you do not bet you are making a mistake and your opponent gains. He is also correct to call since by folding he would give up his entire +ev.
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