#11
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Re: Calculating EV
Pot odds are sometimes insufficient and yet EV > 0. This
is because of implied odds; a good example is from pot limit draw Jacks-or-better: Suppose you have T987 all in diamonds with the deuce of clubs and someone opens for the size of the pot and now are you saying because you have insufficient pot odds (you have 9 flush outs and 6 other straight outs for 15 out of 47 possible combinations but even this is an approximation!), you should fold? No, because of the implied odds on the last betting round! There are implied odds for future betting rounds and pot odds only gives a good approximation for whether you should continue. It is the EV that is the true indicator of whether you should make one decision over the other. Of course, if you are drawing and almost have pot odds in a limit game, you should call because even against a lone opponent, he must pay you off most of the time when you get there. You also sometimes must discount factors such as getting there but you weren't drawing to as many outs as you thought! A practical example comes from PL Omaha high: even if your opponent bets the size of the pot when it is heads up, you only need ten clean outs on the turn (out of 44 cards) to have a profitable call even if your sole opponent plays theoretically correctly. It is because on your ten cards, you can bet the size of the pot and on some bluff cards with optimal frequency and your opponent must call you 1/2 of the time to minimize the amount you make on any bluffs. It is important that none of your outs are "tainted"; i.e., they are truly all outs. Suppose you are in a fixed limit game with one last betting round after you call to draw to your hand with the pot size as P before each of you put in one bet apiece (so now the pot size is P+2 in bets). Your pot odds were 1/(P+2). On the other hand, if you hit and bet, your opponent's theoretical calling frequency is (P+2)/(P+3) so you only need a probability of 1/(P+2 + (P+2)/(P+3)) of completing your hand or (P+3)/((P+2)(P+4)). As you can see, this probability is smaller than the required "pot odds" frequency by a factor of (P+3)/(P+4). Look at an exact example: Suppose the pot has 3.5 BBs on the turn when you have Js Th and the board reads Ks Qh 7d 6c Your opponent bets the turn and now you can call even though you don't have immediate pot odds (1/5.5 = 2/11 = 8/44 > 8/46, this last number being the chance you will make your straight although admittedly your opponent most likely has at least one card in his hand that you don't need to show up on the table!) Also, we are disregarding the extra rake but even that can also be included in the calculations! You actually only need a 26/165 chance or higher of completing your hand to have an EV>=0 here. The EV in this example is 38/46(-1) + 8/46(+4.5+11/13) = 31/299 = 0.1037 BBs. Folding here would be an error of greater than 0.1 BB and would be a "whopper" using backgammon lingo. When you have tainted outs, e.g., when your opponent could have a set and you have nut flush draw in LHE, you must consider how those situations will reflect on your EV. If it is not unlikely your opponent has two pair or a set, you must modify all your EV calculations appropriately, but don't forget it will depend on your opponent and the previous betting action. A good way to think of how pot odds can be applied is to pretend that you will be all-in if the pot you are contesting were to have a fraction (P+2)/(P+3) of a bet (which is almost but not quite 1 in typical pots) added to it. By the way, don't forget to bet when you hit! |
#12
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Re: Calculating EV
I thought it was obvious that I was talking about the equivalence of EV=0 and pot odds in a single bet situation where you are a definite winner if you hit and a definite loser if you dont.
Your EV formulas are quite simplified as well, since a true EV calculation needs to consider probabilities of folds if you raise, the additional chips at risk if you arent a definite winner (a more refined "tainted outs" calculation), as well as the additional bets you may be able to induce from opponents using alternative strategies. For a beginner like the poster, stick with current pot odds, since even the literature does a poor job on what a correct "implied odds" or EV calculation entails. If your hand reading skills arent well developed the latter two are so error prone that considering them will likely hurt your win rate compared to a simple pot odds calculation, using one draw odds unless it is virtually certain that you will be able to stick around for a second draw if you miss. |
#13
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Re: Calculating EV
There is no equivalence if there is a possibility of
future action on the hand; if a player currently has exactly pot odds to draw, his EV is strictly positive. As my previous post has stated, if a player almost (but does not quite) has pot odds in a LHE game, he often will have an EV>0 by calling. As you had pointed out raising may have an even higher EV if you know your opponent folds more than optimally (game theoretically) or if on the end, without position, the alternative strategy of check-raising may have even a higher EV against particular opponents. Nevertheless, the analysis was quite simplified to exhibit the clear reason that for EV>=0, immediate pot odds are not quite necessary. To adjust to the pot odds mindset, you need to adjust the pot by imagining that it is juiced up by (P+2)/(P+3) BBs. |
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