[PokerintheI]

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Although you are correct in your conclusion that getting the other player to call with incorrect pot odds to do so is plus EV over having him fold, you might want to look at how one calculates EV, because you aren't doing it right.

For instance, in scenario 1, you appear to be assuming that your opponent will fold to a 200 dollar bet (if you don't run it twice) 100 % of the time. If this is the case, your ev in that situation is $250 dollars, not $450.

To demonstrate that this is so on an intuitive level, suppose that instead of 200 dollars, you bet a thousand (and again your opponent will fold 100% of the time. ) Your EV has not jumpted to $1250 in the hand, it is still $250. Suppose you bet a million dollars. Your EV is not $1,000,250, it is still $250. Suppose that your opponent was all-in, but for whatever brain dead reason would muck his hand to a $200 bet 100% of the time, your EV if you make the bet is still $250, not $450 (since there is nor functional difference between an inability and an unwillingness to make a call.)


Anyway, and rounding more than you are, I put the EV of not running it twice (for the 100% fold) at $250 and the EV of running it twice (for the 100% call) at $320. So, yes, a plus ev move.

However, since it is obvious that anytime you can induce a call from a player with improper pot odds to call you have a postive EV situation, the only relevant point here would be to show that your EV is unchanged whether you run it once or twice (and thus running it twice is EV if running it once would be, and vice versa). So a more relevant post would have been to demonstrate mathematically, that, in the event of a call, your EV is the same whether you run it once or twice.

Also a much more interesting calculation would assume, that like so much in life, your opponents decision is not 100% either way. If you assume, for example, that if you don't offer to run it twice, that your opponent will still call say twenty percent of the time, and even if you do offer to run it twice your opponent will still decide to fold 10% of the time, your EV caculations get much closer. While still plus EV, (if I've done the math right, which is always a question since I suck at math) and again with a bit more rounding than you were doing, I show the EV of only offering to run it once increasing to $264 (since 20% of the time he incorrectly calls) and the EV of offering to run it twice falls to $313 (since 10% of the time he folds anyway).

As you change those numbers, the plus EV of inducing the call may may make it a more marginal, though still plus EV, play.

--Zetack

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Yeah, I had a feeling I was calculating something incorrectly there.

In my experience with running it twice, it is generally the person calling that asks about running 2 rivers. And it generally involves a relatively large pot with some deepish stacks and regular players. So the determinations with regard to what they might be drawing to or if they might call or fold can be made with the insight gained from playing with the same person for 3 years.

I would agree that in a game with people you don't know as well you would need to factor in the "fold anyway" or "call anyway" percentages to a greater degree.