combining odds of improving to best hand with odds that he's bluffing

[jt1]

Say you are playing hold 'em and you flop top 2 pair. On the turn comes the third spade and a player raises you. You have no read on him and his position doesn't rule out a bluff. Lets say there is 8 big bets in the pot.

You know that you have 11-1 odds to improve to a boat, but the effective odds are only 9-1. However, he may be bluffing, but that only means that you'll have to put 2 big bets into the pot to see the showdown, which makes your effective odds at 4.5-1.

Does this mean that the odds that he is bluffing must be 6.5-1 or better for you to call? 11 minus 4.5 equals 6.5 hence my figure.

[BruceZ]

[ QUOTE ]
Say you are playing hold 'em and you flop top 2 pair. On the turn comes the third spade and a player raises you. You have no read on him and his position doesn't rule out a bluff. Lets say there is 8 big bets in the pot.

You know that you have 11-1 odds to improve to a boat, but the effective odds are only 9-1. However, he may be bluffing, but that only means that you'll have to put 2 big bets into the pot to see the showdown, which makes your effective odds at 4.5-1.

Does this mean that the odds that he is bluffing must be 6.5-1 or better for you to call? 11 minus 4.5 equals 6.5 hence my figure.

[/ QUOTE ]

This is a good question since it is important to be able to analyze this kind of situation, yet I've never seen it done in print. The calculation is simple, but there are plenty of wrong ways to think about the problem. You need to convert the odds to fractions before you subtract them. Don't perform arithmetic on odds. The equation becomes

1/12 + 11/12 *P(bluff) = 2/11.

P(bluff) is the probability that our opponent is bluffing. The left side of the equation is the probability of winning the hand. We have a probability of 1/12 of winning by filling up, and the other 11/12 of the time when we don't fill up, we win the fraction that our opponent bluffs. Since we are multiplying 11/12 by P(bluff), we are assuming that not filling up is independent of our opponent bluffing. The right side of the equation are the overall effective (and implied) odds of 9:2 expressed as a fraction 2/11, since we will win 9 bets when we win, and lose 2 bets when we lose.

Solving for P(bluff) gives P(bluff) = 13/121 = 10.7% = 8.3-to-1. Our opponent must be at least this likely to bluff in order for us to have positive EV to continue in the hand.

Here is a slightly longer method which might be a little easier to understand, and it provides a check. We can write an expression for the overall EV of the hand as

EV = 9*[1/12 + 11/12 * P(bluff) ] - 2*[11/12 * (1 - P(bluff) ].

We want the EV to be positive, so set this to 0 and solve for P(bluff). This gives the same value as before for P(bluff) = 13/121 = 10.7% = 8.3-to-1.