The Fundamental Theorem of Implied Odds

[Collin Moshman]

The Fundamental Theorem of Implied Odds


Hypotheses:

Suppose you are facing a bet B in a no-limit game, and are deciding between calling this bet and folding. Assume further that your move closes the action. Let S = Min {S1, S2} where S1 is your opponent’s stack size and and S2 is your stack size.


Proposition:

Then regardless of the situation, for all B, there exists S such that implied odds justify a call (i.e., calling is +EV).


Proof: Unless your opponent turns his cards face up, you cannot know his hole cards. Therefore there is some percentage P that you will win/split the hand at the time of your decision. For instance, even if you are at the river and cannot beat the board, and your opponent has been betting strong, there exists a low P whereby he is playing the board as well.

Furthermore, there is some chance Q that when you win, you will win a pot of 2 S, representing a gain of S. Therefore your expectation is at least:

P x Q x S

Now we want positive expectation, which means the reward of calling must exceed its cost, so we want:

B < P Q S which will hold iff

S > B / (P Q)

i.e., there indeed exists an S such that calling is profitable, as desired.

Example:

Let us try to plug in numbers for an example. Suppose you are playing full ring $1-$2 no-limit. The player under-the-gun, with a stack of $200, raises to $25. All fold to you in the big blind. You have $200 as well, and look down to see 2s 7c. Fairly obvious fold, right?

But suppose you were both playing very deep. How deep, precisely, would the stacks have to be for you to call? Well, we know B = $25 by hypothesis. As for P, the probability of flopping two-pair or better is around 3%, so let us use P = 0.03 as our rough estimate that you will have the best hand by showdown. Now Q will be a function of S (among many other factors), which is problematic as it suggests that the concluding equation to the proof may require functional analysis to solve for S. But empirically, even as S rises indefinitely, I believe Q has an approximate lower bound, which is not unreasonable since it can only be determined through subjective approximation in the first place. So let us plug in a conservative Q = 0.001, in which case

S > $25 / (.03 x .0001) = $833,333.33

So in this situation, if you’re both sitting on around a million bucks (pretty deep game for $1-$2, admittedly [img]/images/graemlins/smile.gif[/img]), you would have the implied odds to call.

Comments, errors, thoughts?

Best Regards,
Collin