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

[btmagnetw]

this isn't so much a "fundamental theorem" as it is a definition and a mathematical illustration.

[ApeAttack]

Overall seems fine, but 833K is absolutely ridiculous (0.1% is probably a VERY conservative Q).

I think that this is fairly well-known topic though.

[AaronBrown]

First, it is possible for you to know you cannot win. Suppose the board is AAAA2 and you hold a pair of 2's. The other player must have at least one card higher than 2, and therefore must beat you.

Second, you are assuming the other player will be passive for the rest of the hand, only checking or calling. Since she can also raise or fold, there are hands that you should fold. Also you are assuming you are infallible, betting S when you will win or split the pot and checking when you will not. Another reason to fold is you don't always know when you're going to win.

Third, the math is wrong even under your assumptions. I think you define P as the probability that you will win or split, Q as the probability you will win and S as the minimum of the two stack sizes after you call bet B. The other important variable I will call M, the size of the pot at the time of decision (before you call the bet B).

Under your assumptions if you call, you win M + S with probability Q and (M - B)/2 with probability (P - Q). You lose B with probability 1 - P. That makes the EV of calling:

(M + S)*Q + (M - B)*(P - Q)/2 - B*(1 - P)

= (M + B)*(P + Q)/2 + S*Q - B

which is positive if:

S > B/Q - (M + B)*(P/Q + 1)/2

It is true that if Q > 0 you can always make S big enough for this to be true. But the chance of tying is irrelevant to that conclusion. You need a chance to win, not just tie. If we ignore the chance of a tie so P = Q, the formula reduces to:

Q*(S + M + B) > B

That is the expected value of what you win is greater than the amount you have to bet.

[Collin Moshman]

Hi guys,

Thanks for the input. Note that I am not claiming that this result is either original or practical ... just a poker theory idea that interested me.

AaronBrown, excellent example with AAAA2, you hold 22, impossible even to split. Solid point. So let's ignore splits for now, as you suggest.

But when you write:

[ QUOTE ]
I think you define P as the probability that you will win or split, Q as the probability you will win and S as the minimum of the two stack sizes after you call bet B.

[/ QUOTE ]

I actually defined Q to be the conditional probability that you will stack your opponent given that your hand ends up being best by showdown. So regardless of your opponent's actions and/or other factors, Q will be a positive number, even if exceedingly small in certain situations. E.g., you call the $25 bet in the $1-$2 example, flop quads, and your opponent flops ace-high. Then Q > 0 regardless of S since your opponent may make a series of crazy bluffs, turn an ace and refuse to believe you're sitting on the fourth deuce, etc.

So now let us re-define P as per your guidance such that P is the probability that your hand would prove best if dealt to showdown (considered at the time of the call/fold decision), and add on the hypothesis that the situation is one where there is some chance of an outright win, i.e., P > 0.

With these clarifications and added assumptions, I believe the simple S > B / (P Q) is still valid.

Best Regards,
Collin

[Phone Booth]

[ QUOTE ]
Hi guys,

Thanks for the input. Note that I am not claiming that this result is either original or practical ... just a poker theory idea that interested me.

AaronBrown, excellent example with AAAA2, you hold 22, impossible even to split. Solid point. So let's ignore splits for now, as you suggest.

But when you write:

[ QUOTE ]
I think you define P as the probability that you will win or split, Q as the probability you will win and S as the minimum of the two stack sizes after you call bet B.

[/ QUOTE ]

I actually defined Q to be the conditional probability that you will stack your opponent given that your hand ends up being best by showdown. So regardless of your opponent's actions and/or other factors, Q will be a positive number, even if exceedingly small in certain situations. E.g., you call the $25 bet in the $1-$2 example, flop quads, and your opponent flops ace-high. Then Q > 0 regardless of S since your opponent may make a series of crazy bluffs, turn an ace and refuse to believe you're sitting on the fourth deuce, etc.

So now let us re-define P as per your guidance such that P is the probability that your hand would prove best if dealt to showdown (considered at the time of the call/fold decision), and add on the hypothesis that the situation is one where there is some chance of an outright win, i.e., P > 0.

With these clarifications and added assumptions, I believe the simple S > B / (P Q) is still valid.

Best Regards,
Collin

[/ QUOTE ]

Aside from reverse implied odds considerations (as in somehow you'd never pay him off if he gets a better hand than yours), you're ignoring that Q is a function of S. HTH.

[AaronBrown]

I apologize, I read your initial post too quickly. With your definition of Q, your math is correct.

I'm still unconvinced about the theorem as a poker insight. On almost every no-limit hand, I think there's some chance I will stack another player and win. But that doesn't mean I'll call every bet from a big stack when I have a big stack. For one thing, the bigger are stacks are, the more I have to lose as well as win. For another, I'm not trying to maximize my EV every hand, I'm trying to win money for the session. I might feel that I'll have a better chance to take the stack on another hand, with less risk to my stack.

But most important, the intermediate bets go up with the stack sizes. Suppose I tihnk there is a 5% chance on a particular hand that I will stack another player and win $100,000. He bets $1,000 preflop. In your analysis I call, because the stacking possibility is worth $5,000 to me; plus I can win lesser amounts.

But I don't get to see his cards and the board, and then decide whether or not to stack him. He may bet $5,000 or $10,000 on the flop; or go all-in. I have to decide what to do without knowing his cards, or what the turn or river will be. The bigger our stacks, the bigger these intermediate bets are likely to be. The other guy knows as well as I do that we're playing for our full stacks.

[Bad Lobster]

[ QUOTE ]


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)



[/ QUOTE ]

There's a hidden assumption here.

Even if we assume that Q (the chance opponent will call your huge bet) is always greater than 0, your formula requires that either Q is fixed or there is a minimum positive value of Q. But if Q is a function of your bet size B, it's possible for Q to always be positive and still have no minimum.

For example, suppose the probability your opponent will call a bet of B dollars without having the nuts himself is 1/B. There's a 1/1000 chance he'll call a $1000 bet, a 1/1 million chance he'll call a million dollar bet, and so on. Then you can see that no matter how big your bet, your expected profit stays around 1 dollar.

This sort of situation happens in a lot of mathematical proofs: there's a difference between "There is an X such that for every Y..." and "For every Y, there is an X such that..." It seems like an obscure distinction but it can often make the difference between a proposition being right or wrong.

This is more in line with common sense, too--you can't really expect the chance your opponent will call a 1 million dollar bet to be a serious possibility.