Function of pot odds, math related question; theory

[lotus776]

I've stumbled over a theory element of poker that I haven't read about yet and thought I'd post it to see what others who may have realized the same thing say.

When playing No-Limit, the idea of pot odds governs the amount which the bettor bets, not considering all psychological reasons for betting. Primarily pot odds are given to the caller based on how much the bettor decides to bet. For example:

pre-flop the pot contains $100 ([25+25]+50 from the blinds, no callers for simplicity). After the flop the small Blind bets 50, the pot is now $150 and the caller must pay $50 to see the next card. The caller is being offered $150 to $50 or 3 to 1 pot odds.

What I've noticed is that as the bet amount which the bettor places increases (obviously the pot odds for the caller decrease) the pot odds approach 1 to 1, but never quite reach it. This conclusion lead me to believe that there is a simple function that can define that movement of this curve on a graph. For Example:

-if you bet 1/4 of the pot you offer the caller 5 to 1 odds
-if you bet 5/6 of the pot you offer the caller 2.5 to 1
-if you bet 3/2 of the pot you offer the caller 1.667 to 1

this continues on and on for infinity, the bettor is never able to give the caller 1 to 1 odds, but the number approaches an asymptote; all of this leads me to believe that the function looks something like:

f(x)=(x+p)/x, where p is the pot. But I'm not entirely sure.

Has anyone thought of this before? This is really just theory as no one could lay down a bet of infinity times the pot, but for my personal satisfaction I'd like to know if anyone knows this measureable function or if they can help me determine it.

thanks a lot
-Brent

[PaultheS]

You just want a function for pot odds?

f(x) = (1 + x) / x, where x is the fraction of the pot you bet (1/2, 1, 3/2, 2, etc.).

e.g. if you bet 1/2 pot, f(1/2) = (1 + 1/2) / (1/2) = 3. So the player is getting 3:1 on his call (he's calling 1/2 pot to win 3 1/2 pots).

The curve is f(x) = (1/x) + 1, which approaches 1 as x goes to infinity and infinity as x goes to 0 (from the positive side), as you stated.

Hope that helps.

[benkahuna]

[ QUOTE ]
I've stumbled over a theory element of poker that I haven't read about yet and thought I'd post it to see what others who may have realized the same thing say.

When playing No-Limit, the idea of pot odds governs the amount which the bettor bets, not considering all psychological reasons for betting. Primarily pot odds are given to the caller based on how much the bettor decides to bet. For example:

pre-flop the pot contains $100 ([25+25]+50 from the blinds, no callers for simplicity). After the flop the small Blind bets 50, the pot is now $150 and the caller must pay $50 to see the next card. The caller is being offered $150 to $50 or 3 to 1 pot odds.

What I've noticed is that as the bet amount which the bettor places increases (obviously the pot odds for the caller decrease) the pot odds approach 1 to 1, but never quite reach it. This conclusion lead me to believe that there is a simple function that can define that movement of this curve on a graph. For Example:

-if you bet 1/4 of the pot you offer the caller 5 to 1 odds
-if you bet 5/6 of the pot you offer the caller 2.5 to 1
-if you bet 3/2 of the pot you offer the caller 1.667 to 1

this continues on and on for infinity, the bettor is never able to give the caller 1 to 1 odds, but the number approaches an asymptote; all of this leads me to believe that the function looks something like:

f(x)=(x+p)/x, where p is the pot. But I'm not entirely sure.

Has anyone thought of this before? This is really just theory as no one could lay down a bet of infinity times the pot, but for my personal satisfaction I'd like to know if anyone knows this measureable function or if they can help me determine it.

thanks a lot
-Brent

[/ QUOTE ]

As you approach an infinite bet, pot odds approach 1:1 assuming the pot started with any amount.

Yep, sounds right to me. I've thought about this too. This means that someone can bet whatever they want in holdem and I'm not laying down top set on 2 flush, 2 to a straight board when no straight is yet possible when I have a set because Lyle Berman tells me you need 17 outs to be favored against a set. Bring on your overbet.

I've realized this for a long time, because it prevented me from being able to make any hand fold on the flop. Say I have pocket 6s and another player has Ah Kh and the flop has two hearts. The other player has 9 flush draw outs plus 6 Aces and Kings to give him a bigger pair. Therefore, I have the best hand at the moment, but there is no amount that I can bet to make the person fold on the flop, because he is over 50% to improve. This drove me nuts for a while. Try to give a person 1.1:1 pot odds, you have to bet 10 times the pot. You can't give a person 1:1 pot odds but you can give them infinite:1, by giving them a free card.

[BruceZ]

Pardon my temporary brain meltdown, for anyone who saw what I just had up here for a second. [img]/images/graemlins/blush.gif[/img]

[ QUOTE ]
if you bet 5/6 of the pot you offer the caller 2.5 to 1

[/ QUOTE ]

That should be 11 to 5 or 2.2 to 1.


[ QUOTE ]
f(x)=(x+p)/x, where p is the pot.

[/ QUOTE ]

f(x) = (x+1)/x

It's not a function of p. Your opponent must call your bet of xp to win a pot of p+xp, so his pot odds are (p+xp)/xp = (x+1)/x to 1.

Or for an easier way to do it in your head, if you bet a fraction x/y of the pot, then you are making the pot odds x+y to x.