Texas Calculatem and Poker Office Blog

Many players like to protect their hands by betting a large percentage of the pot, in an attempt to making the calling of drawers by other players unprofitable.

However, they will often wrongly assume that, having protected their hand, they're safe to continue betting/protecting in future rounds. Even if scare cards land.

I've done some sums to work out the implied odds one would get, if we assume that a player betting the flop, will make a continuation on the turn. I plan on using this to help me loosen up my play in select situations.

I have assumed that the protecting player either bets the same absolute amount on the turn as on the flop. Or bets the same percentage of the pot on the turn, as they did on the flop. There are therefore two sets of implied odds to use.

Let's start off with some sums:

Let x be the percentage of the pot that the aggressor initially bets.

Let Pf be the size of the pot on the turn.

The aggressor therefore bets x * Pf on the flop. And the pot on the turn is

Pt = Pf + x * Pf + x * Pf
= Pf * (1+2x)

On the flop. Assuming a continuation bet of x% of the pot, the bet will be

Bf = x * Pt * (1+2x)

So the return on the x * Pf flop call will be

R = Pf + x * Pf + Bf
= Pf + x * Pf + x * Pt * (1+2x)
= xPf (1/x + 1 + 2x).

Since the risk is the initial x * Pf call, the odds are:

(2 + 1/x + 2x) : 1.

Similarly, if the continuation bet on the turn is still x * Pf (i.e. the same as the flop bet), then the return is

R = Pf + xPf + xPf,

and the Odds will therefore be

(2+1/x) : 1

Summary:

The implied odds, assuming a continuation bet of x% of the pot on the turn, are:

(2 + 1/x + 2x) : 1. (1)

The implied odds, assuming a continuation bet of x * Pf (i.e. same bet on the turn as on the flop) are :

(2+1/x) : 1 (2)

Tables:

I don't have Excel on this machine, but will produce some odds tables later.

Examples:

Assuming a continuation of x% of the flop on the turn...

x Odds
.1
0.5 3.5
1.0 5.0

Taking this further...

To find the minimum of (1), let

y = 2+1/x+2x

dy/dx = -1/(x^2) + 2

So the turning point is at 1/sqrt(2), which is around 0.71, or 70% of the pot. This gives us that the minimum odds of a constant percentage continuation bet are 4.8:1 !!!!

There is no real solution for a same sized turn bet. But we can see that the implied odds get larger as x gets smaller. So here's a small table

x odds
0.25 6:1
0.33 5:1
0.5 4:1
1 3:1

Anyway, I've just lost my buyin by calling an all in with a weak top pair. What an idiot. Please let me know if what I've written makes sense.