(Immediate) Pot Odds & Cbetting in a Vacuum.

Let's call the pot P.
Let's call the bet B, as the bet is a function of the pot we can also let B = xF where x is a constant. For example if we bet pot x=1 if we bet half pot x=1/2 and so on.

The pot odds we offer when we bet B into a pot P and the pot odds we are offered the other way round can be worked out by the following.

B + P : B

By using the fact B = xP we can neaten this up some more.

xP + P : xP
P (x + 1) : Px
x + 1 : x

As we want pot odds written as something : 1 for ease of use, we can divide by x.

(x + 1) / x : 1

Most of the time we are betting smaller than the pot so I think the most interesting parts to look at are between 1/2 pot and pot, so I will restrict it to that part for now.

When we bet pot, x = 1

(1 + 1) / 1 : 1 = 2:1

When we bet 2/3 pot, x = 2/3

(2/3 + 1) / (2/3) : 1 = 2.5:1

When we bet ¾ pot, x = ¾

(¾ + 1) / (¾) : 1 = 2.3333:1

When we bet ½ pot, x = ½

(½ + 1) / (½) : 1 = 3:1

What we should notice is that the change in pot odds we offer between ½ and ¾ is bigger than the change in pot odds between ¾ and 1.

In the first part the pot odds go from 3:1 to 2.333:1
In the second the pot odds go from 2.333:1 to 2:1

We can see this by drawing out the function (x+1)/x



We can see that the more we bet the less difference it makes to the pot odds and that the upper limit on the pot odds we can offer are 1:1 as we tend our bet size to infinity.

Looking at cbetting in a vacuum.

Pot = P, Bet = B = xP, villain folds F and calls (1-F).

For cbetting to be profitable in a vacuum we need our opponent to fold enough of the time that we pick up the pot that it makes up for the times that he calls and we lose our bet and we assume that when he calls he has always won.

Two outcomes are

1) He folds F of the time and we win P
2) He calls (1-F) of the time and we lose B

The equation therefore is PF – B(1-F) and as we want to know the point at which it happens and we break even we set this to 0.

PF – B(1-F) = 0

PF + BF – B = 0

F(P + B) = B

F = B / (P + B)

Substituting in B = xP

F = xP / (P + xP)

F = xP / P(1 + x)

F = x / (1 + x)

So for cbetting to be profitable in a vacuum we need F to be more than (or equal to, to break even) x/(1+x).

Plugging in the numbers we set that for

pot → x=1 → F > 50%
3/4 pot → x=3/4 → F > 42.9% (3sf)
2/3 pot → x=2/3 → F > 40%
½ pot → x=1/2 → F > 33.3%

Once again we can analyse this by looking at our equation x / (1 + x).



We see that when we bet 0 we need our opponent to fold >0% for it to be profitable (which makes sense, if our opponent folding if we checked to them we'd always profit)

We also see that the more we bet the more the amount that we need our opponent to fold goes towards 100%.