An allin situation ....
and my EV is x and the EV of the villain is 1-x ..... and we decide to run it 3 times and chop at 2 , it gives me an edge if " x " is smaller than 0,5. I understand the logic behind that ..... but i have no idea on how to mathematically prove it.

anyone knows the calculation on how much doing the " 3 times chop at 2 " will improve ur EV ?

ty

-jpp

So you're saying that if after 2 runs you have 1 win each, you'll chop, and otherwise you'll go to 3 runs? Or something else?

If that is what he meant, possible results are:

1. A wins, A wins, A wins => A wins 1, B wins 0
2. A wins, A wins, B wins => A wins 2/3, B wins 1/3
3. A wins, B wins => A wins 1/2, B wins 1/2
4. B wins, A wins => A wins 1/2, B wins 1/2
5. B wins, B wins, A wins => A wins 1/3, B wins 2/3
6. B wins, B wins, B wins => A wins 0, B wins 1

Example: If p(A wins) = 0.6:

1. 0,6³ = 0.216 => A wins 0.216, B wins 0
2. 0.6²*0.4 = 0.144 => A wins 0.096, B wins 0.048
3. 0.6*0.4 = 0.24 => A wins 0.12, B wins 0.12
4. 0.4*0.6 = 0.24 => A wins 0.12, B wins 0.12
5. 0.4²*0,6 = 0.096 => A wins 0.032, B wins 0.064
6. 0.4³ = 0.064 => A wins 0, B wins 0.064

=> A wins 0.584, B wins 0.416 if p(A wins) = 0.6

Was that what you are looking for?

(obv. only applies in spots where p(chop) = 0)

But in general the runouts are not probabilisitically independent like that.

I worked through the simplest case where there is only one card to come.

Let A = number of rivers on which Player A wins

Let B = number of rivers on which Player B wins

Abstracting from other people's cards, assuming NLHE and no chops, then A+B=44.

Working through the algebra:

The expected fraction of the pot for Player A is: E = (A^3 - 66*A^2 + 3677*A) / 119,196.

At A=44, E=100% (as expected)
At A=0, E=0% (as expected)
At A=22, E=50% (as expected)

For 22 < A < 44, 0.5 < E < A/44

For 0 < A < 22, A/44 < E < 0.5

which I guess are pretty obvious.

How'd you get this E = (A^3 - 66*A^2 + 3677*A) / 119,196 ?

You write down the six cases (see post #3).

For each case you calc A's payout (this is easy).

For each case you calc the prob of this case occurring (this is a little tricky as if A wins the first runout, then he is less likely to win the second runout as one of his winning cards is no longer available to appear on the second river, etc).

Then it is a bunch of algebra to combine them all into one EV equation.