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Different EV reference points giving different answers
http://poker-ai.org/phpbb/viewtopic.php?f=24&t=3667		 Page 1 of 1
Author:  namnipeels [ Tue May 30, 2023 9:31 pm ]
Post subject:  Different EV reference points giving different answers
For calculating the EV of an action at a certain decision point here are 3 common reference points used:

1. Change in stack size relative to current decision point (this gives the ubiquitous EV(fold) = 0 ).
2. Change in stack size relative to beginning of hand (EV(fold) = final stack - starting stack ).
3. Stack size at end of hand (EV(fold) = final stack ).

All of these referece points will produce different EVs.
But if we stick to one reference point for a given decision point, and compare the EVs of all the possible actions, the difference between EVs will remain the same.

That is a confusing sentence so an example will help: ​
HU $0.5/$1 NLHE​
Starting stack sizes of SB and BB = $10​
SB posts $0.5​
BB posts $1​
SB first to act with AA, has 2 options fold or shove.​
BB calls a shove 30% of the time.​
BB's calling range has 20% equity against AA.​

Reference 1:
EV(fold) = $0 EV(shove) = $3
Reference 2:
EV(fold) = -$0.50 EV(shove) = $2.50
Reference 3:
EV(fold) = $9.50 EV(shove) = $12.50

When comparing the difference in EVs within a reference point, shoving is always expected to make $3 more than folding.

What confuses me is when I try to apply these different reference points to the indifference principle in a Kuhn poker situation.
https://en.wikipedia.org/wiki/Kuhn_poker

Lets assume both p1 and p2 start with stacks s, then they both ante 1 chip.
Lets assume p1 always bets (1 chip) with K, always checks Q, and sometimes bluffs (1 chip) with J.
When p2 is deciding how often to bluff with a J after p1 checks, he wants to make p1 indifferent to folding and calling with bluff catchers (Q).
EV(p1 folds Q) = EV(p1 calls Q)

Reference 1:
0 = P(p2 has J)*P(p2 bets J)*(3) + P(p2 has K)*P(p2 bets K)*(-1)
0 = (1/2)*P(p2 bets J)*(3) + (1/2)*(1)*(-1)
0 = (3/2)*P(p2 bets J) - 1/2
P(p2 bets J) = 1/3

Reference 2:
EV(p1 folds Q) = (s-1) - s = -1
EV(p1 calls Q) = P(p2 has J)*P(p2 bets J)*(s + 2 - s) + P(p2 has K)*P(p2 bets K)*(s - 2 - s)
-1 = (1/2)*P(p2 bets J)*(2) + (1/2)*(1)*(-2)
-1 = P(p2 bets J) - 1
P(p2 bets J) = 0!?

Reference 3:
EV(p1 folds Q) = s-1
EV(p1 calls Q) = P(p2 has J)*P(p2 bets J)*(s + 2) + P(p2 has K)*P(p2 bets K)*(s - 2)
s-1 = (1/2)*P(p2 bets J)*(s+2) + (1/2)*(1)*(s-2)
P(p2 bets J) = s / (s+2) !?


Why are all of these answers different if I'm sticking to the same reference point throughout the calculation?
Author:  spears [ Thu Jun 01, 2023 7:02 am ]
Post subject:  Re: Different EV reference points giving different answers
Can't say I fully understand the question in the time I can give it, but I hope this helps.

My understanding of indifference is that in a Nash Equilibrium strategy every action at some decision point with some hand will result in the same EV, unless the probability of taking that action according to the NE strategy is one or zero.

Given that definition, I don't see why the difference in EVs between different actions at different decision points would always be the same as I think you are suggesting in your shove fold example

I don't follow your Kuhn poker example, but https://skranz.github.io/gtree/articles ... poker.html seems to support my definition of indifference.
Of course, it must always be the case that a player is indifferent between all moves over which he mixes in an equilibrium.
So you would expect to see the same EV when the strategy is mixed, but different EVs when one action should always be taken and the other never.
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