What's The GTO Solution To This Simple Game?e
04-29-2017
, 03:19 PM
04-30-2017
, 11:22 AM
04-30-2017
, 12:15 PM
04-30-2017
, 10:48 PM
05-01-2017
, 12:42 AM
05-01-2017
, 01:26 AM
05-01-2017
, 01:52 PM
05-02-2017
, 02:52 AM
I think that what I wrote above is correct.
For the two street game... Here's a graph of the EV for each action for each card, when playing against this strategy.
When playing against this strategy:
* Betting is always an action that maximizes EV.
* Check/betting maximizes EV for all cards <= 3/4
* Check/checking maximizes EV for all cards <= 5/12
Note that the strategy is optimal against itself. Therefore it's an equilibrium/GTO.
For the two street game... Here's a graph of the EV for each action for each card, when playing against this strategy.
When playing against this strategy:
* Betting is always an action that maximizes EV.
* Check/betting maximizes EV for all cards <= 3/4
* Check/checking maximizes EV for all cards <= 5/12
Note that the strategy is optimal against itself. Therefore it's an equilibrium/GTO.
05-03-2017
, 09:26 PM
That's an interesting little game to solve! I haven't really seen a game-theoretically sound approach here, so here's some comments and my solution.
(On a sidenote, it's always painful to read "GTO" in serious contexts, that's a meaningless term -- if you are looking for equilibrium play, why not call it that?)
(Because the distribution of the cards over the intervall was not stated specifically, I assume a uniform distribution, that makes the math easier and allows nice graphical representations.)
For sake of simplicity, I will tackle the simpler version with just 1 stage. Some comments on extensions at the end.
The Nash equilibrium is: Both players bet if their card is >= 2/3.
Derivation follows.
First off, some remarks:
* No mixing is needed. In fact, it is easy to show that mixing strategies will always be (at least weakly) dominated. Intuitively, mixing never makes sense, because instead of mixing over a certain intervall of your "card", you would rather bet the top values of that range, and check the bottom ones.
E.g., rather than betting/checking with probabilities 0.25 / 0.75 on the intervall [0, 0.66] as suggested below, it is at least as good (against any opponent strategy, and often strictly better) to check on [0, 0.5] and bet on [0.5, 0.66].
* The same reasoning shows that no "bluffing" will take place, i.e. ranges are linear.
* This allows us to look only at "cutoff-strategies", i.e. such strategies in which the players pick a number (call them A and B) and bet whenever their number is larger than that threshold. I will call the cards actually drawn "a" and "b", i.e. player 1 bets if a>A.
With that being said, we can write down payoff functions, i.e. a function of how much player 1 earns if he plays "A" and player 2 plays "B".
To do so, we need to assess the likelihood of the different events - both check, both bet and a>b, etc. These are illustrated by the graphic below.
Note that drawing from 2 independent standard uniform distribution means randomly picking a point in the unit square -- the probability of each event is therefor given by the size of the accordingly colored area.
One thing complicates the analysis: The way in which the top right rectangle is split depends on whether A>=B (left) or B>A. We therefore must make a case distinction when formulating payoffs (and from there best replies).
The probabilities are multiplied with the payoffs for each event (0.5 in case both fold, 2 if both bet and player 1 wins, -1 for betting and losing, 1 for betting if the other checks, and 0 for checking when the other bets).
__________________________________________________ _________________________________
Thus, if A >= B:
U1(A,B) = 0.5 AB + 1 (1-A)B - 1 A(1-B) + 2 (0.5 (1-A)^2 + (A-B)(1-A))
Staying with this case, we can find the best response by player 1 to any strategy by maximizing this function, i.e. setting its derivative w.r.t. A to 0:
dU1/dA = -3A - 1.5B + 1 = 0
<-> A = 0.5 B + 1/3
(Sidenote: the second derivative = -3 < 0, i.e. it's indeed a maximum).
That's what player 1 should play, as long as A >= B.
[Another footnote: If B were > 2/3, this best response function would prescribe to play a strategy A with A<B -- but in that case, the assumption A>=B is violated, and we would have to look below.]
Similarly, we could derive B = 0.5 A + 1/3 for player 2.
Both equations hold simultaneously for A = B = 2/3, which conveniently also fullfills our assumption that A >= B.
That's our Nash equilibrium!
[Another footnote: If B were > 2/3, this best response function would prescribe to play a strategy A with A<B -- but in that case, the assumption A>=B is violated, and we would have to look below.]
__________________________________________________ _________________________________
For completeness, let's also analyze what holds for A < B (there might be additional equilibria). Then:
U1(A,B) = 0.5 AB + 1 (1-A)B - 1 A(1-B) + 2 (0.5 (1-B)^2 + (B-A)(1-B))
dU1/dA = -1.5B + 1 != 0
Note that this is never 0 - player 1, in these areas will pick a corner solution.
If B > 2/3, the derivative is negative, and player one would play A = 0, i.e. always bet. (Makes sense - 2 folds too often, and player 1 is +EV by picking up the dead money).
If B < 2/3, this first order condition would advise player 1 to play A = 1 - but then, A>B, and we would switch back to the reaction function given in the first case.
By sYmmetry, the same follows for player 2. In neither of these cases, a pair exists in which both players respond optimally to each other - no other Nash equilibrium exists. (At least when one restricts strategies to cut-off strats. Sidenote: stuff like: {bet if a >= 2/3 or a = 0.183746} is actually a Nash equilibrium as well, but clearly degenerate...)
__________________________________________________ _________________________________
That concludes the analysis of the one-stage game in the basic version. 2 further comments:
a) Without formal analysis, I believe the equilibrium two stage-version should closely resemble this. There is no reason to actively try to reach stage 2, if one is confident that one can win in stage 1.
I conjecture that the players would still play A=B=2/3 in stage 1, and then A=B=(2/3)^2 in the second stage [basically, both players end up in stage 2 with capped ranges, and the original game is repeated in the grey square in the graphic above].
b) Also without analysing, I don't think the "both check = higher card gets dead money" rule has much influence on the equilibrium. Analysis could be done the same way as above, but would be more messy, as one would have to cut apart the bottom left area similarly to the top right one.
(On a sidenote, it's always painful to read "GTO" in serious contexts, that's a meaningless term -- if you are looking for equilibrium play, why not call it that?)
(Because the distribution of the cards over the intervall was not stated specifically, I assume a uniform distribution, that makes the math easier and allows nice graphical representations.)
For sake of simplicity, I will tackle the simpler version with just 1 stage. Some comments on extensions at the end.
The Nash equilibrium is: Both players bet if their card is >= 2/3.
Derivation follows.
First off, some remarks:
* No mixing is needed. In fact, it is easy to show that mixing strategies will always be (at least weakly) dominated. Intuitively, mixing never makes sense, because instead of mixing over a certain intervall of your "card", you would rather bet the top values of that range, and check the bottom ones.
E.g., rather than betting/checking with probabilities 0.25 / 0.75 on the intervall [0, 0.66] as suggested below, it is at least as good (against any opponent strategy, and often strictly better) to check on [0, 0.5] and bet on [0.5, 0.66].
* The same reasoning shows that no "bluffing" will take place, i.e. ranges are linear.
* This allows us to look only at "cutoff-strategies", i.e. such strategies in which the players pick a number (call them A and B) and bet whenever their number is larger than that threshold. I will call the cards actually drawn "a" and "b", i.e. player 1 bets if a>A.
With that being said, we can write down payoff functions, i.e. a function of how much player 1 earns if he plays "A" and player 2 plays "B".
To do so, we need to assess the likelihood of the different events - both check, both bet and a>b, etc. These are illustrated by the graphic below.
Note that drawing from 2 independent standard uniform distribution means randomly picking a point in the unit square -- the probability of each event is therefor given by the size of the accordingly colored area.
One thing complicates the analysis: The way in which the top right rectangle is split depends on whether A>=B (left) or B>A. We therefore must make a case distinction when formulating payoffs (and from there best replies).
The probabilities are multiplied with the payoffs for each event (0.5 in case both fold, 2 if both bet and player 1 wins, -1 for betting and losing, 1 for betting if the other checks, and 0 for checking when the other bets).
__________________________________________________ _________________________________
Thus, if A >= B:
U1(A,B) = 0.5 AB + 1 (1-A)B - 1 A(1-B) + 2 (0.5 (1-A)^2 + (A-B)(1-A))
Staying with this case, we can find the best response by player 1 to any strategy by maximizing this function, i.e. setting its derivative w.r.t. A to 0:
dU1/dA = -3A - 1.5B + 1 = 0
<-> A = 0.5 B + 1/3
(Sidenote: the second derivative = -3 < 0, i.e. it's indeed a maximum).
That's what player 1 should play, as long as A >= B.
[Another footnote: If B were > 2/3, this best response function would prescribe to play a strategy A with A<B -- but in that case, the assumption A>=B is violated, and we would have to look below.]
Similarly, we could derive B = 0.5 A + 1/3 for player 2.
Both equations hold simultaneously for A = B = 2/3, which conveniently also fullfills our assumption that A >= B.
That's our Nash equilibrium!
[Another footnote: If B were > 2/3, this best response function would prescribe to play a strategy A with A<B -- but in that case, the assumption A>=B is violated, and we would have to look below.]
__________________________________________________ _________________________________
For completeness, let's also analyze what holds for A < B (there might be additional equilibria). Then:
U1(A,B) = 0.5 AB + 1 (1-A)B - 1 A(1-B) + 2 (0.5 (1-B)^2 + (B-A)(1-B))
dU1/dA = -1.5B + 1 != 0
Note that this is never 0 - player 1, in these areas will pick a corner solution.
If B > 2/3, the derivative is negative, and player one would play A = 0, i.e. always bet. (Makes sense - 2 folds too often, and player 1 is +EV by picking up the dead money).
If B < 2/3, this first order condition would advise player 1 to play A = 1 - but then, A>B, and we would switch back to the reaction function given in the first case.
By sYmmetry, the same follows for player 2. In neither of these cases, a pair exists in which both players respond optimally to each other - no other Nash equilibrium exists. (At least when one restricts strategies to cut-off strats. Sidenote: stuff like: {bet if a >= 2/3 or a = 0.183746} is actually a Nash equilibrium as well, but clearly degenerate...)
__________________________________________________ _________________________________
That concludes the analysis of the one-stage game in the basic version. 2 further comments:
a) Without formal analysis, I believe the equilibrium two stage-version should closely resemble this. There is no reason to actively try to reach stage 2, if one is confident that one can win in stage 1.
I conjecture that the players would still play A=B=2/3 in stage 1, and then A=B=(2/3)^2 in the second stage [basically, both players end up in stage 2 with capped ranges, and the original game is repeated in the grey square in the graphic above].
b) Also without analysing, I don't think the "both check = higher card gets dead money" rule has much influence on the equilibrium. Analysis could be done the same way as above, but would be more messy, as one would have to cut apart the bottom left area similarly to the top right one.
Last edited by Johnny Nashville; 05-03-2017 at 09:36 PM.