Just saw this 3-player game being used for a competition on a TV-show. The rules are as follows:

* There are nine numbers up for grabs: 10, 20, 30, 40, 50, 60, 70, 80, 100
* Each player picks two numbers. (without knowing what the others have picked)
* If a number is only picked once, then that amount of points gets awarded to the player who picked that number. If a number is picked more than once, it doens't give any points.
* The player with the most points wins the game.

Example:
Player A picks 100 and 80.
Player B picks 80 and 70.
Player C picks 100 and 60.
Both 100 and 80 was picked twice so these numbers don't give any points. Player B gets 70 points, and player C gets 60 points, since these numbers were only picked once. Player B wins the game.

What are the optimal frequenzies of each possible double-pick? (I don't have the answer)

Also, forgot to mention, lets say 2nd place is worthless. You play solely to win. 2nd place is just the first loser anyway

i got this on a simulation:

PAIR FREQUENCY [%]

(100, 80) 34.0277777778
(80, 70) 16.6666666667
(100, 70) 16.6666666667
(100, 60) 7.02160493827
(70, 60) 7.02160493827
(80, 60) 7.02160493827
(60, 50) 2.31481481481
(70, 50) 2.31481481481
(100, 50) 2.31481481481
(80, 50) 2.31481481481
(80, 40) 0.462962962963
(60, 40) 0.462962962963
(100, 40) 0.462962962963
(50, 40) 0.462962962963
(70, 40) 0.462962962963

you should never pick below (50,40), that should be clear without any calculation. there are 36 ways for one player to pick 2 numbers and 36 more ways for other player to pick his numbers, so 1296 possible combos. for example, if first picks (50,10) and second (80,70), you should pick (100,60). you would do this for all 1296 combinations and arrive at the result above.

People aren't picking randomly.

yeah... this game would be all about exploiting, you wouldn't need GTO. just get a big enough sample on the pairs that get picked and assume both human players are the same.

but i'm curious how would someone calculate GTO for this? there are 1296 possible opponent selections, each of them with its own probability and they all add to 1, that's all i know lol.

@OP: what i posted is optimal only against 2 people who pick every pair with probability 1/36, it's not an equilibrium solution.

What about just a two player case,

You would just pic top 2 and either win or push. I can't think of any way to improve/exploit that.

The three player case is a bit tricky because you need to pick optimal play given they both use optimal play

If you opponents pick randomly from top 6 that can be exploited, so no one would do that.
If your opponents randomly picked from top 3, 4 or 5, they can also be exploited, they won't do that either.

If I knew my opponents were randomly picking from top 3, what would I do?

Neither player could ever score over 100, i pick 2 of 70/60/50

If my opponents randomly picked from top 4,
Then 5/6 of the time, 60,50 wins.

If they picked randomly from the top 5, me always choosing 100+ another wins at least 46% (46% is just chance neither picks 100)

I'm Getting nowhere....

If they continued trying to out do each other in sequential moves, then

if player a chooses random, then player b chooses 100,90, then C picks random (he doesn't care)

Then a chooses random (as long as it isn't 100,90) as he can only win if c's random choice was 100,90.

B, knows he wins as long as c didn't randomly chose 100,90, so he doesn't change.

Akkopower1, 90 doesn't exist! This actually adds some more finesse to the game. Double-picks totaling 90 become a lot stronger than they would have been otherwise. So picks like (50,40) or (60,30) may not be so unreasonable. I'll think more about the game and come back with a post later, probably. Keep up the discussion

Ohhh, no 90. That sux

What are those numbers?

Percentage of your winning if they choose those numbers?

no, how often you should pick each pair against someone who chooses randomly.

Nash proved that every finite game has at least one equilibrium strategy in mixed strategies. Let's start easier. Suppose that the three players had to choose between X, Y, and Z with the same rules governing the payoffs. In this case, it is fairly straightforward to analytically derive the symmetrical Nash equilibrium.

Here are a few solutions.

[10,9,8] -> [.4904, .2900, .2196]

[10,9,7] -> [.4994, .3295, .1711]

[10,8,7] -> [.5098, .2888, .2014]

The case in OP is analogous, but more complicated since there are undoubtedly more than three choices in each player's optimal strategy (with positive probability). My first guess is that there would be C(6,2) = 15 choices played at equilibrium, but that is speculation on my part.

In any event, solving for such an equilibrium analytically is likely to be next to impossible. However, I am sure that solvers exist which would spit out the solution in no time.

So itll be if they pick randomly, you see what they pic, its how often you would win picking those pairs.

Im pretty sure though, that you should always pic 100,80 vs two people who pic randomly.

In your example, "for example, if first picks (50,10) and second (80,70), you should pick (100,60)."

if you pic 100,80 or 100,70 , you still win too.

Whats the interpretation of those percentages?

Those are the percentages you should choose each of the respective choices in your optimal mixed strategy.

For example, if in the three-person game the choices are $10, $8, and $7, then the symmetrical Nash equilibrium strategy is for each player to "play" $10 50.98% of the time, "play" $8 28.88% of the time, and "play" $7 20.14% of the time.

If that does not answer your question, let me know.

In your game is it correct that each player only has one single choice?

If they are only given one choice;

It doesn't seem obvious to me why a players optimal frequency varies with the differing choice structures.

if we compare two games; choices in game X= [1000,100,1] , choices in game Y = [10,8,7]

in both X and Y if you choose the top pay off you only win if neither player chooses top
in both X and Y if you choose middle you only win if both players choose the same (as long as they both dont pick middle)
in both X and Y if you choose smallest you only win if both players choose the same (as long as they both dont pick smallest)

its quite easy to see why you should choose highest more often. But, why does the optimal frequency depend on the relative sizes of the selections?

Do you know how expected values work?

I have an idea, sum_i prob(x_i)*x_i

in your game, does each player choose one or two numbers?


im assuming in your game players only choose one number. So, it should be irrelevant on the actual size of the numbers, i.e. optimal choices from [100,10,1] should be the same as from [10,8,7]

Yes, in the simplified 3-person 3-choice game, each person chooses just one number and then the payouts follow the rule posted by OP above.

Maybe it would be easier to see that the values surely affect the optimal probabilities if we contrast Game A = {10,9,8} with Game B = {11,9,8}.

Suppose you are told the optimal probs for Game A. Now you get to play Game B. Since Game B is identical to Game A except you get 11 points rather than 10 points in some situations, isn't it clear that you should want to play 11 more often than you played 10?

If you are still not convinced, why don't you write down the payouts to Player 1 if he plays H, M, or L (high, mid, low) given the choices of Player 2 and Player 3. Then write down the expected value of each choice for Player 1.

And then ask yourself what it means for a strategy to be an equilibrium strategy. This is all pretty basic game theory, but I am guessing that you haven't had a course in game theory (that is obviously okay of course).

I still dont think it matters, your not seeking to maximise your EV, but simply score more points (player with most points wins). Perhaps you are seeking to maximise your EV, but your EV doesnt depend on the size of the numbers but simply 1pt for winning a game and 0 for not winning.


from what I posted before;
-----------
if we compare two games; choices in game X= [1000,100,1] , choices in game Y = [10,8,7]

in both X and Y if you choose the top pay off you only win if neither player chooses top
in both X and Y if you choose middle you only win if both players choose the same (as long as they both dont pick middle)
in both X and Y if you choose smallest you only win if both players choose the same (as long as they both dont pick smallest)
------------

if the choices are just [high, med, low], you still win in the same manner as above.

The ways in which you can win are identical, it doesnt matter if the winner has 10pts or 1000pts, a win is a win.

if you then considered the game where choices are [10^10000000,1.1,1], your EV method would assign almost 100% to choosing the biggest number, if I knew two players were doing that it would be pretty simple to exploit.

Correct. All that matters is to beat the other two players. It does not matter whether you rack up 180 points more than them, or just 10 points more than them. It's all the same.

Sorry, I thought the player won the value of his winning bid and he was trying to maximize his expected winnings.

With the clarification (that is, correcting my misunderstanding) that the "prize" for winning is independent of the winning bid, the 3-person 3-choice single-bid game does not depend upon the values of the choices. As the poster(s) above correctly pointed out.

Call them High, Mid, Low. Then the optimal probabilities to play each choice in the symmetrical Nash equilibrium can be analytically found, if I did if correctly, to be the following.

High: sqrt(12)-3 = .46410

Mid: 2-sqrt(3) = .26795

Low: 2-sqrt(3) = .26795

The solution of the more complicated problem presented in OP still seems beyond purely analytical methods, but solvers would have little difficulty with it.

How did you get those

Can you fill in the modelling steps, I don't need any cacculations

Sure (and I apologize again for my earlier posts). At a mixed strategy equilibrium, each choice that has a positive probability of being played must have the same EV.

This is probably obvious but if any single choice had the highest EV, then you'd play that choice 100% of the time. If any choice had a lower EV, then you'd never play that choice.

It is easy to write down (as you did above) all the winning scenarios for each possible choice. Then it is a matter of simple arithmetic to equate the three EV's and solve for the three respective probabilities. In this case, once the winning scenarios are written down, it is almost immediate what the probabilities need to be.

Edit: The simplicity of the formulas makes me wonder if even OP's case can be analytically solved pretty easily. If I have time tomorrow, maybe I'll take another look at this.

my equations are;

EV(high)=EV(mid)
EV(mid) = EV(low)
h+m+l=1

where h is prob you choose high, m prob you choose mid and so forth.

h*(prob h wins) = m*(prob m wins)
m*(prob m wins) = l*prob(l wins)
h+m+l = 1

h*(l^2+2*m*l+m^2) = m*(h^2+l^2)
m*(h^2+l^2) = l*(h^2+m^2)
h+m+l = 1


Were yours along those lines?