Without "subject", again

[Fioravante Patrone1]

$(( I \ \\ \ II \ \vdots,A,B,C),(\ldots,\ldots,\ldots,\ldots),(\ \ \ A \ \ \ \vdots,40 \ 40,60 \ 10,10 \ 40),(\ \ \ B \ \ \ \vdots,10 \ 60,10 \ 10,60 \ 40),(\ \ \ C \ \ \ \vdots,40 \ 10,40 \ 60,50 \ 50))$

You are player $I$.
Assume that choices are simultaneous, and that everything (strategies, payoffs, rationality, intelligence) is Common Knowledge.
And that the game is played just once. You will never meet again player $II$.

If you were to play this game (as player $I$) what would you choose?

Please use the spoiler for your answer and for comments

[Cheguevilla]

Another face of prisoner's dilemma?

[Fioravante Patrone1]

I make public the question by Cheguevilla: "Another face of prisoner's dilemma?", since the answer is "no".

There is one thing that the example introduced here has in common with the PD: the possibility that the choice(s) made by the players do not fit well with the theoretical prediction.

But this is a general phenomenon. Typical of many "interesting" games. That maybe would not be so interesting if they were showing perfect coincidence between theory and "practice" (or experiments).

For those who ae willing to answer, just look at the game and choose which is the choice you would make. That's it.

[kinder1]

I'm sure it is quite easy to forecast my answer, after the discussion we have had about the traveler's dilemma. My choice is C, without doubt. This is enforced by the lack of dominant strategies in this game.
I look forward to know the theoretical prediction.

[Fioravante Patrone1]

@kinder
thank you for the answer
I wait for more, before starting discussing them (otherwise I should say "it")

[kinder1]

waiting for the answers of other guys, I'd like to submit another approach I've just "built", that could furthermore justify the choice I've done, with C.
My thought is the following. Due to the CK and the symmetry of the game, we are sure that the players will choose the same strategy (another "one player game"). If it is true, we can state that the possible game solutions belong to the main diagonal of the table. In this game the best outcome is in the intersection of strategies C.
Make it sense this?

[Fioravante Patrone1]

@kinder

The point is that there is a broad agreement on the fact that this symmetry argument is not valid. That is, even if players face a symmetric situation, there is no guarantee that they will choose the same strategy.
Consider, for example, the "Battle of the Sexes", which is a symmetric game (I hope you agree on this):
$(( I \ \\ \ II \ \vdots,t_2,t_3),(\ldots,\ldots,\ldots),(\ \ \ s_2 \ \ \ \vdots,0 \ 0,2 \ 1),(\ \ \ s_3 \ \ \ \vdots,1 \ 2,0 \ 0))$
Not so easy to be sure that players will be "on the diagonal".
Even if there is a Nash equilibrium in mixed strategies which is symmetric, it is anyhow a (Pareto) inferior choice.

[Cheguevilla]

Without any knowledge about the opponent, the best choice for me is the strategy

C

But

if I can imagine that my opponent has low taste in risk, I can choice the strategy B, assuming that hi will choose the strategy B.

Otherwise

if I can imagine that my opponent believes I choose the lowest risk strategy (C), I also guess that he chooses strategy B, then my best choice should be the strategy A.

Obviously, with the current perception and the current payoff, i will play the strategy

C

without anu doubt.

[maximus2]

In my opinion,

if player I want minimize his risk then the best solution should be strategy A,
else if player I want maximize his payoff the best solution should be strategy C.

[Cheguevilla]

Hi maximus, are you sure you didn't invert the strategies?

[Fioravante Patrone1]

thee is some misunderstanding for what concerns the interpretation of the numbers that appear in the (bi-)matrix

these numbers are values of von Neumann - Morgenstern utility functions

so, risk aversion (or love for) is already bought in!

[marco vicari]

my choise is A, because is my best reply to the player $II$ strategy.
Is it the only Nash equilibrium?

[Fioravante Patrone1]

@marco vicari

"my choise is A, because is my best reply to the player $II$ strategy."
Said in this way is meaningless. $II$ has 3 strategies. And $A$ is best reply to $A$ and $B$ but not to $C$.
If you meant "the strategy that $II$ will choose", of course you have the burden to explain why $II$ will use a specific strategy.

"Is it the only Nash equilibrium?"
Yes, in pure strategies it is the only one.

[marco vicari]

I try to explain.
Player $II$ don't use strategy C because it's not the best reply to my B and C strategy, but only for my A choise.
If I think that player $II$ plays C I play B, but if my choise is B player $II$ plays A and my payoff is $10$.
If player $II$ play A, my best reply is A.
I think this is a backwards induction process.

[maximus2]

...could we use iterated elimination of not dominated strategies (eliminazione iterata di strategie strettamente dominate) ?

[maximus2]

..sorry: iterated elimination of strictly dominated strategies

[Fioravante Patrone1]

@maximus

no, nothing can be eliminated. I will prove this for $I$, then by symmetry...

One can see by direct inspection that $A$ is not even weakly dominated, neither by $B$ ($40 > 10$, nor by $C$ (60 > 40). Similarly for $B$ ($60$ is strictly greater than $50$ and $10$). And for $C$ no hope.

One could wonder whether there could be some kind of domination fo the mixed extension of the game. The answer remains negative.

Since each of the strategies is best reply to at least one of the strategies of $II$ ($A$ is best reply to $A$ and to $B$; $B$ is best reply to $C$; $C$ is best reply to $A$), it is impossible to find any kind of strong domination.

Moreover, since $A$ and $B$ are the unique best reply to some strategy of $II$ (this will hold also for the mixed extension, due to linearity), they cannot be even weakly dominated.

So, it remains to check that $C$ is not weakly dominated.
Is there a convex combination of strategies $A$ and $B$ that will yield a weakly better result for $I$, for whatever strategy played by $II$?
The answer is clearly no.
Take the strategy $\sigma_{\lambda}$ that puts weight $\lambda$ on $A$ and $1 - \lambda$ on $B$.
If $II$ plays $A$, $I$ will achieve a strictly better result using $C$ than $\sigma_{\lambda}$, except when $\lambda = 1$. On the other hand, choosing $\lambda = 1$ is the same as using the pure strategy $A$, a case that we have already considered.

[kinder1]

I'm still sitting on the side of the river, and the cigarette's box is almost empty.

[Fioravante Patrone1]

and what should I say? I don't even have a box!
I stopped smoking years ago, unfortunately (it clashes with the philosophical background of aikido)

users of this forum are extraordinarily lazy
should we fork and create a new forum?

[kinder1]

"Fioravante Patrone":and what should I say? I don't even have a box!
I stopped smoking years ago, unfortunately (it clashes with the philosophical background of aikido)

Great Fioravante.