Next week we'll be looking at stochastic/Markov games (C13 of my course) but this week we took a look at incomplete information in extensive form games (C12 of my course).

We played the following game (a generalization of the matching pennies game that I've blogged about before):

"Player 1 picks Heads or Tails, a coin is flipped and Player 2 is aware of the result of that coin (but not the choice of Player 1), Player 2 then picks Heads or Tails. The is akin to a matching pennies game except that if Player 2 loses whilst picking the same as the random coin then the outcome is $(0,0)$."

Here is a game tree that describes the game:

On a slight tangent our School operates a mid-module feedback system which is not 'official' but allows us to get some feedback from students with regards to how the course is going so far. One piece of feedback I got was something like:

*"The chocolates are making me fat bring fruit instead."*

I'm pretty sure this was tongue in cheek but I brought in the following as a prize for the winner of a 3 player round robin (using the above game):

(A melon and some fizzy cherries)

As I said, I got 3 students (referred to as S, A. R) to play a round robin tournament where each would play each other twice (swapping being player 1).

The results are here (thanks to +Jason Young for collecting them):

In that photo we see the following match ups:

- Game 1 and Game 2: A versus S

- Game 3 and Game 4: R versus S

- Game 5 and Game 6: A versus R

What we see is that R actually won with a final cumulative score of 3, A finished with -1 and S with -2. Looking through the details we see that A and S both went "against" the coin both times whereas R went "against" the coin once.

I asked the class if anyone had an idea of how one

**should**play the game. One student (who I think was from the team Roy crowd) yelled out that one should always go with the coin.

This is exactly correct, the way to 'prove it' it is to obtain the corresponding normal form game for the extensive form game. We do this by using

**expected values over the random nodes:**

**We have $S_1=\{H,T\}$ and $S_2=\{HH, HT, TH, TT\}$. The strategy set for player 2 indicates what player 2 should do for each possible flip of the random coin: so $HT$ indicates playing $H$ if the coin flips Heads and playing $T$ if the coin flips Tails (so $HT$ corresponds to agreeing with the coin, $TH$ corresponds to going against the coin etc...).**

**Using the above strategy ordering we get the following normal form game:**

**$$**

\begin{pmatrix}

(1/2,-1/2)&(-1/2,1/2)&(0,0)&(-1,1)\\

(-1,1)&(-1/2,1/2)&(0,0)&(1/2,-1/2)

\end{pmatrix}

$$

I will skip the analysis of this game (which we did in class) but it can be shown that the Nash equilibrium is for player 1 to play Heads with probability between 1/3 and 2/3 and player 2 to agree with the random coin.

**Importantly,**at the end of the game the victor: R gave the melon to A. I hope he enjoys it and also that my students found this exercise slightly useful and perhaps slightly more memorable.

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