2. (15 pts) Two fair coins are flipped and player I is allowed to view the outcome, player II is not. player II must try to guess ach tail Player I can fold and pay player II S2, or stay. If player I stays, then the number of tails showing. If player II guesses correctly, player I must pay him $2 for e showing. If player II guesses incorrectly, then he must pay player I $2 for each head showing (a) Write out the game in extensive form (that is, write out the game tree), labeling the nodes with the position of the game from the point of view of the player who must make the next move, labeling chance moves with their probabilities, and terminal nodes with the payoff to player I. (b) Carefully describe the set of pure strategies available to each player and write down the payoff matrix for the game with the rows and columns labeled by the strategies. (c) Eliminate all strictly dominated strategies, then determine and u+ for the reduced game. Does the reduced game have a solution in pure strategies? If so, what is it?

Probability of both head and tail in a fair coin is 0.5.
As we know that two fair coins are tossed, the out comes could be HH, HT, TH and TT
where H=heads and T=tails
a) Extensive form of the game
As player one has to make the next make the next move that wheather he has to fold or stay. game tree should be drawn in his respect. Kindly refer to below tree to understands more.
Player I has two choices, either to fold or stay. If he folds, in all the outcomes he has to pay 2$ to player II.
If he stays, Player II can guess 0 tails, 1 tail or 2 tails.
Probability of each outcome i.e. HH, HT, TH and TT is 0.25 each
Thus if Player I stays and Player II guesses 0 tails. So according to diffenrent outcomes, the payoff to player I is explained below:
IF HH comes, then Player II is correct in guessing, so player I must pay him 2$ for each tail. But as no tail showed up, he has to pay 0$
IF HT comes, then Player II guessed wrongly, so player I gets 2$ for each head. Thus player I gets 2$
IF TH comes, then Player II guessed wrongly, so player I gets 2$ for each head. Thus player I gets 2$
IF TT comes, then Player II guessed wrongly, so player I gets 2$ for each head. Thus player I gets 0$ has no heads showed up
Thus the net payoff taking probability into consideration is 0.25*0+0.25*2+0.25*2+0.25*1= 1$
Similarly the entire table can be filled up.
https://d2vlcm61l7u1fs.cloudfront.net/media%2F9d9%2F9d930ecd-2955-4b8d-b4b2-06e742af6f63%2FphpOAF90j.png
b) A pure strategy is all the available moves that a player can adopt during the game.
Pure strategy for Player I: Stay or fold ( 2 pure strategies)
Pure strategy for Player II: Guess 0 tails, 1 tail or 2 tails (3 pure strategies)
It is depicted in table below:
https://d2vlcm61l7u1fs.cloudfront.net/media%2Fd73%2Fd733a575-e070-466d-bf3b-f987af2f0727%2Fphp6zRn40.png
c) Finding the Nash equilibrium in table below:
https://d2vlcm61l7u1fs.cloudfront.net/media%2Fd2b%2Fd2b1b331-52ff-4fd7-865b-c21a11fe242a%2Fphpl5jSqX.png
We see that we find a solution to game i.e. Player I should Stay in the game and Player II should guess 1 tail.
 
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