Consider a game inspired by the long-running television program The Price Is Right. Three players have to guess the price x of an object. It is common knowledge that x is a random variable that is distributed uniformly over the integers 1, 2, . . . , 9. That is, with probability 1/9 the value of x is 1, with probability 1/9 the value of x is 2, and so on.
Thus, each player guesses a number from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. The game runs as follows: First, player 1 chooses a number n1 ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9}, which the other players observe. Then player 2 chooses number n2 that is observed by the other players. Player 2 is not allowed to select the number already chosen by player 1; that is, n2 must be different from n1 . Player 3 then selects a number n3 , where n3 must be different from n1 and n2 . Finally, the number x is drawn and the player whose guess is closest to x without going over wins $1000; the other players get 0. That is, the winner must have guessed x correctly or have guessed a lower number that is closest to x. If all of the players' guesses are above x, then everyone gets 0
(a) Suppose you are player 2 and you know that player 3 is sequentially rational. If player 1 selected n1 = 1, what number should you pick to maximize your expected payoff?
(b) Again suppose you are player 2 and you know that player 3 is sequentially rational. If player 1 selected n1 = 5, what number should you pick to maximize your expected payoff?
(c) Suppose you are player 1 and you know that the other two players are sequentially rational. What number should you pick to maximize your expected payoff?