All these games are symmetric. Therefore all players may expect the same amount as payoff. But since this is a non-zero-sum game with more than two players, we may have several Nash equilibria, some of them nonsymmetric, maening that even though the game is symmetric, the resulting payoff may not be the same for all. We abbreviate by Vn,m(T) (or again just Vn,m if T is understood) any possible Nash equilibrium payoff for a player. Therefore this is not just one value, but could be a set of values.
Any of these games is recursive in the sense that Q3,3(T), for instance, contains Q3,2(T), Q3,1(T), Q2,2(T), Q2,1(T), Q1,2(T), and so on as subgames. These recursiveness is heaviliy used in solving the games. We solve the simpler versions first and base the others on the simpler ones.
|B answers||B waits|
|A answers||((2m-1)T - 8m(m-1))/(2m^2), ((2m-1)T - 8m(m-1))/(2m^2)||(T-4(m-1))/m, T(m-1)/m|
|A waits||T(m-1)/m, (T-4(m-1))/m||V2,m-1, V2,m-1|
................. The conclusion here is: Always wait until the last round. The Nash equilibria have payoffs of V2,m=T/2.
Of course playing one round of Q3,m results in games Qnn,mm(w,l), depending on the decision of the players.
Let is analyze the game played with 3 players and started with three possible answers, where giving a wrong answer costs 4 units. I would like to keep the amount reserved for the winner variable and abbreviate it as w.
We will see that Q(3,2,w,4), Q(3,1,w,4), Q(2,2,w,4), Q(2,1,w,4), Q(1,2,w,4), Q(1,1,w,4) are all subgames of our game. We need to analyze these subgames first.
The games Q(1,2,w,4) and Q(1,1,w,4) are obvious: Since there is only one player left, this player will wait until the last round and then win the amount of w. Therefore the expected value of the player is w.
Q(2,1,w,4) and Q(3,1,w,4) are also very simple: There is only one possible answer, therefore all remaining candidates try this (obviously correct) answer and share the prize of w, getting w/3 each in Q(3,1,w,4) and w/2 each in Q(2,1,w,4).
Q(2,2,w,4) is already slightly more complex. Each player can either try some answer or pass, so we can model this game as a simultaneous game with two moves for each player. If both try, there are four equally likely cases: Both could get the right answer, or only Ann, or only Beth, or none of them. Therefore the expected payoff for both is 1/4·w/2+1/4·w-1/4·4-1/4·4 = 3w/8-2. If Ann tries and Beth passes there are two equally likely cases of Ann getting the correct answer or not. In the second case, Beth has an easy task in the nest round and can win the prize. Therefore the expected payoff for Ann in that outcome equals 1/2·w-1/2·4, and the expected payoff for Beth 1/2·0+1/2·w. The outcome of Beth trying and Ann passing is just the reverse of this case. In the remaining outcome of both passing, both play the subgame Q(2,1,w,4) in the next round and therefore win w/2. The normal form of Q(2,2,w,4) is as follows:
|try||3w/8-2, 3w/8-2||(w-4)/2, w/2|
|pass||w/2, (w-4)/2||w/2, w/2|
This subgame has a dominant move equilibrium, for any w > 4, of passing-passing. Therefore a player facing the subgame Q(2,2,w,4) will pass, wait until the last round, and expect to win w/2 units.