Analysis of SIMULTANEOUS QUIZ SHOW

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 V_n,m(T) (or again just V_n,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 Q_3,3(T), for instance, contains Q_3,2(T), Q_3,1(T), Q_2,2(T), Q_2,1(T), Q_1,2(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.

Two Players

• If both players wait, the payoff is just V_2,m-1.
• If one answers, this answer is correct with probability 1/m and false with probability (m-1)/m. Therefore the expected payoff for the player answering is (T-4(m-1))/m, and T(m-1)/m for the other player (since this other player can then wait until the last round to answer).
• If both answer in this round, then with probability 1/m · 1/m both are right, getting a payoff of T. With probability 1/m · (m-1)/m just Ann is right, with probability (m-1)/m · 1/m just Beth is right, and with probability (m-1)/m · (m-1)/m none of them is right. The expected payoff for each of them in that case is (T/2 + (m-1)(T-4) - 4(m-1)^2)/(m^2) = ((2m-1)T - 8m(m-1))/(2m^2)

................. The conclusion here is: Always wait until the last round. The Nash equilibria have payoffs of V_2,m=T/2.

Three Players

Of course playing one round of Q_3,m results in games Q_nn,mm(w,l), depending on the decision of the players.

• If all players wait, we get the subgame Q_3,m-1.
• If one of them tries an answer, with probability 1/m the answer is correct and the game is over with payoffs of T, 0, and 0 for the three players, and with probability (m-1)/m the answer is wrong and the payoff is -4 for the player answering, and the game V_2,m-1 for the two remaining players. A1 = T - 4(m-1)/m, .... W2 = (m-1)V_2,m-1/m.
• If two players try an answer, then with probabiliy 1/m · 1/m both answers are correct. In this case the payoffs are T/2, T/2, and 0. There are two cases with probability 1/m · (m-1)/m each where the answer of one player is correct and the other one incorrect with payoffs of T, -4, 0 respectively -4, T, 0. With probability (m-1)/m · (m-1)/m both answers are incorrect. In that case the third player will eventually win, by waiting until the last round. Therefore the payoffs are -4, -4, T in that third case. Overall the expected payoffs are A2 := (T + 2(m-1)(T-4) -8(m-1)^2)/(2m^2) for the players attempting an answer, and W1:= T(m-1)^2/(m^2) for the third player.
• Finally, if all three players try an answer in that first round, then the game is finished. With probability 1/m · 1/m · 1/m all three answers are correct, yielding a payoff of T/3 for everybody. There are three cases with probability 1/m · 1/m · (m-1)/m each where two answers are correct. The payoffs are T/2, T/2, -F. There are also three cases with probability 1/m · (m-1)/m · (m-1)/m each where one answer is correct. The payoffs are T, T, -F. Finally, with probability (m-1)/m · (m-1)/m · (m-1)/m all three answers are incorrect. The expected payoff for each player is A3 = (T/3 + (T-4)(m-1) + (T-8)(m-1)^2 - 4(m-1)^3)/ (m^3).

Winner gets 6, wrong answer costs 4

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:

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.