In a sequential game, where there are 8 pizzas.

Player 1 decides number of pizzas he wants. Let's call it S1 (strategy of player 1), and S1 = 5 means player one decided to get 5 pizzas.

Then player 2 decides number of pizzas he wants, and player 2 knows what player 1 decided. Let's call it S2 (strategy of player 2), and S2 = 5 means player two decided to get 5 pizzas.

If the S1 + S2 > 8 then both players get 0 pizzas.

Else, the players get the number of pizzas that they decided to get, also player 1 gets S1 pizzas and player two gets S2 pizzas.

In "solutions" to exercises to an assignment about strategies, they have written the following:

S1* and S2* are "Nash equilibrium" strategies of the player 1 and player 2.

Find all the Nash equilibrium in pure strategies in this sequential game.

(1) S1* ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8} and
S2*(S1) = 8 - S1   if   S1 = S1*,
S2*(S1) > 8 - s1   if   S1 > S1*,
S2*(S1) ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8}   if   S1 < S1*

What does it mean that S1 > S1* or S1 < S1*

It means The player 1 decides a number of pizza which is greater than 8?

What does the notations above mean?

  • $\begingroup$ @Giskard ok I accepted it l now, I forgot this feature here. But can you answer the question or? $\endgroup$
    – Coder88
    Mar 14, 2019 at 9:31
  • $\begingroup$ I'm also confused by the notation you listed, but I've provided a solution with (hopefully) less confusing notation below. I'm happy to elaborate. $\endgroup$
    – dlnB
    Mar 16, 2019 at 1:55

2 Answers 2


The above post is pretty close (+1), but has not mentioned the subgame perfect Nash Equilibrium and the notation $s_2^*=8-s_1^*$ might be misleading, as it implies that player 2's Nash Equilibrium strategy is a function of $s_1$, which would mean the best response for player 1 could not be anything less than 8.

First consider the action set for each player (each has the same action set): $$a_1 \in \{0,1,2,...,8\}$$ $$a_2 \in \{0,1,2,...,8\}$$

First consider the strategy set for each player: $$s_1 \in \{0,1,2,...,8\}$$ $$s_2(s_1) \in \{f: \{0,1,2,...,8\} \rightarrow \{0,1,2,...,8\}\}.$$ Notice that player 1's strategy set is the same as her action set, since player 1 cannot condition her action on player 2's action before moving. On the other hand, player 2's strategy set is the set of all functions mapping player 1's action set into player 2's action set.

If we consider the definition of a Nash Equilibrium,and carefully consider the strategy sets for each player, we can see there are at least 10 Pure Strategy Nash Equilibria: $$(s_1^*,s_2^*) \in \{(0,8),(1,7),(2,6),(3,5),(4,4),(5,3),(6,1),(7,1),(8,0),(8,8-s_1)\}.$$ In the first 9 equilibria listed (i.e. those the first poster was describing), firm 2's equilibrium strategy is simply a number, not a function that depends on player 1's strategy (e.g. $s_2(s_1)=4$). Notice these would be the equilibria if this were a static game (meaning player 2 did not observe player 1's choice before choosing an action).

The last equilibrium listed $(8,8-s_1)$ is the subgame perfect Nash Equilibrium. To see why, we must use backward induction. Starting with player 2's move, we have 9 possible subgames (one for each of the possible actions chosen by player 1). For the arbitrary subgame in which $s_1=C$, player 2's best response is to choose the action corresponding to $8-C$. We can generalize this across the 9 subgames and deduce that the strategy $s_2(s_1)=8-s_1$ forms a Nash Equilibrium in each of these 9 proper subgames.

So we have determined that in the subgame perfect Nash Equilibrium, player 2 must choose strategy $s_2(s_1)=8-s_1$. We must now find player 1's best response, which can easily be seen to be $s_1=8$. Thus the 10th Nash Equilibrium listed above is the subgame perfect Nash Equilibrium.

You might be inclined to argue that $(8,0)$ and $(8,8-s_1)$ are equivalent outcomes, but they are not. The former cannot be a subgame perfect Nash Equilibrium since player 2's strategy $s_2(s_1)=0$ does not form a Nash Equilibrium in every proper subgame (e.g. in the subgame where player 1 had chosen $s_1=7$, player 2 would still have chosen $s_2(7)=0$.


Note that Nash equilibrium means no one can benefit from unilateral deviation. In this case, if $s_1 = s_{1}^*$ (i.e. Nash Equilibrium), then it must be that $s_1+s_2=8$ since otherwise, player 1 or 2 can deviate and benefit (for example, if $s_1+s_2<8$, p1 can order more to let $s_{1}^{\prime}+s_2=8$ and benefit).

So, back to your question, any pair that makes $s_{1}+s_{2}=8$ is a Nash equilibrium, so, there are 9 NEs, with $s_{1}^{*}\in\{0,\cdots,8\}$, and given a $s_{1}^{*}$, $s_{2}^{*}=8-s_{1}^{*}$. I guess the notation $s1>s^{*}$is confusing, it should mean that given a fixed $s_{1}^{*}$, if p1 deviates to $s_{1}<s_{1}^{*}$, then p2's best response is to increase his number from $s_{2}^*$ to $s_2$ and make $s_2+s_1=8$ again.

  • $\begingroup$ You've listed only the equilibria that occur in the static version of the game. In the sequential version, player 2's strategy set is more complex than $\{0,1,...,8\}$, and hence there are more equilibria than just those you mentioned. Take a look at my solution below. $\endgroup$
    – dlnB
    Mar 16, 2019 at 1:57
  • $\begingroup$ @dlnB You are right, I didn't read the question very carefully, sorry about that. $\endgroup$
    – The R
    Mar 16, 2019 at 2:22

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