I am buying an item (which I value at, say, a million dollars) for \$$100$. My friend has a special offer that expires today, and which he can never use again unless used by me, which gets me the item for \$$90$. However, he initially demands that I give him $x \in (0,10)$ dollars to activate this offer. In principle, I can reject and counter-offer, and he can reject my counter-offer and re-offer, without discounting, ad infinitum. What model can be used to resolve this and find an equilibrium? Obviously, both of us strictly prefer reaching an agreement. I was thinking perhaps there is some discounting towards the very end of the time of expiry, but it still doesn't all quite make sense... [Real-life situation between rational, utility-maximizing risk-neutral players.]


2 Answers 2


With discounting, the situation is a classic Rubinstein bargaining game, in which two players make alternating offers to split a shrinking pie.

Without discounting, I'm not sure an equilibrium necessarily exists. Perhaps you can impose restrictions on the offers made in each round, e.g. an offer in round $t$ must be in a range set by the offers made in round $t-1$.


Discounting does not work like that. Discounting means how much you discount the future relative to the present---e.g., receiving \$1 tomorrow is the equivalent of receiving \$0.90 today.

What you are describing is some notion of "importance of reaching an agreement" because the deadline is approaching. However, in terms of modeling this requires finite horizon, which contradicts your ad infinitum assumption. So, I guess one question is whether you have finite or infinite horizon.

I understand the real-life scenario: you guys have to decide by tomorrow (the deadline), but in the meantime you can alternate making offers to each other. In this case, the horizon is finite, albeit the interval between offers is short.

In any case, I think this game is a slight variation of the standard alternating offers game (a'la Stahl-Rubinstein) since the seller does not have a use for the good---as I understand from your description. This matters in the finite horizon model with even periods, where the buyer makes the last offer (see below).

Below I describe the finite horizon version.

  • Players and Utilities: You are the buyer, and the item is worth \$10 to you, and \$0 to your friend (the seller). You both have a common discount factor $\delta$ (note: if $\delta=1$ then you value the future and today the same). If you reach an agreement at price $P$ in period t, then

    • Buyer's payoff in present terms is $V_B = \delta^{t-1} (10-P)$,
    • Seller's payoff in present terms is $V_S = \delta^{t-1} P$.
    • NOTE: I assume that the seller does not value the item, because from your description this seems to be the case.
  • Actions/Offers: The game starts at t=1, and lasts T periods. The seller (your friend) makes the first offer. The buyer either accepts or rejects. If the buyer accepts, the game ends. If the buyer rejects, s/he makes a counter-offer. The seller then either accepts or rejects... and so on... until either an agreement is reached or period T is reached without an agreement.

Since horizon if finite, we can solve the game by backward induction.

  • Odd T

In the last period the seller (your friend) is making the offer. She will just ask $P_T=\\\$10$. And you will accept giving payoffs $V_{B,T}=0, V_{S,T}=10$.

In period T-1, the buyer (you) anticipates this and offers $P_{T-1}$ such that the seller is indifferent between receiving $P_{T-1}$ today and $P_T$ tomorrow. Thus, $P_{T-1}=\delta P_T$. Payoffs $V_{B,T-1}=10(1-\delta), V_{S,T-1}=10\delta$.

In period T-2, the seller anticipates this and offers $P_{T-2}$ such that the buyer is just indifferent between accepting $P_{T-2}$ and offering $P_{T-1}$ next period which will be accepted. Thus, $10-P_{T-2}=\delta(10(1-\delta)$ and $P_{T-2} = 10(1-\delta(1-\delta))$. Payoffs: $V_{B,T-2}=10\delta(1-\delta), V_{S,T-2}=10(1-\delta(1-\delta))$.

By analogous reasoning, at T-3 the buyer offers $P_{T-3}=10\delta(1-\delta(1-\delta))$, which is accepted giving payoffs $V_{B,T-3}=10(1-\delta(1-\delta(1-\delta))), V_{S,T-3}=10\delta(1-\delta(1-\delta))$

And so on until we reach period 1, where the seller offers $P_1=10(1+\sum_{i=1}^{T-1}(-\delta)^i)=10(1-\delta+\delta^2+...+\delta^{T-1})$ and it is accepted by the buyer. Payoffs: $V_B=10-P_1, V_S=P_1$.

If $\delta=1$, the seller offers $P_1=10$ which is accepted immediately. Notice that since T is odd, T-1 is even and $P_1=10(1-\delta+\delta^2+...+\delta^{T-1})=10$ since all $\delta$s cancel out.

(In general, in odd periods T-t, when the seller makes the offer, the seller offers $P_{T-t} = 10(1+\sum_{i=1}^{t}(-\delta)^i)=10(1-\delta+\delta^2-\delta^3+....)$.

In even periods T-t, the buyer offers $P_{T-t}=10\delta(1+\sum_{i=1}^{t-1}(-\delta)^i)$, which is accepted.)

  • Even T

Since the seller makes the offers in odd periods, and the buyer in even periods, the buyer gets to make the last offer.

At t=T, the buyer offers $P_T=0$, which is accepted giving payoffs $V_B=10, V_S=0$. NOTE: Here the assumption that the Seller does not care about the good bites. If the seller has value for the item, then $P_T$ would equal to the seller's value.

At t=T-1, by analogous reasoning to above, the seller offers $P_{T-1}$ to make the buyer indifferent between $P_{T-1}$ today and $P_{T}$ next period. So, she solves $10-P_{T-1}=10\delta$, which gives $P_{T-1}=10(1-\delta)$. The payoffs are $V_B=10\delta, V_S=10(1-\delta)$.

At t=T-2, the buyer offers $P_{T-2}=\delta P_{T-1}=10\delta(1-\delta)$ to make the seller indifferent between accepting $P_{T-2}$ today and offering $P_{T-1}$ tomorrow (which would be accepted). Thus, the payoffs are $V_B=10(1-\delta(1-\delta)), V_S=10\delta(1-\delta)$.

And so on until we reach period 1, where the seller makes the first offer. Remember that this game ends with the buyer making the last offer.

The seller offers $P_1=10(1+\sum_{i=1}^{T-1}(-\delta)^i)=10(1-\delta+\delta^2+...+\delta^{T-1})$ and it is accepted by the buyer. Payoffs: $V_B=10-P_1, V_S=P_1$.

Again suppose $\delta=1$. Now since $T-1$ is odd, we have $P_1=0$.

  • Conclusion: Finite horizon and $\delta=1$ for both players, we have last mover advantage, since both players are infinitely patient.

  • Infinite Horizon: $\delta=1$ is a limit case where you both split it equally.

Useful readings on Stahl-Rubinstein:

(Finite horizon) https://cs.uwaterloo.ca/~klarson/teaching/W06-886/Rubinstein.pdf

(Finite horizon) https://sites.duke.edu/niou/files/2011/05/Lecture-6-Bargaining.pdf

(Infinite horizon) https://web.stanford.edu/~jdlevin/Econ%20203/RepeatedGames.pdf


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