<p>Based on your description, I think players are indifferent between .89 now and .89 later so we can think of money and utils as the same thing. Time (the round) is important only in that it shrinks the total pie and that who makes the offer alternates. This is assuming their utility functions are based on maximizing their money gained--there's no preference for spite or punishing the other player.</p>
<p><strong>Multiple Equilibria Because of Discrete Action Space</strong></p>
<p>If we weren't constrained to working in pennies, so that an offer of .8900000001 were possible, I would say an offer of .89 gives the only equilibrium. Why not .8900000001 to make sure the other player accepts? Because why not .89000000000000001? You could always find a smaller number.</p>
<p>Because we're working with pennies, we've discretized the action space. Now, I think it's fair to say in the two-round game there are multiple equilibria. But the above continuous argument kind of sets a default for how to think about the discrete cases, which is why I think in the referenced course they probably keyed in on an offer of 0.89.</p>
<p>One equilibrium is one where Player 2's strategy is to accept any offer <span class="math-container">$x \geq 0.89$</span>. Then Player 1 offers 0.89 and the equilibrium action profile is (offer 0.89, accept).</p>
<p>Another equilibrium is Player 2's strategy is to accept any offer <span class="math-container">$x &gt; 0.89$</span> (being equivalent to <span class="math-container">$x\geq .90$</span> in this action space) . Then player 1 offers 0.90 and the equilibrium action profile is (offer 0.90, accept).</p>
<p>Both strategies are credible off-path (supposing we want a subgame perfect equilibrium), meaning that if Player 2 does reject .88 or .89, we go to the last round where they are guaranteed at least .89. They do at least as well by rejecting the offer, which is all that is required for equilibrium.</p>
<p><strong>So what should Player 1 do?</strong>
It depends on what equilibrium we're in. With equilibrium knowledge of 2's strategy, Player 1 can offer either .89 or .90.</p>

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2021-10-09T18:48:04.393Z
2021-10-09T18:48:04.393Z
Pburg
<p>I was recently watching <a href="https://oyc.yale.edu/economics/econ-159" rel="nofollow noreferrer">Yale's open course on game theory</a>. One game presented involved the following: the game starts with 1 dollar &quot;on the table.&quot; In each round, a player offers the other player a certain amount of the dollar, which the other player can either accept or reject. If the game continues another round, the amount on the table decreases by 10% (due to the time preference of the players).</p>
<p>Each player must offer the other player some of the money in each round. For simplicity, offers are only in whole pennies (no fraction of penny offers permitted). Obviously, they must be at least 1 and at most the amount of money that's currently on the table - so, for example, in the first round Player 1 could offer Player 2 anywhere from 0.01 to the entire dollar. In the second round, there's only 0.90 on the table, so Player 2 could offer Player 1 anywhere from 0.01 to 0.90.</p>
<p>In a one-round game, player 1 should clearly offer player 2 0.01 and keep 0.99 for themselves.</p>
<p>In a two-round game, I'm a little more confused about the exact amount to offer. Clearly if Player 1 makes Player 2 a bad offer, the game will go to round 2, in which case player 2 will offer player 1 0.01 and keep 0.89 for themselves. So, player 1 should offer player 2 just enough for them to accept the offer and keep the rest for themselves.</p>
<p>This may be a nit-picky point, but should Player 1 offer 0.89 or 0.90? Would player 2 accept the 0.89 because 0.89 now is better than 0.89 later, or would offering 0.89 just make player 2 indifferent because the discounting already accounts for time preference? In the first case, clearly Player 1 should offer 0.89; in the second case, they should offer 0.90 to guarantee that Player 2 will accept the offer.</p>
<p>Can someone set me straight in this area?</p>

<game-theory>
Do I need to offer 0.89 or 0.90 in the following game?
EJoshuaS - Stand with Ukraine
# Do I need to offer 0.89 or 0.90 in the following game?

(Asked by: username_0 on 2021-10-08)

I was recently watching [Yale's open course on game theory](https://oyc.yale.edu/economics/econ-159). One game presented involved the following: the game starts with 1 dollar "on the table." In each round, a player offers the other player a certain amount of the dollar, which the other player can either accept or reject. If the game continues another round, the amount on the table decreases by 10% (due to the time preference of the players).

Each player must offer the other player some of the money in each round. For simplicity, offers are only in whole pennies (no fraction of penny offers permitted). Obviously, they must be at least 1 and at most the amount of money that's currently on the table - so, for example, in the first round Player 1 could offer Player 2 anywhere from 0.01 to the entire dollar. In the second round, there's only 0.90 on the table, so Player 2 could offer Player 1 anywhere from 0.01 to 0.90.

In a one-round game, player 1 should clearly offer player 2 0.01 and keep 0.99 for themselves.

In a two-round game, I'm a little more confused about the exact amount to offer. Clearly if Player 1 makes Player 2 a bad offer, the game will go to round 2, in which case player 2 will offer player 1 0.01 and keep 0.89 for themselves. So, player 1 should offer player 2 just enough for them to accept the offer and keep the rest for themselves.

This may be a nit-picky point, but should Player 1 offer 0.89 or 0.90? Would player 2 accept the 0.89 because 0.89 now is better than 0.89 later, or would offering 0.89 just make player 2 indifferent because the discounting already accounts for time preference? In the first case, clearly Player 1 should offer 0.89; in the second case, they should offer 0.90 to guarantee that Player 2 will accept the offer.

Can someone set me straight in this area?

**Tags:** game-theory


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## 1 answers

### Answer by username_1 on 2021-10-09. Score: 3 | Accepted answer

Based on your description, I think players are indifferent between .89 now and .89 later so we can think of money and utils as the same thing. Time (the round) is important only in that it shrinks the total pie and that who makes the offer alternates. This is assuming their utility functions are based on maximizing their money gained--there's no preference for spite or punishing the other player.

**Multiple Equilibria Because of Discrete Action Space**

If we weren't constrained to working in pennies, so that an offer of .8900000001 were possible, I would say an offer of .89 gives the only equilibrium. Why not .8900000001 to make sure the other player accepts? Because why not .89000000000000001? You could always find a smaller number.

Because we're working with pennies, we've discretized the action space. Now, I think it's fair to say in the two-round game there are multiple equilibria. But the above continuous argument kind of sets a default for how to think about the discrete cases, which is why I think in the referenced course they probably keyed in on an offer of 0.89.

One equilibrium is one where Player 2's strategy is to accept any offer $x \geq 0.89$. Then Player 1 offers 0.89 and the equilibrium action profile is (offer 0.89, accept).

Another equilibrium is Player 2's strategy is to accept any offer $x > 0.89$ (being equivalent to $x\geq .90$ in this action space) . Then player 1 offers 0.90 and the equilibrium action profile is (offer 0.90, accept).

Both strategies are credible off-path (supposing we want a subgame perfect equilibrium), meaning that if Player 2 does reject .88 or .89, we go to the last round where they are guaranteed at least .89. They do at least as well by rejecting the offer, which is all that is required for equilibrium.

**So what should Player 1 do?**
It depends on what equilibrium we're in. With equilibrium knowledge of 2's strategy, Player 1 can offer either .89 or .90.


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