Game theory: inheritance distribution

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An old man is dying and wishes to split his $2^n$-dollar fortune between his two sons. It shall be distributed this way:

(1) The older brother should propose a way to split the money. If the younger brother agrees, then it should be settled that way.

(2) If the younger brother disagrees, half of the money goes to charity, then he should make his proposal on how to split the rest.

(3) If the older brother agrees, then it's settled. If not, again, half of the money goes to charity, and he should make another proposal.

(4) This process goes round and round until there's a plan that's approved by both brothers, or there's only 1 dollar left. In the latter case, no one will inherit any money, and all the legacy will be donated for a better world.

If both brothers are smart enough to make the right decision, and their only goal is to inherit as much money as possible(with no regards to how much the other brother will get), then how will the inheritance be distributed?

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We can build up the answer from the bottom up, by creating a table of "given optimal play from both sides, how much can the starting brother expect to take home when there's $2^n$ dollars on the table", starting with small $n$. This will decide how little he would have been willing to accept in the $2^{n+1}$ round, which again determines how much the starting brother in that round should offer.

However, we then immediately run into a problem: If your brother offers you exactly the amount you can expect to end up with if you decline and move on to the next round, what do you do? You can either decline out of spite, or accept out of a wish to keep the most of the fortune inside the family.


Let's first suppose both brothers have a secondary goal of keeping money within the family and each brother trusts the other to have that goal.

Then in the \$2 round you should offer to split the money with both dollars to you and nothing to your brother. He then has a choice between letting both dollars go to you, or letting both dollars go to charity, and we're assuming he will prefer you having them, so you end up with \$2.

Now, in the \$4 round you know that if you offer less than \$2, the offer will be rejected and you end up with nothing. But if you offer \$2, your brother will end up with \$2 no matter whether he accepts or declines, so we assume he will prefer you to have the \$2 rather than them going to charity. So you should offer \$2, and you end up with \$2.

In the \$8 round, it is the same! If you offer less than \$2, your offer will be rejected. On the other hand if you offer \$2 your brother can either take the \$2 or become the starting brother in a \$4 round, where he will take home \$2. Again we'll assume he is nice, so if you offer \$2 your offer will be accepted and you take home \$6.

In the \$16 round, don't offer less than \$6, but if you offer exactly \$6 that will be accepted. So that's what you do, and you take home \$10.

In the \$32 round you offer \$10, which is accepted and you take home $22 ...

Now a pattern begins to show, and it looks like the general rule is that you should offer one third of the pool, rounded up or down to the nearest even number. I'm confident this can be shown by induction.


On the other hand if the brothers don't trust each other they will assume that whenever someone gets an offer where it doesn't matter for their end result whether they accept or decline, the offer will be declined out of spite.

Then in the \$2 round you should offer \$1, since that is the lowest offer that is striclty better for your brother to accept than to decline.

In the \$4 round, offer one more than the \$1 your brother will get by declining -- that is, \$2. You get \$2 yourself.

In the \$8 round, offer \$3, and you get \$5 yourself.

In the \$16 round, offer \$6, and you get \$10 yourself.

In the \$32 round, offer \$11, and you get \$21 yourself.

In the \$64 round, offer \$22, and you get \$42 yourself.

Here it seems the general rule is to offer one third of the pool rounded up to the next integer.


The difference between the two outcomes is at most \$1, so it is probably rational to spend an extra dollar on not needing to assume mutual trust.

On the other other hand, if you don't trust your brother at all, perhaps you should spilt the pool in half from the beginning, to eliminate every possible chance that he can consider declining to be a rational choice.

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This is an example of Rubinstein's bargaining game. See Osborne and Rubinstein , A course in game theory, for a detailed discussion. For Nash equilibrium, there are many solutions, because one brother may say, give me X or I will never agree and we will both get nothing, and then the other brother will agree. For subgame perfect equilibrium, there is a unique equilibrium that balances preferences and the shrinking of the inheritance each round of disagreement. In this case the older brother suggests he gets 2/3 and the younger gets 1/3, and the younger accepts. Though given that this is a discrete case, there may be a subtlety I missed.