There are 8 coins to be divided between A and B. A first makes an offer of how to divide the 8 coins between them. If B refuses A’s offer, then now it is B’s turn to make the offer, but this time 4 coins are taken away so that only 4 coins can be divided between them. Again, if A refuses B’s offer then now A can make an offer again, but again the stake is halved to two coins. Once a single coin remains, whoever makes the offer takes the coin. What offer should A make at the beginning? (assuming B is rational, and A knows that B is rational)
closed as off-topic by LeppyR64, Gamow, Rand al'Thor, Mithrandir, JonMark Perry Dec 4 '16 at 4:49
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I believe the answer is
A gets 5 coins, B gets 3 coins.
Because, working backwards:
B is guaranteed one coin, 8A->4B->2A->1B
A and B have no reason not to collaborate, both are rational actors, so won't spite the other if they would not lose anything. This means A can safely offer a 1:1 split at step 2A.
Given that A has a path to earning a coin, he will reject a split that doesn't offer him a coin at 4B. The best B can do is offer 1:3, claiming three for himself.
Now that B has laid claim to 3, A can offer a 5:3 split, B has no reason not to accept, and they both go home happy.
This problem can be extended iteratively upwards
B would be able to offer 11:5 in a 16 coin situation, A would be able to offer 21:11 in a 32 coin situation, and the sequence would continue with each person claiming f(x)=2^x-f(x-1)
The answer (assuming that they are trying to get as many coins as they can) is:
A offers a 4-4 split,
Because: B knows that he can't get more than three if the coins split, and A knows if it splits he will only get one coin.
if the coins split all the way to 2 B knows no matter what A offers, (unless he gives both coins to B) that he should refuse A's offer, (so he gets a coin, and A doesn't). but when there are 4 coins, B will offer a 3-1 split, because A knows he won't get a coin if he refuses, and B knows if A refuses, he will only get one, (so it's best for both of them).
Now. if A offers a 4-4 split they both know they get the most coins they can.
Of course if there are different rules to what each others goals are then this answer changes.
What about A offering a
If B refuses, he will need to make an offer splitting 4 coins,
and he'll never be able to get more than 2 coins out of that.
So the best action for him is to accept A's initial offer, leaving A with
If A instead offers a
6-2 split, B could refuse and offer the 4 coins in an even split. So that would be suboptimal for A.
If it gets to the stage when there's only 1 coin left, then it's B turn to make an offer, so
If it gets to the stage when there are 2 coins left, then it's A's turn, but because of the above,
whatever offer A makes, B will simply refuse it and then win. So A loses.
If it gets to the stage when there are 4 coins left, then it's B's turn, and because of the above,
A should accept any offer which gives A any coins at all. (If A refuses the offer, then we're down to the 2-coin case and A loses.) So B wins, simply by making the offer "all four coins to me, none to you".
At the start, when there are 8 coins left and it's A's turn to make an offer,
it doesn't matter what offer A makes, because B can refuse it and win by the above strategy.
A makes an offer for 8 coins. B refuses it and makes a non-offer for 4 coins (the "I take them all" kind of offer). A refuses it and makes an offer for 2 coins. B refuses it, makes a final offer for 1 coin, and takes it.
If A is not sure about B strategy/ goals/rationality, and we assume A cannot communicate with B (apart by doing moves in-game).
Let's first note one important thing:
B is in advantage position, no matter what A does, B can finally make the gain for A equal to 0
Now assuming both players are humans and does not try to just win but to gain maximum coins
I would agree with other answer that A should offer 4-4 so it results in best for both
However B is still in a winning positions, and if winning does count more than earning money
than A cannot afford to offer to B less than 5 coins (of course this is not states in Original post, so it is open to interpretation). The reason for that is that B cannot gain anyway more than 4 coins in his round so a offer of 5 coins is automatically accepted.
But why should B rejects 4-4 and pretends a 3-5?
Again this depends on interpretation: but if winning is more important than coins number then a no-winning position will not be accepted by B that starts in a winning position.
Assumes B rejects 4-4
then B could offer A 1-3 which grants victory to B and gives enough coins to A to accept, in the attempt to still maximize the gain. (if B offers 0-4 then A could just refuse to reduce gain of B to 0-1)
Can we assume there are repeated games?
If A offers 4-4, then B could reject it and offering a 1-3 which in the end cause B to win with triple the coins, A could reject that, but a 0-1 victory means that B has "infinite times more coins than A" (1/0 is infinite) Infact the best offer is offering at start 3-5 that on the long run keeps the coins gained by both players the same on average, but it is at same time the strategy that allow one of them to gain more coins and hence win (they are still players afterall). Also is the choice that has less assumptions possible on what's B strategy. And it is also the choice that gives the winner the minor possible proportiong compared to the loser.