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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)

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closed as off-topic by LeppyR64, Gamow, Rand al'Thor, Mithrandir, JonMark Perry Dec 4 '16 at 4:49

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    $\begingroup$ Unless you have an inventive solution in mind, I think this belongs on Math SE. $\endgroup$ – boboquack Dec 2 '16 at 19:51
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    $\begingroup$ I've heard a similar riddle, and i'm pretty sure that in this example, that their number one priority is to get as many coins as they can, and their second priority is to make the other person get the least coins they can. $\endgroup$ – Sam Harrington Dec 2 '16 at 20:03
  • $\begingroup$ all of these answers could be different depending on what objectives they each have, so I added that comment above to try to narrow the riddle down for all these puzzlers. $\endgroup$ – Sam Harrington Dec 2 '16 at 20:25
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    $\begingroup$ So the players are trying to maximize their winnings, but are they trying to minimize their opponents winnings, or maximize their opponents winnings or indifferent? If they had to choose between 3 -1 and 3-5, where they get 3 which would they do? $\endgroup$ – Dr Xorile Dec 2 '16 at 22:33
  • $\begingroup$ I agree with @DrXorile. While they are trying to maximize their profit, they are trying to minimize their opponents win? $\endgroup$ – Oray Dec 3 '16 at 9:36
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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)

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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.

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  • $\begingroup$ Thank you for putting your answer in a spoiler block.  But why do you subsequently repeat it, fully visible? $\endgroup$ – Peregrine Rook Dec 4 '16 at 18:26
  • $\begingroup$ I actually can't figure out how to make a spoiler block. @PeregrineRook The person who edited my answer added that $\endgroup$ – Sam Harrington Dec 5 '16 at 17:49
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What about A offering a

5-3 split?

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

5 coins.

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.

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  • $\begingroup$ But B could offer 1-3. Because if A reject to win 1 coin and give 3 to B then he automatically have to do a offer that B will reject because next round B will win 1 coin anyway. So A is not going to let be offer 1 and win 3, because actually A could win already 3 by offering 3-5 in first glance (the scores are in A-B format) $\endgroup$ – GameDeveloper Dec 4 '16 at 0:33
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If it gets to the stage when there's only 1 coin left, then it's B turn to make an offer, so

B wins.

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.

Final answer:

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.

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  • $\begingroup$ If OP was going for the game theory "gotcha" answer, this is the correct one (so +1). Unfortunately the objectives of A and B were poorly defined... $\endgroup$ – monoRed Dec 2 '16 at 20:11
  • $\begingroup$ I think we can at least assume that the players are trying to optimize their payment. It's unclear whether they are indifferent to, maximizing or minimizing their opponent - I know rational people who'd do all three... $\endgroup$ – Dr Xorile Dec 2 '16 at 22:31
  • $\begingroup$ Note I actually upvoted your answer. Just saying because I'm getting downvotes for no apparent reason and without explaination, so I try at least to avoid retail downvotes.. here's the upvote screenshot: imgur.com/a/bY17K $\endgroup$ – GameDeveloper Dec 4 '16 at 3:01
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    $\begingroup$ (1) “You shouldn’t both answer and VTC.”  (2) I believe that you’ve misunderstood the game.  As far as I can tell from the incomplete, unclear problem statement, it’s a quantitative win-win game, and you’ve treated it as a binary (qualitative) win-lose game. $\endgroup$ – Peregrine Rook Dec 4 '16 at 18:07
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    $\begingroup$ @PeregrineRook (1) I know, sorry. At the time of posting the answer, I thought the question was perfectly valid, but after seeing several other answers which could all fit, I decided to agree with the close vote. (2) So I've gathered from the downvotes and from checking the other answers :-) $\endgroup$ – Rand al'Thor Dec 5 '16 at 0:55
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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.

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  • $\begingroup$ Any other offer put B in further advantage because basically B could make the coins winned by A become 0, so it is only in the first roudn that A can win some money (and we assume B is rational, so it is not willing to make A lose all money as long as A makes a good offer) $\endgroup$ – GameDeveloper Dec 4 '16 at 1:22
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    $\begingroup$ You seem to have introduced a notion of "winning" which is (a) not part of the question, at all, and (b) apparently different in your mind from maximizing coins won—that is to say, you seem to think "winning" is more important than "winnings", to the extent that you propose 1-3 is a better result for B than 4-4. That is, bluntly, irrational, and if you're still wondering why people are downvoting your answer I think this is the biggest reason why. $\endgroup$ – Rubio Dec 4 '16 at 7:40

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