| StarDrive July 24 2012 11:59. Posts 90 | Profile # |
There are two players, Alice and Bob. There is an initial pool of 100 dollars. Alice proposes an allocation of the dollars (real numbers, not necessarily integers), and Bob can either accept or reject. If he accepts, the allocation stands. If he rejects, half of the money is destroyed and Bob proposes an allocation. If Alice rejects, half the money is again destroyed. This continues until one of them accepts. What allocation should Alice propose?
Example play through of the game:
1) Alice proposes (100 - pi, pi)
2) Bob rejects and proposes (e, 50 - e). Since half the money was destroyed, only 50 remains.
3) Alice rejects and proposes (20, 5). Since half the money was destroyed, only 25 remains.
4) Bob rejects and proposes (6.25, 6.25). Alice accepts and each walks away with 6.25.
My solution that I'm not sure is correct:
+ Show Spoiler +
Let the final allocation of the game be (p, 100 - p). Say Alice proposes (x, 100 - x). If Bob accepts, he gets 100 - x. If he rejects, the game recursively transitions to the exact same game except Bob is the first player and there is half as much money, hence he gets p/2.
So Bob accepts if 100 - x >= p/2. Not sure if this next step is legit. Alice will propose the minimum she can to end the game on the first turn, and her payoff will be x. The game will end and the final allocation will be (x, 100 - x), hence x = p. So 100 - x = x/2 and x = 200/3.
Last edit: 2012-07-24 12:24:06 |
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| Antisocialmunky United States. July 24 2012 12:05. Posts 5070 | Profile Blog # |
Rejecting at 50/50 makes no sense because you can never get more in subsequent rounds.
0_o What is the incentive in that? So the question is how far can you get above 50/50 before they reject out of hand.
You don't have enough math to come up with a good answer since there's no way to gauge likeness of rejection.
The optimal solution is probably around 50:50.
The fun solution is to say "Give me X or I will reject until there is no money".Last edit: 2012-07-24 12:09:50 |
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| StarDrive July 24 2012 12:05. Posts 90 | Profile # |
| But can Alice squeeze more money out of Bob than (50, 50)? I'm not sure of the answer. |
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| publicenemies July 24 2012 12:06. Posts 254 | Profile Blog # |
| I don't know anything about the game theory but I have a question about what you're trying to ask. Do you mean that should Alice propose an amount to benefit her or can she just come up with a random number? |
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| StarDrive July 24 2012 12:07. Posts 90 | Profile # |
| Alice and Bob are both trying to maximize the money they walk away with. |
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| CaptainPlatypus United States. July 24 2012 12:10. Posts 698 | Profile Blog # |
On July 24 2012 12:05 StarDrive wrote:
But can Alice squeeze more money out of Bob than (50, 50)? I'm not sure of the answer.
Yes. The best deal Alice can get is definitely better than 50/50. I'm too lazy to do the math, but I'm betting she can get somewhere between 65 and 70 (assuming perfect play by both participants). Non-iterated, of course. |
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| StarDrive July 24 2012 12:10. Posts 90 | Profile # |
On July 24 2012 12:10 CaptainPlatypus wrote:
Show nested quote +On July 24 2012 12:05 StarDrive wrote:
But can Alice squeeze more money out of Bob than (50, 50)? I'm not sure of the answer.
Yes. The best deal Alice can get is definitely better than 50/50. I'm too lazy to do the math, but I'm betting she can get somewhere between 65 and 70 (assuming perfect play by both participants). Non-iterated, of course.
Yes, what we need is answer and proof. |
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| iTzSnypah United States. July 24 2012 12:11. Posts 1219 | Profile Blog # |
| 66 2/3... that way bob can never have more than 33 1/3.... |
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| StarDrive July 24 2012 12:11. Posts 90 | Profile # |
On July 24 2012 12:11 iTzSnypah wrote:
66 2/3... that way bob can never have more than 33 1/3....
That is what I got, what is your solution?
My solution is not rigorous and I'm not 100% sure it is right.Last edit: 2012-07-24 12:12:23 |
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| MrZentor United States. July 24 2012 12:13. Posts 1541 | Profile # |
| Alice can definitely get away with proposing a 62.5/ 37.5 split. Last edit: 2012-07-24 12:14:48 |
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| caradoc Canada. July 24 2012 12:13. Posts 2892 | Profile Blog # |
Alice should propose the maximum possible for Alice such that Bob could not, under any reasonable allocation in the future, expect to make as much.
But I reject the premise of it outright as it applies to things like this-- If alice proposed 63-37, even though I could not get any more than 37 under any counterproposal scenario (assuming Alice acts rationally), I would reject it cuz Alice has no reason to be so greedy, and I value money falling out of the sky less than exploding the myth of economic rationality as a foundation for theory.
Therefore 50-50. Or 0-100, and I'll give her 50 cuz I'm nice. Or we just say fuck the question and lets go get pizza and beer together and get to know each other better....
Last edit: 2012-07-24 12:14:47 |
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| TyrantPotato Australia. July 24 2012 12:14. Posts 1231 | Profile Blog # |
well assuming both players are perfectly rational, and that alice MUST offer something to bob, alice will offer 99 for herself and 1 for bob. Bob being rational would still see this as a positive as he will have more money then he would other wise, and if he rejects the deal the total surplus from this arrangement will be cut in half. only when the notion of "fairness" and rationality is thrown out the window will bob refuse the offer.
yawn i think that makes sense, never liked nash in economics anyway  |
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 NeMeSiS3 Canada. July 24 2012 12:14. Posts 2969 | Profile Blog # |
| Is this homework or what? Seems randomly placed by a low post user, but the questions cool. To be frank, I have absolutely no idea. |
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| Tewks44 United States. July 24 2012 12:15. Posts 1831 | Profile # |
| I would probably offer 50/50. Sure, there's essentially no risk, but with no risk comes a guaranteed return. |
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| StarDrive July 24 2012 12:16. Posts 90 | Profile # |
On July 24 2012 12:14 NeMeSiS3 wrote:
Is this homework or what? Seems randomly placed by a low post user, but the questions cool. To be frank, I have absolutely no idea.
No, talking about random questions with guys at work and I couldn't remember a question precisely. While trying to formulate it, came up with this by accident.
Actually the original question had 3 people but I wanted to solve the 2 person case first. |
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| kdgns United States. July 24 2012 12:16. Posts 1447 | Profile # |
On July 24 2012 12:14 TyrantPotato wrote:well assuming both players are perfectly rational, and that alice MUST offer something to bob, alice will offer 99 for herself and 1 for bob. Bob being rational would still see this as a positive as he will have more money then he would other wise, and if he rejects the deal the total surplus from this arrangement will be cut in half. only when the notion of "fairness" and rationality is thrown out the window will bob refuse the offer. yawn i think that makes sense, never liked nash in economics anyway
but wouldn't bob just reject and propose 49-1 for the same reason you suggested? |
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| iTzSnypah United States. July 24 2012 12:18. Posts 1219 | Profile Blog # |
On July 24 2012 12:11 StarDrive wrote:
Show nested quote +On July 24 2012 12:11 iTzSnypah wrote:
66 2/3... that way bob can never have more than 33 1/3....
That is what I got, what is your solution? My solution is not rigorous and I'm not 100% sure it is right.
If Bob doesn't accept the offer Alice can just decline the next one unless bob's offer is 1/2 the pot. But bob can't reject the first offer if Bob wants to maximize his money. All Alice has to do is offer 2/3 of the pot every time and she will automatically always walk away with at least half of the remaining pot.
EDIT: I CHANGE MY MIND:
74/26. As bob can never have more money. I reread the problem and realized the burn was half the whole pot, not just alice's offer.Last edit: 2012-07-24 12:23:25 |
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| starfries Canada. July 24 2012 12:18. Posts 2617 | Profile Blog # |
| I think the answer is pi > 100/3. |
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| TyrantPotato Australia. July 24 2012 12:19. Posts 1231 | Profile Blog # |
On July 24 2012 12:16 kdgns wrote:
Show nested quote +On July 24 2012 12:14 TyrantPotato wrote:well assuming both players are perfectly rational, and that alice MUST offer something to bob, alice will offer 99 for herself and 1 for bob. Bob being rational would still see this as a positive as he will have more money then he would other wise, and if he rejects the deal the total surplus from this arrangement will be cut in half. only when the notion of "fairness" and rationality is thrown out the window will bob refuse the offer. yawn i think that makes sense, never liked nash in economics anyway
but wouldn't bob just reject and propose 49-1 for the same reason you suggested?
only if he wasnt rational. because he would be throwing away 50 total utils in surplus.
nash theory if i remember dictates that each person will work to achieve their own maximum benefit AND that of the group (not exact line but its along those lines)
therefore considering complete rational decision making and ceterus paribus bob will accept ANY OFFER first given to him.
the question really lies in whether or not alice is willing to give bob more as an incentive for rationality, but that line of thought then sways from the original question. |
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| StarDrive July 24 2012 12:19. Posts 90 | Profile # |
On July 24 2012 12:18 iTzSnypah wrote:
Show nested quote +On July 24 2012 12:11 StarDrive wrote:
On July 24 2012 12:11 iTzSnypah wrote:
66 2/3... that way bob can never have more than 33 1/3....
That is what I got, what is your solution? My solution is not rigorous and I'm not 100% sure it is right.
If Bob doesn't accept the offer Alice can just decline the next one unless bob's offer is 1/2 the pot. But bob can't reject the first offer if Bob wants to maximize his money. All Alice has to do is offer 2/3 of the pot every time and she will automatically always walk away with at least half of the remaining pot.
Hmmm, not fully convinced by it. I arrived at the same answer but don't know if it's correct. |
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