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JocK
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10 bananas for 3 gorillas
« on: Dec 12th, 2004, 2:19pm » |
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Pleasing 3 gorillas with 10 bananas
You are one of the several individuals who are allowed once a day to enter the gorilla cave in the zoo. Like any of the other visitors, each time you enter, you have to distribute 10 bananas over three bags, and put the three bags in a corner of the cave. Subsequently you watch the same ritual that happens each day and with each visitor:
1) the leader of the pack, a huge black gorilla, approaches the three bags, looks in all three bags, and takes the bag that contains the most bananas. If the number of bananas in that bag exceeds what he got from the previous visitor, the gorilla is happy and throws a gold coin at you. If the number of bananas is the same, the gorilla will be in neutral mode: nothing happens and the gorilla disappears. However, if the number of bananas is less than what he received from the previous visitor, the gorilla gets angry and approaches you with clear intentions. Only when you throw a gold coin at him he will disappear and won’t attack you.
2) These events repeat itself with the next gorilla, a large brown specimen: he approaches the two remaining bags, looks in both bags, and takes the one with the most bananas. If the number of bananas in that bag exceeds the number he got from the previous visitor, the gorilla is happy and throws a gold coin at you. If the number of bananas is the same, nothing happens and the gorilla disappears. However, if the number of bananas is less than what he received from the previous visitor, the gorilla gets angry and you have to throw a gold coin at him to prevent him from attacking you.
3) The last gorilla, a big elderly member of the group approaches the last bag, takes this bag and counts the number of bananas in it. If the number of bananas in that bag exceeds the number he got from the previous visitor, he throws a gold coin at you. If the number of bananas is the same the gorilla disappears. However, if the number of bananas is less than what he received from the previous visitor, the gorilla gets angry and you have to throw a gold coin at him to prevent him from attacking you.
Obviously, in this daily routine your intention is to gain as many gold coins as you can, and lose as few as possible. But you can’t please all three gorillas at the same time. You don't know the other cave visitors and have no means to communicate with them. Also, the order in which the various visitirs are allowed to enter the cave is completely random.
What is your strategy? How will you distribute the 10 bananas over the three bags?
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xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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towr
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Re: 10 bananas for 3 gorillas
« Reply #1 on: Dec 12th, 2004, 2:57pm » |
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That's an interesting problem..
Quote:| You don't know the other cave visitors and have no means to communicate with them |
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Does that mean the other visitors can't come back each day?
Otherwise you communicate to them through the gorillas' behaviour (How they react to your offering will say something, though not much, about the offerings of the previous person; and about yours to the next.)
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JocK
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Re: 10 bananas for 3 gorillas
« Reply #2 on: Dec 12th, 2004, 5:07pm » |
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'Communicating' via variations in the offerings to the gorillas is OK, but remember: there are many cave visitors each day, and they do come in in random order. Each day some visitors might decide to stop with this daily routine and be replaced by others...
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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towr
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Re: 10 bananas for 3 gorillas
« Reply #3 on: Dec 13th, 2004, 2:12am » |
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{10, 0, 0} seems to be the worst thing you can do, as it doesn't beat any of the other combinations.
If the other people played randomly, {4,4,2} would be best. But for all intends and purposes, without further communication, it's seems like playing rock-paper-scissors.
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rmsgrey
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Re: 10 bananas for 3 gorillas
« Reply #4 on: Dec 13th, 2004, 2:16pm » |
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The only thing that beats {4,4,2} is {5,5,0}, but that loses against a random population.
Without putting in rather more work than I plan to, I'm not going to be able to find an equilibrium strategy - one where the expected return from any other strategy playing next is always non-positive - in fact, I'm not even able to convince myself there is one.
I'd expect that, unless you have a reasonable chance of being the next visitor after yourself, assuming all visitors to be rational and greedy, that the expected return would be 0 - each player greedily picks an "optimal" (non-losing) strategy, but, since, for each loop with positive expected gain among all players, there's the reverse loop with an equal expected loss, the net expected return amongst all players is zero, so if any strategy has a positive expected gain, there must be at least one strategy with an expected loss, which contradicts our assumption that no player chooses a losing strategy... Note that the expected value of a strategy is calculated without knowing the actual strategies chosen by others, so depending on the specific choices, one player may have an advantage in the actual field.
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towr
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Re: 10 bananas for 3 gorillas
« Reply #5 on: Dec 13th, 2004, 2:31pm » |
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It's interesting to note that it's only possible to either gain one coin, loose one coin, or neither loose nor gain. And so if people could communicate it doesn't matter if they're greedy or generous, since there is a strategy in which everyone always gains 1 if they know what the previous person did.
The trick is of course to find such a situation without that information. Which due to symmetry can't exist unless there is more information, and a low turnover in the visitor population.
For instance if there is one visitor every hour, you can eventually synchronise behaviour (even without explicit communication).
For instance you could work towards a repeating sequence of {4,3,3}, {5,4,1}, {6,2,2} (or a similar cycle) It might take days or weeks to reach concensus without communication (and every newcomer would screw up every now and then before grasping the emerged norm), but I think it would work eventually.
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rmsgrey
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Re: 10 bananas for 3 gorillas
« Reply #6 on: Dec 13th, 2004, 2:45pm » |
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So what happens when someone plays {6,4,0} for second place in that cycle? He wins consistently, but the third place guy is now breaking even instead of gaining... I guess things will eventually settle - anyone who's consistently earning has no reason to rock the boat - unless they turn out to be a jerk who values making others miserable...
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towr
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Re: 10 bananas for 3 gorillas
« Reply #7 on: Dec 13th, 2004, 3:19pm » |
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There may be a better cycle. But otherwise a bastard is pretty similar to the newbie, he screws things up a little every now and then, but if order changes randomly each day no one should suffer them too often (unless there are too many bastards and newbs).
And since you get the maximum gain either way (once there is some semblance of concensus), there is no incentive not to cooperate. Whereas everyone cooperating helps everyone win.
I suppose from a game theory p.o.v. that's not too good an argument. There may not be a reason not to further the common good if there is no price, but that neither gives a reason to do do it.
It might be interesting to see what a genetic algorithm would make of this. Either for the original problem, or the variation with indexed time slots.
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rmsgrey
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Re: 10 bananas for 3 gorillas
« Reply #8 on: Dec 13th, 2004, 3:36pm » |
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An interesting side question is how long it takes to find a cycle if you try looking for one - it seems fair enough to assume that each individual repeats their behaviour if they gain from it, and changes if they lose - so the question is how complex "behaviour" can be - if the same person visits at different points in the cycle, do they recognise the difference? And what frequencies can they pick up? (If they're looking for a 4-cycle or 5-cycle, then a 3-cycle is going to pass them by completely)
And, of course, how fragile the cycle-forming process is - if you have 1 clueless schmuck who always plays {10,0,0} - or a scheming jerk who always plays {4,4,2} - can a near-cycle form? (assuming they don't always visit at the same point in the hypothetical cycle) And how much longer does it take to find?
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Re: 10 bananas for 3 gorillas
« Reply #9 on: Dec 14th, 2004, 4:04pm » |
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I'm a long-time forum lurker, and was interested by the concept of this puzzle. I think that it may be changed into a potentially more interesting one as follows. Let me know what you guys think.
The same gorillas with the same rules are present. Rather than random visitors, there are 2 teams of N people who alternate turns visiting the gorillas. Each team has 10*N bananas total, and each team is trying to maximize its profit. Each team member visits the gorillas once and distributes an arbitrary number of their team's remaining bananas to the 3 gorillas.
Team A's first member visits first, and Team B's last member visits last. After each person's visit, both teams observe the outcome for all 3 gorillas - whether they give or take a coin (or neither).
For the boundary case of Team A's first player, the non-existent previous visitor can be assumed to have given (0,0,0).
What is each team's best strategy to maximize their worst case profit? Does either team have a stragety which is guaranteed to yield a positive profit?
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Hippo
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Re: 10 bananas for 3 gorillas
« Reply #10 on: Mar 19th, 2007, 6:00am » |
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Denote bag choices by triples in decreasing order. Than if your triple agrees in a cordinate with previous one, you gain/loose nothing.
Some triples cannot gain at all ... (10,0,0),(9,1,0),(8,2,0).
Graph of gaining/loosing does not make some vertex superior to others providing others are usefull. But if you choose (4,4,2),(6,4,0),(6,2,2) each with probability 1/3. You don't loose. Against this strategy choosing (6,3,1), (5,4,1) or (5,3,2) is bad play ... I will continue later..
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Hippo
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Re: 10 bananas for 3 gorillas
« Reply #11 on: Mar 19th, 2007, 7:48am » |
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Let us continue ... (8,1,1) looses to this strategy in long run, too.
(7,3,0), (7,2,1), (5,5,0), and (4,3,3) had equal chances against it, but any their combination is beaten by combined strategy of (5,3,2), (6,3,1) and (5,4,1). This is why I will play 1/3((4,4,2),(6,4,0),(6,2,2)).
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srn437
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Re: 10 bananas for 3 gorillas
« Reply #12 on: Sep 1st, 2007, 10:54pm » |
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Couldn't you have modified the last post? It's impossible to use a distribution that works for all other distributions having been done previously. If it was 10, 0, 0, you need to also do that or you'll get beaten by the first one. If you do that, the other person(the previous one)could have done 8,1,1 and you'll get beaten by 2 or 3. It can't be done.
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