Pirate Game - A Variation [solutions]

A forum for good logic/math puzzles.

Moderators: jestingrabbit, Moderators General, Prelates

Posts: 225
Joined: Tue Mar 29, 2011 1:41 pm UTC

Pirate Game - A Variation [solutions]

Postby sfwc » Thu Apr 06, 2017 7:19 pm UTC

Original thread

To help distinguish them grammatically, I’m assuming that pirates A, C and E are male but pirates B and D are female.

For a given pirate p, I’ll use the phrase `scenario p’ to describe what would happen if, in the final voting, pirate p (and therefore also all pirates of higher rank) were killed. So for example scenario C is what the final outcome would be if the proposals of pirates A, B and C were all rejected. It is clear that this scenario only depends on the proposals of the lower ranked pirates (in this case D and E). Furthermore, in deciding what his proposal should be, pirate C only needs to take into account scenario C, rather than details of exactly what pirates D and E proposed. Then scenario B is completely determined by C’s proposal and scenario C in the following way: if at least half the pirates are better off according to C’s proposal than scenario C then scenario B will just be C’s proposal, otherwise scenario B will be the same as scenario C.

Now let’s consider what pirate A should do. He wishes to survive, and so must ensure that at least two other pirates are better off according to his proposal than according to scenario A. Subject to this, he wishes to get as much gold as possible. So he should pick two of the pirates who are worst off according to scenario A and offer each of them one more coin than in scenario A, taking the rest for himself. He can always do this, since the gold obtained by the two worst off pirates according to scenario A can total at most 500 coins, so he has to pay out at most 502 coins.

On the basis of this, we can see that pirate A’s proposal will always be accepted, and we know exactly how his proposal is determined by scenario A. Let’s now turn our attention to pirate B. She is confident that she will survive, since she knows that pirate A’s proposal will be accepted. So she just wants to maximise how much gold she gets. She can do this by using her proposal to tailor scenario A in such a way that pirate A will offer her as much gold as possible. This means that she would like to be one of the two pirates who are worst off in scenario A, or to put it another way she would like there to be at least two other pirates who do at least as well as her according to scenario A.

Ideally, she should hope that in scenario A she will get exactly 332 coins, whilst 2 other pirates get at least 333. That way, pirate A will offer her 333 coins in his proposal (if she gets 333 coins in scenario A then there will be at most one other pirate who is strictly better off than her in that scenario and so she gets at best a 1 in 2 chance of being offered money in pirate A’s proposal). To achieve this, she certainly needs to survive in scenario A and so she must make a proposal which will be accepted. She can certainly do this with a proposal of the kind stated above: since there are only 1000 coins, there is some pirate who gets at most 333 coins in scenario B, and so pirate B could, for example offer that pirate and some other unfortunate pirate each 334 coins and herself 332 coins.

In summary, pirate B will choose at random amongst proposals satisfying the following conditions:
- she gets 332 coins
- 2 other pirates (who I’ll call the unlucky pirates) each get at least 333 coins
- at least one pirate other than her is better off than in scenario B

I’ll call the other pirate, the one to whom pirate B offers less than 300 coins, the lucky pirate.

Then pirate A will propose to give pirate B 333 coins and the lucky pirate one more coin than in B’s proposal and to keep the remaining gold for himself. Pirates A and B and the lucky pirate will vote in favour of this proposal and it will be accepted.

In light of all this, what sort of proposal should pirate C make? Obviously, he would like to end up as the lucky pirate. But he cannot guarantee this, because it is possible that he will end up being the `some other unfortunate pirate’ from a couple of paragraphs ago. So the best he can do is to maximise the chances that he will be the lucky pirate.

Let’s analyse this more carefully. We know that there are no more than 6 possible proposals in which he is the lucky pirate, corresponding to the 6 ways of distributing the extra 1000 - 332 - 2*333 = 2 coins amongst 3 pirates. In three of these he is offered nothing, and so ends up with just a single coin in the final distribution. In two of them he is offered one coin and so ends up with two, and in one of them he is offered 2 coins and so ends up with 3. On the other hand, whatever scenario B might be, there are always at least 6 possible proposals for pirate B in which pirate C is not the lucky pirate. If some pirate gets at most 332 coins in scenario B then this is clear, and it can also be checked that there are at least 8 such proposals when two of the pirates get 333 coins and the other one gets 334 in scenario B. So the expected takings of pirate B are at most (3*1 + 2*2 + 1*3)/(6 + 6) = 5/6 of a coin.

In order to maximise his takings, he must keep the number of possible proposals for pirate B in which he is not the lucky pirate down to 6. That is, he should aim to make scenario B such that he and at least one other pirate (who I’ll call the successful pirate) each get at least 335 coins and the remaining pirate gets at least 2 coins. He can always achieve this by, for example, offering 335 coins to himself, 663 coins to whichever of pirates D and E gets least in scenario C, and the other 2 coins to the remaining pirate.

After he has made a proposal of this kind, it is equally likely that he or the successful pirate will become the lucky pirate when pirate B comes to make her proposal. The other pirate has no chance of becoming the lucky pirate. So it is imperative to pirate D that she should become the successful pirate. She can guarantee this by proposing to give at least 663 coins to pirate E and to take the rest herself.

We now have a complete understanding of what will happen. Pirate E will propose to take all the gold. Then pirate D will choose a random number from 663 to 1000 and propose to give that many coins to pirate E, taking the rest herself. Pirate C will then pick a random proposal in which he and pirate D each get at least 335 coins and pirate E gets at least 2, and will make that proposal. Pirate B will choose pirate C or pirate D at random to be the lucky pirate, then pick a random proposal in which she gets exactly 332 coins and each of the unlucky pirates gets at least 333. Finally, pirate A will propose to give pirate B 333 coins and the lucky pirate one more coin than in B’s proposal and to take the rest for himself, and this proposal will be accepted.

So each of the distributions 664:333:3:0:0 and 664:333:0:3:0 has a 1 in 12 chance of occurring (here the amounts given to the pirates are listed in order of decreasing seniority from A to E). The distributions 665:333:2:0:0 and 665:333:0:2:0 each have a 1 in 6 chance of occurring and the distributions 666:333:1:0:0 and 666:333:0:1:0 each have a 1 in 4 chance of occurring. The expected takings of the pirates are, in order from A to E, 665 and a third, 333, 5/6, 5/6 and 0.

Return to “Logic Puzzles”

Who is online

Users browsing this forum: No registered users and 10 guests