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5 pirates namely A, B, C, D and E (really interesting names) found a chest containing 1000 diamonds...

A being the leader is given the task of distribution of the diamonds...A's proposal is voted by the other pirates...if majority of them are against his proposition he will be shot dead and the task of distribution falls upon B and so on...

The priority order of every pirate is as follows-

1. To live

2. To get as many diamonds as possible

3. To get as many of their colleagues dead

Your Task- How should A distribute the diamonds such that he ensures his survival and also manages to get as many diamonds as he can..

I hope i was clear..

Oh Note: If the votes turn out to be 50-50 the proposition is still taken as accepted i.e. the pirate wont be shot...

Have fun ^_^

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Since his first priority is to live (and this trumps all other priorities), he should:

Take no diamonds for himself and distribute the remainder to 2 of the 4 pirates (doesn't matter which). 50% of the diamonds is the best the other two will possibly get, so 2 or the 4 will accept the proposal and the other 2 will reject it. 50% of the vote is positive so it will pass.

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I think it's better to go from the end to the beginning. If it gets to D's turn, he's dead, cause E will vote no and he's 100% of the other pirates, which means D will vote yes on C's turn, even if he gets nothing, which means C won't give anything to D or E, so B can buy both their votes giving each a diamond. Since C will get nothing from B, A can buy him with one diamond. D and E will get one diamond from B, so they would rather see A die, so A must give either one 2 diamonds. Let me know if I'm missing something.

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I think it's better to go from the end to the beginning. If it gets to D's turn, he's dead, cause E will vote no and he's 100% of the other pirates, which means D will vote yes on C's turn, even if he gets nothing, which means C won't give anything to D or E, so B can buy both their votes giving each a diamond. Since C will get nothing from B, A can buy him with one diamond. D and E will get one diamond from B, so they would rather see A die, so A must give either one 2 diamonds. Let me know if I'm missing something.

I see what you are getting at..

I think you're right.

Starting Backwards:

E - Obviously if it gets here he gets everything. He will always vote no and people will die (priority #3).

D - NO chance, on his turn E will always vote NO (agreed). On all other turns he will vote yes if he gets SOMETHING.

C - D will always vote yes, as long as he gets SOMETHING, because he will die if it gets to the next turn. E will vote no because he will get the jackpot if it fails. C's best option is to keep 999 and give D 1

B - B need two votes. D will vote yes if he gets 1 diamond. E will always vote no. C will vote yes if he gets ANYTHING. So B should give D and C 1 diamond and keep 998.

A - needs two votes, if C and D get 1 diamond they will vote yes. So he should keep 998 and give C & D one.

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I see what you are getting at..

I think you're right.

Starting Backwards:

E - Obviously if it gets here he gets everything. He will always vote no and people will die (priority #3).

D - NO chance, on his turn E will always vote NO (agreed). On all other turns he will vote yes if he gets SOMETHING.

C - D will always vote yes, as long as he gets SOMETHING, because he will die if it gets to the next turn. E will vote no because he will get the jackpot if it fails. C's best option is to keep 999 and give D 1

B - B need two votes. D will vote yes if he gets 1 diamond. E will always vote no. C will vote yes if he gets ANYTHING. So B should give D and C 1 diamond and keep 998.

A - needs two votes, if C and D get 1 diamond they will vote yes. So he should keep 998 and give C & D one.

I don't think C needs to give D anything, because if C dies, D dies, so D will vote yes anyway to stay alive. E may vote yes if he gets something, since he know D will vote yes, that's arguable, but irrelevant. C will give E nothing because he doesn't need his vote. So on B's turn, D and E know that if he loses, they get nothing, so B gives each of them 1 diamond and they'll vote yes. He can't buy C because C gets everything if he loses, so he won't bother. So on A's turn, C know that he'll get nothing if A loses, so A gives him 1 and he votes yes. D and E know that they'll get 1 on B's turn, so A needs to give 2 to get their vote. But since A already has C's vote, he doesn't need both D's and E's votes, only one of them, so he chooses one and gives him 2 diamonds, keeping the other 997. Again, A can't buy B unless he gives him 999 diamonds, because B gets 998 if A loses.

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A needs three votes including his vote..so he need only two more votes to get max. diamonds..

Here's the strategy..

A gives B nothing because B will anyway not vote in favor of A..

C will accept anything A gives him because he knows he'll get nothing if B becomes the leader..so A gives C one diamond

Now only one vote left for A..he chooses D and gives him 2 diamonds, D agrees because he'll get only one diamond if the leaders are either B or C..

Now A has three votes and he gives E nothing

Thus, A gives 3 and keeps 997 diamonds to himself..

I hope I'm correct..

Nice puzzle

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Split it three ways, giving B and C each 334 for good measure, and himself 332. I say B and C because they'll be likely to accept, considering after A is dead anything B or C decides is easily vetoed by D and E. So actually, I'd argue for A gets 998, and B and C each get 1. If this becomes vetoed, B would have no choice but to split the diamonds 500-500 between either 2 remaining, as with any other deal the remaining pirates would be able to hold out for more.

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Imagine only 2 pirates, A and B. Since A only needs 50% of the votes, he'll take all the diamonds for himself and vote for himself. A:1000, B:0.

Now take 3 pirates, A,B,C. A needs one extra vote to live. C knows that if A dies, he'll get nothing (see above with 2 pirates, and remember that if A dies, C in this case becomes B above). So C will vote for A if A gives him anything. Therefore, A gives C 1 diamond and keeps the rest. A:999, B:0, C:1.

Take 4 pirates, A,B,C,D. A needs one extra vote to live. C knows that if A dies, he'll get nothing (see above with 3 pirates, and remember that if A dies, C in this case becomes B above). So C will vote for A if A gives him anything. A:999, B:0, C:1, D:0.

Now 5 pirates, A,B,C,D,E, like the puzzle asks for. A now needs two votes to live. C and E both know that if A dies, they get nothing (see above with 4 pirates, and remember that if A dies, C and E in this case become B and D above, respectively). So C and E will vote for A if A gives them anything. A:998, B:0, C:1, D:0, E:1.

Edited by sacohen0326
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