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Hard riddle--for da pros only

Burberry London

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So I was browsing on mathstackexchange (a website to help people with their math problems) since I am a actuarial math student and needed help with some problems. I stumbled into a very interesting riddle.

Here goes:

There are 100 pirates on an island. They just found a treasure consisting of 200 gold coins. Now every pirate is ranked from #1 all the way to #100. The point of this is because the #1 ranked pirate decides how to split the "loot". Obviously every pirate is extremely greedy (they want the most gold for themselves), but they are also extremely smart (they have perfect logic and are able to figure everything out). Now if the #1 ranked pirate decides on a way to split the loot and the other crew member doesn't agree on it (meaning if more than 50% of the pirates disagree with his way of sharing the loot) then we kill the #1 pirate and the #2 becomes #1, #3 becomes #2 so on and so forth. This also means that if the votes are tied then the #1 pirates's decision is put into effect. Also every pirate loves blood meaning that if by killing the #1 pirate they would get the same amount of gold as if they used the #1 pirate's plan to split the gold, then they would much rather kill the #1 pirate.

You are #1 pirate, what would be your decision, so that you keep the most gold for yourself?

Basically you have to do this by starting with 1 pirate. Obviously he would keep 200 to himself.

Now 2 pirates he would also keep 200 for himself since if the votes are 50-50 he wins.

If there's 3 pirates then he would have to keep 199 gold and give 1 to the #3 pirate since #3 pirates knows that if he votes against him, then he would get 0 gold since they would kill the #1 leaving 2 pirates left.

Similarly, when there's 4 pirates, we give 1 gold to #3 and keep the rest since the #3 knows that no matter what, he won't get any gold if #1 dies, so by giving him 1 gold there will be a 50-50 vote which means that #1 will not die.

Now with 5 pirates we give 1 to #3 and 1 to #5 because if #1 dies, then obviously #5 won't get anything (because of the previous example with 4 pirates) neither will #3.

Basically the general rule for this is to give 1 gold to every odd # pirate and keep the rest, it will ensure that you win the vote and you keep as the most gold.
 
How would you know if the other pirates are going to agree with you or not.
 
Heidy
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You said I am pirate #1 and all pirates have perfect logic, I do not have perfect logic. Therefore riddle broken. If I were the first pirate in the proposed scenario that would invalidate the scenario because I would have perfect logic and automatically know the answer to the scenario.
/meta
 
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This answer seems pretty poor. No pirate would be happy with 1 gold.
 
This answer seems pretty poor. No pirate would be happy with 1 gold.

Obviously they won't be happy, but they know that if they vote against pirate #1, then they won't get any gold at all, thus they just have to settle with 1 gold.
 
Chess Master
The odd numbered pirates will get 1 each, with pirate 1 getting the rest.

Why? Use backward induction. Two pirates left (1 and 2) -- 1 will keep everything. Three pirates (1,2,3) -- 1 will want to get 3's vote (because if he doesn't get it, pirate 2 will just kill pirate 1 and keep everything for himself. But 3 knows that he will not get anything if he kills pirate 1, but he is bloodthirsty. Therefore, offering pirate 3 1 gold coin will make pirate 3 vote for pirate 1.

And so on.
 
How about I tell them I'm not really a pirate but a banker that will invest their gold and create a joint account between them so they can watch it grow?
 
How about I tell them I'm not really a pirate but a banker that will invest their gold and create a joint account between them so they can watch it grow?

bankers didn't exist back in 2500BC, so they would think you are a scammer and would kill you
 
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