The Math Factor Podcast

3 Comments »

1. jlundell said,

   September 3, 2007 at 4:46 pm

   The solution to the proposed version of the problem seems straightforward enough (I await its publication), as does the alternative in which ties lose.

   However (and this is where I must be missing something), I don’t see how the penalty for losing a vote (exclusion vs execution) makes a difference. I initially assumed that there must be a situation where some probability of return would be involved (say a 50% chance of getting all the loot), but I don’t see such a situation arising, since all ambiguous cases appear to be resolved by suitable application of the toady principle.

   (Was there a Labor Day edition?)

2. strauss said,

   September 4, 2007 at 5:27 pm

   There’s only a slight difference between losers being executed and losers being ejected: a pirate who fears execution will gladly sell a vote for 0 gold pieces, but a pirate who fears ejection won’t sell for less than 1 gold piece. But this distinction turns out not to affect to many of the pirates.

   (Next segment will be coming up soon…)

3. B Boom said,

   September 6, 2007 at 12:34 pm

   I’m going to go by the rules that a tie wins, and that you are killed if your proposal isn’t accepted.

   Now, for this problem it’s easiest to work backwards, so I’ll call the last person in line “pirate 1”, the next to last person “pirate 2”, etc.

   An ideal situation for pirate 1 is if he is the last one left, because then he can get all of the loot, but if it gets down to pirate 2, he could also vote to give himself all the loot. Pirate 1 knows this, so he will want to accept a proposal before it gets down to pirate 2. Now, pirate 3 needs to get a majority of the votes, so he will try to make a plan to get pirate one to vote with him. Of course, pirate 1 will vote for a proposal that will give him any loot, because if pirate 3 gets killed, pirate 2 will get all of the loot, so the best thing for pirate 3 to do will be to give pirate 1 a single gold piece and to take the rest for himself. Pirate 2 knows this, so he will try to accept a proposal before it gets down to pirate 3, so pirate 4 should make a proposal to give pirate 2 1 gold and to take the rest.

   As you can see, this becomes the even pirates against the odd pirates, no matter who is making the proposal. Each even pirate knows that if an odd pirate makes a proposal, he won’t get any loot, so all even pirates will vote for any other even pirate’s proposal as long as he gets any loot at all. It’s the same for the odd pirates.

   So, the best thing for pirate 100 (the first in line) to do is to give himself all of the loot, except give 1 gold each to all of the other even pirates.

   Now, the most senior pirate gets to decide what order they stand in, so the best order would obviously be from most to least senior, because then he gets most of the loot

   Ben N
   Minnesota

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