09-18-2012, 04:48 PM
My thoughts...
Rules:
1) Captain must ensure that he gets the votes needed to remain as captain. Therefore, in a 50:50 situation (i.e. the other pirates are no better, no worse), he must guarantee that the voter will vote for him (i.e. offer 1 more laser than the scenario if he loses).
2) If the captain loses, then the next scenario occurs with 1 less pirate (i.e. 6 becomes 5, and etc).
3) The captain must maximize as many lasers as possible.
Nomenclature:
The #1 pirate is the current captain, #2 is the next in command, and etc to #6 as the least senior pirate. Lets work backwards to get the most lasers for the #1 pirate (aka captain) and avoid having to give equal share (which means less for the captain).
This following logic is to figure out what is the minimum number of lasers for pirate #6.
The #6 pirate will never be alone. If you offer him at least 1 laser, he'll most likely take it because he can never vote anyone off and expect to get more than 1 laser.
If there are 2 pirates remaining, then the #5 pirate could take all the booty, as he wouldn't have to share with #6. In this scenario, #6 would receive zero lasers.
If there are 3 pirates left, then the #4 pirate would have to contend with the greed of #5. If he offered 1 laser to #6, then #6 must agree. In this case, though, #5 gets zero lasers.
If there are 4 pirates left, then the #3 pirate could never get #4 to vote with him. However, if you give #5 1 laser, he must agree with you as he'd get zero lasers with a 3 pirate scenario. Thus, in a 4 pirate scenario, #4 and #6 never gets any lasers (important for later).
Now, how do you maximize the number of lasers for the captain with 6 pirates?
Lets look at what happens with 5 pirates first:
The #2 pirate would offer #4 one laser and #6 one laser. Why would #4 and #6 take this? Because they know with a 4 pirate scenario, #3 would offer #5 one laser and win the tied vote.
Take note: in this case, the #3 and #5 pirates would get no lasers.
Thus, with 6 pirates, the captain would offer #3 and #5 one laser each, as in a 5 pirate scenario, they would each get zero lasers. Thus, the captain would receive 28 lasers, with #3 and #5 receiving 1 laser each.
Greed is Good! :-)
Rules:
1) Captain must ensure that he gets the votes needed to remain as captain. Therefore, in a 50:50 situation (i.e. the other pirates are no better, no worse), he must guarantee that the voter will vote for him (i.e. offer 1 more laser than the scenario if he loses).
2) If the captain loses, then the next scenario occurs with 1 less pirate (i.e. 6 becomes 5, and etc).
3) The captain must maximize as many lasers as possible.
Nomenclature:
The #1 pirate is the current captain, #2 is the next in command, and etc to #6 as the least senior pirate. Lets work backwards to get the most lasers for the #1 pirate (aka captain) and avoid having to give equal share (which means less for the captain).
This following logic is to figure out what is the minimum number of lasers for pirate #6.
The #6 pirate will never be alone. If you offer him at least 1 laser, he'll most likely take it because he can never vote anyone off and expect to get more than 1 laser.
If there are 2 pirates remaining, then the #5 pirate could take all the booty, as he wouldn't have to share with #6. In this scenario, #6 would receive zero lasers.
If there are 3 pirates left, then the #4 pirate would have to contend with the greed of #5. If he offered 1 laser to #6, then #6 must agree. In this case, though, #5 gets zero lasers.
If there are 4 pirates left, then the #3 pirate could never get #4 to vote with him. However, if you give #5 1 laser, he must agree with you as he'd get zero lasers with a 3 pirate scenario. Thus, in a 4 pirate scenario, #4 and #6 never gets any lasers (important for later).
Now, how do you maximize the number of lasers for the captain with 6 pirates?
Lets look at what happens with 5 pirates first:
The #2 pirate would offer #4 one laser and #6 one laser. Why would #4 and #6 take this? Because they know with a 4 pirate scenario, #3 would offer #5 one laser and win the tied vote.
Take note: in this case, the #3 and #5 pirates would get no lasers.
Thus, with 6 pirates, the captain would offer #3 and #5 one laser each, as in a 5 pirate scenario, they would each get zero lasers. Thus, the captain would receive 28 lasers, with #3 and #5 receiving 1 laser each.
Greed is Good! :-)