I need some help with this puzzle

[space-time]

I can't figure this one out. Anyone? Thanks.

Six (M=6) desperate newly proclaimed laser pirates, kicked out of a corrupt lab because of too honest laser profiling
results, have obtained “as a compensation” N=30 beam profilers and have to divide up the loot. The pirates are all
extremely intelligent, greedy and selfish. They value goods in this priority order:

1. Their career

2. Getting beam profilers

The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if at least half the crew or
more go “Aye”, the loot is divided as proposed, as no pirate would be willing to take on the captain without superior
force on their side.

If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all
pirates will turn against him and make him lose the career forever (walk the plank), using past results of dishonest
beam profiling against him. The pirates then start over again with the next senior pirate as captain.

What is the maximum number of beam profilers the captain can keep without risking his career?

[mikebw]

Are you getting fired?

[bik]

I guess I'm too old. I don't understand a "puzzle" that is based on human traits (and variables) like greed, loyalty, and courage.

That said, of the captain is safe by getting two other pirates to side with him, I'll say 10 each for the captain and two coconspirators, and zero for the other three.

[space-time]

I'll say 10 each for the captain and two coconspirators, and zero for the other three.

can it be that simple?

[wowzer]

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! :-)

[gabester]

Where on earth do you find a puzzle as bizarre as this?

I'm going to guess the captain, wanting to keep his career first and foremost, offers two of the crew 15 "beam profilers" each to go along with his plan. But it really makes no sense... can this be rephrased in a more conventional way?

However, that solution doesn't make sense because the 3 who get nothing would just badmouth the captain and the other two, thus destroying their careers... so I'm going to go against my initial answer and suggest the captain keeps none and splits the beam profilers evenly distributing 6 amongst the other 5, to buy everyone's silence.

This puzzle just seems to be missing an element.

Or rather, maybe the missing element is the question - it's not how it's split overall, it's how many the CAPTAIN can keep. I'm assuming the agreement made by the pirates is binding, i.e. the captain can't get 2 guys to agree to take 15 each then the captain just keeps 30 himself?

In either case I think the captain ends up keeping 7. That's my final answer. I can't explain why.
g=

[gabester]

wowzer wrote:
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.

This makes sense when you assume there is a RANK (although the numbers are effectively useless meaningless...)

But I am assuming that unless a pirate gets SOMETHING he will work to undermine the group's career (after all, he has nothing in his possession indicating he was part of the corrupt lab.) So I would think everyone might need to keep at least 1 laster to buy their silence.

Still, my head hurts, this problem is silly, and I'm going to do something else now. Wowzer, I wish you'd posted 20 minutes earlier!
g=

[bik]

Now, Wowzer has way too much time on his hands, but his answer seems logical.

The problem I have with the puzzle, as I said before, is the humans elements.

The question presented is, "what is the maximum number of beam profilers the captain can keep without risking his career?" Risk, by definition, is a variable. In a purely logical exercise, how is risk assessed?

It would seem to me that the captain's career is at some level of risk if he pisses off more than three other pirates.

If I was one of the the two pirates getting offered only 1 each while the captain keeps 28, I would be just as pissed as the guys getting none.

I can see how Wowzer's scenario would be the correct answer, but it seems to require six ROBOT pirates with all logic and no emotion.

But he's probably right.

[space-time]

Where on earth do you find a puzzle as bizarre as this?

it's a newsletter I get every two months, and if you get the right answer, you are entered into a sweepstake to win a certain tablet (yes, w/retina display). I usually try to solve it myself, but this one seems either very simple or very hard. I am not sure which way to go. I haven't spent much time on it, will give it some thought tonight.

[wowzer]

If you win the retinal display, drop a picture to me with whatever you have on it that looks cool...or you could give me the display. :-)