Jonny
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- Joined
- Sep 8, 2009
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He should offer one coin each to 1 & 3 and offer nothing to the others.
Incorrect.
He should offer one coin each to 1 & 3 and offer nothing to the others.
Incorrect.
In a Pirate’s Game of N players–and gold pieces outnumber players–if the players are labeled 1, 2 … N, in order of of first to last proposer, player 1 should make a nominal offer to the odd-numbered players 3, 5, 7, etc., and keep the remaining wealth for himself.
Okay, I would light the first rope on both ends, that means it would burn down in 30 minutes. Then I cut the second rope in two and do the same thing or light it in the middle, that's another 15 minutes, giving me a total of 45 minutes.
This is not an answer, this is a link to a site. Your previous answer was incorrect, and of this I am certain. Try again.
Correct.Right. But the 30 minute thing should still work despite the irregular density.
Can I split a rope half length?
Actually it's a fairly easy puzzle to solve.
The link shows the logic of the puzzle. Granted, it does not serve you the answer on a plate - but I assumed for a brief moment that you are capable of making simple substitutions.
I apologise, Johhnyboy. I got that wrong.
A man has a lighter and two lengths of robe with varying densities. The only thing the man knows about each length of rope is that if he lights an end of a rope, it will burn for exactly one hour; however, it may be that 90% of a rope burns up in 1 minute, and it takes 59 minutes for the remaining section of rope to burn (this is a product of the varying densities).
How can this man accurately measure 45 minutes?
You do not need to apologize, and have yet to solve the problem. The solution you gave above is incorrect, and from my brief glance at the website, it isn't applicable to this problem.
Try presenting the correct answer with your reasoning, in your own words.
Edit:
Truth be told, I find your initial incorrect answer and your citing of this website to be distasteful, when taken together. I cannot help but assume you care more about being perceived as intelligent and right than having fun thinking about the problems.
http://en.wikipedia.org/wiki/Pirate_gameIt might be expected intuitively that Pirate A will have to allocate little if any to himself for fear of being voted off so that there are fewer pirates to share between. However, this is as far from the theoretical result as is possible.
This is apparent if we work backwards: if all except D and E have been thrown overboard, D proposes 100 for himself and 0 for E. He has the casting vote, and so this is the allocation.
If there are three left (C, D and E) C knows that D will offer E 0 in the next round; therefore, C has to offer E 1 coin in this round to make E vote with him, and get his allocation through. Therefore, when only three are left the allocation is C:99, D:0, E:1.
If B, C, D and E remain, B knows this when he makes his decision. To avoid being thrown overboard, he can simply offer 1 to D. Because he has the casting vote, the support only by D is sufficient. Thus he proposes B:99, C:0, D:1, E:0. One might consider proposing B:99, C:0, D:0, E:1, as E knows he won't get more, if any, if he throws B overboard. But, as each pirate is eager to throw each other overboard, E would prefer to kill B, to get the same amount of gold from C.
Assuming A knows all these things, he can count on C and E's support for the following allocation, which is the final solution:
* A: 98 coins
* B: 0 coins
* C: 1 coin
* D: 0 coins
* E: 1 coin[1]
Also, A:98, B:0, C:0, D:1, E:1 or other variants are not good enough, as D would rather throw A overboard to get the same amount of gold from B.
Did you see that I edited my question? I obviously meant from one end to another, assuming that the density would vary over the length of the rope but not over its width.
(English isn't my native language, I was searching for the English word for längsseits)
Well, if I thought we were going to cast our feelings into words, I'd have quoted the Song of Solomon.
And what you find distastful is neither here nor there.
Before you post some old hat game theory, check out the possibility that there may be answers you cannot or will not understand.
Or check out wiki...
http://en.wikipedia.org/wiki/Pirate_game
Sorry - it's not my "words".
Have a good day.
Suppose there exist 1000 gold coins which are to be split up among ï¬ve pirates: 1,2,3,4, and 5 in order of rank. We assume that the coins must remain wholly intact and that the pirates have the following characteristics: a pirate is inï¬nitely knowledgeable; a pirate values his own life above wealth and above his need to kill, and will thus always vote for his own proposal; a pirate will always choose a greater amount of wealth over killing another pirate; caeteris paribus, a pirate will kill another pirate if given the chance; a pirate is risk neutral.
Starting with the highest numbered pirate, he can make a proposal as to how the coins will be split up. This proposal can either be accepted or the pirate making the proposal is killed. A proposal is accepted if and only if a majority of the pirates accept it. If a proposal is accepted the coins are split up in accordance with the proposal. If a proposal is rejected and the pirate making the proposal is killed, the next ranking pirate makes a proposal, so on and so forth.
What proposal should pirate 5 make?