# Thread: Word problems for fun!

1. 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)

2. Originally Posted by Jonnyboy
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.
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...

It 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.
http://en.wikipedia.org/wiki/Pirate_game

Sorry - it's not my "words".

Have a good day.

3. Originally Posted by Red Herring
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)
Interesting thought. Given that I did not specify, I will accept this answer. However, in the spirit of the original riddle, let us assume that the density varies in all directions, such that the rope cannot be divided in such a way. Have a go at this 'clarified question'.

4. Originally Posted by matmos
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.
This is incorrect. I hope someone else comes along and provides an answer with his or her reasoning, so that you can read what the correct answer is. Me telling you why it is incorrect will amount to me solving it for you. I have written a solution myself, but want to give others a chance first.

5. I light rope A on one end and rope B on both ends. By the time rope B is burned down (i.e. after 30 minutes) there will be exactly 30 minutes left of rope A, no matter how long or short. THEN I light the other end of rope A as well, which reduces the process to 15 minutes.

6. YES!!! How'd you like it?

Edit: I felt the same way afterwards... Riddles like these seem so simple once you know them, but so complicated if you aren't familiar with the process. I remember wracking my brain on this one.

7. I asked for distraction from work and I got it

It is nice because it is elegant. I like puzzles where you need to find an idea rather than do a lot of calculations.

This is a classic some of you might already know (verbatim from an old Columbo episode ):

In a room are several sacks of gold pieces, as many sacks as you like. Each sack contains several of these gold pieces -- again, as many as you like. One sack, however, is full of artificial gold pieces, and they weigh differently. The solid gold pieces weigh, let's say, a pound each. And the artificial pieces weigh, let's say, a pound and an ounce.

Now you have a penny scale. You put the penny in, and you get a card, and that tells you how much the weight of the gold is. But, you only have one penny. You have one reading on the weight.

Which sack has the artificial gold pieces?

8. Let us assume that all sacks have the same number of bars. We take 1 bar from sack 1, 2 bars from sack 2, 3 bars from sack 3, etc. The amount that the scale is off, 1 ounce, 2 ounces, 3 ounces, will indicate the sack with the fake gold.

I liked this one.

Edit: I don't think my solution would work if the number of sacks was greater than the number of bars in each sack. Am I missing something?

9. So the pirate riddle still stands. There was an attempt made by one person, who based his solution on a slightly different version of the this problem and thus, got it wrong. I would encourage all who attempt it to avoid looking at internet solutions (even solutions to other similar problems) as that defeats the fun. I have quoted the problem below for your convenience; read it carefully! Good luck and have fun!

Originally Posted by Jonnyboy
Suppose there exist 1000 gold coins which are to be split up among ﬁve 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 inﬁnitely 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?

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