Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**wl1****Member**- Registered: 2009-01-21
- Posts: 2

The puzzle:

"5 pirates of different ages have a treasure of 100 gold coins.

On their ship, they decide to split the coins using this scheme:

The oldest pirate proposes how to share the coins, and all pirates remaining will vote for or against it.

If 50% or more of the pirates vote for it, then the coins will be shared that way. Otherwise, the pirate proposing the scheme will be thrown overboard, and the process is repeated with the pirates that remain.

Assuming that all 5 pirates are intelligent, rational, greedy, and do not wish to die, (and are rather good at math for pirates) what will happen?"

from http://www.mathsisfun.com/puzzles/logic-puzzles-index.html

has got me confused.

The solution given didn't maske sense, due to "50% or more of the pirates vote for it", not "at least 50% vote for it".

So I think it would go to the second eldest, who would give all the coins to any of the other 3, to save his life. That should secure 50% of the vote.

Or am I way off?

Wayne

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Yes, I think the solution is wrong. It's very important what happens when the votes are tied.

The question says that the sharing scheme will be used in that case, but the answer assumes that the pirate gets thrown overboard.

I think the solution will still be something similar to the original one though. In particular, the eldest pirate will make a proposal that he knows will be voted for, since he doesn't want to die.

Edit: Following similar reasoning to the posted solution, I get that the eldest pirate will propose a split of 98:0:1:0:1.

Why did the vector cross the road?

It wanted to be normal.

Offline

**wl1****Member**- Registered: 2009-01-21
- Posts: 2

Mmm - I think I have it. Eldest gives 2nd eldest 1 coin, and gives 99 to the third eldest. Secures the vote.

Offline

**chesler****Member**- Registered: 2009-08-15
- Posts: 2

Old thread, I was cleaning out last year's school papers and found this. Agreed the tie-breakers make a big difference. Otherwise it comes down to working backwards and gaining votes by offering someone more than he will get if the vote is no. The solution seems to ascribe a value to not getting thrown overboard beyond not getting any gold. It is solvable, and is not "You will be hanged within the next week, on a day when you don't expect it."

Offline

**Jevandtieriel****Member**- Registered: 2009-10-01
- Posts: 1

I am confused by the answer as with this logic anyone could vote to toss the oldest. If 2 get less than they think they deserve then the oldest is out.

But folowing some sort of logic...

If you give any three pirates 30 gold coins each, and keep 10 gold coins for yourself (you have to keep some treasure for yourself as captain) then the three pirates should vote for you and only one against.

Or you could give all the gold shared evenly by any three. Hence ruling age out of the equasion.

You could give the three oldest pirates the money and forget about the youngest vote.

Does this make sense to anyone?

Cheers

Jev.

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

If a pirate is offered 30 coins, he still won't accept it if he thinks he can get more.

On the other hand, if he's offered 2 coins, he still might accept it if he thinks he wouldn't get that much otherwise.

Here's how I got my solution in detail:

Why did the vector cross the road?

It wanted to be normal.

Offline

**phillip1882****Member**- Registered: 2013-12-25
- Posts: 2

I also tend to think that a fairer distribution is far less likely to get you killed.

within the parameters of the puzzle, I would offer to split the gold evenly between the four other pirates and take 4 gold for myself.

I'm greedy in the sense I want to stay alive as the oldest. that to me is the most important parameter.

I highly doubt a bloodthirsty pirate would be satisfied with only 1 gold, even if he is intelligent.

if i was bloodthirsty i can't imagine being willing to take only 1 gold for myself from a loot of 100 gold. even if i thought it might mean getting no gold instead.

Offline

**zoya****Member**- Registered: 2013-12-26
- Posts: 1

I think the solution is wrong but i could not understand the problem properly.

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 102,853

Hi zoya;

Welcome to the forum.

I think the solution is wrong but i could not understand the problem properly

So it could be right, yes?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

Offline

Pages: **1**