5 pirates, 100 coins, wrong answer...
Feb. 5th, 2012 09:13 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
rabidsamfanSo, I was wandering the 'net and came across the "Greedy Pirate Problem". You've got 5 pirates who have 100 gold coins and agree that they'll divide the loot by letting the senior-most pirate propose a divvy and ask for votes. All the pirate (including the senior) vote, and if half or more agree, that's the divvy. If they don't agree, then the seniormost pirate gets fed to the sharks and the new senior pirate makes his proposal.
The assumptions are that the pirates are sane, and they all
1) would rather stay alive and
2) want as much gold as possible and
3) are bloodthirsty. They'd just as soon end the process with fewer pirates aboard, all other things being equal.
So, if you follow the link you'll see that the standard logic says the "right" solution is for the seniormost pirate of 5 to offer the juniormost and middle pirate 1 coin apiece to get their votes and then keep 98 coins.
I disagree.
That analysis pretty much looks at the problem from the top down. But let's try it from a different angle.
It will be easiest if we name our pirates. Alfred is senior, followed by Byron, then Cuthbert, then Dennis, and finally we have our junior pirate Eric the Sanguinary.
Eric has different considerations than the others. First of all, he's not going to get fed to the sharks. By the time the crew is thinned down to him, there's no one left to throw him in. So he wants lots of gold and he'd actually enjoy seeing a fellow pirate playing lunchmeat for the local fishies. Now let's look at the scenarios.
If Eric is the only pirate left, he gets all the gold. Doesn't he wish!
If Dennis and he are the only ones left, Dennis will say, "I get 100 coins and you get zip, but since we each have one vote, my vote gives me half the votes and I win." So Eric doesn't want Dennis being left in charge.
If Cuthbert is deciding, he needs either Eric's vote or Dennis's vote. Dennis has the counterweight of wanting to be the pirate making the decision so he can get all the gold, so Cuthbert is better off bribing Eric. What Eric wants is the heftiest bribe he can get from Cuthbert.
Byron and Alfred's scenarios are both not great for Eric if he's trying to maximize his gold, but if they're following the standard logic of only offering single coin bribes, he's got a chance. So Eric, if he's offered a bribe by Alfred will turn it down if it is too small (saying so) and watch Alfred get eaten as he cheers lustily, making it clear that he likes gold, but he likes blood even better. Byron, if he's not smart enough to bribe the other two (who may be valuing themselves higher than a single coin now) is going to also go overboard. And with any luck Cuthbert will raise the bribe to Eric in order to get that vote and stay alive.
Right?
Now Alfred, who is a canny old pirate and has been around the deck more than once, knows that Eric will get more joy out of seeing him get eaten than gaining a tiny bit of gold, so his best bet is to ensure he gets the votes he needs to stay alive. If he claims that seniority counts and splits the coins 40 for Alfred, 30 for Byron, 20 for Cuthbert and 10 for Dennis, he's got a pretty good chance of getting four votes, leaving Eric to fuss and fume and wait for the time when Ferdinand joins the crew... (Or better yet, he splits it 39, 29,19,9 and 5 and sleeps at night.)
One, one and ninety eight my eye...
The assumptions are that the pirates are sane, and they all
1) would rather stay alive and
2) want as much gold as possible and
3) are bloodthirsty. They'd just as soon end the process with fewer pirates aboard, all other things being equal.
So, if you follow the link you'll see that the standard logic says the "right" solution is for the seniormost pirate of 5 to offer the juniormost and middle pirate 1 coin apiece to get their votes and then keep 98 coins.
I disagree.
That analysis pretty much looks at the problem from the top down. But let's try it from a different angle.
It will be easiest if we name our pirates. Alfred is senior, followed by Byron, then Cuthbert, then Dennis, and finally we have our junior pirate Eric the Sanguinary.
Eric has different considerations than the others. First of all, he's not going to get fed to the sharks. By the time the crew is thinned down to him, there's no one left to throw him in. So he wants lots of gold and he'd actually enjoy seeing a fellow pirate playing lunchmeat for the local fishies. Now let's look at the scenarios.
If Eric is the only pirate left, he gets all the gold. Doesn't he wish!
If Dennis and he are the only ones left, Dennis will say, "I get 100 coins and you get zip, but since we each have one vote, my vote gives me half the votes and I win." So Eric doesn't want Dennis being left in charge.
If Cuthbert is deciding, he needs either Eric's vote or Dennis's vote. Dennis has the counterweight of wanting to be the pirate making the decision so he can get all the gold, so Cuthbert is better off bribing Eric. What Eric wants is the heftiest bribe he can get from Cuthbert.
Byron and Alfred's scenarios are both not great for Eric if he's trying to maximize his gold, but if they're following the standard logic of only offering single coin bribes, he's got a chance. So Eric, if he's offered a bribe by Alfred will turn it down if it is too small (saying so) and watch Alfred get eaten as he cheers lustily, making it clear that he likes gold, but he likes blood even better. Byron, if he's not smart enough to bribe the other two (who may be valuing themselves higher than a single coin now) is going to also go overboard. And with any luck Cuthbert will raise the bribe to Eric in order to get that vote and stay alive.
Right?
Now Alfred, who is a canny old pirate and has been around the deck more than once, knows that Eric will get more joy out of seeing him get eaten than gaining a tiny bit of gold, so his best bet is to ensure he gets the votes he needs to stay alive. If he claims that seniority counts and splits the coins 40 for Alfred, 30 for Byron, 20 for Cuthbert and 10 for Dennis, he's got a pretty good chance of getting four votes, leaving Eric to fuss and fume and wait for the time when Ferdinand joins the crew... (Or better yet, he splits it 39, 29,19,9 and 5 and sleeps at night.)
One, one and ninety eight my eye...
![[identity profile]](https://www.dreamwidth.org/img/silk/identity/openid.png){kind=small}
(no subject)
Date: 2012-02-06 02:33 am (UTC)I salute you, sir or madam, and hope that if I'm ever on a pirate ship with you the captain deals with divvying up the gold and does not give you a say.
(no subject)
Date: 2012-02-06 05:45 am (UTC)(no subject)
Date: 2012-02-06 03:19 am (UTC)But #3 is where the five-pirate idea really breaks down. If they'd really "just as soon end the process with fewer pirates aboard, all other things being equal," then no matter what the first proposal, the others would nay it just to have fewer shares to divvy up. In this scenario, I think the only safe proposal would be 20 pieces to each all round, unless they really are bloodthirsty, and this gets rejected so the next lot is 25 pieces all round. (and perhaps another round, except it's hard to divide 100 by 3)
What am I missing?
(no subject)
Date: 2012-02-06 03:45 am (UTC)They say 50/50. This is clearly wrong. It's 50/50 per child, but a family that rolls 50% and gets a boy has no other children, where a family that has a girl rolls again and gets a boy. Um, China, anyone? There will be way more boys than girls in this scenario, not quite 2:1, because a family could have multiple girls, but it's going to skew male.
(no subject)
Date: 2012-02-06 05:43 am (UTC)(no subject)
Date: 2012-02-06 05:45 am (UTC);D
(no subject)
Date: 2012-02-06 03:28 pm (UTC)(no subject)
Date: 2012-02-06 07:31 am (UTC)Drink up, me hearties...
(no subject)
Date: 2012-02-06 03:29 pm (UTC)