Pirates Puzzle! (no Googling)

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Pirates Puzzle! (no Googling)

Post by Jon Corby » Tue Jan 22, 2008 1:42 pm

Ten pirates find a treasure chest containing 100 gold coins. After squabbling for some time, they agree on the following process to decide how to divvy up the booty: The pirates are ranked in order, from 1 (highest) to 10 (lowest). The lowest-ranked pirate (10) will offer up a division, for example he might suggest that #1 takes 20 coins, pirates #2 thru #8 take 9 coins each, and he gets 8. The other pirates then each vote "yes" or "no" to his suggestion; if the majority vote "yes", then they divide up the loot as suggested, but if not then that pirate walks the plank to his death, and the next lowest ranked pirate then offers his suggestion, and the process is repeated. Each pirate wants as many coins as he can get for himself, and does not care for any other pirates - he would rather see them walk the plank than not if it makes no difference to the number of coins he receives.

Assuming the pirates are perfectly logical, how much does each pirate end up with? And how many pirates die?

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Tue Jan 22, 2008 4:57 pm

Well, the first pirate is really only fighting for his life. He needs to get 5 of the remaining 9 to agree with his idea, and he's unlikely to think too much about giving himself gold. All the pirates know that when it comes to their go, they will be in the same position, so want it to end quickly.

On the other hand, if they let the first one walk the plank, they will all know that the 9th ranking pirate can offer a deal pretty much the same as what the 10th ranking pirate does except with only 9 to think about. Without even thinking about details, that might appeal. But on yet another hand, they will all know when it gets to their turn, the pirates ahead of them might do to them what they did to the lower down pirates.

Also, do they all know all the other pirates are logical?

Maybe if the first pirate gave himself 95 coins and each of the next five (so ranks 5 to 9) 1 each, they would all vote in his favour knowing that they might otherwise die. No-one would actually die.

If it is logical for those five to vote for his deal then fine for them. But if it is not logical, then it would likely be illogical for higher up pirates to vote for their deal, pretty much whatever it is. Given this, ranks 5 to 9 would probably just decide it's logical to go with it.

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Tue Jan 22, 2008 5:07 pm

But then, maybe not. It might be logical for him to look at pirates 5-9 but offer them a more convincing deal than that. They could easily vote him off and use the same principle themselves but not in a 95 and then the rest sense, but a more even spread. But whichever way you look at it, even if they are more generous than the first pirate, the other pirates could still expect a better deal when it gets to the next tuen and so on, whether it be a by a small amount or a big one. They will also know that the principle will always apply when it gets to their turn, however generous they are.

So the only way for it to be logical for them (priates 5-9) to be kept alive is for it to be logical for the first pirate to be kept alive. Whether waiting would get them a small increase or big increase, as long as they are offered something to start with, it doesn't matter. They need to nip it in the bud and vote for the 95, 1, 1, 1, 1, 1.

Is this close, Corby?

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Re: Pirates Puzzle! (no Googling)

Post by Jon Corby » Tue Jan 22, 2008 5:11 pm

Gevin-Gavin wrote:Also, do they all know all the other pirates are logical?
Every pirate is entirely logical, and knows all other pirates will be entirely logical.

Every pirate cares only about his own haul, so while he would gain enjoyment from the deaths of others, he wouldn't sacrifice even a single coin for it.

You're on the right-ish sort of lines, but the answer is precise, not a vague "yeah, that pirate might vote yes to that I suppose" :)

I'll refrain from saying anything else at the moment until others have had a go.... :D

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Tue Jan 22, 2008 5:20 pm

How does the voting work when there are an even number of voters and it's a draw? Is there a clue in this (i.e. it will be settled in the first round)?

That aside, I would say that it must be possible for the first pirate to survive or all the others would be in exactly the same position and also die. So the first pirate, being entirely logical, knows that whatever he goes for would be whatever the others would do when it gets to them, and he knows the others must vote for it or they will also die at their go! So it's 100 to him and none to anyone else!!!

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Re: Pirates Puzzle! (no Googling)

Post by Charlie Reams » Tue Jan 22, 2008 5:25 pm

I've seen this puzzle before and it's a good'un.

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Re: Pirates Puzzle! (no Googling)

Post by Joseph Bolas » Tue Jan 22, 2008 5:34 pm

Charlie Reams wrote:I've seen this puzzle before and it's a good'un.
Also if anyone has heard of Perplex City before, you can find this puzzle on card #213 of Season 1.

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Re: Pirates Puzzle! (no Googling)

Post by Paul Howe » Tue Jan 22, 2008 5:45 pm

Corby wrote:Ten pirates find a treasure chest containing 100 gold coins. After squabbling for some time, they agree on the following process to decide how to divvy up the booty: The pirates are ranked in order, from 1 (highest) to 10 (lowest). The lowest-ranked pirate (10) will offer up a division, for example he might suggest that #1 takes 20 coins, pirates #2 thru #8 take 9 coins each, and he gets 8. The other pirates then each vote "yes" or "no" to his suggestion; if the majority vote "yes", then they divide up the loot as suggested, but if not then that pirate walks the plank to his death, and the next lowest ranked pirate then offers his suggestion, and the process is repeated. Each pirate wants as many coins as he can get for himself, and does not care for any other pirates - he would rather see them walk the plank than not if it makes no difference to the number of coins he receives.

Assuming the pirates are perfectly logical, how much does each pirate end up with? And how many pirates die?
Just had a quick go at this, and I can't be bothered typing out my reasoning in detail, but the answer I get is that pirate 10 gives himself 95 coins, and 1 coin each to pirates 9,7,5,3 and 2, so none of the pirates die. I also think that if you play the game with 9, 7 or 5 pirates, there's no way for the first pirate to avoid death!

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Re: Pirates Puzzle! (no Googling)

Post by David O'Donnell » Tue Jan 22, 2008 5:53 pm

It seems like any offer the first pirate makes is bound to be rejected since the others can divide the booty amongst fewer and fewer pirates each time. If this process continues I don't see why you just don't end up having two pirates at the end with the one who is doing the voting having complete control over the other one. If the first pirate is logical then surely he is going to realise that his only chance of survival is to co-opt at least five of the others in voting with him on some proposal. He may point out that with the first plan pirates 3-10 will all die and only pirates 1 and 2 will get any gold (probably a 99-1 split!). If I were the lowest ranked pirate I would offer to divide some of the treasure with pirates 5-9. From a staunchly logical point of view you could get away with offering them a coin each since if they don't accept they will end up dead. However, it would depend on pirate 5 since he holds the balance of power. Maybe you have to offer pirate 5 more than pirate 6 etc. In short, I don't know :lol:

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Re: Pirates Puzzle! (no Googling)

Post by David O'Donnell » Tue Jan 22, 2008 5:55 pm

Paul Howe wrote:
Corby wrote:Ten pirates find a treasure chest containing 100 gold coins. After squabbling for some time, they agree on the following process to decide how to divvy up the booty: The pirates are ranked in order, from 1 (highest) to 10 (lowest). The lowest-ranked pirate (10) will offer up a division, for example he might suggest that #1 takes 20 coins, pirates #2 thru #8 take 9 coins each, and he gets 8. The other pirates then each vote "yes" or "no" to his suggestion; if the majority vote "yes", then they divide up the loot as suggested, but if not then that pirate walks the plank to his death, and the next lowest ranked pirate then offers his suggestion, and the process is repeated. Each pirate wants as many coins as he can get for himself, and does not care for any other pirates - he would rather see them walk the plank than not if it makes no difference to the number of coins he receives.

Assuming the pirates are perfectly logical, how much does each pirate end up with? And how many pirates die?
Just had a quick go at this, and I can't be bothered typing out my reasoning in detail, but the answer I get is that pirate 10 gives himself 95 coins, and 1 coin each to pirates 9,7,5,3 and 2, so none of the pirates die. I also think that if you play the game with 9, 7 or 5 pirates, there's no way for the first pirate to avoid death!

Please explain your reasoning because I had something similar (without offering money to odd numbered pirates!) and it's bothering me that I haven't got it fully worked out.

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Re: Pirates Puzzle! (no Googling)

Post by Paul Howe » Tue Jan 22, 2008 5:58 pm

Will do in half an hour or so David, after I've had my dinner (one of the nice things about this forum is that I don't feel guilty about emailing posts like this to hundreds of people!)

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Re: Pirates Puzzle! (no Googling)

Post by David O'Donnell » Tue Jan 22, 2008 6:00 pm

:D

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Tue Jan 22, 2008 6:09 pm

Hope you don't mind me having multiple goes at this! I've just reread it and seen that the majority need to say yes for the deal to go ahead, so in a draw he walks the plank. So if it gets as far as #3, even if #2 goes for it, #1 won't and it will be the plank. So at #4's turn, #2 and #3 will have to vote for it, whatever it is! #4 could happily give everything to himself. So even if they are just offered 1 coin, #1, #2 and #3 would do well to vote for #5's bid if it reaches him. So what can #6 do to stop it going further? If he just offers 2 coins each to #s 1-3 and the rest to himself, they will vote for it, knowing that #5 could offer them just 1 coin.

What about #7? What does he do? He could offer 3 coins to #s 1-3 which would be better for them than it going further and just 1 coin to #4. #8 could offer 4 coins each to the same #s 1-3 and 2 to #4. #9 would offer 5 coins to #s 1-3, 3 to #5 and 1 to #6. Pirate #10 would offer 6 coins to #s 1-3, 4 to #5 and 2 to #6.

However, it does not specify that the pirates do not want to die, so this could change things. #4 may have to give a coin each to #s 2 and 3. So running through it again:

#5 - 2 to #s 2,3 and 1 to #1
#6 - 3 to #s 2,3; 2 to #1
#7 - 4 to #s 2,3; 3 to #1; 1 to #4
#8 - 5 to #s 2,3; 4 to #1; 2 to #4
#9 - 6 to #s 2,3; 5 to #1; 3 to #4; 1 to #5

#10 - 7 to #s 2,3; 6 to #1; 4 to #4; 2 to #5 - 74 to himself! This is my answer.

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Re: Pirates Puzzle! (no Googling)

Post by Paul Howe » Tue Jan 22, 2008 6:51 pm

Well, whilst I was having dinner I realised that there was an error in my reasoning that I have now (hopefully) corrected. My new conclusion is that pirate 10 takes home 95 coins, pirate 9 gets nothing, and that 5 of pirates 1-8 get one coin each (it doesn't matter which 5). Also, none of the pirates now have to die if you change the number of pirates.

Something that gets hammered into you by doing degree level maths is that for puzzles like this, it's instructive to look at small cases first, so lets start with 2 pirates. Well, if pirate 2 wants to live, he's got no choice but to give 100 coins to pirate 1. With 3 pirates, things aren't that much different. If pirate 3 offers pirate 1 anything less than 100 gold coins pirate 1 will vote no and we're back to the 2 pirate case where pirate 1 takes all. With 4 pirates, things are a little different, because pirate 1 can be outvoted by pirates 2 and 3, so pirate 4 has to offer them more than they will get in the 2 and 3 pirate cases. Since they're not getting anything in these cases, 1 gold each will be sufficient, leaving 98 gold for pirate 4. So we now have coin allocations for 2,3, and 4 pirates that will be accepted by rational pirates.

Now, suppose there are N pirates, how will a rational pirate behave? Well since the N-1 pirate deal will be accepted, a pirate will only reject the N pirate deal if the N-1 pirate deal offers him more money. To get a majority in the N pirate deal, we need (N-1)/2 (rounded up to the nearest integer) pirates to accept it, which means offering this number of pirates the same amount of coins as they will get in the N-1 pirate deal. Since pirate N-1 gets a massive lump of coins in the N-1 pirate deal, we might as well offer him 0 coins as he's going to reject pretty much anything. The rest of the pirates get no more than 1 coin from the N-1 pirate deal so it suffices for pirate N to offer (N-1)/2 (rounded up) pirates 1 coin each and keep the rest for himself.

With 10 pirates, this is (10-1)/2 = 4.5 so round up to 5 pirates that have to be given 1 coin each, and 100-5 = 95 coins to pirate 10. So, no-one has to die, and furthermore, the pirates have redistributed their wealth to the least fortunate of their number. Perhaps we have misjuded these lowly seadogs?

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Re: Pirates Puzzle! (no Googling)

Post by Jon Corby » Tue Jan 22, 2008 8:08 pm

Paul Howe wrote:With 4 pirates, things are a little different, because pirate 1 can be outvoted by pirates 2 and 3, so pirate 4 has to offer them more than they will get in the 2 and 3 pirate cases. Since they're not getting anything in these cases, 1 gold each will be sufficient, leaving 98 gold for pirate 4. So we now have coin allocations for 2,3, and 4 pirates that will be accepted by rational pirates.
This isn't quite true for the parameters I've given - each pirate will happily watch others walk the plank provided it doesn't reduce their share. Therefore #2 & #3 know they will die if it gets as far as pirate #3 suggesting, so will happily accept a 0-0-0-100 split offered by pirate 4 (ie pirate 4 keeps the lot) as it's the only way they'll preserve their lives.

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Tue Jan 22, 2008 9:10 pm

Corby wrote:
Paul Howe wrote:With 4 pirates, things are a little different, because pirate 1 can be outvoted by pirates 2 and 3, so pirate 4 has to offer them more than they will get in the 2 and 3 pirate cases. Since they're not getting anything in these cases, 1 gold each will be sufficient, leaving 98 gold for pirate 4. So we now have coin allocations for 2,3, and 4 pirates that will be accepted by rational pirates.
This isn't quite true for the parameters I've given - each pirate will happily watch others walk the plank provided it doesn't reduce their share. Therefore #2 & #3 know they will die if it gets as far as pirate #3 suggesting, so will happily accept a 0-0-0-100 split offered by pirate 4 (ie pirate 4 keeps the lot) as it's the only way they'll preserve their lives.
Oh really? Then my answer will have to change. I don't think it was made clear that a pirate would preserve his life over and above the rule that says he would rather see another pirate die if it made no difference to number of coins received. So:

#4 - All for himself
#5 - 1 to #s 1,2,3
#6 - 2 to #s 1,2,3
#7 - 3 to #s 1,2,3; 1 to #4
#8 - 4 to #s 1,2,3; 2 to #4
#9 - 5 to #s 1,2,3; 3 to #4, 1 to #5

#10 - 6 to #s 1,2,3; 4 to #4, 2 to #5, 76 to himself and he doesn't die. Looking back, I had this slightly different, but I don't think it really matters if #7 gives 1 coin to #4 or #5 and various other changes could be made.

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Re: Pirates Puzzle! (no Googling)

Post by Paul Howe » Tue Jan 22, 2008 9:23 pm

Corby wrote: This isn't quite true for the parameters I've
given - each pirate will happily watch others walk the plank provided
it doesn't reduce their share. Therefore #2 & #3 know they will die
if it gets as far as pirate #3 suggesting, so will happily accept a
0-0-0-100 split offered by pirate 4 (ie pirate 4 keeps the lot) as it's
the only way they'll preserve their lives.
Hmm, you're right. I was assuming each pirate would keep his fellows alive if he didn't stand to gain financially, but in fact the Jon Corby definition of rationality appears to sanction murder as long as it doesn't hit you in the pocket. Truly, you're an inspiration to us all.

Reasoning as I did before, for pirate N's proposal to get accepted in the N pirate case now requires improving the reward of the majority of the pirates in the N-1 pirate case. Starting from 0-0-0-100 for 4 pirates, we get something like:

0-0-0-100
1-1-1-0-97
0-2-2-1-0-95
1-0-3-2-1-0-93
2-1-0-0-2-1-0-94
0-2-1-1-0-2-1-0-93
1-0-0-2-1-0-2-1-0-93

You could make some different choices at some points, but I think you will always end up with pirate 10 getting 93 coins, pirate 9 getting nowt, two of the rest getting 2 coins and 3 getting 1.

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Tue Jan 22, 2008 10:02 pm

You might have got me there Paul. I was always including the same pirates in the majority, but that doesn't have to be the case.

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Re: Pirates Puzzle! (no Googling)

Post by Jon Corby » Tue Jan 22, 2008 10:15 pm

Paul Howe wrote:Hmm, you're right. I was assuming each pirate would keep his fellows alive if he didn't stand to gain financially, but in fact the Jon Corby definition of rationality appears to sanction murder as long as it doesn't hit you in the pocket. Truly, you're an inspiration to us all.
They're bloodthirsty pirates of the high seas! I don't make the rules!



(Except these rules, which I made)

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Tue Jan 22, 2008 10:16 pm

Paul Howe wrote:Reasoning as I did before, for pirate N's proposal to get accepted in the N pirate case now requires improving the reward of the majority of the pirates in the N-1 pirate case. Starting from 0-0-0-100 for 4 pirates, we get something like:

0-0-0-100
1-1-1-0-97
0-2-2-1-0-95
1-0-3-2-1-0-93
2-1-0-0-2-1-0-94
0-2-1-1-0-2-1-0-93
1-0-0-2-1-0-2-1-0-93

You could make some different choices at some points, but I think you will always end up with pirate 10 getting 93 coins, pirate 9 getting nowt, two of the rest getting 2 coins and 3 getting 1.
Here is another way of doing it using your method up to pirate 6:

0-0-0-100
1-1-1-0-97
2-2-0-1-0-95

Now using your own pirate 7 method we have:

1-0-3-2-1-0-93

Pirates 3,4 and 5 are better off in the case of 7 than using my own pirate 6 system. 1, 2, and 6 are not. But is this relevant? Yes, I think it is. You see, if it gets to the point where pirate 6 makes his decision, he is not required at all to look back into the past and see what has been done. He would still be making a decision that is just as logical in its own right as your own 6 pirate decision. So the majority will not be guaranteed an increase under this system.

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Tue Jan 22, 2008 11:18 pm

So, I've been thinking about it a bit more, and here's what I've come up with. 4 and 5 are still as before:

4: 0-0-0-100
5: 1-1-1-0-97

But then there are different possibilities that will give the majority a favourable outcome for the 6 case over the 5 case.

6: 0-2-2-1-0-95
6: 2-0-2-1-0-95
6: 2-2-0-1-0-95

Since the 5 case is set, any of these will work for the 6 case. So, moving onto the 7 case:

6: 0-2-2-1-0-95 leads to

7: 1-0-3-2-1-0-93 or
7: 1-3-0-2-1-0-93

6: 2-0-2-1-0-95 leads to

7: 0-1-3-2-1-0-93 or
7: 3-1-0-2-1-0-93

6: 2-2-0-1-0-93 leads to

7: 0-3-1-2-1-0-93 or
7: 3-0-1-2-1-0-93

We have six distinct possibilities here and if any of these come up, the pirates will know that there is more than one possibility for the next round so will look at the likelihood of a better deal next time. Let's look at:

7: 1-0-3-2-1-0-93

The next one could still be any of:

6: 0-2-2-1-0-95
6: 2-0-2-1-0-95
6: 2-2-0-1-0-95

#1 has a 1 in 3 chance of losing out by one by continuing, but a 2 in 3 chance of gaining by one so will vote against. #2 can only gain or stay level so will vote against. #3 loses out so votes in favour. #4 also loses out. And #5 loses out. #6 will gain but it's 3 votes all and pirate 7 walks the plank. The other possibilities for seven are just anagrams of this one, so the same will happen, so it doesn't work! This post is long enough. I will gather my thoughts before the next one.

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Wed Jan 23, 2008 12:00 am

So I'm still with these as the "correct" possibilites for six pirates.

6: 0-2-2-1-0-95
6: 2-0-2-1-0-95
6: 2-2-0-1-0-95

We now need a 7 case where 4 out of the 6 voters will go with it. #s 5 and 6 will increase by one and #7 will just get whatever is left. But we need two out of #s 1-3 to vote in favour. They know that they've got a 2 in 3 chance of getting 2 next time. But by just giving 2 of them 2, it's a guarantee so they will vote for it!

7: 0-2-2-2-1-0-93
7: 2-0-2-2-1-0-93
7: 2-2-0-2-1-0-93

And we haven't increased the possibilities either so it's easy to keep track of. OK, so #8 needs four to go for his system. Again just increase the last two by one to keep them happy. Then you can give a guaranteed 2 to any two of #s 1,2, or 3 and give #4 nothing.

8: 0-2-2-0-2-1-0-93
8: 2-0-2-0-2-1-0-93
8: 2-2-0-0-2-1-0-93

Again just the three possibilities. #9 needs to keep five happy. Again, giving two of the first three two each will be fine for them. Give #4 one, none to #5 and increase by one for the back two and insert a zero as usual.

9: 0-2-2-1-0-2-1-0-92
9: 2-0-2-1-0-2-1-0-92
9: 2-2-0-1-0-2-1-0-92

And finally #10, who still needs five votes. Seems easy enough. Keep two out of the first three happy again. Give nothing to #4, one to #5, nothing to #6, increase #s 7 and 8 to two and one and nothing to #9.

10: 0-2-2-0-1-0-2-1-0-92
10: 2-0-2-0-1-0-2-1-0-92
10: 2-2-0-0-1-0-2-1-0-92

In fact you could give two to #4 and none to #7 giving:

10: 0-2-2-2-1-0-0-1-0-92
10: 2-0-2-2-1-0-0-1-0-92
10: 2-2-0-2-1-0-0-1-0-92

And that's it. Paul came up with 93 for the end, so it would be interesting to see what is correct.

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Re: Pirates Puzzle! (no Googling)

Post by Paul Howe » Wed Jan 23, 2008 12:03 am

Gevin, I think you are right. For N=6 and beyond the allocation is not uniquely determined, so you have to offer the pirates a situation in which their expected return in the current round is higher than their expected return in the next one. A bare minimum would be to ensure that the number of coins shared between pirates 1 up to N-2 in round N is greater than the number of coins shared between the same pirates in round N-1, so that pirate N gets one fewer coin each time N is increased. You'd also have to divide the coins in a "fair" way (obviously giving one pirate all the coins and the rest nothing would get voted down). Whether this fair division would involve pirate N sacrificing more coins, I'm not sure. I see you've been busy churning out various combinations which I don't have the stamina to read through at the minute, so maybe you've cracked it.

Anyway, having done all that, I looked up the puzzle on the internet and discovered that when posed with slightly different conditions, it has a much simpler and more elegant solution, which is perhaps what Jon was aiming for in the first place?

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Re: Pirates Puzzle! (no Googling)

Post by Joseph Bolas » Wed Jan 23, 2008 12:27 am

Paul Howe wrote:Anyway, having done all that, I looked up the puzzle on the internet and discovered that when posed with slightly different conditions, it has a much simpler and more elegant solution, which is perhaps what Jon was aiming for in the first place?
Could there possibly be then, more than 1 answer to this puzzle?

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Re: Pirates Puzzle! (no Googling)

Post by Jon Corby » Wed Jan 23, 2008 12:59 am

Paul Howe wrote:Anyway, having done all that, I looked up the puzzle on the internet and discovered that when posed with slightly different conditions, it has a much simpler and more elegant solution, which is perhaps what Jon was aiming for in the first place?
Not really, I didn't bother to check for an exact wording which people could then look up. The puzzle works however you word the subtleties, the conclusions are just slightly different. It doesn't make it unsolvable or nonsensical, so no I wasn't aiming for whatever answer you found.

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Wed Jan 23, 2008 11:56 am

I was thinking about this in bed last night (yes, it's quite sad), and I decided that there would be more solutions than I came up with. This is still fine for six I think:

6: 0-2-2-1-0-95
6: 2-0-2-1-0-95
6: 2-2-0-1-0-95

But seven would change. For some reason, I always stuck with two out of the first three getting two coins each. But instead of just giving #4 two, he can join their club. It would be three out of four getting two and one getting none.

7: 0-2-2-2-1-0-93
7: 2-0-2-2-1-0-93
7: 2-2-0-2-1-0-93
7: 2-2-2-0-1-0-93

So, onto eight, and the next one joins the two rotation club.

8: 0-0-2-2-2-1-0-93
Without writing them all out, any three from the first five get two and the other two get nothing.

9: 0-0-2-2-2-2-1-0-91
Here it is any four from #s 1-6.

Finally:

10: 0-0-0-2-2-2-2-1-0-91
Any four from the first seven pirates to get two coins.

Am I any closer?

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Re: Pirates Puzzle! (no Googling)

Post by Jon Corby » Wed Jan 23, 2008 1:29 pm

Gevin-Gavin wrote:10: 0-0-0-2-2-2-2-1-0-91
Any four from the first seven pirates to get two coins.

Am I any closer?
I agree completely with this answer :)

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Re: Pirates Puzzle! (no Googling)

Post by David O'Donnell » Wed Jan 23, 2008 1:37 pm

Corby wrote:
Gevin-Gavin wrote:10: 0-0-0-2-2-2-2-1-0-91
Any four from the first seven pirates to get two coins.

Am I any closer?
I agree completely with this answer :)
Please post the answer Corby before I go insane.

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Re: Pirates Puzzle! (no Googling)

Post by Paul Howe » Wed Jan 23, 2008 3:06 pm

Corby wrote:
Gevin-Gavin wrote:10: 0-0-0-2-2-2-2-1-0-91
Any four from the first seven pirates to get two coins.

Am I any closer?
I agree completely with this answer :)
Wahey, at last! At first, I thought Jon's rules led to a needlessly complicated solution (I'm not just bitter because I arsed it up three times. No siree), but actually I think they result in some very interesting behaviour when you have large numbers of pirates.

Clearly, Gevin's solution can be extended to deal with up to 100 pirates, but what happens when there are 101 pirates? What about 102? 103? I think something interesting and different to the basic pattern that Gevin outlined happens in each case, although having made a bit of a mess of this puzzle already, I'm not completely convinced of my reasoning.

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Re: Pirates Puzzle! (no Googling)

Post by Joseph Bolas » Wed Jan 23, 2008 4:15 pm

Corby wrote:
Gevin-Gavin wrote:10: 0-0-0-2-2-2-2-1-0-91
Any four from the first seven pirates to get two coins.

Am I any closer?
I agree completely with this answer :)
Ooh. The numbers I have then must be wrong.

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Wed Jan 23, 2008 8:22 pm

I just spent ages on a reply but it turned out I wasn't logged in when I submitted and it lost it! But I think I was logged or the quote thing doesn't come up.

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Wed Jan 23, 2008 8:52 pm

Paul Howe wrote:Wahey, at last! At first, I thought Jon's rules led to a needlessly complicated solution (I'm not just bitter because I arsed it up three times. No siree), but actually I think they result in some very interesting behaviour when you have large numbers of pirates.

Clearly, Gevin's solution can be extended to deal with up to 100 pirates, but what happens when there are 101 pirates? What about 102? 103? I think something interesting and different to the basic pattern that Gevin outlined happens in each case, although having made a bit of a mess of this puzzle already, I'm not completely convinced of my reasoning.
I'll try again. I think I said something like: I arsed it up more than three times. But anyway, pirate #100:

Two each to any 49 of the first 97, one to #98 none to #99 and one to himself.

#101: He cannot survive the plank because he needs 51 votes. He can give one coin to #99 but still needs to give two each to 50 others and doesn't have enough.

#102: He automatically has #101 on his side and can also give one to #99. Two each would go to 49 of the rest of them (#s 1-98 and #100) and he'd have one for himself. This would get him 51 votes.

#103: He needs 52 votes. #101 can be given just one, and interestingly #s 1-98 and #100 averaged less than one each in the #102 scenario (98 between 99 of them), so he can dish out just one to any 52 of #s 1-101 (apart from #99) and they would gain from it. He keeps 48!

#104: He can now give out one each to any 52 of #s 1-102 and he also takes 48.

#105: He gives out one coin each to 53 of #s 1-103 and takes 47. And so on...

#199: He gives out 1 coin each to 100 of #s 1-197 and keeps none. He wins by 100 votes to 98.

#200: He gives out 1 coin each to 100 of #s 1-199 and keeps none. He gets the 100-99 majority he needs.

#201: He walks the plank. Not enough coins for the 101 votes he needs.

#202: He also needs 101 votes. He gets 100 coin votes and the guaranteed vote from #201. He gives none to himself but survives.

#203: Walks the plank. As do #s 204-205

#206: Gets three votes from #s 203-205 for them to survive and 100 coin votes, giving him 103 votes and a majority.

#s 207-213 walk.

#214: Gets seven survival votes (#s 207-213) and 100 coin votes which gives him his 107 votes.

#s 215-229 walk.

#230: 15 survival votes and 100 coin votes give him the 115 he needs.

I think it just carries on with the gap doubling between each survivor - the next being #262.

What were your thoughts, Paul?

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Re: Pirates Puzzle! (no Googling)

Post by Paul Howe » Wed Jan 23, 2008 9:45 pm

I got the same as you for numbers 101, 102, 103. Beyond that, I hadn't really bothered reasoning anything out but can't see anything wrong with what you've written.

I think it's interesting that once you increase the number of pirates enough you get completely different behaviour (the gap between surviving pirates doubling), that also runs counter to the intuitive guess that once you increase the number of pirates beyond a certain point the first pirate always has to die.

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Thu Jan 24, 2008 3:03 pm

OK, I think I've done this to death now, so I think I've earnt a "Googling".

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Sat Jan 26, 2008 3:01 pm

David O'Donnell wrote:
Corby wrote:
Gevin-Gavin wrote:10: 0-0-0-2-2-2-2-1-0-91
Any four from the first seven pirates to get two coins.

Am I any closer?
I agree completely with this answer :)
Please post the answer Corby before I go insane.
It's in the post you quoted. :?:

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Re: Pirates Puzzle! (no Googling)

Post by Rhys Benjamin » Sun May 29, 2011 3:53 pm

Wow. I can't do this!
The forum's resident JAILBAKER, who has SPONDERED several times...

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Re: Pirates Puzzle! (no Googling)

Post by Jon Corby » Thu Aug 27, 2020 12:39 pm

Gavin Chipper wrote:
Wed Jan 23, 2008 11:56 am
I was thinking about this in bed last night (yes, it's quite sad), and I decided that there would be more solutions than I came up with. This is still fine for six I think:

6: 0-2-2-1-0-95
6: 2-0-2-1-0-95
6: 2-2-0-1-0-95

But seven would change. For some reason, I always stuck with two out of the first three getting two coins each. But instead of just giving #4 two, he can join their club. It would be three out of four getting two and one getting none.

7: 0-2-2-2-1-0-93
7: 2-0-2-2-1-0-93
7: 2-2-0-2-1-0-93
7: 2-2-2-0-1-0-93
I was just doing this puzzle with somebody else, and they agreed - up until this point. They say that 7 can go:

7: 1-1-1-0-1-0-96

#5 will vote yes as he's getting nothing on the next vote.
#1,#2,#3 will also vote yes, as any of them realise they may get nothing on the next vote, so would take a single coin now.

I make him right? I can't see any flaw in the logic.

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Re: Pirates Puzzle! (no Googling)

Post by Ian Volante » Thu Aug 27, 2020 1:36 pm

I still don't like this.
meles meles meles meles meles meles meles meles meles meles meles meles meles meles meles meles

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Thu Aug 27, 2020 9:40 pm

Jon Corby wrote:
Thu Aug 27, 2020 12:39 pm
Gavin Chipper wrote:
Wed Jan 23, 2008 11:56 am
I was thinking about this in bed last night (yes, it's quite sad), and I decided that there would be more solutions than I came up with. This is still fine for six I think:

6: 0-2-2-1-0-95
6: 2-0-2-1-0-95
6: 2-2-0-1-0-95

But seven would change. For some reason, I always stuck with two out of the first three getting two coins each. But instead of just giving #4 two, he can join their club. It would be three out of four getting two and one getting none.

7: 0-2-2-2-1-0-93
7: 2-0-2-2-1-0-93
7: 2-2-0-2-1-0-93
7: 2-2-2-0-1-0-93
I was just doing this puzzle with somebody else, and they agreed - up until this point. They say that 7 can go:

7: 1-1-1-0-1-0-96

#5 will vote yes as he's getting nothing on the next vote.
#1,#2,#3 will also vote yes, as any of them realise they may get nothing on the next vote, so would take a single coin now.

I make him right? I can't see any flaw in the logic.
I've just read through the thread, but I haven't gone through it in enough detail to get back into the same thought process as back then. Looking at that though, #1, #2 and #3 are getting an average of 1.5 coins under my solution (3 of them get 2 and 1 gets 0), whereas they are getting exactly 1 under this new proposal. Is a bird in the hand worth two in the bush? Well, their expected monetary gain is higher under my solution, but you could also look at utility. With utility there is generally diminishing returns - getting a million pounds when you have nothing is worth more to you than another million when you already have a million. So 1 coin is worth 1 utility point, 2 might be wirth less than 2, but you'd still expect at these small numbers for it to be worth pretty much 2.

None of this is utility stuff is defined in the question of course, but in terms of expected monetary gain, it's higher under my solution and it's probably the best way to go. Obviously you'd say a 99% chance of getting all the coins is better than a guaranteed 1, so if you take that as your starting point, the least arbitary thing to do in the absence of any other specified criteria in the question is to go for expected number of coins.

But maybe the question just isn't rigorously defined enough.

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Re: Pirates Puzzle! (no Googling)

Post by Jon Corby » Fri Aug 28, 2020 7:49 am

Gavin Chipper wrote:
Thu Aug 27, 2020 9:40 pm
Jon Corby wrote:
Thu Aug 27, 2020 12:39 pm
Gavin Chipper wrote:
Wed Jan 23, 2008 11:56 am
I was thinking about this in bed last night (yes, it's quite sad), and I decided that there would be more solutions than I came up with. This is still fine for six I think:

6: 0-2-2-1-0-95
6: 2-0-2-1-0-95
6: 2-2-0-1-0-95

But seven would change. For some reason, I always stuck with two out of the first three getting two coins each. But instead of just giving #4 two, he can join their club. It would be three out of four getting two and one getting none.

7: 0-2-2-2-1-0-93
7: 2-0-2-2-1-0-93
7: 2-2-0-2-1-0-93
7: 2-2-2-0-1-0-93
I was just doing this puzzle with somebody else, and they agreed - up until this point. They say that 7 can go:

7: 1-1-1-0-1-0-96

#5 will vote yes as he's getting nothing on the next vote.
#1,#2,#3 will also vote yes, as any of them realise they may get nothing on the next vote, so would take a single coin now.

I make him right? I can't see any flaw in the logic.
I've just read through the thread, but I haven't gone through it in enough detail to get back into the same thought process as back then. Looking at that though, #1, #2 and #3 are getting an average of 1.5 coins under my solution (3 of them get 2 and 1 gets 0), whereas they are getting exactly 1 under this new proposal. Is a bird in the hand worth two in the bush? Well, their expected monetary gain is higher under my solution, but you could also look at utility. With utility there is generally diminishing returns - getting a million pounds when you have nothing is worth more to you than another million when you already have a million. So 1 coin is worth 1 utility point, 2 might be wirth less than 2, but you'd still expect at these small numbers for it to be worth pretty much 2.

None of this is utility stuff is defined in the question of course, but in terms of expected monetary gain, it's higher under my solution and it's probably the best way to go. Obviously you'd say a 99% chance of getting all the coins is better than a guaranteed 1, so if you take that as your starting point, the least arbitary thing to do in the absence of any other specified criteria in the question is to go for expected number of coins.

But maybe the question just isn't rigorously defined enough.
Yeah, it's definitely fair to say that the question isn't rigorously defined enough. When he first asked the question of "how risk-averse are the pirates? Would they risk a guaranteed coin for a chance of 2 later? (if "on average" they would get more, but might also end up worse off)" I didn't really have to think, and said it was all about guaranteeing themselves the most coins. But yes, when I looked back at the rules, it doesn't explicitly say precisely that. It's fairly easy fixed just by specifying that they want to maximise their own guaranteed haul, but yes it's certainly a reasonable question to ask and it definitely isn't covered off well enough as it stands. I don't think the utility stuff is relevant once you remove the 'gamble' aspect.

Also though, your (and it was mine, remember, I'm not criticising you!) solution doesn't work even if it is factoring in 'expected gain'. #6's proposal sees pirates #1, #2 and #3 getting an 'expected gain' of 1.3333 coins each, and #7s tops this by offering those same three pirates an expected gain of 1.5 coins each, so they would indeed vote for this. However, #8s subsequent proposal then offers these same three pirates an expected gain of only 1.2 coins each. Therefore they wouldn't vote for this. So the solution we all accepted isn't even the "best expected value" solution.

Edit: no wait. Of course it works, they would vote "yes" because they're literally being offered 2 coins at that point, not an expected 1.2 coins 🤦

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Fri Aug 28, 2020 8:02 pm

Jon Corby wrote:
Fri Aug 28, 2020 7:49 am
Edit: no wait. Of course it works, they would vote "yes" because they're literally being offered 2 coins at that point, not an expected 1.2 coins 🤦
So you're saying our original solution is still right (based on expected gain anyway)? So I don't have to go through it all again?

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Re: Pirates Puzzle! (no Googling)

Post by Jon Corby » Sat Aug 29, 2020 8:55 am

Gavin Chipper wrote:
Fri Aug 28, 2020 8:02 pm
Jon Corby wrote:
Fri Aug 28, 2020 7:49 am
Edit: no wait. Of course it works, they would vote "yes" because they're literally being offered 2 coins at that point, not an expected 1.2 coins 🤦
So you're saying our original solution is still right (based on expected gain anyway)? So I don't have to go through it all again?
Yes, I believe so. But I have to admit I don’t think I ever considered it as “expected gain”. But then I am also struggling to see what other rationale I could have used :?

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Re: Pirates Puzzle! (no Googling)

Post by Gavin Chipper » Sat Aug 29, 2020 9:08 am

Jon Corby wrote:
Sat Aug 29, 2020 8:55 am
Gavin Chipper wrote:
Fri Aug 28, 2020 8:02 pm
Jon Corby wrote:
Fri Aug 28, 2020 7:49 am
Edit: no wait. Of course it works, they would vote "yes" because they're literally being offered 2 coins at that point, not an expected 1.2 coins 🤦
So you're saying our original solution is still right (based on expected gain anyway)? So I don't have to go through it all again?
Yes, I believe so. But I have to admit I don’t think I ever considered it as “expected gain”. But then I am also struggling to see what other rationale I could have used :?
Me too actually, so I'm not sure what went on.

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Re: Pirates Puzzle! (no Googling)

Post by Jon Corby » Sat Aug 29, 2020 1:06 pm

By the way, I'm far happier with the alternative solution being correct though (which is the pirates maximising their guaranteed gain, not 'gambling') and I would always specify this in the rules in future.

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