Tricky Logic Puzzle

Miscellaneous and Personal Stuff I Must Share
21

Robot Arm – if a pirate will choose better or equal probability over a sure thing, then I totally agree with you.

And now that I’m thinking more about it, if the pirate will pick a sure thing over the unknown (albeit good probabilities) – then I see no reason why that bribe of 2 gold even be offered. Offer 1 gold piece to all your bribees every time – because only 1 of them would be guaranteed a bribe at the P-1 case, and they wouldn’t know who it is.

So under my current theory (which relies on a pirate picking a known amount of gold over an unknown amount of gold), the gold needed would be the same as the minimum votes needed – 1 gold piece to each of the bribed.

If pirates choose good probability of more gold over certainty of gold, then this breaks down.

22

Actually, I may need to scratch that. Clearly at P=5, somebody (#1 or #2) must be offered a bribe of 2 gold pieces, or else they will just let it revert to P=4 (in which case #1 and #2 both get a bribe of 1 Gold).

23

You guys have come to precisely the same analysis I have… the problem is undefined without knowing how pirates treat possible things vs. sure things, but with the most obvious behavior (2 guaranteed is better than a chance of 2, but no EV calculations) it always ends up with the guy dividing getting a bunch, the next guy getting 0, then 1, then a bunch of guys some of whom get 2 and some of whom don’t.
The next interesting question is how it works out for very small G. For isntance, what if G = 1:

1 Pirate: 1 takes it
2 Pirates: whatever #2 proposes, he dies, 1 takes it
3 Pirates: 3 takes it
4 Pirates:whatever he does, 3 votes down, and 1 or 2 vote down, so he dies, 3 gets it
5 Pirates: whatever he does, 4 votes up, 3 votes down. So he has to bribe 1 or 2. He ends up alive with no gold (interestingly, the first person to do so)
6 Pirates: no matter who he offers it to, that person votes up and everyone else votes down, he dies (then 1 or 2 end up with it)
7 Pirates: 6 autovotes up, he can bribe any one other pirate to vote up, he votes up, still dies 4-3 (then 6 dies, then 1 or 2 end up with it)
8 Pirates: 7 and 6 autovote up, he bribes one other person , vote is 4-4, he dies, 7 and 6 die, 1 or 2 end up with it
9 Pirates: 8, 7, and 6 autovote up, he bribes one of 1-5, and lives with no gold
10: dies
11: dies
12: dies
13: dies
14: dies
15: dies
16: dies
17: lives with no gold
18 through 32 die
33: lives with no gold

etc., where (assuming I did the math right) only pirates #1 and 3 actually end up with the gold, and all the others die, except for 5, 9, 17, 33, and so forth, where the number for each one doubles -1.
Next up: G = 2! (I excpect that it will end up with a basically identical pattern, except that the interval between guys who live will be slightly smaller.)

24

The obvious answer to me was that they do use expected value calculations. These problems always rely on the participants having flawless reasoning, and a good chance (better than 50-50) at two gold pieces is better than one sure thing.

25

Are you sure you have the problem right? I remember a very similar problem. The only difference is that the seniormost pirate gets to propopose first, and he is killed if the proposal is rejected; then, the next seniormost goes.

26

I forgot; another difference was that the proposal is accepted in the event of a tie.

I found the original statment of the problem I saw.
http://godel.theory.cs.cmu.edu/GTI/archive/2003spring/Materials/Assignments/assignment03/assignment03.pdf

27

I actually registered just to weigh in on this little matter.

Here’s how I see it. The pirates, being infinitely logical and wise, must act in the following ways. (All of this, obviously, works for P values greater than 3, wherein the entire system falls down.)

  1. The juniormost pirate will always propose a scheme where he lives.
  2. This scheme, having satisfied 1, will always give the juniormost pirate the greatest gold possible, in accordance with pirate quality B.
  3. Each pirate knows that the other pirates will act in a same manner, that they will invariably propose a scheme that passes and that will simultaneously give themselves the largest share possible.

The following scenarios assume vast quantities of gold, so that the juniormost pirate will always keep more than he doles out in bribes.

With a solo pirate, the motion obviously passes, as the only guy left gets all the gold.

P=1:
P1- gets G, votes yea
1-0, Scheme passes

With one more pirate, pirate 1 always will vote nay, as after axing the neck of 2, he is the only one left with all the gold. All things being equal, the doomed pirate 2 gives himself all the gold. It matters not. Even if he gave 1 all the gold, 1 will still kill him as he gets the gold anyway and wants bloodshed.

P=2:
P1- gets 0, votes nay
P2- gets G, votes yea
1-1, Scheme fails

-or-

P=2:
P1- gets G, votes nay
P2- gets 0, votes yea
1-1, Scheme fails

With three pirates, pirate 3 votes to save his life, and will automatically get pirate 2’s vote to avoid the above situation. 3 therefore gives himself all the gold, relying on 2’s need to save his own life above all else.

P=3:
P1- gets 0, votes nay
P2- gets 0, votes yea
P3- gets G, votes yea
2-1, Scheme passes

Four pirates complicate things, as all pirates now know that pirate 3’s scheme will pass, and number four knows that 3 will always vote against his scheme, as if it fails, his scheme passes and he gets all the gold. Numbers one and two also know that, if this motion fails, they will get nothing under 3’s scheme. But, to vote for 4’s idea, as they still enjoy bloodshed, they need a small gold bribe. Therefore, pirates 1 and 2 vote for the following scheme, as it’s more than they will get under 3’s scheme, which will invariably pass.

P=4:
P1- gets 1, votes yea
P2- gets 1, votes yea
P3- gets 0, votes nay
P4- gets G-2, votes yea
3-1, Scheme passes

The fifth pirate’s situation is the following. Everybody knows that the above scheme is the only scheme that number 4 can submit. Five will always vote for his own plan to save his own life, and number 4 will always vote against it to get G-2 gold, with the assurance that his own plan will pass. Thus, to get the majority needed, 5 needs to bribe two others that his plan is better than 4’s. But which two? Among pirates 1, 2, and 3, two of them need to be offered a better deal than 4 gives them. In scheme 4, (which will always pass,) the take of 1 and 2 are interchangable, so they can be bribed in similar fashions. The two options are, therefore, to bribe both 1 and 2, or to bribe 3 and either 1 or 2. For either of these two options, 5’s life is spared, so he then turns to the problem of maximizing wealth. To bribe both 1 and 2 requires upping both their takes to 2 gold each for a max of 4 gold. To bribe 3 and either 1 or 2 requires upping one of their takes to 2 and one of them to 1, for a max of 3 gold.

Pirate 5 chooses the second option, maximizing his own take. Whether he bribes pirate 1 or 2 is incidental, as each will guarantee his max take. He bribes pirate 1 for no good reaso, or 2. It really doesn’t matter. There is no use giving gold to someone who will invariably vote against your plan, so he gives nothing to pirate 1 or 2, whichever doesn’t get his first bribe.

P=5:
P1- gets 2, votes yea
P2- gets 0, votes nay
P3- gets 1, votes yea
P4- gets 0, votes nay
P5- gets G-3, votes yea
3-2, Scheme passes

-or-

P=5:
P1- gets 0, votes nay
P2- gets 2, votes yea
P3- gets 1, votes yea
P4- gets 0, votes nay
P5- gets G-3, votes yea
3-2, Scheme passes

Pirate six is the first with some real thinking to do. He knows that pirate 5 will vote against his plan. He knows that he, himself, will vote for it. Thus, he needs to get three more on board for the 4-2 decision. Anything that won’t guarantee that 4-2 will give him the possibility of death, which is intolerable. His first motivation is life. Alas, he doesn’t know whether number 4’s scheme will bribe 1 or 2. One of them will get 2, the other will get 0. Therefore, to ensure one of their votes for 5’s life, he will need to ensure that he bribes them enough that they won’t be enticed by the possibility of getting more than 4. This means that, to be assured of either 1 or 2’s vote, he will need to bribe at least 3 gold, which is more than they will ever get with 4. (Pirates want to maximize their take.) 5 can get away with bribing 3 with two gold, which is more than he will get under any scheme 4 comes up with. And, of course, bribing number 4 is easy, and worth only 1 gold piece.

Therefore, pirate 5s scheme involves bribing the cheapest other pirates (3 and 4) and bribing pirate 1 or 2 with enough money to assure survival. Which one is a matter of choice. Those who will alawys vote against his plan will recieve nothing. (If you’re going to die, you might as well give yourself lots of gold in the process.)

P=6:
P1- gets 3, votes yea
P2- gets 0, votes nay
P3- gets 2, votes yea
P4- gets 1, votes yea
P5- gets 0, votes nay
P6- gets G-6, votes yea
4-2, Scheme passes

-or-

P=6:
P1- gets 0, votes nay
P2- gets 3, votes yea
P3- gets 2, votes yea
P4- gets 1, votes yea
P5- gets 0, votes nay
P6- gets G-6, votes yea
4-2, Scheme passes

Pirate 7’s scheme is complicated yet again. He will vote for his own life, and number 6 will vote against, to recieve the optimal cut of G-6 gold under his own plan. Securing a majority vote of 4-3 will require the bribery of three other pirates. Number 7 does not know if 6’s scheme will bribe pirate 2 or 1 with the 3 gold, and neither do they. The chance of each getting bribed under 6’s scheme is 50/50, which are bad odds for anyone to risk their own life. Therefore, to assure a vote, 7 would have to bribe either of them 4 gold pieces, more than they would get under 6’s plan. But wait! Number 7 doesn’t need to bribe either of them, as he can secure his life cheaper by bribing pirates 3, 4, and 5. (All of them know exactly how much they will get under 6’s plan.) Thus, number seven will always do the following, securing his own life and maximizing his own wealth. There is only one alternative that will maximize wealth and guarantee his own life.

P=7:
P1- gets 0, votes nay
P2- gets 0, votes nay
P3- gets 3, votes yea
P4- gets 2, votes yea
P5- gets 1, votes yea
P6- gets 0, votes nay
P7- gets G-6, votes yea
4-3, Scheme passes

Pirate 8 knows exactly what 7 will do, so his job is a tad easier. He needs a 5-3 vote, knowing that he will always vote to save his own life, and 7 will always vote against, to get a G-6 cut of the gold. Pirate 7 needs to bribe four others. The cheapest others to bribe are pirates 1, 2, 5, and 6. These four will always respond to this cheap bribe as all pirates know exactly what number 7’s scheme will give them.

Note that this particular situation puts pirate 8 in the happy situation of giving away less gold than 7 needed to give away. Strange how these things work…

P=8:
P1- gets 1, votes yea
P2- gets 1, votes yea
P3- gets 0, votes nay
P4- gets 0, votes nay
P5- gets 2, votes yea
P6- gets 1, votes yea
P7- gets 0, votes nay
P8- gets G-5, votes yea
5-3, Scheme passes

The ninth pirate will vote his own life, and pirate 8 will vote against, to get the cut of G-5. The vote needs to be 5-4, so pirate 9 needs to bribe 4 others. As pirate 8’s scheme is assured, pirate 9 knows whom he needs to bribe. He bribes 3, 4, and 7, as they are cheap bribes. He forgets bribing 5, as he’s expensive. He therefore needs to bribe number 1, 2, or 6 with 2 gold pieces, which is more than they get under 8’s plan.

P=9:
P1- gets 2, votes yea
P2- gets 0, votes nay
P3- gets 1, votes yea
P4- gets 1, votes yea
P5- gets 0, votes nay
P6- gets 0, votes nay
P7- gets 1, votes yea
P8- gets 0, votes nay
P9- gets G-5, votes yea
5-4, Scheme passes

-or-

P=9:
P1- gets 0, votes nay
P2- gets 2, votes yea
P3- gets 1, votes yea
P4- gets 1, votes yea
P5- gets 0, votes nay
P6- gets 0, votes nay
P7- gets 1, votes yea
P8- gets 0, votes nay
P9- gets G-5, votes yea
5-4, Scheme passes

-or-

P=9:
P1- gets 0, votes nay
P2- gets 0, votes nay
P3- gets 1, votes yea
P4- gets 1, votes yea
P5- gets 0, votes nay
P6- gets 2, votes yea
P7- gets 1, votes yea
P8- gets 0, votes nay
P9- gets G-5, votes yea
5-4, Scheme passes

And so on and so forth. The permutations get more complicated, but not much more. Each successive pirate needs to get a majority of votes, and therefore bribes whoever is cheapest. The trouble comes when you have a finite amount of gold. For instance, when you have only 5 pieces of gold to divy up, pirate 7 cannot possibly assure his own safety, as he needs bribes amounting to six gold. Pirate 7 will be axed, as the other pirates are greedy. And without gold, they are bloodthirsty. Pirate 7 cannot possibly offer a better deal to enough people to save his skin, so if the route comes down to him, he will get axed. This makes pirate 7 vote for pirate 8’s plan, even if he wouldn’t with an infinite amount of gold, to save his own skin. Assured of pirate 7’s vote no matter the price, pirate 8 no longer needs to bribe pirate 5 the high price of 2 gold, and the scheme looks like this.

P=8, G=5:
P1- gets 1, votes yea
P2- gets 1, votes yea
P3- gets 0, votes nay
P4- gets 0, votes nay
P5- gets 0, votes nay
P6- gets 1, votes yea
P7- gets 0, votes yea
P8- gets G-3 (2 gold), votes yea
5-3, Scheme passes

With this knowledge of pirate 8’s guaranteed proposition, pirate 9 will have a different offer than he would without that knowledge.

P=9, G=5:
P1- gets 0, votes nay
P2- gets 0, votes nay
P3- gets 1, votes yea
P4- gets 1, votes yea
P5- gets 1, votes yea
P6- gets 0, votes nay
P7- gets 1, votes yea
P8- gets 0, votes nay
P9- gets G-4 (1 gold), votes yea
5-4, Scheme passes

Putting a limit on the gold changes votes and incentives eventually down the line of pirates, which changes the needed bribes up the line, which changes everything completely. The infinite gold formulas and logics break down.

Of course, in these schemes, I’m assuming that the pirates whose necks are not on the line differ in their weighings of possibly money in the future, and that they cannot be relied upon unless their bribes outweigh anything they’d get in the future. Therefore, no pirate who proposes a plan can hazard less than a 100% guarantee that he lives, and that the proposing pirate is willing to give up whatever is needed (but not a dime more) to ensure his personal safety.

That took two hours to write. Whew!

28

See, my first interpretation was that pirates would choose a sure result over an uncertain result, probabilities be damned, ye scurvy dogs!

But it goes to show that the problem as listed is not precisely formulated.

I googled a little, and I found this kind of puzzle formulated in a little more specific way (i.e. not an unknown number of pirates P, for instance – that is pretty general, you must admit, and there’s no one clear solution for all possible P).

There are two variations (both more specific, although still not answering the “sure thing” vs. “probabilities” question – although it may not matter since P=5 and P=6 in these problems) listed on this page

AND

29

Nevermind, I just realized that the junior vs. senior thing is irrelevant. In that case, the first guy to make the proposal should suggest that the pirates who go after him should alternatingly get 0 or 1 gold coin and he should get the remainder.

30

I address this problem by assuming that the pirate who’s neck is on the line wants to avoid such confusion by giving away enough bribes to ensure his own safety.

Here’s a similar problem that I ran across.

There’s a teacher and his math class that meet Monday through Friday. One Friday, he tells his class that they will have a pop quiz on a certain day the following week, on Monday, Tuesday, Wednesday, Thursday, or Friday.

He proceeds to tell them that they will not know which day the quiz is given until the quiz is handed out. (It is administered during the middle of the class, so there will be no outward signs of an impending quiz. Papers on desk, that sort of thing.) He also says that, if anyone can walk into the room and can logically 100% prove that the quiz is on that day, and if the student is correct, then he will give the entire class perfect marks on the quiz, without even administering it. To be fair, the teacher promises to not change the day of the quiz.

After class, a student walks up and confronts the teacher, saying that she can prove that the quiz is inadministerable.

Student: “Teacher, the entire class should get full marks on the quiz, because I can prove that you can never actually give the quiz. Imagine we walk in on Friday without having taken the quiz yet. We will absolutely be assured that the quiz will be given that day, because it is the last possible quiz day. Therefore, you will not give the quiz on Friday. Likewise, Thursday won’t work. When we walk into the room on Thursday, we will know the test is that day, because, as I have already shown, it cannot be on Friday, and Wednesday, Tuesday, and Monday will have passed quizless, and therefore Thursday is the only day remaining for the quiz to be given. Therefore, you cannot give the quiz on Thursday. Having nixed Friday and Thursday, Wednesday won’t work either for similar reasons. Likewise, Monday and Tuesday are out. Therefore, you cannot possibly administer the quiz without us knowing the day!”

What is the teacher’s response?

I enjoy brain teasers. :smiley:

31

Here’s how the teacher responds:

Smile sweetly and say nothing. Then administer the test on whichever day (s)he likes. When the students protest, tell them by that their logic, any day would be unexpected!