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Unraveling the Enigma of Hidden Pirate Treasure: A Captivating Escale into the Unknown

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I frequently benefit from the weekly puzzles from FiveThirtyEight’s riddler, Oliver Roeder. Right here’s a superb one from this week:

The Puzzle Of The Pirate Booty:

Ten Completely Rational Pirate Logicians (PRPLs) discover 10 (indivisible) gold items and want to distribute the booty amongst themselves.

Pirate Puzzle POSTThe pirates every have a novel rank, from the captain on down. The captain places forth the primary plan to divide up the gold, whereupon the pirates (together with the captain) vote. If a minimum of half the pirates vote for the plan, it’s enacted, and the gold is distributed accordingly. If the plan will get fewer than half the votes, nevertheless, the captain is killed, the second-in-command is promoted, and the method begins over. (They’re mutinous, these PRPLs.)

Pirates at all times vote by the next guidelines, with the earliest rule taking priority in a battle:

  1. Self-preservation: A pirate values his life above all else.
  2. Greed: A pirate seeks as a lot gold as attainable.
  3. Bloodthirst: Failing a risk to his life or bounty, a pirate at all times votes to kill.

Underneath this technique, how do the PRPLs divide up their gold?

Further credit score: Resolve the generalized downside the place there are P pirates and G gold items.

___________

I feel I received it—click on under to develop and skim my resolution:

[expand title=”SOLUTION“]

Consider it from the view of the lowest-ranked pirate and work up from there.

The bottom ranked pirate (we’ll name him P10 and work our method as much as the captain, who’s P1) would like to have each different pirate die so he can take all 10 gold items for himself. He additionally has no risk to his life, as a result of by the point he’s promoted to captain, it means everybody else is lifeless.

P9 would additionally like to have all pirates above him killed, as a result of if it will get down to only himself and P10, he can allot all 10 gold items to himself and none to P10. The vote will find yourself 1-1 (P9 will vote in favor and P10 will vote towards), and since half of the members voted for the plan, it’ll be enacted and P9 is not going to be killed. So he, too, has no risk to his life.

P8 is aware of all this. He is aware of that if it will get all the way down to the ultimate two pirates, it’ll find yourself as:

P9: 10 items
P10: 0 items

And he is aware of that P10 and P9 know this, so if it finally ends up attending to the purpose the place there solely three pirates left and he’s captain, he’ll have that in thoughts. He might allot the gold items like this:

P8: 10
P9: 0
P10: 0

That’s a lot worse for P9 than the result if P8 is killed, so P9 will vote towards the plan. However P10 is in the identical place as he’ll be if P8 is killed: zero gold items. That is when rule #3 kicks in: the pirates are bloodthirsty. If it gained’t put both a pirate’s life or bounty at stake, “a pirate at all times votes to kill.” On this case, P10 will vote to kill as a result of it gained’t have an effect on his life or bounty both method. So if P8 desires to win P10’s vote, he wants to offer him one gold piece. Then rule #2 kicks in—greed. P10 will perceive that he ought to vote in favor of the plan, as a result of if he doesn’t, he’ll get nothing after P8 is killed, so he’ll vote for the plan and it’ll be enacted. P8’s life might be spared and he’ll make off with an superior 9 gold items. So P8 will allot the items like this:

P8: 9
P9: 0
P10: 1

Within the case the place there are 4 pirates left and P7 is captain, P7 is aware of the entire above and he is aware of that the opposite three pirates all know the entire above—and he solely wants one of many three different pirates to vote in favor of his plan for it to be enacted with a view to spare his life. So he’ll allot them like this:

P7: 9
P8: 0
P9: 1
P10: 0

He is aware of P8 and P10 will vote towards it, as a result of if he dies, they’ll be within the above scenario the place P8 will get 9 items and P10 will get one, and his plan has them each getting zero. However he’ll get P9’s vote, as a result of P9 is aware of that if he votes towards it, P8 will enact his plan and P9 will get nothing. One piece is healthier than zero items, so P7’s plan might be enacted.

You’ll be able to proceed this logic upwards via the ranks. In every case, if the highest-ranked pirate allots one piece to all pirates who might be getting zero items ought to he be killed, his plan might be enacted, so it’ll go like this:

If it will get all the way down to the ultimate 5 pirates, the plan might be:

P6: 8
P7: 0
P8: 1
P9: 0
P10: 1

If it will get all the way down to the ultimate six pirates, the plan might be:

P5: 8
P6: 0
P7: 1
P8: 0
P9: 1
P10: 0

If it will get all the way down to the ultimate seven pirates, the plan might be:

P4: 7
P5: 0
P6: 1
P7: 0
P8: 1
P9: 0
P10: 1

If it will get all the way down to the ultimate eight pirates, the plan might be:

P3: 7
P4: 0
P5: 1
P6: 0
P7: 1
P8: 0
P9: 1
P10: 0

If it will get all the way down to the ultimate 9 pirates, the plan might be:

P2: 6
P3: 0
P4: 1
P5: 0
P6: 1
P7: 0
P8: 1
P9: 0
P10: 1

However nobody will die in any respect, as a result of the unique captain will allot them like this:

P1: 6
P2: 0
P3: 1
P4: 0
P5: 1
P6: 0
P7: 1
P8: 0
P9: 1
P10: 0

He’ll get votes from P3, P5, P7, P9, and himself to create a 5-5 break up—and his plan might be enacted.

So the answer is: 6, 0, 1, 0, 1, 0, 1, 0, 1, 0

___________

Observe about collusion: Collusion at first looks like it might present an fascinating wrinkle. For instance, P9 and P10 might collude and determine that as a substitute of P9 voting to enact P1’s plan (which might give him one gold piece and P10 zero gold items), they might each vote no, after which proceed to vote no time and again, stopping a majority vote till they’re the one two left. This might be nice for P9, as a result of he’d get ten gold items as a substitute of 1, and likewise good for P10 as a result of, regardless of ending up with zero items in each circumstances, he’d have killed extra individuals within the latter case, and as per rule #3, pirates are bloodthirsty, so he’d fairly try this than P1’s plan. However this fails, as a result of P10 would fairly have one gold piece than kill these different pirates (since rule #2, greed, trumps rule #3, bloodthirst). So in the event that they went with this collusion, as soon as they received to the ultimate three, P10 would break the settlement and vote sure to P8’s plan to grab his final likelihood at one gold piece. Figuring out this may occur, P9 would vote sure to P7’s plan. This might proceed upwards till the highest, so P9 would finally vote sure to P1’s plan, as described within the authentic resolution.

P9 might attempt to thwart this downside by promising P10 two gold items if he agrees to vote no every time till they’re the one two left—however that fails too, since as soon as there are solely the 2 of them left, P9 has no incentive to honor the plan and can take all ten for himself. For this reason all collusion inevitably fails.

___________

Further Credit score

The additional credit score problem is: Resolve the generalized downside the place there are P pirates and G gold items)

The second in command at all times will get 0, the third in command at all times will get 1, and that alternates—0, 1, 0, 1…—all the best way down. Then the primary in command will get no matter’s leftover.

This works high-quality so long as there are sufficient gold items for the primary in command to dole out one piece to only beneath half of the opposite pirates. If not, he’s in hassle, and so might be each pirate under him till sufficient pirates are killed off that the primary in command has sufficient gold items to make use of to purchase the votes he wants. Or, till G ≥ P/2 – 1 (if P is even) or G ≥ (P-1)/2 (if P is odd).

Further Credit score add-on:

Shit. As commenter gpsimms identified, the additional credit score is extra sophisticated than I gave it credit score for. Within the case that there’s sufficient gold to cowl the requisite votes (G ≥ P/2 – 1 (if P is even) or G ≥ (P-1)/2 (if P is odd), as described above), what I stated within the first additional credit score part is right. However when there’s not sufficient gold, issues get extra sophisticated.

Let’s say there’s just one gold piece. Right here’s what occurs (I bolded those that will vote “sure”):

If there’s just one pirate (P = 1):

P10: 1 (He takes the gold piece.)

End result: Nobody dies.

If P = 2:

P9: 1
P10: 0

End result: Nobody dies.

If P = 3:

P8: 0
P9: 0
P10: 1

End result: Nobody dies.

If P = 4:

P7: 0
P8: 0
P9: 1
P10: 0

End result: Nobody dies.

If P = 5:

P6: 0
P7: 0
P8: 0
P9: 0
P10: 1

End result: P6 dies, plan fails, strikes onto the following spherical.

If P = 6:

P5: 0
P6: 0
(votes sure as a result of he’ll die within the subsequent spherical if he doesn’t)
P7: 0
P8: 0
P9: 0
P10: 1

End result: Nobody dies.

So the fascinating factor is that due to what would occur right here when P = 6, P5 and P6 are protected. They haven’t any likelihood of dying—one thing they’ll concentrate on when voting in P ≥ 6 conditions.

If P = 7:

P4: 0
P5: 0 – votes no as a result of they’ll survive within the P = 6 spherical and he’d fairly kill P4 first.
P6: 0 – votes no as a result of they’ll survive within the P = 6 spherical and he’d fairly kill P4 first.
P7: 0
P8: 0
P9: 1
P10: 0

End result: P4 dies.

If P = 8:

P3: 0
P4: 0 – is aware of he’ll die within the subsequent spherical if P3 is killed
P5: 0 – is aware of he’ll be saved when P = 6
P6: 0 – is aware of he’ll be saved when P = 6
P7: 0
P8: 0
P9: 1
P10: 0

End result: P3 dies.

If P = 9:

P2: 0
P3: 0
– is aware of he’ll die when he’s captain if P2 is killed
P4: 0 – is aware of he’ll die when he’s captain if P2 and P3 are killed
P5: 0 – is aware of he’ll be saved when P = 6
P6: 0 – is aware of he’ll be saved when P = 6
P7: 0
P8: 0
P9: 1
P10: 0

End result: P2 dies.

If P = 10:

P1: 0
P2: 0
– is aware of he’ll die when he’s captain if P1 is killed
P3: 0
– is aware of he’ll die when he’s captain if P1 and P2 is killed
P4: 0 – is aware of he’ll die when he’s captain if P1, P2 and P3 are killed
P5: 0 – is aware of he’ll be saved when P = 6
P6: 0 – is aware of he’ll be saved when P = 6
P7: 0
P8: 0
P9: 1 – votes sure as a result of if he doesn’t, it’ll whittle all the way down to P = 6 and he’ll get no gold items when that plan is enacted
P10: 0

End result: Nobody dies.

As soon as we get to P > 10, the entire above 10 pirates know they’re protected. P1 – P5 will all be saved when P = 10. In order that they’ll cease voting sure until there are gold items allotted to them. So P11 – P17 are all screwed till we get to the purpose the place P = 18. Then, P11 – P18 will all vote sure to save lots of themselves, together with one decrease pirate who P18 provides the gold piece to.

So when there’s one gold piece, nobody dies when P = 1, 2, 3, 4, 6, 10, 18, 34, 66…

The sample, beginning at P = 4, is the final protected quantity instances two, minus two (2n – 2).

That’s for G = 1. Rising the variety of gold pieceschanges the “protected sample.” You may create a generalized formulation for P and G, however that sounds very icky, so I’ll go.

[/expand]

_______

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Extra puzzles on Wait However Why:

The Stick Determine puzzle

The Three Jellybean Drawback

The Two Envelopes Drawback

The Infinite Checkerboard Quandary


And one very massive puzzle:

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