Can you solve the pirate riddle? - Alex Gendler

Selfishness does not a great captain make.

bro just give 20 each

Daniel would obviously keep all the gold. LIKE IM SORRY HUH?

why did the answer make me happy hahaha

I thought it would be 33 for A, C and E... never thought of this solution!

Yawa lisuda sabton oy

what genius. i would have just gone nay nay nay nay 1 coin or 0? id take 0 and nay

1st are you so damaged that you feld obliged to violate historical facts (no women pirates) just to avoid BS critisism ?
2nd experimental psycology found that humans have a very strong sense for injustice against themselfs that are willing to tolerate only small percentages of unjustice, up to 50%.
after that they choose to be doomed together with the oppressor. So, .... 99 to 1 , definetely is not the correct answer......

Thanks Ted.

Why not everyone takes 20 gold coins.

Says you can't communicate or bribe... immediately suggests to attempt to bribe. Besides how silly and nonsensical the logic is here, it is a funny game.

Female pirates? Yeah, right! 😂😂😂😂

98 1 1 is the answer

Charlotte got the GYAT

Never let them know your next move: Amaro votes nay after proposing the 98-0-1-0-1 distribution

Maybe A B C D gets 25 gold while E gets 0 :)😊

The Nash Equilibrium is a nice theory for theoretical situations; the assumption that because of shared knowledge everyone will come to the same conclusion or solution is very one-dimensional thinking.

You forgot the most important piece of logic "We are pirates and we don't follow the rules"

I’ll throw out my guess: A gives pirates C and E one coin each and keeps the rest.

If pirate E becomes captain, he will get all the coins

If pirate D becomes captain, he will get all of the coins (his best-case is proposing 100 for himself and 0 for E, they will vote, it will be a tie, and proposal D will be accepted.

If pirate C becomes captain, his best bet is offering E 1 coin to win his vote. C’s proposal will win 2-1 and C will keep 99, D will keep 0, E will keep 1.

If B becomes captain, His best bet is offering D 1 coin. D knows that he will get nothing if C becomes captain so he accepts the deal. Proposal B wins 2-2. B keeps 99, C keeps 0, D keeps 1, E keeps 0.

If A becomes captain, his best bet is offering C and E one coin each. They know they will get nothing if B becomes captain, so they take the single coin. Proposal A wins 3-2

I got it right but me not being a pirate offered C and E 33 coins each and A keeps 34. If the game is to repeat, A is better off giving 2 coins each to C and E and 1 coin each to B and D, keeping 94 for A. Otherwise he won't have enough support for the next mission!

In the scenario where Amaro offers Eliza one coin, why doesn't she vote nay? She gets the same amount if Charlotte becomes captain. It would make more sense for Eliza to get 2 coins in that scenario
Feel free to correct me, but I think that's true?

Nope. Read some of Thayler and other behavioral economists. In practice, people make irrational decisions. Also, this “riddle” is fundamentally flawed because it doesn’t take into account what happens in the following time periods. So many flaws with this “riddle”. Poorly done.

he said game thoery

I don't think it is necessary to give the coins to C and E. He can give it to any 2 of the pirates and get their votes

Nice, I had the same exact thought process. Eliza is in worst position. Daniel can keep all gold and give her none if they are the only remaining, then I thought that Charlotte can have 90 coins and give 10 to Eliza, Eliza would say yes. Then with Bart, he would give Daniel 10 coins and he'll keep 90. Then with Amaro, he has to get 2 votes in so I thought he'd have to give Daniel and Eliza 1 coin. Isn't this another way to win the riddle, because, if the vote gets to Bart, he would give the lowest amount possible to Daniel anyway so getting 1 coin from Amaro or 1 coin from Bart is the same. Also Eliza always gets 1 coin in best case.

But Eliza would also get 1 coin if she rejects Amaro's plan and goes with Charlotte's. So, I think it should be 97-0-1-0-2 for full insurance

good solution:
"EVERYBODY GETS 20 COINS"

А могли бы каждому по 20 монет дать

may be not nash equilibrium but it needs to use backward induction

I thought and found the answer to be A=98, D=1 and E=1 coin. Wouldn't this be also a right answer since in this case also, D and E will support A as that is the best they can get

😀i like ted ed

I should get all the gold, since i have green eyes

Good job Amaro, now try that solution with a real pirate crew.

I almost got this, I forgot that we need to consider the weakest voters on their then (their turn as captain). Position 3 & 5 are weak in a sense that they need to follow another’s actions for their own to be successful.

I was initially thinking to give position 2 & 3 one coin each to secure their votes. But was thinking that position 2 would probably collaborate with position 3 to secure a larger booty for himself.

In summary, tap the weakest knowing that they know they are weak themselves. Position 5 is the weakest followed by position 3.

I feel like this explanation is completely incorrect. Nobody is gonna be cool with taking only 1 coin while another person gets 99.

What should do ?
Me: kill the others

2nd last guy after voting out 5th guy well I will get more by myself if I vote out fourth guy and can make the last girl say yes by giving her 2 coins, so I WIN! LAUGHS!

1st observation (I suspect this will require working from the end up, lol, this type of problem often does...well, maybe not, but this problem feels to me like it will): if there are only two pirates left, 4 and 5 (easier than names for me, lol), then 4 can propose to get all 100 coins and the other one can't "object", lol, so 4 would would like it to get to that point, and 5 would not, so 5 would not vote in a way that would get things to that point, lol...I have suspicion, without working it out, that Amaro will have to get very few coins, lol, to stay alive...like give a large amount to the others...perhaps not equal, lol, one to the other...at any rate, lol...

pirates are so smart these days

Charlotte looking DAYUMMMMMM

If all the rules are the same but ties do not pass (and everyone knew), here's how it would go
Eliza would deny the proposal as she'd get all the gold and see Daniel perish.
Charlotte would keep all the gold for herself, and Daniel would vote yarr to ensure he's never the pirate captain.
Bart would give Daniel and Eliza one coin each to secure their votes as he needs at least 3 votes (including himself)
98:0:1:1
Amaro needs to make his scenario beneficial for 2 people to secure their votes, so 2 people (to save coins) can be reduced to zero. Obviously no point giving Bart any coins
So Amaro should propose something like
97:0:1:2:0 or
97:0:1:0:2
To secure daniel or elizas vote, they'd need at least 2 coins each. No point giving them each 2, as Amaro can win over charlottes vote with 1 coin.
This scenario basically means that ties are allowed but the captain cannot vote.

However, what if the captain still cant vote but ties still don't pass? Tricky
Eliza would vote to keep all the gold and there'd be nothing daniel could do.
Charlotte already has daniels vote secure but she still needs elizas. Eliza would obviously vote nay to ensure she receives all the gold and daniels would not be enough.
If either charlotte or daniel are the captain, their deaths are immediately guaranteed.
So Bart would choose to keep all the gold, and to save their own lives, charlotte and daniel would agree.
If Bart's plan is accepted, charlotte, daniel and eliza still get nothing and the plan would pass.
Amaro needs a whopping 3 votes out of 4 (as Amaro cannot vote).
So Amaro should choose to distribute the coins 97:0:1:1:1

Therefore, even when votes dont pass and the captain cannot vote, Amaro is paradoxically only loosing 1 coin over having the captain vote and ties pass

Сделать бы сайт с решением задач про пиратов.
1)С пользовательским количеством пиратов
2)Может ли сам пират, предлагающий разделение денег голосовать?
3)Какая доля нужных голосов? (>=x, >x)

this is not nash equilibrium.
if "Each pirate is an expert in logical deduction", this is not solution. contrary to the ordinary course of life

the solution is 20-20-20-20-20 or 22-21-20-19-18.

What about 6 pirates, or 7 pirates,...... or generally n pirates?

No one accepts 1 or 0 coin while the 1st captain gets 98 coins

this is what the politicians do to get elected, most of us are Eliza!

What a bout pirate a 34gold c 33gold and e 33gold?

historically pirates should trust each other quite a lot, i mean, their whole industry basically was maintained over trust. also since they used numbers most of the time for intimidation tactics to get their booty, killing each other for fun was probably not a good option

if we assume that bart is the quartermaster, the real distribution on that crew would probably be:

A:28
B:21
C:14
D:14
E:14

and 9 in expenses to fix the boat and the like