Riddles and puzzles

[Loki]

Spoiler:

The idea of risk doesn't exist with perfect information. Risk is something that can't be measured. We could measure every relevant variable here. Solution 1 works only if the pirates aren't sure if the other pirates are rational. Solution 3 requires P6 to be irrational. There's no element of risk because each pirate knows how the others will respond to every possible move.

Spoiler:

I guess we can think of it this way. Let's say 6 people were getting to play the 6 crew scenario over and over. Suppose they play it once per second for every second that passes for a year. if you could have a 50/50 chance of getting 2 dollars every second or to get 1 dollar garaunteed every second, what would you pick. They're statistically identical in the total money earned. So we live eitherway, we statistically make the same money eitherway, so let's go for the kill. That's one way to think about it... However, they're only equal for you in the long run, so since we're playing this game once (you don't constantly find chests of 100 gold), I think you're better off taking the 1. My ultimate vote is, the solution you pointed out, after doing the correct maths of 50%(0) + 50%(2), is the correct one. So I think P6: 97 p5: 0, P4: 1, P3: 0, P2: 1, P1: 1 is correct. Again, I know I would take it.

Spoiler:

If I'm rational, I'm indifferent between the two outcomes. The expected return is identical. Regarding your other point, there aren't multiple equilibria, which means it doesn't really matter if we play this game multiple iterations.

[RandBlade]

Spoiler:

I guess we can think of it this way. Let's say 6 people were getting to play the 6 crew scenario over and over. Suppose they play it once per second for every second that passes for a year. if you could have a 50/50 chance of getting 2 dollars every second or to get 1 dollar garaunteed every second, what would you pick. They're statistically identical in the total money earned. So we live eitherway, we statistically make the same money eitherway, so let's go for the kill. That's one way to think about it... However, they're only equal for you in the long run, so since we're playing this game once (you don't constantly find chests of 100 gold), I think you're better off taking the 1. My ultimate vote is, the solution you pointed out, after doing the correct maths of 50%(0) + 50%(2), is the correct one. So I think P6: 97 p5: 0, P4: 1, P3: 0, P2: 1, P1: 1 is correct. Again, I know I would take it.

Spoiler:

No its wrong. P6 won't offer it as he can't risk his own life in order to get 1 more gold. Life takes priority over gold so he won't offer this one in the first place as he can't guarantee his own life.

Plus 1 gold + a death > 1 gold alone.

[Loki]

It's not a matter of risking his own life. Based on the rules of the game, we know (and he knows) with 100% certainty that he will be killed if he makes that offer.

[RandBlade]

Spoiler:

The idea of risk doesn't exist with perfect information. Risk is something that can't be measured. We could measure every relevant variable here. Solution 1 works only if the pirates aren't sure if the other pirates are rational. Solution 3 requires P6 to be irrational. There's no element of risk because each pirate knows how the others will respond to every possible move.

Spoiler:

If I'm rational, I'm indifferent between the two outcomes. The expected return is identical.

Spoiler:

Solution 3 is only irrational for P6 if he knows that the other 2 pirates won't risk their gold to kill another. Given that information, it also makes sense that Solution 1 doesn't work for the exact same reason - P6 refuses to offer it as he won't risk his life for gold.

Without that certainty Solution 3 is the rational solution as its the only solution where 3 voters are guaranteed to be better off.

[Loki]

Just what is he uncertain about (I really don't think there's any more point to spoiler tags)?

[RandBlade]

It's not a matter of risking his own life. Based on the rules of the game, we know (and he knows) with 100% certainty that he will be killed if he makes that offer.

Only if a guarantee of 1 gold is valued the same as an average of 1 gold.

Many theories would say that's not true.

[Loki]

Those many theories assume irrationality.

Are you assuming that the pirates don't know n (the probability that the 2 coins are given to P2 by P5)?

[RandBlade]

Those many theories assume irrationality.

No they don't, diminishing returns (or diminishing utility) is a real concept. If 0 gold means not being able to afford grog, 1 gold gives grog and 2 gold gives more grog then they might prefer the guarantee of grog over the possibility of no grog.

[Lebanese Dragon]

I disagree with loki to simply check statistical long run equivalency in expected returns, in a game you're only playing once. Even if the expected returns are equal for each distribution function you should take the one with the least variance (since you're only playing once).

Also Rand, P6 knows exactly how the game will turn out, he never risks his life.. He is the same person, intellectually as P5, P4, P3, P2, and P1. He knows when he'll die. If he'll get one more gold and not die, he'll do it. It's a matter of us determining if he'd get that one more gold, of us determining do P2, and P1 accept. It's clear they accept in the case we all agree on, it's not as clear if they accept in the other case. Loki's saying it's irrational to accept

Spoiler:

97 p5: 0, P4: 1, P3: 0, P2: 1, P1: 1

but I disagree if you're only playing once, becuase the other one is only equal in the long run.

[Loki]

No they don't, diminishing returns (or diminishing utility) is a real concept. If 0 gold means not being able to afford grog, 1 gold gives grog and 2 gold gives more grog then they might prefer the guarantee of grog over the possibility of no grog.

Sure, their utility function can have a diminishing returns component. We were not given that information though.

Let's even assume that only P5 knows what n is. P2 and P1 don't know what n is. There's a huge range of outcomes they can each expect, and in only one of those outcomes (where n equals to either 1 or 0) are they better of rejecting the offer of 2 coins. They are thus best off accepting P6 offer with a certain probability (which is somewhere near 1). P6 knows this. I suppose if he values his life at infinite coins, he'd make the more generous offer in solution 3. That information is not available to us, however.

[Loki]

I disagree with loki to simply check statistical long run equivalency in expected returns, in a game you're only playing once. Even if the expected returns are equal for each distribution function you should take the one with the least variance (since you're only playing once).

Also Rand, P6 knows exactly how the game will turn out, he never risks his life.. He is the same person, intellectually as P5, P4, P3, P2, and P1. He knows when he'll die. If he'll get one more gold and not die, he'll do it. It's a matter of us determining if he'd get that one more gold, of us determining do P2, and P1 accept. It's clear they accept in the case we all agree on, it's not as clear if they accept in the other case. Loki's saying it's irrational to accept

Spoiler:

97 p5: 0, P4: 1, P3: 0, P2: 1, P1: 1

but I disagree if you're only playing once, becuase the other one is only equal in the long run.

That's only true if you're risk-averse. I'm also not quite sure you chose a good example for having a game with many iterations when a pirate can get killed in any game, and that death is final.

[RandBlade]

That's only true if you're risk-averse. I'm also not quite sure you chose a good example for having a game with many iterations when a pirate can get killed in any game, and that death is final.

Or if you don't have diminishing marginal utility.

Either way its irrelevant as the first pirate won't risk his life on it.

EDIT: The fact that life is valued at infinite coins is given to us.

[Loki]

I'm not doubting that you can; I'm saying this information was not provided.

Why do you assume that the pirate values his life at infinity? He's a pirate. Surely if his life meant that much to him, he'd choose a safer profession.

[RandBlade]

I'm not doubting that you can; I'm saying this information was not provided.

Why do you assume that the pirate values his life at infinity?

That information is given to us:

These particular pirates are all perfect logicians, and want to stay alive, get as much gold as possible, and see as many of the other pirates die as possible, in that order of priority. They'll never sacrifice from one of the higher priorities for a lesser priority.

Life is a higher priority, and they'll "never" sacrifice from a higher prioirty to a lower.

He's a pirate. Surely if his life meant that much to him, he'd choose a safer profession.

Well if you want to think like that, then the next question I suppose would be that in any of the solutions certain pirates are guaranteed to get nothing - in which case why bother sailing at all? If they were perfectly rational they'd surely not have gotten themselves into this situation in the first place?

[Lebanese Dragon]

Either way its irrelevant as the first pirate won't risk his life on it.

EDIT: The fact that life is valued at infinite coins is given to us.

He knows because it's the highest priority listed above all else. Furthemore P6 knows exactly when he'll die, the other pirates never randomly pick yes or no. They will either pick yes or no every time, and he knows exactly what they'll pick, becaues he thinks exactly like them.

That's only true if you're risk-averse

I simply think in this case, as these pirates are logicians, they would know it's better in playing one game to not gamble getting gold or no gold. (even though long run is the same amount of gold).

It's hard for me to argue that point (about the risk-aversion) with a hole proof argument. I'm content with saying

Spoiler:

P6: 95, P5: 0, P4: 1 and 2 gold is given to any two between P1-3

is a clear accepted solution and the next most money for P6 possible other than the situation in question.

[RandBlade]

He knows because it's the highest priority listed above all else. Furthemore P6 knows exactly when he'll die, the other pirates never randomly pick yes or no. They will either pick yes or no every time, and he knows exactly what they'll pick, becaues he thinks exactly like them.

I simply think in this case, as these pirates are logicians, they would know it's better in playing one game to not gamble getting gold or no gold. (even though long run is the same amount of gold).

I think you're wrong. The key words were these:

These particular pirates are all perfect logicians, and want to stay alive, get as much gold as possible, and see as many of the other pirates die as possible, in that order of priority. They'll never sacrifice from one of the higher priorities for a lesser priority.

Priority 1: Stay alive
Priority 2: Get as much gold as possible
Priority 3: Kill

Getting as much gold as possible is a priority, now they can gamble their 1 expected gold on getting more gold. That's not sacrificing from one priority to a lesser priority so is perfectly valid. In fact you could argue that taking 1 guarantee would be rejected as its not "as much as possible" as 2 was possible [even if they end up getting nothing]

[Lebanese Dragon]

I guess in the list of priorities given, risk-aversion, even if it seems very clear isn't listed as a factor. In this case the following, which ironically i came to using the wrong expected return is what we all agree on.

Spoiler:

P6: 95 gold, P5: 0 gold, P4: 1 gold and 2 gold is given to any two between P1, P2, and P3

[RandBlade]

Time for a new puzzle and this one is so stupidly easy its ridiculous. Saw the following on Facebook and the really scary thing is how many people are getting it wrong

[Wraith]

Spoiler:

7

[Ominous Gamer]

Yeah, this isn't really a puzzle. Its grade school math Order of operations.

[Wraith]

I think you're wrong. The key words were these:

These particular pirates are all perfect logicians, and want to stay alive, get as much gold as possible, and see as many of the other pirates die as possible, in that order of priority. They'll never sacrifice from one of the higher priorities for a lesser priority.

Priority 1: Stay alive
Priority 2: Get as much gold as possible
Priority 3: Kill

Getting as much gold as possible is a priority, now they can gamble their 1 expected gold on getting more gold. That's not sacrificing from one priority to a lesser priority so is perfectly valid. In fact you could argue that taking 1 guarantee would be rejected as its not "as much as possible" as 2 was possible [even if they end up getting nothing]

They're perfect logicians, trading 1 gold for a 50% chance at 0 or 2 gold is a wash. Therefore, they go to their next priority, and vote to kill somebody.

P=5 case with >50% voting can be reduced to P5=97 P4=0 P3=1 P2=1 P3=1. You can go from there to get the solution to P=6.
I think this solution is correct:

P6: 95 gold, P5: 0 gold, P4: 1 gold and 2 gold is given to any two between P1, P2, and P3

[Loki]

Only if we assume P5 will choose which pirate to give money to with a 50% probability. That information is not given (I would imagine it would be in a formal game).

[RandBlade]

Yeah, this isn't really a puzzle. Its grade school math Order of operations.

Agreed but sadly most people are saying 1 or 5.

[Loki]

Not everyone is smarter than a first grader.

[Loki]

I have a bunch of examples from my old textbook. Don't have the answers to most of them.

The game is played between two players, Player A and Player B. Each player has a penny and must secretly turn the penny to heads or tails. The players then reveal their choices simultaneously. If the pennies match (both heads or both tails) Player A keeps both pennies, so wins one from Player B (+1 for A, -1 for B). If the pennies do not match (one heads and one tails) Player B keeps both pennies, so receives one from Player A (-1 for A, +1 for B). What's the optimum strategy for the two players?

[DecoyMilk]

Easy one:

You're in a room with no windows and one door. There are 3 lightswitches on the wall, A, B & C

Upstairs, in a room that you cannot see into other than being in the room, is a lightbulb connected to the electricity supply.

One of the lightswitches turns the bulb on and off. How can you find out which one, using only one trip to the room with the lightbulb in?

[RandBlade]

Not everyone is smarter than a first grader.

The frustrating thing is how cocky and arrogant the dumb idiots can be. Comments like

You idiots there's no parantheses so you need to go from left to right. Anything times zero is zero so the answer is 1. I used my calculator and it says 1.

Or even better:

Don't you remember PEMBAS (then something about Aunt Sally). It should be done as
6 - 1 x 0 + 2 / 2 = 6 - 0 + 2/2 = 6 - 0 + 1 = 6 - 1 = 5

How do these people remember to breath properly? And they get to vote.

[Ominous Gamer]

Don't you remember PEMBAS (then something about Aunt Sally). It should be done as
[...]
6 - 0 + 1 = 6 - 1 = 5

That math has nothing to do with PEMDAS, please excuse my dear aunt sally Thats just....bad math.

[Loki]

Parentheses, exponents, multiplication, division, addition, subtraction. It does apply there and it gives you the proper result.

[RandBlade]

Sorry PEMDAS. I never learnt that mneumonic (I just learned how to do maths properly). The Aunt Sally types are saying as A comes first ...