Shamelessly stolen from you-know-where.
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The ball grid problem
Albert knows that Albert doesn’t know where the ball is, and also knows that Bernard doesn’t know where the ball is, which means the ball is in a row where every ball has another ball in its column, which narrows it to C or D
The information that the ball must be in C or D is enough to exactly select the ball given knowledge of the column, which makes it either C3, D2 or D4
The information that Albert knows which cell it is once Albert knows that Bernard knows means that it must be C3 because if it was a D cell then Albert would still not know
C3 is the ball that was pulled from door 1, thus there was a gold ball behind door 1
Monty hall problem
Which means we dont know whether there were 6 or 3 gold behind door 1, which means its essentially a slightly different Monty Hall problem
The first door either has 2 gold balls (and 3 silver) behind it or 5 gold balls, and the second door either has 3 gold balls and 3 silver or 6 gold balls
The second door has a 66% chance of having 6 gold balls (guaranteed gold) and a 33% chance of having 3/3
The first door has a 66% chance of 2/3 and a 33% chance of 5 (guaranteed gold)
Thus door 1 has a 2/3x2/5 + 1/3 = 3/5 chance of you pulling a gold ball
Door 2 has a 2/3 + 1/3x1/2 = 5/6 chance of pulling a gold ball
You should switch doors to maximise your chance of being allowed to swap tracks
Knowing that the ball was gold gives you Bayesian knowledge about the boxes behind the door, since the prior probability of the host pulling a gold ball from a 6-gold door is different than from the 3/3 door. So you have to multiply Monty Hall probabilities and Bayesian probabilities together.
That assumes the host pulled a ball at random, of course, and not a deliberately gold ball.
Hmm, you’re quite right. My intuition is that the Bayesian portion would exactly offset the Monty hall portion. I think, at a glance, Bayes would give door 1 a 2/3 probability of having 6 gold, but Monty Hall would give door 2 the same probability, so we can effectively cancel these out and just consider a raw probability
You either have 5 gold or 2 gold 3 silver behind door 1, and 6 gold or 3 and 3 behind door 2, which gives door 2 a very slight edge. Does that check out?
Yes! Cancels out, leaving only a very slight edge on door 2. All that work for only… 2.77% edge over picking at random. What a troll problem, huh?
Heh what a trolly problem indeed
I don’t have that many family members, so I don’t know who all those people are. Anyway I’m just gonna pick at random because I am too old and too tired for this shit.
Uh, deal? I walk up to the third door and pull a gold ball from it. I shrug at the madman.
Also, it doesn’t say I can only open one box, and all three doors should still have at least one gold ball remaining.
Doesn’t say the trolley is a runaway. Don’t even need to make deals, you can safely walk away.
It also doesn’t say that the track isn’t already headed away from the people.
Smullyan would be proud.
I don’t see no question.
Commenting before reading other comments
Solution to grid puzzle
The henchmen’s discussion implies that the letter row and number column both have at least two balls in them (required for “I don’t know, but I know you don’t know)”. Bernard’s statement to Albert makes it clear to Albert that the letter must be either row C or D depending on the number he knows.
If it was row D the answer would still be ambiguous to Bernard so it must be C3 and the ball is gold
Solution to overall puzzle
I’ve been successfully nerd sniped and my family is dead.
I had the grid solution wrong, but your explanation is sound! Thanks!
I’m so confused by the added complexity here. If there’s a gold ball behind every door, the choice of door doesn’t matter.
We were asked to pull a ball from behind one of three doors, not figure out what two dudes know about balls in a grid.
I choose the following option:
Kick the madman in the nuts, open all the doors and take all of the balls, switch the trolly track myself, save the people, sell the balls, then go back and tie the madman to the tracks and ask him to pick the door with the black and blue balls. If he doesn’t choose his own fly as the door, TROLLEY.
I’m think it’s implied that you can only take pull out one ball at random from behind a door.
We don’t even get that far. Everytime we try, the dude does something before we get the chance and then changes the rules lol
I'm pretty sure the ball landed in
C3.
Albert is very sure that Bernard doesn’t know either. Bernard would know the location if it was in 5 or 6, indicating to all of us that Albert was told a row that isn’t A or B.
Now that Bernard can also deduce that it’s not A or B, he’s narrowed it down to one possibility. That means all of us now know it can’t be column 1 either, because if it were, he wouldn’t have gotten anything from that new fact.
Finally, now that column 1 is eliminated, Albert has deduced the location. Row D would’ve left two more possibilities, but row C leaves just one. Albert must know it is in row C.
For the rest, well, there isn’t even actually a question, I suspect you’d open a door and pick a box and hope that you’ve got a gold ball to pick, and it’s not clear that he’s following Monty Hall rules and always opening a bad door, but I think knowing which ball got thrown would make the rest of the odds fall into place.
C3.
Thanks for the explanation, but I cannot follow on this line
Since he knows that Bernard doesn’t know given just the row, each ball in that row is in a column that contains more than one ball.
Why is that? Why couldn’t it be A2 or A3? In this case neither Albert nor Bertrand could tell what row/column this was either, because it would be in a row/column with another ball. How can you exclude any row with overlap with any single-ball columns?
It couldn’t be row A or B because there would then be a possibility the ball was in A6 or B5 and that Bernard actually did know where the ball was. Albert says he knows Bernard doesn’t know where the ball is, which means he is 100% confident, and the only rows that allow for that confidence are rows C and D, because each ball in each row is not alone in its column.
Scenario 1: The madman tells you the ball is in row A. Then the madman whispers the column to Bernard. Can you really be sure that Bernard knows or doesn’t know where the ball is?
Scenario 2: The madman tells you the ball is in row C. From this, you can be completely sure that Bernard doesn’t know where the ball is, because any column he would be given doesn’t narrow it to a single row.
Yes and because C3 is a golden ball, you should confidently switch to the second door. Because now it’s just the Monty Hall problem with balls instead of goats. When the madman chose a door to opened, he deliberately chose a bad (mixed) door, otherwise he would have given away the correct location. The fact, that he opened the third instead of the second gives you new information, that you can take advantage of by switching, increasing your chances. Had the ball been silver, it might have been revealed to come from a bad door.
Edit: damn I just realized he picked the ball from the first door, not the second. Okay in that case we might actually have to calculate probabilities, but I’m too lazy for that.
I have no idea what the ball thing is about. I just assumed that since he was a madman, he was just doing madman things.
Can I try again until I get a silver ball? Please.
Yeah, problem doesn’t specify whether you want your family to live. Common mistake.
That’s like the old joke of the CIA loyalty test.
A CIA recruit reached the final stages, at the last step he found his mother tied to a chair and gagged. His prospective boss hands him a gun and says "To prove you are ready for this job and will follow any orders, You’ll need to kill your mother, I’ll be outside.
10 seconds after leaving the recruiter hears the bang of a gun… followed by other bangs, clangs of metal.
Prospective recruit walks out the door… “The god damn gun you gave me had blanks in it, so I had to beat her to death with the chair”
Sounds to me like you’re allowed to see the color of the ball before you take it out. This means it doesn’t matter which door you open, a gold ball will be available to pull and you win. Checkmate.
Doesn’t say it’s your family tied to the tracks either.
It does in the first line?
Your family was kidnapped and is now safely eating ice cream backstage. Tied to the tracks are 5 infinite worlds Hitlers.
Sorry fam, I don’t negotiate with game theorists.
So the only winning move is not to play? Caught you, game theorist.
I think the ball is at C3.
Albert knows that Bernard doesn’t know where it is exactly, that means the ball can’t be in the first two rows, because if it was that would mean there’s a chance for the ball to be in column 5 or 6, in which case Bernard would known right away where it is exactly (as those columns have only one ball). This means the ball must be in row C or D and column 1, 2, 3 or 4. We know Bernard was able to deduce the exact position of the ball from this information, which means it can’t be column 1 (as there are 2 balls still remaining there). That means it must be in column 2, 3 or 4. We know Albert was able to deduce the exact position if the ball from this, and since row D has 2 balls still in play (collumn 2 and 4), meaning Albert wouldn’t be able to guess which one it was, that means it’s in row C, as row C has only one ball left in play.
Now idk what the correct choice would be statistically. If I remember correctly, the Monty Hall problem states that there’s 2/3 of a chance you’ll get it correctly if you switch doors, but since a gold ball was pulled out from behind the first door (the ball at C3 is gold), I think that means there’s a 2/3 chance of the first door (the one from which the ball was pulled out of) being the one with all gold balls (since it contains 2/3 of all gold balls). In that case I think the probabilities cancel out? Which means it doesn’t increase your chances whether you switch or not.
What I don’t understand is how Albert got the information from Bernard.
Written as it is, we can deduce the C3.
But how did Albert understand from Bernards sentence that it’s in the 3rd column?
Beautifully written!
I think the probabilities cancel out?
My conclusion as well. Except that since Jigsaw has taken one gold ball out, door 2 must have slightly better chance of gold ball remaining on average (75% vs 72.2%?).
Hoping to select the door with a goat, that will headbutt the switch for me.
