1781

This is a terrible riddle. Good riddles have simple, fore-head smacking answers. The answer to this one seems to involve really heavy and incredibly boring math. I actually think I’ll tell this riddle to people but use Rich’s answer as the actual answer.

1782

How can you hear a croak unless at least one of the pair is male?

1783

Well, this is a good puzzle in the sense that there’s an obvious answer which is right for the wrong reason, a seemingly clever answer which is just plain wrong, and a slightly more subtle correct answer.

In both the frogs and the CWB puzzles, we start with four equally likely events: FF, MF, FM, or MM. And both puzzles would like us to rule out FF without providing any other information, leaving MF, FM, and MM equally likely. But, being puzzles, they both give us some event which rules out FF but which also generally changes the relative chances of the remaining possibilities. The wrong answer to CWB implicitly invokes the bizarre assumption that girls in boy-girl families never stand in windows, and the wrong answer to the frogs puzzle implicitly invokes the bizarre assumption that two male frogs croak at the same rate as one male frog.

1784

I guess to me “puzzle” is different than “riddle” though. I hear riddle and I expect to feel dumb after I hear the (now very obvious) answer.

For example,

We hurt without moving.
We poison without touching.
We bare the truth and the lies.
We are not to be judged by our size.

What are we?

Words.

Though seeing the very first page of this thread, the first post to me is a puzzle and not a riddle, so I am almost certainly entirely wrong.

1785

how can you have any pudding if you don’t eat your meat!

1786

I thought the stackexchange guy was implying in the answer that neither of the frogs could be male, given his FF0 probability in the chart but I was wrong. I just misinterpreted things. The video posted was a better explanation IMO, the stackexchange guy does 2 things: lays out the probability space in text clearly, and confuses the issue by adding the probability that a male will croak §. I feel like it’s splitting hairs a bit with this riddle to go that far.

Although I still can’t come to terms with the answer. I mean, what if there were 21 frogs and you know 20 of those frogs have just croaked (because you’re a wizard or something)? How does the presence of males next to the maybe-female change the actual probability at all of the remaining frog actually being a female. It does not follow via common sense. Stuff like the Monte Hall problem I get after thinking it through, but this one I do not get. “Get” from a common sense perspective, obviously the math is fine.

1787

arrendek, don’t forget the explanation posted in the video is very clear but completely wrong.

It’s not that hearing one frog croak changes the sex of the other frog, it’s that hearing one frog croak and not hearing the other frog croak gives us information that we should generally use to update our beliefs about whether there’s a female in the pair.

First consider the sole frog. Before we learn anything, we think the frog is a girl with probability 0.5. Then we listen for a bit and we don’t hear it croak. If we know we’d hear the frog croak for sure if it’s a male, then we would conclude that it must be female. At the other extreme, if we knew that neither male nor female frogs croak, then not hearing a croak tells us nothing and we’d still think the frog is a girl with probability 0.5. Generally, a bit of math tells us that the probability the frog is female given we do not hear it croak is 1/(2-p), where p is the probability we’d hear a male croak. (This is why we need to know the probability we’d hear a male frog croak.)

Now consider the pair. What’s the probability that the other frog is female, given we hear one of them croak? The bit of math in the stackexchange post shows that it’s still 1/(2-p) if we assume that frogs croak independently. So in that case your intuition is correct: hearing one frog croak and not hearing the other croak is just like looking at one frog and not hearing it croak.

If croaks are dependent, then we’d get a slightly trickier answer. For example, suppose that male frogs croak to attract a female, and if there’s already a female present, a male frog doesn’t croak. Then if we heard a croak from a pair of frogs we’d know with certainty the other frog is also male. On the other hand, suppose that male frogs never croak if there’s another male present, because that would cause a fight. Then if we heard a croak from the pair, we’d know with certainty that the other frog must be female.

1788

And the thread is a math thread again. ;)

1789

Definitely. It’s a deductive consequence of the premise that only male frogs croak.

Only male frogs croak. (Or, all croaking frogs are male.)
A frog croaked.
Therefore, the frog that croaked is male. (Or, there exists a male frog.)

Probability literally doesn’t come into it when it can be concluded from modus ponens. Phrasing typically matters more than anything in these puzzles.

1790

Here’s another puzzle, the Blue-eyed Islander problem. It only requires logic, although it is advanced logic. There’s no trouble from probabilistic vagueness at least.

https://www.math.ucla.edu/~tao/blue.html

1791

That’s good. My answer is that all the blue-eyed people kill themselves on the 100th. The logical flaw with the other argument is that everyone already knew that there was at least one blue-eyed person, but the fact there is at least one blue-eyed person was not common knowledge before the speaker’s announcement. That is, I knew there was at least one blue-eyed person, and so you did you, but I did not know that you know, and so on.

1792

Correct. It’s similar to this old joke: http://spikedmath.com/445.html

Or if you want to get into a hardcore analysis of the actual logic involved (which is easier to understand intuitively), there’s Terry Tao’s own explanation, in long form.

1793

I admit I’ve always struggled with this particular problem. So I don’t understand your explanation. Every native knows that if there are 99, 100, or 101 blue eyed natives, then every native can see at least one blue eyed native. And every native knows one of the following:

(1) There are 99 or 100 blue eyed natives
or
(2) There are 100 or 101 blue eyed natives

Therefore, every native knows that every native knows that there is at least one blue eyed native. What does the explorer add?

EDIT:

Consider a variant, in which the explorer instead says:

“So, according to your custom, if there were one blue eyed person on this island and he discovered his eye color today, he would kill himself tomorrow at noon?”

That’s not a statement of fact at all, just a restating of the rules which are common knowledge. But the following day, nobody would kill himself, and therefore n=1 would be excluded. Would the rest of the induction follow the same way?

1794

Truly the riddle of the ages.

1795

Ok, I think I finally understand why the blue eyed native problem bugs me so much. The premises are self contradictory. You cannot have a stable population of perfect logicians ready to kill themselves once they learn their eye color, they should already be dead.

Suppose instead of logicians, we had blue eyed zombies. They are not logicians but they perfectly obey your instructions:

  1. If you see more than two blue eyed zombies, you should conclude that everyone sees at least one blue eyed zombie.

  2. Once you have concluded that everyone sees at least one blue eyed zombie, count the number of blue eyed zombies and add one. Wait that number of days, and then if there are still any blue eyed zombies left, kill yourself.

This heuristic has identical output to the perfect logicians.

Now every Sunday, you put one blue eyed zombie on an island. I think it’s clear that every third Wednesday, all the zombies kill themselves. There is no way to reach a population of 100 in the first place. And if they magically appeared on the same day the explorer arrived, it wouldn’t matter what he said.

1796

Think of a couple very simple examples:
There is 1 villager total who has blue eyes. He now knows he has blue eyes and kills himself.
There are 2 villagers, the first with blue, the second with brown. The one with blue sees the other has brown, knows it is him and kills himself on the first day. The second one sees that the first has blue eyes so thinks it might not be him and when the first killed himself the second knows that he must not be blue too.
There are 2 villagers, both blue. The first sees the other is blue so he waits. The second sees the other blue so he waits. When neither of them kill themselves the first day then they both know that they have blue eyes otherwise the other would have killed themselves.

The thing that is interesting (to me) is that on day 101 all of the brown eyed people would kill themselves and it is only stopped because the 100 beat them to it.

Before the person talks, they know of themselves that they have one of many eye colors. Once the person notes a specific eye color then he creates two groupings, people that see all X of the existing people with that color and people that see X-1 of that eye color. And from that, due to the killing rules, he separates the two groups into one that thinks the blue people should kill themselves on the Xth day and one that thinks blue people should kill themselves on X+1 day.

1797

The explorer’s statement is just there to support the n=1 case for the induction to work. In the general case, I think magnet’s right that there’s no stable state where n>1.

You could start the induction at n=2, as magnet did, and the explorer become superfluous.

1798

There’s more than just the possibility of blue or brown eyes. In an xkcd variant of the blue-eyed islander problem, a new villager with green eyes shows up & tells the others that he sees someone with blue eyes. The green-eyed villager & the other brown-eyed villagers have no idea (each individually) what their own eye color is.

Let’s do a variant that you can play at home. Everyone picks a card from the standard deck & without looking at it sticks it on their forehead. Everyone else can see your card, but you can’t. Let’s say everyone happens to have drawn a red card. Now someone else, someone not playing the game, comes up & says “I see a red card.”

Do you know what color card you have? Of course not, unless you’re the only one playing. You know everyone else has a red card, but you haven’t heard anything to indicate that your card matches theirs. (And your situation is identical to everyone else’s, so it doesn’t matter whose perspective we choose.)

This problem is logically similar to the blue-eyed islander problem.

Now the outsider asks us if we know what color card we have. If there is only one person playing (n=1) then we immediately say yes. We have the red card. But for n>1, the first person says “I don’t know.”

If n=2, the first person saying they don’t know tells us, what we did not know before, that they also see a red card. Then the second person knows they have a red card.

For n=3, we know that if there were only 2 red cards, the second person would have known their card color, so after 2 “I don’t know” answers, the 3rd person knows their card must be the 3rd red card.

This continues by induction. If there are n red cards, then after n-1 “I don’t know” answers, the nth person, who can only see n-1 other red cards & knows that the (n-1)th person would have known their card if there were only those n-1 red cards, now knows that their own card must be the nth (& last) red card.

In the case of the blue-eyed islanders, each day that passes with nothing happening is equivalent to one “I don’t know.” The inaction itself is a statement of epistemic uncertainty. Since each blue-eyed islander can see n-1 blue-eyed islanders, they know that it would take n-1 days for them to reach certainty if there were n-1 blue-eyed islanders. But when nothing happens after n-1 days, they know for certain that there must be more than that. Since they can see everyone but themselves, they conclude, inescapably, that they themselves must be the additional nth blue-eyed islander.

The xkcd version of the problem has islanders who discover their own eye color leave by ferry instead of killing themselves. It is rather pleasanter than the traditional version.

1799

I should add that, in the case of the blue-eyed islanders (though not in the case of the red cards), for some lower bound the outsider probably becomes unnecessary to start the chain reaction. I wanna say n=3, but I’d have to think about it. Three guarantees that everyone knows that everyone sees at least one blue-eyed islander.

So if there’s one green-eyed islander, 100 blue-eyed islanders, and 900 brown-eyed islanders, only the green-eyed islander won’t eventually discover their own eye color. But again, I’m just conjecturing on this point right now.

1800

So you guys actually think that this:

is somehow “better” than the probability-based puzzle?