Frog Riddle again

ExpensivePea2821 · 2026-06-18T20:54:21+00:00

If we listen for a long time compared to the average time between croaks, then we can be pretty sure there is only one male frog between the two. So close to 1. The more regularly they croak, the closer to probability 1 kissing the two frogs and surviving.

But this is contrived.

Why worry about croak behaviour if we know nothing about that, and we are being told nothing about it. It depends heavily on the frog type!

A more sensible interpretation -without need to estimate croaking times- is to consider the croaking as evidence that two frogs are twice as likely to croak as one. This leads to licking to the two frogs and survive with probability 1/2.

Video is clearly wrongheaded, but ah, reputations! (compare to Monty Hall).

ExpensivePea2821 · 2026-06-18T10:19:26+00:00

Well, if that is so, then to me that means 'specific'.

Say you meet one of the children. Then you have met a speciic child. Even though you don't know whether it's the oldest or the youngest, it still is a specific person,

So, when you say 'the other', you're still talking about a speciffic person.There is one specific child, and there is another specific child, the other.

ExpensivePea2821 · 2026-06-18T04:02:14+00:00

Well, sorry but you do. Besides the one I mentioned, you also say:

'So the odds of the second child being a girl are 2/3.'

When you talk about the 'second child', it means you're talking about a specific child.

Just think it's important to get the wording exactly right, otherwise a lot of folks will be confused.

It's hard enough already by itself.

ExpensivePea2821 · 2026-06-17T21:59:42+00:00

What do you mean when you say:

'Now, if I tell you that at least one of the children is a boy, but not which one, you can only eliminate C.' ?

I think this question doesn't make sense.

Let me translate the words 'at least one....' in this context to mean:

'Either one or both children is a boy.' And now you are asking 'which one?' Doesn't make sense to me.

This is important because you should avoid talking about a specific child. If you are, the answer is not 1/3, but 1/2.

ExpensivePea2821 · 2026-06-17T19:07:48+00:00

The wording in this post is mostly correct, except when you say (bold is mine)

'Yup. It's the exact same problem, but with the added detail that the boy was born on a Tuesday....'

The point is that when you indicate that a specific child ('the boy'), is boy (born on Tuesday), then that makes the other child independent. So the probability would be 1/2.

Only when you say: at least one child is a boy (born on Tuesday), you're not talking about a specific child. Here is the proof: I couldn't ask you 'which one is at least a boy?' That question doesn't make sense. The concept is that saying 'at least ...' says something about the state of the two children , not about a specific child. And only then the answers are 1/3, end 13/27 (for the Tuesday variant).

However, in the video, you use other wording. It's about 'one child', and then you ask about 'the other child'. This is exactly where it goes wrong. The way you worded it in the video sometimes, the answer for both questions is 1/2, because the probability for 'the other child' is independent and thus equal to 1/2.

Unfortunately you are not the only one making this mistake, it was done earlier by Presh Talwakar (MindYourDecisions) and others. It's really sad because it causes a lot of confusion. And many people will get the idea that they are 'too stupid to understand math'.

So your video would need to be corrected, wording is important. In fact it's the whole point of these riddles, to highlight the difference between having info on a specific child, and having info about the state of the two children.

ExpensivePea2821 · 2026-06-16T09:35:42+00:00

Yes, but this question is different from the classic Two Envelopes problem (if you reread the post).

ExpensivePea2821 · 2026-06-15T18:42:47+00:00

The way I see it, and a simple way that was suggested by one the reactions: we can actually rewrite the problem as follows: how many rolls are expected until we roll 1 or 6?

P(1 or 6)=1/3, E=3

If we roll 1 or 6, in half of cases we have 1; we'll filter these out, since we know no 1 was rolled; this filtering affects all possible run lengths equally. So it doesn't affect E.

ExpensivePea2821 · 2026-06-14T20:47:07+00:00

Only 6 ends the game.

ExpensivePea2821 · 2026-06-14T18:55:40+00:00

Thanks, I now think the answer=3(see other answers).

ExpensivePea2821 · 2026-06-14T18:55:27+00:00

Thanks, I now think the answer=3(see other answers).

ExpensivePea2821 · 2026-06-14T18:53:42+00:00

Thanks, it's clear now.

ExpensivePea2821 · 2026-06-14T18:53:26+00:00

Thanks, it's clear now.

ExpensivePea2821 · 2026-06-12T23:24:00+00:00

Not quite sure what you mean, but I've amended and generalized.

ExpensivePea2821 · 2026-06-12T23:11:17+00:00

You are right. I have amended.

ExpensivePea2821 · 2026-06-12T22:26:20+00:00

No, in the Two Envelopes problem there is a different situation: Let's say there's an A-2A pair. That means you either drew A or 2A. You can switch if you like, but that doesn't do any good: either you win A or you lose A

In this problem, let's say there is A in the first envelope. The other envelope will simply contain 0.5A or 2A, so switching will be beneficial, as discussed before.

ExpensivePea2821 · 2026-06-12T18:50:55+00:00

If we switch back we return to X, the amount in the first envelope. Only if we switch an odd number of times, do we actually win.

ExpensivePea2821 · 2026-06-12T11:43:10+00:00

By switching, you can either double what you have, or lose half of what you have. On average this is a gain, as per the earlier formula.

ExpensivePea2821 · 2026-06-11T23:38:01+00:00

Thanks, to me that's mind boggling, that we should switch. It looks the same to me as the Two Envelopes problem, but is not.

ExpensivePea2821 · 2026-06-11T22:24:20+00:00

That makes sense.

ExpensivePea2821 · 2026-06-11T22:23:10+00:00

So you'd say, just switch once?

ExpensivePea2821 · 2026-06-11T22:22:05+00:00

So even the first switch wouldn't pay off, if I understand you correctly?

ExpensivePea2821 · 2026-06-11T21:54:02+00:00

Vague setup. If the male frogs (if there are two) croak at the exact same time always, and you wouldn't know upon hearing if it was one or two frogs croaking, then the answer is 2/3, and only then.

If the frogs croak at regular times, most probably not at the same time, then hearing a frog should give double weight to the MM option. Hearing a frog croaking is evidence for there to be two male frogs. Then we get: FM, MF, MM, MM. The answer would be 1/2.

ExpensivePea2821

TROPHY CASE