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Do you think the solution to this riddle is wrong?

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clem84

clem84

Gold Member
Can you solve the frog riddle?

This is why I think his conclusion is wrong. When you cross out possibilities from the sample space, knowing that there is at least one male lets you cross out two possibilities. FF can be eliminated. MF or FM makes no difference because you KNOW that at least one of them is wrong. So you're left with MM and (MF or FM. Either way there's at least one female in there). If he goes to the clearing with two frogs, his odds of survival are 50% so regardless of where he goes, he still has 50% chance of survival.
 
So you're saying the MF and FM possibilities are redundant? Cause I was kinda thinking that. If you imagine a hypothetical that you know which of the two frogs croaked, you'd know which one was male. Let's say it was the first one. Therefore, that would eliminate the possibility of it being a FM because you know it can only be either MM or MF. Although nothing has changed in reality in light of the hypothetical, it makes sense for the possibility now to be 50% instead of 66%.

When I was first prompted to make a choice I thought either was fine because it made sense that they were both 50/50. Then when the video introduced the sample space I conceded that that makes sense and I was wrong. But then now I feel like the sample space is flawed in how it allows for both of the MF and FM possibilities in in light of my hypothetical.
 
MF and FM are two possibilities for the type of frog though. Just because they are both "the same" in terms of frog count does not reduce their likelihoods or conditional probabilities. This makes sense intuitively, as there should be more than a half chance if there are two frogs right?
 
I was thinking that the male frog was the one with the antidote leaving me very confused.

But no, I agree with the solution. Four possibilities, one is impossible, two of the options that are left make sure that you survive.
 
Trigonometrize. said:
So you're saying the MF and FM possibilities are redundant?
Click to expand...

Well, kind of. We know one of those 2 is wrong so we're left with 2 possibilities out of 4 originally. So 50%. I don't know much about conditional probability so I'd like for someone to tell me if I'm wrong.
 
My way of thinking is that the male
calling out could very well have attracted a mate, explaining the second frog. Males would likely fight with each other, although that's in no way guaranteed. Still, that's why I would choose the clearing regardless of fancy probability shenanigans.

Then it occurred to me that we were only told that we heard the croak from that direction. So it could very well be that there's a third frog hiding behind the two, like in some bushes or something, meaning the chance that the two might save my life could be even higher.

Yeah I know, that's not what this was about. Sorry. I'm the type that would not eat a poisonous mushroom anyw*gets clubbed*
 
CTLance said:
My way of thinking is that the male
calling out could very well have attracted a mate, explaining the second frog. Males would likely fight with each other, although that's in no way guaranteed. Still, that's why I would choose the clearing regardless of fancy probability shenanigans.

Then it occurred to me that we were only told that we heard the croak from that direction. So it could very well be that there's a third frog hiding behind the two, like in some bushes or something, meaning the chance that the two might save my life could be even higher.

Yeah I know, that's not what this was about. Sorry. I'm the type that would not eat a poisonous mushroom anyw*gets clubbed*
Click to expand...
I thought these too, but it wouldn't count purely on outcome probability.
 
clem84 said:
Well, kind of. We know one of those 2 is wrong so we're left with 2 possibilities out of 4 originally. So 50%. I don't know much about conditional probability so I'd like for someone to tell me if I'm wrong.
Click to expand...

But there aren't two possibilities wrong.

FM and MF are both possibilities.

According to your logic there is only 33% chance since you know only one of the three options can be right.
 
Took a bit of thought, but I ended up agreeing with the video's solution.
It's almost intuitive to assume that MF and FM are redundant choices, but it ignores the actual sample space where the frogs have distinct genders even if you as the observer don't know which one is correct. That leaves a total of three distinct possibilities: MF, FM, or MM. This gives a 66% chance of survival.
 
Cyan said:
Is there a text version of this?
Click to expand...

You are in a forest and have been poisoned. The only cure for the poison is to lick a particular species of frog. Unfortunately for you, there are two types of frogs, male and female, and they are identical except for the fact that male frogs make a croaking noise.

You see a frog to your right, and you hear a croaking noise, which leads to you seeing two frogs to your left. You can only go in one direction. What are the odds of survival in each case?
 
I remember this. It's permutations vs combinations. One is where the order matters ( MF is different from FM) and the other is where order doesn't matter.
 
Cyan said:
Is there a text version of this?
Click to expand...

you need to find a female frog. They occur at a 1:1 rate of male frogs in the wild. You can identify male frogs only by their sound. You see two frogs sitting someplace and you hear one of them make a sound. What are the odds that the other is a female?
 
Murugo said:
Took a bit of thought, but I ended up agreeing with the video's solution.
It's almost intuitive to assume that MF and FM are redundant choices, but it ignores the actual sample space where the frogs have distinct genders even if you as the observer don't know which one is correct. That leaves a total of three distinct possibilities: MF, FM, or MM. This gives a 66% chance of survival.
Click to expand...

I don't see how that's not a flawed sample space. The positioning of the F and the M in the MF versus FM is arbitrary, just like in real life when you go to lick the frogs because you are, in fact, licking them both. It IS redundant.
 
clem84 said:
Well, kind of. We know one of those 2 is wrong so we're left with 2 possibilities out of 4 originally.
Click to expand...

Huh? Either frog could be either gender. Each frog is independent of each other and we don't know which frog is male until we do the test so MF or FM are both valid possibilities.

With your logic, out of the 3 possibilities we know two of them will be wrong so it should actually be 33%.

Unless I'm not following what you're trying to suggest.

PeskyToaster said:
I remember this. It's permutations vs combinations. One is where the order matters ( MF is different from FM) and the other is where order doesn't matter.
Click to expand...
I don't have the vocabulary to describe it since I never studied this in school or anything but this is what I was thinking and where OP is getting confused.
 
siddhu33 said:
You are in a forest and have been poisoned. The only cure for the poison is to lick a particular species of frog. Unfortunately for you, there are two types of frogs, male and female, and they are identical except for the fact that male frogs make a croaking noise.

You see a frog to your right, and you hear a croaking noise, which leads to you seeing two frogs to your left. You can only go in one direction. What are the odds of survival in each case?
Click to expand...

Krejlooc said:
you need to find a female frog. They occur at a 1:1 rate of male frogs in the wild. You can identify male frogs only by their sound. You see two frogs sitting someplace and you hear one of them make a sound. What are the odds that the other is a female?
Click to expand...

Ok, these are actually two different problems, which might help explain why people are confused. For the problem as siddhu outlines it, the answer is 2/3 as other people have explained. For the problem as Krejlooc outlines it, the answer is 1/2.
 
Oreoleo said:
Huh? Either frog could be either gender. Each frog is independent of each other and we don't know which frog is male until we do the test so MF or FM are both valid possibilities.

With your logic, out of the 3 possibilities we know two of them will be wrong so it should actually be 33%.

Unless I'm not following what you're trying to suggest.
Click to expand...

What if we DID know that frog 1 was male and only frog 2's gender was unknown. Why would having that knowledge change the chance of survival? It wouldn't, so something is wrong elsewhere in the logic, and that is that the sample space doesn't support arbitrary changes in ordering when ordering has no effect on reality when you're going to lick both.

Cyan said:
Ok, these are actually two different problems, which might help explain why people are confused. For the problem as siddhu outlines it, the answer is 2/3 as other people have explained. For the problem as Krejlooc outlines it, the answer is 1/2.
Click to expand...

Ha, this is an interesting scenario considering the nature of the probability at play here.

My thesis on this is that reality is one way. Having information does not change reality, so you need to look at how you're constructing your sample space and the way it most reflects reality. Information does not change that.
 
Prophet Steve said:
I was thinking that the male frog was the one with the antidote leaving me very confused.

But no, I agree with the solution. Four possibilities, one is impossible, two of the options that are left make sure that you survive.
Click to expand...

Me too. I was really confused at first.
 
I overthought it a bit, but I came to the right conclusion of
going left
. My thinking was that
if you hear a male croak from the left, and if you assume that you have equal numbers of males and females, then that automatically gives you a better chance of the other being female because you now have uneven numbers.
 
Cyan said:
Ok, these are actually two different problems, which might help explain why people are confused. For the problem as siddhu outlines it, the answer is 2/3 as other people have explained. For the problem as Krejlooc outlines it, the answer is 1/2.
Click to expand...

Why is it half in the other description? Is it because his description specifically asks for the odds of the other frog being female?
 
serenewarfare said:
I overthought it a bit, but I came to the right conclusion of
going left
. My thinking was that
if you hear a male croak from the left, and if you assume that you have equal numbers of males and females, then that automatically gives you a better chance of the other being female because you now have uneven numbers.
Click to expand...

Eh, that would only apply if the eveness of the males and females was in effect for the sides of the forest separately. If that were the case, yes, the left would have a better chance, but they are all apart of the same pool, or at least there was no indication that they weren't, so that logic didn't apply.
 
I wasn't thinking about math at all. I thought the tree stump frog was the female and the two frogs in the clearing were males trying to attract it. Makes more sense.
 
In the possibility space MF and FM are not as likely to occur as MM because there are actually two MM's. Let M be the male that you heard and m be the male you're assuming, then the possibility space is [Mf Mm fM mM], thus F is only 50% chance, like basic instinct implies.
 
siddhu33 said:
Why is it half in the other description? Is it because his description specifically asks for the odds of the other frog being female?
Click to expand...

The odds of the other frog being female and the odds of him surviving after licking the two frogs are the same.
 
liliththepale said:
In the possibility space MF and FM are not as likely to occur as MM because there are actually two MM's. Let M be the male that you heard and m be the male you're assuming, then the possibility space is [Mf Mm fM mM], thus F is only 50% chance, like basic instinct implies.
Click to expand...

Yes! That is a good way of inversely saying what I've been trying to say. Thank you!
 
I watched the video and came to the same conclusion as the narrator.

Basically, since both frogs's sexes are independent of the others, you have to count both FM and MF as separate situations.
 
Trigonometrize. said:
I don't see how that's not a flawed sample space. The positioning of the F and the M in the MF versus FM is arbitrary, just like in real life when you go to lick the frogs because you are, in fact, licking them both. It IS redundant.
Click to expand...

The way I see it is that the information you're given matters as much as the frog's actual genders. The reason they're not arbitrary is because an individual frog's gender is not arbitrary. It's just information that you're lacking.

All you're given is that you have is that you have two frogs, and at least one of the frogs is male, but you don't know which one. You have to consider both FM and MF because the gender of an individual frog is distinct, and changing the order of the frogs does change the order of their genders. Among the four possible combinations FF, FM, MF, and MM, you can only cross out one, which is FF, since this combination violates "at least one is male". You have absolutely no information on exactly which one is male, so the combinations MM, MF and FM still stand.

Given #M >= 1, you reduce your sample space to {MM, FM, MF}. Then P(#F >= 1 | #M >= 1) = 2/3.
 
Murugo said:
The way I see it is that the information you're given matters as much as the frog's actual genders. The reason they're not arbitrary is because an individual frog's gender is not arbitrary. It's just information that you're lacking.

All you're given is that you have is that you have two frogs, and at least one of the frogs is male, but you don't know which one. You have to consider both FM and MF because the gender of an individual frog is distinct, and changing the order of the frogs does change the order of their genders. Among the four possible combinations FF, FM, MF, and MM, you can only cross out one, which is FF, since this combination violates "at least one is male". You have absolutely no information on exactly which one is male, so the combinations MM, MF and FM still stand.

Given #M >= 1, you reduce your sample space to {MM, FM, MF}. Then P(#F >= 1 | #M >= 1) = 2/3.
Click to expand...
This one makes the most sense. I think. You wouldn't count MM twice, since it is the same outcome.
 
FiggyCal said:
This one makes the most sense. I think. You wouldn't count MM twice, since it is the same outcome.
Click to expand...

MF and FM are also the same outcome when you're licking them both. It's the same as Mm and mM.

Starfish Hero said:
But we don't. We hear the croak and we see two frogs. Which is male is unknown.
Click to expand...

Playing shrodinger's cat with genders does not affect reality. There is a way that you create the sample space that doesn't fuck with alternate realities, and that is by being consistent with either including redundancies or not: [FM MF Mm mM] or [MF MM]
 
liliththepale said:
In the possibility space MF and FM are not as likely to occur as MM because there are actually two MM's. Let M be the male that you heard and m be the male you're assuming, then the possibility space is [Mf Mm fM mM], thus F is only 50% chance, like basic instinct implies.
Click to expand...

Trigonometrize. said:
MF and FM are also the same outcome when you're licking them both. It's the same as Mm and mM.



Playing shrodinger's cat with genders does not affect reality. There is a way that you create the sample space that doesn't fuck with alternate realities, and that is by being consistent with either including redundancies or not: [FM MF Mm mM] or [MF MM]
Click to expand...

It doesn't work like that.

Start with two frogs that you have no information of.

What are the chances there is one male and one female?

What are the chances there are two males?

What are the chances there are two females?

Now apply the information.
 
clem84 said:
Can you solve the frog riddle?

This is why I think his conclusion is wrong. When you cross out possibilities from the sample space, knowing that there is at least one male lets you cross out two possibilities. FF can be eliminated. MF or FM makes no difference because you KNOW that at least one of them is wrong. So you're left with MM and (MF or FM. Either way there's at least one female in there). If he goes to the clearing with two frogs, his odds of survival are 50% so regardless of where he goes, he still has 50% chance of survival.
Click to expand...

Your reasoning is wrong. MF and FM are 2 unique possibilities out of 3. Just because you know one of them is wrong doesn't reduce their weight to 50%.

Their answer is correct. 2/3 possibilities have a female in the pair. So 67%.
 
Trigonometrize. said:
MF and FM are also the same outcome when you're licking them both. It's the same as Mm and mM.
Click to expand...

You heard one frog croak. Either frog can be male (FM or MF). Or both frogs can be male (MM).

I don't know why we're adding what this imaginary person is assuming, since that was never part of the question.
 
Trigonometrize. said:
MF and FM are also the same outcome when you're licking them both. It's the same as Mm and mM.
Click to expand...

Uhhhh...what?

You're doing it wrong.

Code: 
Frog 1  |  Frog 2
M           M
M           F
F           M
F           F

Those are the four possibilities. You don't count MM twice.

It doesn't matter that MF and FM give you the same outcome. That's actually the point of the riddle...what's the probability of getting that outcome, which is 2/3 after eliminating 1/4.
 
Korey said:
Uhhhh...what?

You're doing it wrong.

Code: 
Frog 1  |  Frog 2
M           M
M           F
F           M
F           F

Those are the four possibilities. You don't count MM twice.
Click to expand...

Why count MF twice as FM when you're licking them both but then also not count the ordering of the double males? Once someone explains that to me I'll concede defeat. It's like a disingenuous perspective of what's going on. I don't see why we should be constructing our sample space inconsistently for no reason.
 
Regardless of whether you lick them at the same time, you get the probability based on which frogs exist!
 
Trigonometrize. said:
Why count MF twice as FM when you're licking them both but then also not count the ordering of the double males? Once someone explains that to me I'll concede defeat. It's like a disingenuous perspective of what's going on.
Click to expand...

Because order has nothing to do with it.

Frog 1 can be male or female.
Frog 2 can be male or female.

If you follow that those are the possibilities. We aren't switching them around.
 
Korey said:
Those are the four possibilities. You don't count MM twice.
Click to expand...

No it's not. That's only the table if you don't know either. You can't just scratch off FF and assume the rest of the table remains valid. Since we know at least one of the frogs is male, that changes it so there's definite-male and possible-male, ergo MM gets counted twice.

dM pM
dM pF
pM dM
pF dM

It's not that complicated.
 
I think it's still 50%, since you don't know which frog croaked, only that one of them did.
If the left frog croaked, the possibilities would be MM and MF.
If the right frog croaked, the possibilities would be MM and FM.

Either way, this could probably be tested.
 
Trigonometrize. said:
Why count MF twice as FM when you're licking them both but then also not count the ordering of the double males? Once someone explains that to me I'll concede defeat. It's like a disingenuous perspective of what's going on.
Click to expand...

Because you're looking at the gender of each frog individually, when determining the # of possibilities.

Imagine if there was such thing as 3 genders, M F A.

Frog 1 could be M F or A. Frog 2 could be M F or A.

Code: 
Frog 1  |  Frog 2
M           M
M           F
M           A
F           M
F           F
F           A
A           M
A           F
A           A

There would be 9 possible combinations of what those frogs could be. They are two distinct individuals.

Swap out frogs for two people named Jesse and Casey or something.

This is basic probability...
 
serenewarfare said:
I overthought it a bit, but I came to the right conclusion of
going left
. My thinking was that
if you hear a male croak from the left, and if you assume that you have equal numbers of males and females, then that automatically gives you a better chance of the other being female because you now have uneven numbers.
Click to expand...

Isn't that the gambler's fallacy?
 
Dongs Macabre said:
I think it's still 50%, since you don't know which frog croaked, only that one of them did.
If the left frog croaked, the possibilities would be MM and MF.
If the right frog croaked, the possibilities would be MM and FM.

Either way, this could probably be tested.
Click to expand...
This is based on Monty Hall problem which has been empirically tested.

Monty_problem_monte_carlo.svg
 
liliththepale said:
In the possibility space MF and FM are not as likely to occur as MM because there are actually two MM's. Let M be the male that you heard and m be the male you're assuming, then the possibility space is [Mf Mm fM mM], thus F is only 50% chance, like basic instinct implies.
Click to expand...

No because order doesn't matter. You're double counting. In this setup Mm and mM should have half the likelihood of whatever MM had.
 
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