A zoologist is traveling through the rainforest when he is bitten by a venomous snake...

A zoologist is traveling through the rainforest when he is bitten by a venomous snake. He knows that only the female of a certain species of frog native to this rainforest secretes the antidote to this venom, on its back. He also knows that males and females of this species are in equal proportion, look exactly the same, but only the male croaks.

Luckily he sees such a frog a few meters in front of him. But then he hears a croak from a male behind him and turns around to see two frogs of this species a few meters away in the other direction.

Knowing he only has enough time to run in one direction and lick the pair of frogs or the lone frog before the venom kills him, which should he choose and what is his chance of surviving?

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en.wikipedia.org/wiki/Boy_or_Girl_paradox
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I've always been shit at probabilities, but I think it shouldn't matter whether he chooses the first frog or the pair of frogs.

Regardless of which option he chooses, he still has the knowledge that one of the frogs from the pair is a male. At this point, there should(?) be an equal chance that either of the remaining frogs is female. With a total population of 100 frogs, we know know that either choice has a 50/99 chance of being a female.
Although,
>I've always been shit at probabilities

This is good so far, but there's also the chance that he doesn't catch any frogs at all. In which case, it would be better to run toward two frogs, in case the second one is close enough if he can't catch the first one.

You're correct about boyh choices giving the sane chance of survival, but for the wrong reason. And your calculation is wrong. You can't assume there are only 100 frogs, plus you're missing something big.

Nope, he will catch the frogs.

Partial credit is about as much as I had hoped for. I chose a population size of 100 as a cop-out for doing actual math.

Guess I'll wait for a smarter person to come along

Thinking apart from statistics I would go for the lone frog.

He doesn't know which of the two frogs croaked, so he could grab a male frog when he turns around. So the probability of grabbing a male frog would be 75%, assuming he doesn't know which one is male (I think). But going after the one right in front of him would be a 50% chance of grabbing either male or female.

amiright?

Also on top of this, knowing that there is one male frog behind him means that there is a higher chance of a female frog being in front of him.

2 frogs.
Possibilities are
Mm mf fm

For the single frog is a 50/50 chance.

You have a 2/3 chance to survive licing 2 vs 1

Why do people get so mindfucked by basic applications of Bayes' rule

it's not that hard to understand

Keep in mind that you know one of the two frogs is male, so at least one of the frogs is irrelevant in the pair.

So if you turned around just a moment sooner and saw which of the two frogs made the croak, thereby eliminating, your odds of survival would drop to 1/2 if you went over and licked the remaining frog?

Frogs in a pair likely indicate some kind of mating behavior, and he knows one is a male so he could lick the other one.

you mean 1/3, males are the ones you DONT want
Meaning that your odds, understandably, increase to 1/2.

No, it's 2/3 because you can lick both.

If the problem stated you could only lick 1 of the pair then it would be 50/50 in all cases

Not 50/50 sorry. You d lick the one in front because there's a 1/3 chance that neither are male

Male frogs croak to attract females, so it has to be the pair of frogs because the male has attracted a female. Probably. So higher chance is with the females, not so much of math but more knowledge on frogs.

Holy fuck Veeky Forums, how have none realized it's the exact same problem as the one with 3 doors and 1 car

if he licks the wrong frog he might get HIGH

They didn't study enough isomorphisms, or category theory in general

I may be adding complexity to the question, but if two frogs are together, isn't the chance of them being opposite gender more than 50/50 due to mating instinct?

This sounds like a retelling of the Monty Hall problem, but there are some differences, such as there could be two female frogs.

The question seems to assume that there is no chance of the frogs escaping the zoologist. I'd go for the two frogs, but I think the chance is 50/50.

It's not the same as Monty Hall, because in the game show it was guaranteed that exactly one door had the prize. Here there could be 2,1, or 0 females.

It's also not the same as Monty Hall because for MH to work it HAS to be that the host is explicitly going to open a door THAT IS NOT THE ONE THE PLAYER ORIGINALLY CHOSE

This is an essential part of MH. A lot of people who expose the problem forget to state this because they don't understand the MH problem, which results in more confusion than there should be.

American education everyone.

take the female frog out of your pocket and lick that

if you left your froggo at home, lick the pair because of this then lick the last frog too

this is an example of how observation can change probability

So basically, a guy encountered 3 frogs and 1 is male. Okay, then he is equally likely to get a female frog from whichever of the two remaining frogs.

He can lick both frogs behind him. Also, your probabilities are not correct.

How do you know this species croaks to attract females? It could be a mechanism to assert dominance over a territory to other males, which would make the other frog male. Don't assume information not given to you.

All of you have failed. I'll give you a hint: the fact that he only heard one croak is very important.

All you know is that only the males croak. The frog in front has a 50/50 chance of being male or female. There's definitely at least 1 male behind him. That means his chance of getting a male in front is 0.5 because of the equal sex ratio but it's 0.75 for the two behind him because he knows at least one is male, and one is 50/50.

If you average the probabilities behind him (0.50+1)/2 = 0.75.

There's no point really to averaging because he knows at least one behind him is male so that brings up his chances of accidentally picking one dramatically.

>The frog in front has a 50/50 chance of being male or female.
Incorrect. This would only be true if male frogs fail to croak as often as females, which is impossible.

>That means his chance of getting a male in front is 0.5 because of the equal sex ratio but it's 0.75 for the two behind him because he knows at least one is male, and one is 50/50.
That's just so wrong I don't even know where to begin...

Actually, he does math without irrationals. He's got no problem with real numbers, as long as they're also rational.

If you're op then you're changing the game. You said only males croak, you did not say how often or that they need to croak.

Also, in front is 50/50. Behind is 2/3rds.

MM MF FM FF are the possibilities of behind pre-croak

MM MF FM are the possibilities behind post croak. MF and FM are distinct cases, therefore 2/3rds, therefore go back.

You're trying to pull some clever bs like "o haha you know they're both female since they didn't croak!" but you didn't write the question correctly to have that be the case.

Its like that ball problem in boxes but the opposite.
1/3 chance of finding a female from the croak direction. 1/2 of it being from the not croak direction.

Clearly he should sit down and calculate what he should do

The male frog doesn't matter to the zoologist. The croak means that one of the frogs behind identified itself as of no interest. There is one frog of potential interest behind him and one in front of him, each having a 50% chance of being female, so he's got the same chance of survival with either choice.

That's not how probability works

>If you're op then you're changing the game. You said only males croak, you did not say how often or that they need to croak.
You don't need to know how often they croak to see that is impossible for a frog which you didn't hear croak to have the same chance of being male as female. That would imply males are as likely to croak as females (i.e. never), which we know cannot be true since we heard a male croak.

>MM MF FM are the possibilities behind post croak. MF and FM are distinct cases, therefore 2/3rds, therefore go back.
You're assuming these are all equally likely to produce a croak. If the chance of a single male croaking while you were listening is x then the chance of hearing a single croak from two males is 2x(1-x). The chance of hearing a single croak from a male and female is x. Given that x cannot be 0 since we heard a male croak, these are only equally likely when x = 1/2.

But if x = 1/2, then the chance that the lone frog which we did not hear croak is female is 0.5*1/(0.5*1+0.5*0.5)= 2/3.

So running to the pair of frogs confers no advantage!

The true probability of survival is 1/(2-x) regardless of which direction you run.

Yes I did write the question to fool both naive internists AND people familiar with the boy girl paradox. And it worked. You assumed things instead of thinking about all the information given.

*naive intuitionists

You never stated the likelihood of males croaking, you just stated the possibility of croaking. A male may never croak or may constantly croak, these are things that you do not know. It does NOT mean that every second a frog is not croaking makes it more likely to be a female. After all, maybe male frogs only croak when they think of the song Closer by The Chainsmokers and only the frog who croaked has heard it and the other one hasn't and therefore will never croak.

>x cannot be 0
You've just made an assumption that all males croak at the same rate. x does not have to be a single variable, as you never stated "only males croak, and they croak at a rate (or likelihood) of x"

Therefore we can go ahead and discount x since making any decisions about it relies on assumptions that we don't have proof for, and just treat it as a boy girl problem.

>You never stated the likelihood of males croaking, you just stated the possibility of croaking.
Yes. Therefore x > 0 and chance of survival cannot be 1/2.

>A male may never croak or may constantly croak, these are things that you do not know.
Wrong. Since we heard a male croak the probability is nonzero.

>It does NOT mean that every second a frog is not croaking makes it more likely to be a female.
It literally does. Since females never croak while males sometimes croak, a non-croaking frog is always more likely to be female.

>After all, maybe male frogs only croak when they think of the song Closer by The Chainsmokers and only the frog who croaked has heard it and the other one hasn't and therefore will never croak.
This simply proves it is possible to not hear a male frog croak. But we are talking about the probability of male vs. female. Females never croak and there female is more likely.

>You've just made an assumption that all males croak at the same rate.
No I did not. I specifically said that x is the chance of a make frog croaking, not a rate. This chance exists regardless of the variability in rates across the frog population.

To math it out more:

chance of mm with one male croaking: x+y
chance of mf croaking: w (Since the change of a male croaking may be reliant on the presence of a female)
chance of lone male not croaking: z

Where none of these variables are related without assumptions.

So guess what, your problem, which intended to fool others only fooled yourself.

No, because x for male 1 != x for male frog 2. You'd have to make an assumption that x is constant across all male frogs. You didn't even say all male frogs croak, you just said only male frogs croak. Only men have gone to the moon, that does not mean all men go to the moon.

>Wrong. Since we heard a male croak the probability is nonzero.

Wrong since we don't know it's a constant probability.

>It literally does. Since females never croak while males sometimes croak, a non-croaking frog is always more likely to be female.

We do not know that all males croak, only that some males croak. Therefore the probability that it is a male AND croaks decreases, but the probability that it is a male does not decrease.

>This simply proves it is possible to not hear a male frog croak. But we are talking about the probability of male vs. female. Females never croak and there female is more likely.

Not if the probability of all males except that one male that croaked is 0, which is totally possible. Maybe this male had a mutation and he's the only male who croaks.

>No I did not. I specifically said that x is the chance of a make frog croaking, not a rate. This chance exists regardless of the variability in rates across the frog population.

Sorry, should have written, "You've just made an assumption that all males have the same chance of croaking"

This.

>chance of mm with one male croaking: x+y
>chance of mf croaking: w (Since the change of a male croaking may be reliant on the presence of a female)
>chance of lone male not croaking: z
None of this contradicts anything I said. Regardles of whatever assumptions you add to the problem, there is still a chance x of a random male croaking, which cannot be 0. And the chance of a random male not croaking is 1-x.

>No, because x for male 1 != x for male frog 2.
By definition it is. You don't seem to understand probability. If you had information to distinguish between the two frogs in that manner, then you would be right. But you don't. You inky have the infuriating given in the problem, which is the only information I used.

>You didn't even say all male frogs croak, you just said only male frogs croak. Only men have gone to the moon, that does not mean all men go to the moon.
Why do I have to keep repeating this? The chance that a man has gone to the moon is nonzero, unless you have information that this man does not hasten to be one of the astronauts who did. A random male frog has nonzero chance of croaking since at least one male croaked. Not to mention that your original answer was based on the assumption that x HAD TO BE 0, not simply that it could be 0.

>Wrong since we don't know it's a constant probability.
This is nonsense. Probabilities are measures of how much information you have. Probabilities change when you gain new information. What new information is in the problem?

>We do not know that all males croak, only that some males croak. Therefore the probability that it is a male AND croaks decreases, but the probability that it is a male does not decrease.
Again whether specific males croak is irrelevant since the probability describes the entire population.

>Not if the probability of all males except that one male that croaked is 0, which is totally possible.
This possibility is included in the probability of a random male croaking. You don't understand how probability works.

>Sorry, should have written, "You've just made an assumption that all males have the same chance of croaking"
All males have the same chance of croaking as a random male until you get more relevant information. Please tell me what relevant information I missed in the problem (not scenarios you add to the problem) or admit defeat.

Here is the problem:

> If the chance of a single male croaking while you were listening is x then the chance of hearing a single croak from two males is 2x(1-x).

This is ONLY true is the probability that male 1 croaks is the probability that male 2 croaks. Before either had croaked you could say "hey it is equally likely that these frogs will croak" without making any assumptions, but once on frog croaks, all bets are off. We cannot assume that the probability of croaking is the same between a male frog that has croaked and a male frog that has not croaked, as they may be different.

>The information we have to distinguish between them is that one has croaked and one has not croaked. They (may) now belong to two different populations, and as such you cannot treat their probability the same. The one that croaked belongs to the population of known croakers, the other one belongs in the greater population of unknown croaking ability.

>A random male frog has nonzero chance of croaking since at least one male croaked.

/A/ random male frog does have a non-zero chance. /TWO/ random male frogs do not both have nonzero chances.

If you look at frog A, there is a nonzero chance since we know at least one frog can croak.

Now if you have two frogs the chance that both can croak may very well be 0 since you do not know how many can croak.

>Had to be 0
No, I'm saying since the probability CAN be zero you must discount the croaking altogether and simply treat it as a boy-girl, since you don't have enough information.

You turn back duh, faggots.

The probability of surviving is 100%, as you can lick the PAIR of frogs, and you know at least one of the pair is male.

>This is ONLY true is the probability that male 1 croaks is the probability that male 2 croaks.
This is true since you have no information to distinguish between the two frogs. The probability is not inherent to the frogs, it describes your knowledge of the frogs.

>Before either had croaked you could say "hey it is equally likely that these frogs will croak" without making any assumptions, but once on frog croaks, all bets are off.
You don't know which frog croaked so this is all irrelevant.

>We cannot assume that the probability of croaking is the same between a male frog that has croaked and a male frog that has not croaked, as they may be different.
Again, the probability of a random frog croaking encompasses all possible situations and permutations involving the frogs. One frog could be a mute. This is included in the chance of a random frog croaking. Until you have intonation which tells you this frog is mute, this changes absolutely nothing.

I asked you to provide me with information in the problem which changes my analysis. You have failed.

>/A/ random male frog does have a non-zero chance. /TWO/ random male frogs do not both have nonzero chances.
This is simply nonsense. Prove it mathematically.

>Now if you have two frogs the chance that both can croak may very well be 0 since you do not know how many can croak.
The probability of interest here is the chance of two males producing a single croak while you are listening. This is 2x(1-x). I have no idea what you think you're arguing here.

->No, I'm saying since the probability CAN be zero you must discount the croaking altogether and simply treat it as a boy-girl, since you don't have enough information.
The probability cannot be zero given the indignation you have. It's as simple as that. I priced it to you in the first few replies using bayes theorem. You made a mistake and instead of admitting it you simply keep repeating vague descriptions of the problem without disproving my proof.

Here is the problem:

Let x be the chance of a random male croaking while you are listening.

Since you heard a male frog croak while you were listening, x > 0. If x was 0, it would be impossible for us to hear a male croak. Yes or no?

The chance of a random frog being female given we did not hear it croak is 0.5*1/(0.5*1+0.5(1-x)) = 1/(2-x) by Bayes' theorem. Yes or no?

The chance of a pair of random frogs being both male given we heard a male croak is 0.25*2x(1-x)/(0.25*2x(1-x)+0.5x) = (1-x)/(2-x) by Baye's theorem. Which means the chance that the pair contains a female given we heard a male croak is 1/(2-x). Yes or no?

You want to lick a female frog, smart guy.

playing the game, i'd say 0.5 for either choice.

tryna think out of the box i'd say the couple of frogs.
- opposite sex mating probability
- there clearly are more frogs over there
- nothing asserts that the croak came from one of the two frogs. could be an hidden 3rd
no idea of the proba then ofc

Ah, fuck me.

Then it's irrelevant which side you pick..

i forgot: also lick both frogs' bellys. if they're both males, there's a chance they've copulated with a female. as male frogs climb onto females backs, there could be antidote left....

Imagine you have two boxes. One with one marble and one with two marbles. The marbles were randomly placed into those boxes from a big container that is half blue marbles, and half red marbles.

If you are given the information that the box with two marbles contains at least one blue marble, which box should you take if you want a red marble?

It can't be 0.5, that would imply females are as likely to croak as males, which is not true.

The box with two marbles has a 4/3 higher chance of containing a red marble.

no
ill posed. boy or girl paradox.

no
"your statement would imply (insert random falsehood here)" is not a convincing argument

If females never croak while males sometimes croak, then a frog which didn't croak has a higher chance of being female. Moron.

Oh, you break it down as P(male croaks in time). This turns it into some Poisson process Bernoulli random variable type thing. I think its a Poisson random variable but my nomenclature is terrible.

Then you compare the chance of 1 Process having 0 hits vs 1 or 2 processes having 1 hit to know what he should do.

I cant latek so bear with me.
P\left( x \right) = \frac{{e^{ - \lambda } \lambda ^x }}{{x!}}

We don't know what lambda (how often the frogs croak) but we dont need it to make a decision.

you just compare e^{-\lambda } vs \frac{\lambda e^{-\lambda }}{\lambda e^{-\lambda} + 2\lambda e^{-2\lambda}}

My reasoning:
Let A be the single frog in front, and BC the frogs behind him, and denote A male, B female and C male as MFM as an example.
The following constellations are possible and equally likely pre-croak:
MMM
FMM
FFM
FMF
MFM
MMF
FFF
MFF

He knows he hasn't encountered the last two since one of BC croaked, hence a male has to be present between them.
Therefore, in the remaining 6 options, he would have a 50% chance of licking a female if he picks A, and a 2/3 chance of licking a female if he picks BC (in 4 out of 6 cases there is a female among BC).

Nope. Your assumption that each permutation is equally likely is incorrect. Read the thread.

Not this question again.
Stop fapping to undergrads failing at probabilities. Go away. And before you ask; I don't care about answering your question.

If we're going to play the unjustified assumptions game, I have three more that everyone in this thread has ignored so far:

1. The zoologist is deontically implied to choose the option that maximizes his probability of survival. He could be suicidal, or evil (in which case his non-survival would be justified under utilitarian ethics)
2. The real numbers exist, so that probability is meaningful as a measure of the chance of survival.
3. The words used by OP have their standard meanings, so e.g. 'the venom kills him' refers to the event of non-survival, and not some arbitrary bullshit like 'the venom causes him to become depressed and enter an existential crisis, thus killing him while he is still surviving in a technical sense'.

Tl;dr changing the parameters of the problem after looking at the results is fraudulent, inconsistent and unscientific

But there can be two cars here if the frog in front of you is female and the frog that didn't croak is female

>it's not that hard to understand

Never took probability, explain it to me please.

first, (random vomit) has nothing to do with (arguable statement that i believe to be false)
second, you're assuming that the fact it didn't croak yet actually gives information
I concede this is not an easy question.
you're basicly saying that p(female | didntcroak) = p(didntcroak | female) * p(female) / p(didntcroak) = 1 * 0.5 / p(didntcroak)
then than p(didntcroak) is strictly lower than 1, proving your idea.

I say you have no idea about p(didntcroak)
(like 's x being nonzero is an unproven assumption)
why would you?
> we heard a croak, so at least one has croaked at least once
are you implying that "#propitiouscases / #possiblecases is hence nonzero"? why would this apply to a continuous situation / how the fuck do you count the non-croak events?
if not, what's the idea then?

>calling people morons
impressive reasoning point

Answer can be anything depending on the tendency of frogs to congregate.
For example, suppose the species is characterized by a heterosexuality parameter 0

>he has no problem with R, just 100% of its elements

The two frogs that are chilling are probably about to fugg. Id go for them senpai.

This shit is wrong because people are assuming that the order of the back frogs matters, which it doesn't.

these are all of the possibilities of frog combinations

front:m back:m f
front:m back:m m
front:f back:m f
front:f back:m m

as you can see, because order of back frogs doesn't matter, the likelihood of having a female frog is 50% on each side, assuming that those who didn't croak have a 50% chance of male or female

> drops mic

en.wikipedia.org/wiki/Boy_or_Girl_paradox

>mops dric

>first, (You) # (random vomit) has nothing to do with (You) # (arguable statement that i believe to be false)
They are equivalent statements. To say that a frog which you didn't hear croak is equally likely to be male as female is to say that males and females are equally likely to not croak, which is to say that they have the same chance of croaking. The chance of croaking = 1 - chance of not croaking.

>second, you're assuming that the fact it didn't croak yet actually gives information
I didn't assume it. I proved it.

>I say you have no idea about p(didntcroak)
>(like (You) #'s x being nonzero is an unproven assumption)
I proved it in that post. Can't you read?

If x = 0 then a random male will never croak. We heard a random male croak, so by contradiction x is not 0.

>are you implying that "#propitiouscases / #possiblecases is hence nonzero"?
x is the chance of a random male croaking while you were listening. I have no idea what you're going on about.

Anyway even if you ignore the proof that x>0, your hidden assumption that x=0 is unfounded.

You're right about why his reasoning is wrong but the problem is not equivalent to the BG paradox due to the croak giving information about a single frog.

>moron
>can't you read?

...y...you're right...

...you... you were absolutely, completely right... the whole time...

...i get it now... i'm just a moron...

...

... ...you know... i was thinking that, maybe... sometimes... events with probability exactly 0 could actually happen...

... it's foolish, i know, i'm an idiot for even considering this...

...

all i can say now is that i am sorry...

i'm sorry i made you lose your time and energy, all because i didn't make the effort to understand that your statements were all provided with unquestionable and complete evidence.

i'm sorry i upset you to the point that hatred and insults had to be expressed in order to unlock such a tough situation.

i'm sorry, above all, that i wasn't cautious enough not to use fallacious reasoning as hidden assumptions.


... I'm so sorry. Please forgive my behavior.

I hope one day we discuss something again, and that day i hope i'm less of an idiot than today.

He should switch his door

[pedant] Don't use the word "inuitionism" in math unless you really mean intuitionism as stated by Brouwer. [/pedant]

maybe it croaks to ward off normies?

1 frog ahead of him, no evidence:

50/50 chance.

2 frogs behind. Possible combinations of frog sexes (F: Female, M: Male)

FF
FM
MF
MM

Evidence: One of them croaks, leaving us with the probabilities:

MM
FM
MF

There's a 2/3 chance that one of the frogs behind him is a female.

The 2 frogs are either male-male, male-female or female-male

Therefore he has a 2/3 chance to get a female if he runs to the two frogs, while he has a 1/2 chance to get a female if he runs to the lonely frog.

Events with probability 0 can happen if the number of potential events is infinite. But there aren't infinite frogs.

Wrong. You're ignoring the fact that not hearing a frog croak is informative.

See

Why not lick all three?

Is this way the right line of thinking? Just add bayes
P(F|0 croaks) vs 1 - P(M&M| 1 croak)

P(F|0 croaks) = P(0 croaks|F)*P(F)/(P(0 croaks|F)*P(F) + P(0 croaks|M)*P(M)
P(M&M| 1 croak) = P(1 croak | M&M)*P(M&M)/(P(1 croak| M&M)*P(M&M) + P(1 croak| M&F))*P(M&F))

the P(croaks|gender) are the Poisson random variables with lambda multiplied by the number of males.

Who says that the two frogs behind him qere the source of the xriak?

Could be 2 females with a nale creepin on dey asses

the one in front of him
0% chance of surviving because everyone dies eventually

There could be infinite frogs. "the rainforest" could be an infinite rainforest

Here is what we know based on the information
>male and female frogs are indistinguishable from mere sight
>only male frogs croak
>a single frong is in front, has not croaked
>croaking sound came from behind
>two frogs behind
>female frog is the goal

what we do not know
1. if one of the frogs behind him croaked
2. quantity of frog goo needed to cure the bite

therefore,
the probability of any of the frogs being female is assumed 50%. one frog in front of us has a 50% chance to be female. of the two frogs behind us, there is AT A MINIMUM a 50% chance for one frogs to be female.

Because we did not witness a frog croaking, there is a greater than 0% chance that either of the frogs are female. It could be both are male, BUT, it is also possible that both are female.

Due to this, you have a 50% chance of a female at a MINIMUM if you run to the two frogs, and a >0% of the other frog being female. Thusly, you would run to the two frogs, and lick both all over. If a female, you got her goo. If a male, there is a chance there is female goo on him somewhere if he is an alphamale.

prove me wrong

>prove me wrong
here.

You forgot to check your privilege, the species could be [math]\varepsilon[/math]-heterosexual for an arbitrary small [math]\varepsilon>0[/math] which means that the probability of the partner of the croaked frog being female can be made arbitrarily small. (You can even assume [math]\varepsilon=0[/math] if the frogs are parthenogenetic.)

Though I should add that if we let [math]\delta\geq 0[/math] be the probability of a female being foreveralone you can likewise parametrize [math]\delta[/math] to make the survival probabilities whatever you like.

there was no information given on the mating or congregation habits of these frogs. thus, there is an equally likely chance that frogs like to assemble in an all-male triangular pattern as they are in male/female pairings

thus, everything you wrote is pointless and has no effect on this scenario

>there is an equally likely chance that frogs like to assemble in an all-male triangular pattern as they are in male/female pairings
I see, so you're one of those people who go 'all probabilities are 50%, either it happens or it doesn't'.

Also see

nope, I'm one of those people who is able to read.

OP may have believed he wrote a "clever" question, but he never specified that one of the frogs behind you croaked, only that you heard a croak. due to this crucial component, you have a >0% greater chance of getting a female frog from the two behind you.

should have said that one of the frogs croaked behind you but you didn't see which one. In which case it wouldn't matter either way since the one who croaked is effectively removed, leaving you with a frog in front or a frog in back. would still go for the two frogs since you have a higher likelihood of getting residual antidote from two frogs than one.

>but he never specified that one of the frogs behind you croaked, only that you heard a croak.
>But then he hears a croak from a male behind him

>male behind him
remind me again which part of that implies that one of the two frogs on the log is a male, or that one of them croaked?
what if there is a third male out of view behind the log?

fucking uncreative brainlets, I swear to god.
>hurr what do you mean i have to put my name on the paper it didnt explicitly tell me to do so what the fuck

take the bread out of your pocket and tempt all three frogs to come near you