/sci/ - Science & Math

>>11371758
We have one none of the above vote. Care to explain your answer?

The probability that the lone frog is a male is ever so slightly higher than 1/2, since absence of evidence is evidence of absence.

You know for certain that one of the pair is a female. The equally likely possibilities thus are FM, MF, FF. Hence each frog in the pair has 1/3 possibility of being male. Hence it's more advantageous to lick the single male in front of you.

>>11371765
Oh shit, I just realized I misread the question. I take back my answer and choose to lick both of the frogs. (I thought you could only lick one at first)

>>11371765
>The equally likely possibilities thus are FM, MF, FF.
Those are not equally likely. There are four possibilities:

Fc M
M Fc
Fc F
F Fc

If we order according to the female that croaked we have

Fc M
Fc F

This has the same probability of containing a male as the lone frog that didn't croak:

M
F

>>11371769
You're a dumb faggot. Write a computer simulation and you'll see that I'm correct.

>>11371770
Why would you need a simulation when I gave you mathematical proof?

Hint: ignoring information, such as the fact that you only heard one croak, will give you the wrong answer.

So far we have 4 votes that all disagree with each other. They're also all wrong. This board really has gone to shit.

Go for the pair, gay frogs dont exist in the rainforest, so the pair must be a mating pair.

>>11371777
That would mean male and female frogs are distinguishable aside from their croak, when the question explicitly says they aren't.

I wrote a python script and I get very different results depending on what the probability that a female frog croaks is

import random
probabilityofcroak = 0.9
n=0
totalcases=0
N=100000
for i in range(1,N):
#0 male, 1 female
singlefrogMale = random.uniform(0,1) <= 0.5
frog1Male = random.uniform(0,1) <= 0.5
frog2Male = random.uniform(0,1) <= 0.5
singlecroak = False
if singlefrogMale == False:
singlecroak = random.uniform(0,1) < probabilityofcroak
frog1Croak = False
frog2Croak = False
if frog1Male == False:
frog1Croak = random.uniform(0,1) < probabilityofcroak
if frog2Male == False:
frog2Croak = random.uniform(0,1) < probabilityofcroak
if (frog1Croak ^ frog2Croak) and not singlecroak:
totalcases = totalcases + 1
if frog1Male or frog2Male:
n = n+1
probability = n/totalcases
print(probability)

>>11371789
What results did you get?

>>11371789
>the brainlet CS major vs. the chad mathematician

>>11371794
If probabilityofcroak = 0.9, result about 0.9
If probabilityofcroak = 0.5, result about 0.66
If probabilityofcroak = 0.5, result about 0.55

>>11371801
>If probabilityofcroak = 0.5, result about 0.55
meant
If probabilityofcroak = 0.2, result about 0.55

>>11371794
>>11371797
Modify and run the code yourselves here:
https://trinket.io/python/4fe16d9e07

>>11371801
Let x = probability of female croaking

P(M|~C) = (1/2)(1)/((1/2)(1)+(1/2)(1-x)) = 1/(2-x)

>>11371806
>>11371809
BTFO

>6 votes, all for different answers
>only one is correct

>>11371780
You should try to actually read posts before responding to them. You dont need to know which frog is which sex, you just need to observe that they are paired up.

>>11371812
>>11371809
>>11371806
>>11371758
I assume we all agree by now that the correct answer is "doesn't matter, not enough information", right?

>>11371822
>You dont need to know which frog is which sex, you just need to observe that they are paired up.
Are you illiterate? Saying that the two frogs are different sexes is still distinguishing between male and female. And you already know one is female so your point is moot anyway.

>>11371822
Saying that only males stand next to females is adding distinguishing information not in the problem, brainlet.

>>11371758
Too long. Didnt' read. Eat shit nerd

>>11371828
No it doesnt. All you need to see is that two frogs are paired. Gay frogs only exist in American rivers, so if two frogs are paired together in the rain forest, and they cannot be gay, then they must be a mating pair. Lick both frogs and you're fine.

>>11371834
>did read, can't figure out out
ftfy

>>11371831
I'm not adding anything. The problem told us it takes place in the rain forest. That's clearly fair ground for inference.

>>11371837
Saying that the two frogs are different sexes is still distinguishing between male and female.

>>11371842
You heard a croak. One of the pair is female. It doesnt matter that you cannot tell which is female and which is male. The only thing that matters is that if one frog is female, the other must be male. Lick both.

Some questions:
Does the croak come for sure from one of the two frogs you've turned around and seen?
Have we seen the first frog not croak?

>>11371855
Saying that the two frogs are different sexes is still distinguishing between male and female.

>>11371859
Yes to both.

>>11371870
Thanks, sorry if the second one was already clear.

>>11371789
I modified this by calculating m - number of passed cases in which the single frog is male. Then n/m is not dependent on the probability of a croak, and I think it's fair to say that the probability of any of them croaking if they are female is the same so the approach is fine. n/m ~= 1
I don't have the confidence to make the call, I guess I'd stay in place.

>>11371869
Just because someone, somewhere, did the observational studies or autopsies to know that no frog in the rainforest is a gay or lesbian frog, doesnt mean I have to distinguish between the sexes at all. All I need to know is that any pair of frogs are god fearing heterosexuals

>>11371985
Saying that the two frogs are different sexes is still distinguishing between male and female.

I need 2 more values (frequency of croaks and observation time or expected value of number of croaks during observation time) to solve this, so far looks like an alright tricky problem.

>>11372153
Are you doing a Poisson process?

>>11372002
I am not making any conclusion about the sex of either of the frogs. I have no idea which frog is female, and I dont need to. I just need to know that one of them is female, which has already been confirmed for me in the prompt.

>>11372391
Saying that the two frogs are different sexes is still distinguishing between male and female.

>>11371758
The correct answer is to choose from one of the frogs where the croak came from. You have a 50% chance of choosing the one that is a guaranteed antidote, and still a 50% chanve if you choose the one that didnt croak. 75% chance of curing yourself.

>>11372594
Read the question again.

>>11372618
Why would the best option be omitted? The question has various flaws. Is there a downside from licking a from that doesnt contain the antidote? If not, then you are guarabteed to be cured by the pair. If so, then the either decision has the same probability.

>>11371758
looks more like bill burr

>>11372636
Read the question again.

>>11372636
You know one of the pair is a female because it croaked, that probability is fixed (and not in your favor). OP never said either sex was poisonous, so it won't hurt you to lick both of the paired frogs (which are not mating so you don't know that one is male). The other frog has an equal probability of being either male or female, just like the one sitting in front of you. It therefore doesn't matter which option you pick, but I'd advise running to whichever seems closer.

You do NOT need a computer simulation to think this through, anons.

Oh i misread it lol, croakers dont contain the antidote, so both options have a 50% chance of curing you.

Didn't we have this exact autistic garbage thread some two months ago?

>>11373062
>it doesn't matter
Correct
>50% chance
False.
See
>>11371809
>>11371806

>>11372309
No, he's doing a Grenouille process

>>11372309
Yep, it's becomes simple when you just apply it. Something like 1/(1+exp(-z)) for 1 frog vs 1/(1+1/(2z)) for 2 frogs behind you where z is expected number of croaks.

>>11373086
Incorrect. You already have a guaranteed croak from a female from the pair, and since you are not limited to only licking one, you can effectively pretend that whichever frog croaked does not exist. Its 50/50.

>>11373127
The odds that you'll lick a male are equal if you go either way, but the actual chance that you will lick a male is undetermined: it depends on the probability that a female croaks. (Bayes' theorem)

>>11371758
So i'm choosing between a set of two frogs of which one is female (but since i'm able to lick both i'm guaranteed to lick a frog which has not croaked) or a lone frog which has not croaked, thus without trying to infer more information (such as an increased/decreased probability of a male being with a female or the other frog not croaking in response to the croaking frog implying it is more likely male) each option seemingly has the same likelihood of success.

>>11373145
Except evaluating this as a scenario where all females had an equal chance of croaking is a fallacy. It is a foregone conclusion that only one frog croaks, even if it is not the only female. If you were sit and wait for one frog to croak, and discard all scenarios where the single frog croaks first, you would still be left with a 50/50.

>>11373162
This. You beat me to it.

>>11373162
>Except evaluating this as a scenario where all females had an equal chance of croaking is a fallacy.
They did, females croak randomly.

>It is a foregone conclusion that only one frog croaks, even if it is not the only female.
You don't understand conditional probability.

> If you were sit and wait for one frog to croak, and discard all scenarios where the single frog croaks first, you would still be left with a 50/50.
You keep assuming that a silent frog has a 50% chance of being male but that's false. See >>11371809

>>11373584
6 possible scenarios that can produce the conditions of this problem. all females on the board have an equal chance of croaking first. Likelihood of the scario is dependant upon ratio of females in the pair to total number of females.

Is this similar to the monty hall thing

>>11373632
You're ignoring that the croak came from a specific female. So by your argument there are 8 scenarios:

F Fc F
M Fc F
F F Fc
M F Fc
F Fc M
M Fc M
F M Fc
M M Fc

But these are not equally likely since a silent frog is more likely to be male than female.

>>11373637
No.

>>11373096
That's wrong, they should both be 1/(1+exp(-z))

>>11373639
False. The question states that you hear a croak, then turn around and see 2 frogs. It could be either one.

>>11373650
Non sequitur. It could be either one but you know it came from a specific frog. That's why F Fc F and F F Fc are separate.

1/3 + 2/3(1/2)=3/6

Nothing matters

>>11373652
You would just be splitting scenarios 1 and 2 into 2 equally likely scenarios, which does not affect the final result at all.

>>11373669
Yes, the final result is that the probability of survival in either direction is 1/(2-x) as I already showed. I would invite you to do the math yourself but you seem to be incapable.

>>11371758
You dont have to have A 220 iq to understand conditional probability But it helps

>>11373674
No, you've demonstrated nothing. The probability of a croak is irrelevant. There is a guaranteed croak for this situation to occur, and no future croaks can change the effectiveness of a choice.

>>11373691
>No, you've demonstrated nothing. The probability of a croak is irrelevant.
I already showed it's relevant. Your incorrect intuition that it's not relevant is not an argument. Take a freshman course in probability theory and then try again.

>>11373691
Guaranteed croak

>>11373708
>for this situation to occur
Brainlet

>>11373699
Lets hear your explanation of how the frequency of croaks affects which group of frogs is more likely to croak first. I'm interested.

>>11373723
I didnt even read your answer
i just wanted to post the guaranteed grug

>>11373723
A situation which happens to occur is not necessarily guaranteed to occur. A coin which lands on heads is not guaranteed to land on heads.

>>11373726
You're quite dumb if you couldn't figure this out yet. If the probability of a female frog croaking is 1, then this immediately tells you that the lone frog is male and the frog which didn't croak in the pair is male, because if they were females they would not be silent. So if x=1 you are guaranteed to lick a male frog. Now what happens if x is less than 1?

>>11373740
Which is directly accounted for in the scenarios lol the fractions above and below the figures represent the likelihood that they occur. One frog must croak first, and assuming everything in the problem occurs at once, then future croaks mean nothing as it is implied your choice must be made immediately.

>>11373746
>Which is directly accounted for in the scenarios lol the fractions above and below the figures represent the likelihood that they occur.
They don't seem to represent anything coherent since you assume a silent frog is equally likely to be male as it is to be female, which cannot occur. I suggest you try to argue against my proof instead of referring to your flawed from the start argument.

>>11373641
I think it is a modified version of monty hall. Go for the two frogs, if there’s at least one male frog in the group of three (your only hope) then there’s a 2/3 chance he’s lurking in the pair.

>>11373762
>a silent frog is equally likely to be female
You are failing to solve the problem because you are working backwards from the final result to determine the genders of frogs. Before any frogs croaked, there was n number of female frogs. One frog has to be the first to croak. All the female frogs had a 1/n chance to be the one that croaked. All others are silent, but their silence does not affect the fact that the odds of them being female was 1/2 in the first place.

>>11372515
Only insofar as the prompt told us. I'm making no distinctions the prompt didn't already make.

Alright, it's Poisson guy again. Fixed some mistakes in my previous attempt.
1) Let's assume that croaks occur with a constant mean frequency independently of each other (should be true as long as frequency is not too high);
2) We can now apply Poisson's distribution;
3) We need only 1 parameter — expected number of croaks during observation time, let's name it z;
4) [math]P(k)=\frac {z^k e^{-z}}{k!}[/math]
5) For 0 croaks P(0)=e^(-z), for 1 croak it's z*e^(-z).

For 1 silent frog in front we have 2 options: it's male or it's female with 0 croaks, of which male satisfies our problem conditions.
[math]P(surv_1)=\frac {P(male)}{P(male)+P(female_0)}=\frac{1/2}{1/2+1/2*e^{-z}}=\frac {1}{1+e^{-z}}[/math]


For 2 frogs behind you we have 3 options: MF, FM and FF, of which the first two will save us and third won't.
[math]P(surv_2)=\frac {P(mf)+P(fm)}{P(mf)+P(fm)+P(ff)}=\frac{1/3*2ze^{-z}}{1/3*(2ze^{-z}+e^{-2z})}=\frac {1}{1+e^{-z}/2z}[/math]

Putting that shit in Wolfram says that for 0<z<1/2 it is better to lick single frog and after that it is better to lick the pair.

https://www.wolframalpha.com/input/?i=1%2F%281%2Bexp%28-x%29%29-%281%2F%281%2Bexp%28-x%29%2F%282x%29%29%29+%3E+0

>>11374052
Fuck, no, I am retarded, need to redo the second part.

>>11374061
Alright. For 2 frogs behind us we have the following: MF,FM,F(c)F and FF(c), so
[math]P(surv_2)=\frac{1/3*2ze^{-z}}{1/3*(2ze^{-z}+1/2*2ze^{-z})}=\frac {2}{3}[/math]
Which means that for z>ln(2) it is better to lick a single frog. Intredasting.

>36 votes
>only 3 correct answers
the absolute state of /sci/

>>11374068
Fuck, I am super retarded, forgot exp(-2z).
[math]P(surv_2)=\frac{1/3*2ze^{-z}}{1/3*(2ze^{-z}+1/2*2ze^{-2z})}=\frac {1}{1+1/2*e^{-z}}[/math]

>>11373778
Wrong.

>>11373786
>You are failing to solve the problem because you are working backwards from the final result to determine the genders of frogs.
It's the complete opposite. I used the initial probabilities to calculate the conditional probability using Bayes theorem.

>One frog has to be the first to croak.
Or none of the frogs could have croaked. This is a dumb assumption and also gets you no closer top answering the question, which is why you have failed in every single post to even give a formula for finding it.

>All others are silent, but their silence does not affect the fact that the odds of them being female was 1/2 in the first place.
I never said it did. It affects the conditional probability, not the initial probability.

>I'm making no distinctions the prompt didn't already make.
Saying make only make frogs are next to female frogs is a distinction the prompt didn't make.

>>11374076
Still wrong, they're both 1/(1+exp(-z))

>>11374456
This makes a certain amount of sense, yes.

>>11374456
>>11371809
The only correct answers ITT.

>>11374614
And they're from the OP...

>>11371758
as you run to the pair, if one of them croaks, you still have hope that it was the one that croaked in the first place.
in other words, it's harder for the pair to be both female than for the lone one to be male
t. /pol/

>>11371758
>applies antiophid serum from my pocket
Pshh. Nothing personal, mathcuck.

>>11371758
I lay down

>implying a frog could produce anti-venom against a snakebite
probability is 0, it would never happen. And dont go licking random frogs.

>>11375119
The frog which didn't croak in the pair has the same chance of croaking while you are running towards it as the lone frog, so that's wrong.

>>11374412
> Wrong
Are you sure?
From a game theoretical view, the cases where there’s no males and the cases where there are two males are moot (doesn’t matter which way we run). We thus have to optimize for the case where there’s one male in the group of three frogs. There are three configurations where this is true and therefor matter, and two of them are such that the male is in the pair. I think I’m right.

>>11376528
>There are three configurations where this is true and therefor matter, and two of them are such that the male is in the pair.
Sure, but they're not equally likely. You haven't actually done any math, just made a bunch of naive assumptions.

The antivenom apparently has a low enough dosage to cure yourself by licking a frog, so I’ll just bring a small vial with me before I go out into snake territory.

>>11376898
Hindsight is 20/20

>>11376936
/sci/ is still at an elementary school level of probability theory, as I thought.

>>11371758
Wouldn't you have a 100% chance with the pair since there is a 100% chance there is a female among them if you hear a croak?

>>11377917
You want a male.

Is the probability 1/2 with the pair?

MC
FC
CM
CF

>>11377984
No, a non-croaking frog is more likely to be male than female.

>>11371758
Is this right?

A: pair contains a male
B: one croak is heard
Let x be the probability a female frog croaks
P(A|B) = P(B|A)P(A)/P(B)

P(A) = 2/3 since there are three options MM MF FF and all are equally likely given the population exists in equal proportion

P(B) = 1/3 * 0 + 1/3 * x + 1/3 * (x(1-x)) = 2x/3 – (x^2)/3

P(B|A) = 1/2 * 0 + 1/2 * x


P(A|B) = (3x + 4) / (4x – 2x^2)

I don't know how you'd figure out x though

>>11371758
The probability depends on the average time between croaks and how long I spend observing the frogs. Of course, the probability that the silent frog is a male is higher simply because I know at least one of the frogs behind me is female. Effectively, I am being asked to pick a red ball from two boxes; one box contains either a red or blue ball, and the other contains another red or blue ball, but also *definitely* a red ball. Obviously I want to pick from the one that doesn't definitely have a red ball in it.

>>11371770
>faggot
Why the homophobia?

>>11376882
> Sure, but they're not equally likely. You haven't actually done any math, just made a bunch of naive assumptions
They are equally likely and I did do the math, I guess you didn’t notice because I used correct English and spelled numbers lower than ten with letters and included them in the sentences explaining their relevance.
The numbers in your formulas are supposed to be motivated by logic, the only thing I see in this thread is everyone pulling irrelevant formulas from thin air, computing the results and calling it a day.
I went ahead and googled the answer and turns out that of course I am correct and my reasoning was correct.
Your chance of survival if you go for the single frog is 50%, as anyone can figure out with their pinky toe. Your chance of survival if you go for the pair is two thirds, 67%, and the reason this is so is easily googleable or understood from my posts. So there.
I came here from /pol/, is this actually a troll board and nobody here knows math or science at all? If so I guess I fell for it.

>>11378880
>P(A) = 2/3 since there are three options MM MF FF and all are equally likely given the population exists in equal proportion
Wrong. MF is twice as likely as MM. It's 3/4.

>>11378894
>Effectively, I am being asked to pick a red ball from two boxes; one box contains either a red or blue ball, and the other contains another red or blue ball, but also *definitely* a red ball. Obviously I want to pick from the one that doesn't definitely have a red ball in it.
Incorrect. You know the definite red ball is separate from the unknown ball, so it doesn't matter which box you choose.

>>11379058
There are also google results for 1/2 either way. You probably just went for one based on the TedEd talk, which is incorrect.
For the pair, there are 3 different states a frog could be: M, Fs (silent female), and Fc (croaking female). The condition is that there must be 1 Fc.
Therefore, the permutations are:
>M Fc
>Fs Fc
>Fc M
>Fc Fs
That’s 2/4 of the sample space containing a male, or a 50% chance. If you don’t understand the difference between permutations and combinations, or why they matter, then you’re a retard who shouldn’t be browsing a maths board.

>>11379058
>They are equally likely and I did do the math
Nothing you said showed they're equally likely. They can't be, since a frog which didn't croak is more likely to be male than female. And by your logic, the answer could be any number since you can arbitrarily split the possible events up many different ways. For example:

Fc F
Fc M

Oh I guess that means it's 1/2, not 2/3.

>The numbers in your formulas are supposed to be motivated by logic, the only thing I see in this thread is everyone pulling irrelevant formulas from thin air, computing the results and calling it a day.
I've pointed out the flaws in everyone's logic, including your own. You haven't a thing to argue against mine.

>I went ahead and googled the answer and turns out that of course I am correct and my reasoning was correct.
Let me guess, you found a YouTube video that told you you were right? Guess what, the video is wrong for there reasons I explained above. I even contacted Derek Abbott, the guy behind the video and he admitted it was an incorrect answer.

Since you'll probably be in denial, here's what you should have found on your Google search: https://mindyourdecisions.com/blog/2016/03/08/ted-eds-frog-riddle-is-not-entirely-correct-game-theory-tuesdays/

11% correct response rate. Pathetic.

>>11379470
The Popular Mechanics article dismissed your approach.

>>11379472
No I read the Popular Mechanics article, among others, but that was the most interesting as it provided a twist: if you catch a glimpse of which frog croaked, survival chance goes down from 2/3 to 1/2. That is a bit of a mindfuck.

>>11379472
I read your blog link. He begins to explain the problem correctly, arriving at 2/3. Then he claims to have “possibly” found an error in that approach and goes on to bring up croak frequency. This is of course wrong and also shown so in the PM article.
And do I really have to explain why the four combinations of two bits where each bit has a 50:50 chance of being set are equally likely?
Come on, people. This puzzle is super easy, I solved it in my head in seconds. All that is needed is an elementary understanding of probability. Stop making fools of yourselves.

This is such a reverse troll problem. You think the answer would be counterintuitive at first but it is actually exactly what you would think. There are like three people in this thread who are not retarded btw

>>11381301
>This, however, is an incorrect interpretation because it assumes information that we are not given. Specifically, this interpretation assumes that two male frogs are twice as likely to make a croaking sound than a pair of one male and one female. But the problem says nothing about the frequency of croaking, so it may be that it's not uncommon for a male frog to go his whole life and only croak once. We know there was a croak, and therefore we know that one of the frogs is male; we cannot extrapolate further than that without more information.
This is a non-sequitur, regardless of what the frequency of croaking is, two frogs are twice as likely to croak as one. The article claims to use conditional probability but falls into the naive trap of assuming that just because you made a list of events, they are equally likely. You don't need to make any assumption about the frequency of croaking to see that the two choices give the same probability. If we list the possible events by position of the frogs then we have

Fc M
Fc F
M Fc
F Fc

Notice that nowhere is anything said about the frequency of croaking. We simply took into account that a frog with unknown position croaked. According to Popular Mechanics' pseudoargument, this should mean the probability is 1/2.

>>11381311
That's wrong, you don't need to see the frog to know a single croak came from a specific frog. You don't need to know which frog. It also ignores these fact that a non-croaking frog is more likely to be male than female, so it can't be 1/2. You're in denial so instead of making an argument you're just going to keep referring to sources which have already been debunked.

>>11381331
>begins to explain the problem correctly, arriving at 2/3. Then he claims to have “possibly” found an error in that approach and goes on to bring up croak frequency.
No, he claims that the riddle is wrong access you have to take into account all the information given. Stop trying to use weasel words.

>This is of course wrong and also shown so in the PM article.
It's not, the PM article dismisses this approach with handwaving, not an argument. The author clearly has very little knowledge of conditional probability since he doesn't realize the difference between information coming from a specific source and knowing which source it came from.

>And do I really have to explain why the four combinations of two bits where each bit has a 50:50 chance of being set are equally likely?
No you don't. These are not analogous to bits since single bits don't have a mechanism similar to croaking. If bit 0 can be followed by 0 or 1 randomly while bit 1 can only be followed by a 0, then possible combinations of two pairs will be

00 00
00 01
00 10
01 00
01 01
01 10
10 00
10 01
10 10

If I told you that the sum of the secondary bits was 1 then we would have

00 01
01 00
01 10
10 01

Now does this mean the chance of getting a 10 is 1/2? No! Because the chance of getting a 00 randomly is less than the chance of getting a 10 randomly, since 1 is sheats followed by a 0.

The one frog can be simply M or F meaning 1/2.
The pair can be FF, MF or FM so 2/3.
Ridiculously trivial.

>>11371769
>>11371771
>Let's ignore that there are two scenarios where both are female and consider them as one single scenario
Wow, very mathematical, such proof! Retard

>>11382099
>The one frog can be simply M or F meaning 1/2.
Wrong. A non-croaking frog is more likely to be male.

>>11382103
>>Let's ignore that there are two scenarios where both are female and consider them as one single scenario
Where did I ignore it? There are also two scenarios where a frog is male. Fucking retard.

To make this easier for you morons, notice that the 4 scenarios have a symmetry over position. Fc M has the same chance of occurring as M Fc and likewise for Fc F and F Fc. So we can simplify the list by ordering by croaking frog:

Fc M
Fc F

Notice that this is gives the same probability as our single non- croaking frog:

M
F

There, now I have treated the two scenarios as one. Try making an actual argument.

>>11382099
>The pair can be FF, MF or FM so 2/3.
Wrong

Fc M
Fc F
M Fc
F Fc

Reminder that I have made a simulation of the scenario with thousands of iterations, to find that the correct answer is
doesn't matter which direction, probability is indeterminate (depends on the probability of croak)
>>11371806

>>11382112
My bad, I overlooked the symmetry

>>11381832
It's actually a reverse reverse troll.

>>11382115
>simulation is considered a stronger argument than proof
This board is so fucking shit.

I got to the right answer by intuition, am I a genius?

>>11382132
Simulation confirms the proof. If you got anything different, that means your proof (or your assumptions) is wrong.

>>11382165
It confirms nothing except that you got the same results by two different methods
>If you got anything different, that means your proof (or your assumptions) is wrong.
Or the code

>>11382159
What's the right answer?

>>11382173
It doesn't matter, not enough information.
It doesn't matter because you just know one of the two frogs that are together is female, so you can remove him from the equation.
Not enough information because you don't know how often females croak on average. If they croak very often, it is more likely that the two other frogs are male, resulting in high chances of survival. If they almost never croak, then it's basically a 50/50 chance whether each of the two others are male or female, resulting in a chance of survival close to 50%

>>11371985
>>11371837
You don't actually know they're a mating pair. They could be two females that were fighting over food or territory or something, or that just happened to be in the same area.

>>11371758
>Dash to nearest frog
>Lick it
>Quickly use it as a projectile to ricochet the other two frogs into the air
>Jump up and catch one frog in each hand
>Lick them both before you hit the ground

>>11382113
The chance of one of the frogs being the one that croaked and therefore female is 1/2 but that is a different question from the chances of there being a male frog where there is at least one female.

>>11371758
i would re choose and pick the new door. thus, monty hall python solved.

>>11382272
There being a male frog where at least one is female is a different question from there being a male frog where one frog croaked. Explain what's wrong with my list.

>>11382515
If something is Fc then it is also F.
Counting them separately doesn't work

>>11382542
>If something is Fc then it is also F.
Wrong, F is a non-croaking female. Sobering can't have crushed and not croaked while you were listening. Counting then together doesn't work.

>>11382542
>If something is Fc then it is also F.
Wrong, F is a non-croaking female. Something can't have croaked and not croaked while you were listening. Counting them together doesn't work.

>>11382558
It requires that one frog has an equal chance of being a female who did not croak, a female who croaked and a male.
Meaning one frog has a 2/3 chance of being female which can't be.

>>11382584
no, it has a 50% chance of being a female who croaked, since you know one croaked but don't know which one. thus a 25% chance of being a female who didn't croak and a 25% chance of being male
Just because there are multiple outcomes doesn't mean all are equally likely

>>11382584
>Meaning one frog has a 2/3 chance of being female which can't be.
It certainly can be after you hear a frog croak. You fundamentally don't understand conditional probability.

>>11382073
These three cases are equally likely;
FF
MF
FM

Let’s say we run the simulation 300 times, we’ll get roughly 100 cases of each.

The FF cases are then spread evenly into an equal amount of FcF and FFc cases, so roughly 50 of each case.

>>11382099
Congrats on having a brain.

>>11378894
>the probability that the silent frog is a male is higher simply because I know at least one of the frogs behind me is female

????????????????

>>11382073
> Notice that nowhere is anything said about the frequency of croaking
Exactly. What conclusion can we draw from this? Either the frequency is relevant, in which case we have insufficient information to solve the problem. Or it is not relevant, and we do have enough information. The assumption should be that we are interested in a solvable version of the problem, or we’re all wasting our time.

Could the frequency be relevant? Yes. For example, if the females croak on average 50 times per second, the chances of a silent frog being male will quickly approach 1. However, if croaking is rare, it doesn’t affect the probabilities significantly and we end up with the correct answer 2/3 for the SOLVABLE VERSION OF THE PUZZLE!

>>11383071
>These three cases are equally likely;
>FF
>MF
>FM
By what argument?

>Let’s say we run the simulation 300 times, we’ll get roughly 100 cases of each.
By what argument?

If you actually do the math, without ignoring the crucial information that the croak had to come from a single frog, you'll see that FF occurs (1-x)/(2-x) of the time, where x is the probability of a female frog croaking while you were listening. Only when x=1/2 is FF as likely as FM or MF.

Instead of just begging the question and making naive assumptions, why don't you actually do the math?

>>11383075
He's wrong, you're a brainlet.

>>11383093
He's saying that because males and females are in equal number, if you see a male it's more likely the next frog you see is female. However this would only be significant in a very small population of frogs.

>>11383096
>The assumption should be that we are interested in a solvable version of the problem, or we’re all wasting our time.
Idiotic. You're ignoring relevant information to "solve" the problem, which just means you got a wrong answer. You don't need to know the frequency of croaking to know that your survival chance is the same for either choice.

>However, if croaking is rare, it doesn’t affect the probabilities significantly and we end up with the correct answer 2/3 for the SOLVABLE VERSION OF THE PUZZLE!
Completely wrong. If croaking is rare then 1/(2-x) approaches 1/2, not 2/3. You still doing get it, and your fragile ego is keeping you from understanding the problem.

Wow, Reddit really did a number on this place.

>>11371758
Pair:
FM
FF

50% shot that one of them is male

Single:
M
F
50% shot that it is male

it doesn't matter

>>11383806
>50% shot that one of them is male
Wrong. How do you know they're equal?

>>11383809
it didn't croak in that particular time and the problem doesn't state that they croak every x seconds. so the only information we have is there is a 50% chance it is male, based on the population

>>11371806
I'd sooner shoot myself than "run code".

>>11383811
It doesn't have to croak after a certain amount of seconds for it to have a chance of croaking. A frog can have a 99% chance of croaking but still not croak. Saying that a non-croaking frog is equally likely to be male than female is the same as saying that a male frog croaks as often as a female frog. The question explicitly states that's not true.

>>11383815
If female frogs croak at least once a minute and you've been waiting for 10 minutes, it's almost a certainty that it's a male.

>>11383825
Yes, so why do you say it's 50/50?

>>11383825
Because we don't know how often the female frogs croak.

If female frogs croak once a day, but you've only been there for 5 minutes, that doesn't tell you anything. Only that as time goes on, you have a greater and greater chance of it being female.

>>11383830
>Because we don't know how often the female frogs croak.
Doesn't matter, it's impossible for a non-croaking frog to have an equal chance of being male or female.

I'm not surprised people here are wrong, but I am surprised how many people can be so sure in their answer when they are wrong.

>>11383850
It's human nature. People will irrationally defend their intuition. Even if you explain what they're missing, they'll insist what they're missing is irrelevant, even though you already showed them exactly how it's relevant.

This is like the flies in the jar problem. The aim is not to establish the answer to a childishly simple question, but rather to demonstrate that the vast majority of people posting an answer are (1) incapable of reading a question properly, or (2) incapable of understanding the question even if they managed to read the question properly or (3) incapable of reasoning out a correct answer even if they did read and understand the problem correctly.

Therefore most of /sci/ users who post answers to questions like this are retards, morons and just plain pig fuck stupid.

We should not be surprised by these results. After all this board frequently contains threads of trash quality, such as flat earth, moon hoax, and 9/11 conspiracy. These threads are not only just tolerated on this board but they also receive plentiful replies.

Thank you OP for demonstrating this truth, but most /sci/ users with a mental age of above 14 are already aware of the situation..

15% correct response rate, slightly above random chance. Impressive for a bunch of retards.

>>11383843
yeah, the chance is higher of it being a male. But you do not know the exact chance, and thus do not know the probability of your survival

>>11383814

>>11384927
Many of those probably read the thread before answering. But then again, many might not have thought it through before answering and just went with their first hunch, or just straight out ignored the 2nd part of the question and checked the first answer where the first part fits