[a / b / c / d / e / f / g / gif / h / hr / k / m / o / p / r / s / t / u / v / vg / vm / vmg / vr / vrpg / vst / w / wg] [i / ic] [r9k / s4s / vip / qa] [cm / hm / lgbt / y] [3 / aco / adv / an / bant / biz / cgl / ck / co / diy / fa / fit / gd / hc / his / int / jp / lit / mlp / mu / n / news / out / po / pol / pw / qst / sci / soc / sp / tg / toy / trv / tv / vp / vt / wsg / wsr / x / xs] [Settings] [Search] [Mobile] [Home]
Board
Settings Mobile Home
/sci/ - Science & Math

[Advertise on 4chan]


Thread archived.
You cannot reply anymore.


[Advertise on 4chan]


You hear a croak behind you and turn around to see two frogs of the same species. You know only males of this species croak and are otherwise indistinguishable from females. Croaking is random, females and males exist in equal proportion, and males and females congregate randomly. What is the probability both frogs are male?

https://www.strawpoll.me/45216510
>>
>>13062235
100%
>>
>>13062249
I asked for the chance both are male, not the chance the first reply in this thread would be a shitpost.
>>
>>13062235
1/2 fuck niggers who think MF and FM are separate in this context.
>>
>>13062235
There is no probability; either they are both male or they are not. This is the same misconception as with confidence intervals.
>>
>>13062342
0.25
>>
>>13062342
>There is no probability; either they are both male or they are not.
Doesn't follow.
>>
>>13062336
Wrong.
>>
>>13062496
kys
>>
>>13062235
1/2
options include,
FF
MM
MF

Since we heard a croak, that means there must be at least one male, eliminating FF option, leaving just MM and MF, thus a 50% chance both are male.

Now if you think MF != FM then that has 4 options, eliminating FF leaving 3, thus giving 1/3 probability. However MF == FM so if you think otherwise you are a huge faggot.
>>
>>13062235
I can tell you what's 100%, the chance that both of those frogs are going up my ass.
>>
>>13062522
The point isn't that MF == FM. The point is that one male and one female is twice as likely to occur if you find 2 random frogs.

In b4 bump limit, again.
>>
>>13062319
It's 100% retard you said only males croak
>>
>>13062235
oh no, the dog is eating the cookie!
>>
>>13062507
Wrong.
>>
>>13062522
Why do you think those three options are equally likely?
>>
>>13062587
Right, and you heard a croak. That doesn't mean both frogs croaked. Are you OK, anon?
>>
>>13062534
Correct, but the frogs aren't random since you heard a croak.
>>
File: Girls.png (490 KB, 449x401)
490 KB
490 KB PNG
Oof, /sci/ still can't do a basic probability question. How embarrassing.
>>
>>13062859
How are you supposed to know which frog croaked?
>>
>>13063007
You aren't.
>>
If a tree falls in a forest and no one is around to hear it, does it still make a sound?
>>
>>13063010
Then how are you supposed to know if it's a male
>>
>>13063012
Yes.
>>
>>13063019
The frog that croaked is male since only male frogs croaked. But you don't know which frog it is.
>>
>>13063025
How can you prove it if you arent around to hear the tree?
>>
>>13063030
But if both frogs croaked then they would both be male
>>
>>13063046
The question doesn't say both frogs croaked, so how did you get 100%?
>>
>>13062235
>Croaking is random
What is the probability of hearing a croak from a male during the time period you were standing there? If they croak constantly, the probability is as low as 1/3. If they almost never croak, the probability is much higher.
>>
>>13063037
With a recording device. Or by using basic logic.
>>
>>13063090
Because only male frogs croak...
>>
>>13063097
The question doesn't say both frogs croaked.
>>
>>13062867
>MF is twice as likely as MM
>somehow probability of MM and MF is equal
kys
>>
>>13063100
But the one that croaked is a male
>>
>>13063093
>What is the probability of hearing a croak from a male during the time period you were standing there?
Unknown.

>If they croak constantly, the probability is as low as 1/3.
If they croak constantly then you would have heard two croaks instead of one if they were both male. Show your work.
>>
>>13063106
>MF is twice as likely as MM
Before hearing a croak, correct.

>somehow probability of MM and MF is equal
They aren't equal, since females are always silent but males are not always silent. If we assume males almost never croak then they are equal after hearing a croak.
>>
>>13063107
Yes, and the question is whether both are male. So how did you get 100%?
>>
>>13063133
>only male frogs croak
>frogs croak
>>
>>13063139
So how did you get 100%?
>>
>>13063146
because only male frogs croak, if the frogs croak then it is 100% absolutely certain they are male
>>
>>13063152
The question doesn't say that both frogs croaked, so your argument fails.
>>
>>13062996
Can do, its 50%
>>
>>13062391
This is the least wrong answer in this thread. Pathetic.
>>
>>13063161
Why?
>>
>>13063108
>If they croak constantly then you would have heard two croaks
Or one big simultaneous croak and you couldn't tell if it was one or two since you don't know how loud these frogs normally are. Followed by both frogs instantly dying.

>Show your work.
The 1/3 is just the typical conditional probability problem of "You know at least one is male". It gets more complicated when you factor in "How likely were you to even find out that at least one is male", and my intuition is telling me that, if it were very unlikely, the likelihood of both being male increases. But intuition is often wrong with probability, and I don't even remember how to do math with random events with unknown frequencies that with an observation window of unknown length.
>>
>>13063171
>Or one big simultaneous croak and you couldn't tell if it was one or two since you don't know how loud these frogs normally are.
Seems highly unlikely. Also, if the frogs croak constantly then you would hear more than one croak.

>The 1/3 is just the typical conditional probability problem of "You know at least one is male".
But that's wrong. Here, the way you know one is male is from a croak that had to come from a specific frog. So it's the 1/2 interpretation if you ignore the frequency of croaking is different for males and females. The point of this problem is to trip up both naive 1/2ers and more advanced 1/3ers. 1/3ers are actually more wrong than the 1/2ers because they think the problem is equivalent to what they already know.
>>
>>13063207
It's like this. You are given a 4 sided die, and are told to produce random numbers from one to three. How do you do it? You roll the die, and throw out the fours. Then the 1, 2 and 3 are truly random, and each has a prob. of 1/3. In our case, we have four cases, male male, male female, female male, and female female. We throw out female female. We are left with a prob. of 1/3 for male male. Ta.
>>
>>13063230
Wait a sec, how are Female Male, Male female seperate
>>
>>13063230
No, we have 9 cases

M M
Mc M
M Mc
Mc Mc
M F
Mc F
F M
F Mc
F F

We throw out 5

Mc M
M Mc
Mc F
F Mc

By symmetry we can reduce to

Mc M
Mc F

Now the only question is how likely a silent frog is to be male. If x is the chance of a male frog croaking while you were listening then it's (1-x)/(1-x+1) = (1-x)/(2-x).
>>
>>13063281
The fuck
>>
>>13063156
The question doesn't specify whether both frogs croaked, so 100% is a consistent answer
>>
>>13063305
The question specifies you heard a croak, not two croaks. Your answer is inconsistent with question.
>>
>>13063295
What don't you understand?
>>
>>13063329
You wouldn't be able to tell a single croak from two frogs croaking at the same time
>>
>>13063335
Yes you would. And now you're assuming that they croaked at the same time. If your answer is based on making up informatiom not in the problem, it's wrong. Get over it.
>>
>>13063345
Fine jfc, it's 50%
>>
>>13063281
>If x is the chance of a male frog croaking while you were listening then
This is slightly different from how I was interpreting it in >>13063171. For me, based on the scenario presented, you heard at least 1 croak, and at that moment, time stopped and you had to make an evaluation. I was working off of the assumption that, even if both frogs were going to croak, as soon as you heard one, there was no more information collected.

If you go from the perspective of "I listened to some frogs for an amount of time and only heard 1 croak", then things change around a bit, since then suddenly the probability of hearing only 1 croak in the first place gets very low if the croak probability per frog increases.
>>
>>13063129
>Before hearing a croak, correct.
Why would it stop being correct after hearing a croak, you retard?
>>
>>13062495
>Doesn't follow.
It doesn't need to follow, it's true by definition. X or (not X) is always true.
>>
>>13063354
Wrong.
>>
>>13063373
>For me, based on the scenario presented, you heard at least 1 croak,
The problem said you heard a croak, not at least one croak.

>and at that moment, time stopped and you had to make an evaluation. I was working off of the assumption that, even if both frogs were going to croak, as soon as you heard one, there was no more information collected.
It doesn't matter, my argument still applies.
>>
Well, 1 of them hast to ne Male. Other one has 50/50 Chance of being Male. So its 50/50
>>
>>13064196
>Why would it stop being correct after hearing a croak, you retard?
Because hearing a croak gives you information about the frogs. Are you really this dense?
>>
>>13064205
>It doesn't need to follow, it's true by definition
"There is no probability" doesn't follow from "x or not x."
>>
>>13064472
>Other one has 50/50 Chance of being Male.
A silent frog is equally likely to be male or female? No.
>>
>>13064478
We dont know the probability of croaking tho
>>
>>13064491
Doesn't change what I said. If male frogs croak more frequently than females, silent frogs must be more likely to be female.
>>
Assuming a population size large enough, 50%, with a negligible advantage to MF.

They exist in equal proportions, so MF would be a hair more likely, but it approaches 50% as population size increases.
>>
>>13064544
>Assuming a population size large enough, 50%, with a negligible advantage to MF.
You're ignoring that silent frogs are more likely to be female.
>>
>>13064571
If croaking is random, then it's another hair toward MF.

At this point, I'm ready to dissect them.
>>
>>13064586
Not really, the probability can be anywhere between 1/2 and 0 depending on the frequency of croaking.
>>
>>13064477
Yes, that part follows from you referring to one particular, already existing instance. It's like asking what the probability of my hair being red is - there is no probability, I have brown hair. If you formulated your problem as "What's the probability that a random pair of frogs has property XYZ", it'd be a different story.
>>
>>13064669
>Yes, that part follows from you referring to one particular, already existing instance.
How?

>It's like asking what the probability of my hair being red is - there is no probability, I have brown hair.
The probability for you is 1. The probability for me is whatever the probability of a random individual having red hair is. It depends on what information you have. Nothing you said implies "there is no probability."

>If you formulated your problem as "What's the probability that a random pair of frogs has property XYZ", it'd be a different story.
How so?
>>
>>13062235
this is just monty hall
>>
>>13062235
1/3
>>
>>13064768
Wrong.
>>
>>13062249
>>13062587
read the OP, it's written the chance that both are males
>>
>>13062996
don't generalize
>>
>>13064461
The question says males and females are evenly distributed
>>
>>13064695
This is true an may help ppl. Another way to think about it: two coins are flipped, you are NOT allowed to see them. You are told AT LEAST ONE is heads. What is the probability that both are heads? 1/3. H=head, T=Tail, the possibilities are HH, HT, TH, and TT. But TT is out (AT LEAST ONE HAS HEADS), so you are left with HH, HT, TH. prob. of HH is 1/3.
>>
>>13065145
Literal troll. MF != FM and FF is out. Just because a car looks identical to yours in say a parking lot, it ain't yours, it's a different entity. Is crow A male or female, is crow B male or female, so AF BM is NOT AM BF!!! In math land this is called order matters, and order matters in this problem.
>>
>>13065145
I change the answer again since you know that the FF possibility is not true since you heard a croak (at least one male is there)
you have 50% chance that every individual is a male
and you have two possibilities
MM
MF = FM
both with 50% chance each
thus it is 0.5 * 0.5 = 0.25
25% chance of being both males
>>
>>13062235
I'm a bayesian, an event either happens or it doesn't.
>>
>>13065159
yeah didn't take the problem seriously at the beginning, I say 25% or if you consider MF and FM as two different things (but why?) it's 16.5%
>>
File: prob.png (24 KB, 829x615)
24 KB
24 KB PNG
guys this is seriously embarassing, so many people answering 1/2 or 1/3
/sci/ needs some training if we want to look like an actual science discussion community
I hope that most of these answers are shitposts or we're in a bad situation here
>>
I'm the one who wrote >>13065189, maybe it can also be 1/2 after all... I tried solving with a tree diagram and if you consider that you're sure that the first frog you pick is a male (because it croaks) then you have 50/50 of having a male or a female, thus 1 * 0.5 = 0.5 or 50% but idk if this is the answer OP wants
I don't really know, could be 50%, 25% or 16.5% depending on the interpretation
>>
>>13062235
it's random
>>
>>13062235
P(both frogs are male) = 0.5
There are 4 possible outcomes
MM
MF
FM
FF
Given we heard a croak, P(FF) = 0
Croaking is a random event but I presume having 2 males doubles the probability of hearing a croak. I think about it backwards. For each of the remaining 3 outcomes, what are the chances of hearing a croak. Let C be the event that a croak is heard. P(C | MF) = P(C | FM) = (1/2) * P(C | MM). Using bayes rule, P(MM | C) = P(C | MM) * P(MM) / P(C).
P(MM) = P(MF) = P(FM) = P(FF) = 0.25
P(C) = P(C | MM)*P(MM) + P(C | MF)*P(MF) + P(C | FM)*P(FM) + P(C | FF)*P(FF)
Let a = P(C | MF) = P(C | FM)
Then P(C) = 2a*0.25 + a*0.25 + a*0.25 + 0*0.25 = a

Therefore P(C | MM) * P(MM) / P(C) = 2a * 0.25 / a = 0.5
>>
>>13065142
And?
>>
>>13065150
Incorrect. A croak came from a specific frog so it's not the same as just giving information about the pair. Also, the fact that the other frog didn't croak makes it more likely to be female.
>>
>>13066059
>Croaking is a random event but I presume having 2 males doubles the probability of hearing a croak.
Incorrect. If x is the chance of a male croaking then the chance of two males producing one croak is 2x(1-x), which is less than 2x.
>>
>>13065137
Not generalizing, so far no one has given the correct answer in this thread. It's pathetic.
>>
>>13062235
1/3
>>
>>13064469
>It doesn't matter, my argument still applies.
I disagree. The interpretation matters. If you listen for a set amount of time and discard scenarios where you hear two croaks, that's different from including scenarios where you would have heard two croaks but stopped to assess the probability before the second croak would have been heard.
>>
>>13065180
Because in binary the cases are:
00 01 10 11
Where 0 is female and 1 is male.
Case 00 is thrown out. There are 3 other cases, only one is 11. 1/3
> for the last time, 01 != 10
>>
File: f.png (13 KB, 631x590)
13 KB
13 KB PNG
A+B+C+D = events of one croak.
B+C = in which both are male
x = prob of a male croaking
x -> 0, in infinitesimal time frame
(B+C)/(A+B+C+D) -> 1/2
x -> 1, in sufficiently long time frame
(B+C)/(A+B+C+D) -> 0
>>
>>13066131
Wrong.
>>
>>13066243
Wrong.
>>
>>13066131
If x=50%. x is always 50%. So you are correct.
>>
>>13066170
Then calculate the probability.
>>
>>13066228
No, the cases are

M M
Mc M
M Mc
Mc Mc
M F
Mc F
F M
F Mc
F F
>>
>>13066248
See>>13066243
>>
>>13066268
Mc and M are the same.
All the frogs that can be Mc are also M.
>>
>>13066113
>>13066113
I may be wrong here. I assumed the croaks followed a poisson distribution. When you double the rate, you double the expected outcome, but that's not what we're measuring exactly. We're after the probability ratio of croaking between mixed gender pairs and all male pairs.
That being said, if we use a binomial model, then you're incorrect. The probability of at least on out of two males croaking would be x + x(1-x) = x(2-x), which is also less than 2x, so you're right about that.
Using the poisson model, the probability ratio follows the curve 2e^x / e^(2x) where x is the rate at which croaks occur.
>>
File: 3USgt2gh.jpg (28 KB, 500x281)
28 KB
28 KB JPG
>>13066279
>Mc and M are the same
>croaking is the same as not croaking
>a is not a
>>
>>13066255
The average value given a uniform distribution over x is 1-ln2 ~ 0.307
>>
>>13066294
Any frog that is male can either be the frog that croaked or a male who didn't for all we know.
>>
>>13066282
You're confusing the probability of two males producing a croak with the probability of two males producing at least one croak. It's the former that matters here.
>>
>>13066306
OK, and what does that have to do with my post?
>>
File: 1619989223055.png (12 KB, 1758x590)
12 KB
12 KB PNG
>>13066260
Borrowing >>13066242, if you allow the scenario where both males would have croaked but you stopped and looked at them as soon as you heard one, as x approaches 1, the probability approaches 1/3. As x approaches 0, the probability approaches 1/2. That aligns with my initial thoughts in >>13063093.
>>
>>13066327
Mc isn't separate from M.
These letters represent groups of frogs. Any member of M can be a member of Mc therefore all possible members of Mc are all possible members of M so they represent the same frogs.
>>
>>13066336
>At time T1, you hear one fart. What's the chance that both are males.
P(male not farting in T1) = e^(-F)
P(male farting once in T1) = F e^(-F)
P(two males producing one fart in T1) = 2F e^(-2F)
P(2 males | one fart in T1) = (1/4)(2F e^(-2F))/((1/4)(2F e^(-2F))+(1/2)(F e^(-F))) = 1/(e^F+1)
>>
>>13062235
It's 1/3, and you should literally kill yourself and your whole god damn family to remove your retard genes from the gene pool, if you answer anything else.
>>
>>13066402
>Mc isn't separate from M.
Incorrect. Mc is a male that croaked and M is a frog that didn't croak. These cannot be the same by definition.

>These letters represent groups of frogs
No.
>>
>>13066476
Incorrect.
>>
>>13066380
I don't see how. Using Poisson distribution, if F is the frequency of male croaking in the time you were listening, the probability of two males producing one croak is 1/(e^F+1), which ranges from 0 to 1/2.
>>
>>13066504
*two males given one croak
>>
>>13066504
>>13066523
That's why I specified the assumption that you stopped and tried to assess the frogs as soon as you heard any croak, opposed to there being exactly 1 croak during some predetermined listening period (e.g. you were planning on standing outside for X minutes). That's how I interpreted "You hear a croak behind you and turn around".
>>
>>13066536
The listening time isn't predetermined. It doesn't matter whether you waited for some amount of time and only heard one croak or stopped listening immediately after hearing a croak.
>>
>>13066580
>It doesn't matter whether you waited for some amount of time and only heard one croak or stopped listening immediately after hearing a croak.
Why wouldn't it matter. If there are two male frogs and they croak very frequently, the scenario of hearing exactly one croak over a specified time period would be much less likely than hearing at least one croak eventually. Solutions like >>13063281 and >>13066242 explicitly exclude the possibility of both male frogs croaking and come to very different conclusions as a result.
>>
>>13062235
1/3 of course
>>
>>13062530
Hot
>>
>>13066688
>If there are two male frogs and they croak very frequently, the scenario of hearing exactly one croak over a specified time period would be much less likely than hearing at least one croak eventually.
The "specified time" is the same as the time it took you to hear a croak. The probability is exactly the same.

>Solutions like >>13063281 (You) # and >>13066242 # explicitly exclude the possibility of both male frogs croaking and come to very different conclusions as a result.
They come to the exact same result. You have no clue what you're talking about.
>>
File: 1619878913532.jpg (123 KB, 1024x539)
123 KB
123 KB JPG
>>13062235
Ribbit
>>
>>13066795
I think I understand now. The x I was using is "the probability of a male frog croaking during the time I was planning to stand outside", while your x is "the probability of hearing a croak during the time before hearing the croak". If it's known that male frogs croak very frequently, and yet it took a very long time before hearing a croak, then I can see how that would make the probability of there being 2 male frogs very low. I was focused on whether or not a croak was heard at all, without considering the time before hearing a croak being part of the input to the problem.

>They come to the exact same result.
I typed badly. They came to the same result, which was different from mine.
>>
File: brainlet.jpg (8 KB, 231x250)
8 KB
8 KB JPG
>>13063093
>>13066242
>>13066380
That is irrelevant. Lets suppose in the situation you observe the frogs (or you where in their presence enough to hear them, like in the original scenario) until you heard one croak then you immediately stopped considering if there where any subsequent croaks, you instantly start considering the chances of both the frogs being male.

No matter how frequently the frogs croak, you will only hear 1 croak, and the other frog will be silent, If they croak more frequently, you would have observed their behavior a much shorter time until 1 croaked,

>>13066688
>Why wouldn't it matter. If there are two male frogs and they croak very frequently, the scenario of hearing exactly one croak over a specified time period would be much less likely than hearing at least one croak eventually.
that specified time period is how long you where present until one frog croaked which your consideration of following croaks were immediately ceased.

>>13067041
>The x I was using is "the probability of a male frog croaking during the time I was planning to stand outside", while your x is "the probability of hearing a croak during the time before hearing the croak".
yes that changes the question, they are different. The first one would be depended on how frequently the frogs croak. But in the second one, the probability of hearing a croak, given you will observe the frogs until one croaks will be, by definition will be 100% (also given at least 1 is male).


also Pic rel, whats the probability wojak hears a croak
>>
>>13067217
>No matter how frequently the frogs croak, you will only hear 1 croak
Not if both frogs are female. And if both frogs are male you are more likely to hear a croak faster than if only one is male. The rate of croaking determines how much more likely it is. So it's highly relevant to the probability of two males.
>>
>>13064474
You still didn't explain how, retard. How did MM and MF suddenly become of equal probability when FF is removed - which is the only information you get when you hear a croak - you fucking schizo??
>>
>>13067730
>You still didn't explain how, retard.
But I did. You just didn't listen, retard. >>13063129

>How did MM and MF suddenly become of equal probability when FF is removed
They didn't. Can't you read?

>which is the only information you get when you hear a croak
Wrong. Brainlet. You get information that there is a frog that didn't croak which is more likely to be female.
>>
Wait why is everyone arguing

Normally it's 50/50. The possibilities are equally split between:
MF, MM, FM, FF

A frog croaked. The only possible situations where a frog croaks are:
MF, MM, FM

Thus it's 33.3%

I'm high so maybe I missed something? Or is this board incredibly bad at logic?
>>
>>13067477
this is not relevant for the given situation. time isnt a factor

>>13067852
MF == FM
>>
>>13062235
How much time has elapsed and what is the probability density function for the frequency of croaks?
>>
>>13067896
MF =/= FM

Think of it as Frog #1 and Frog #2. Frog #1 can be male, or Frog #2 can be male, or both can be male. Those are the possibilities.
>>
>>13067802
>You get information that there is a frog that didn't croak which is more likely to be female.
LMAO how the fuck does this even work you fucking retard
>>
>>13067896
>MF == FM
Define "=="
>>
>>13067915
The order in which the frogs are doesnt matter in this case since its identical if the frog that croaks is #1 or #2. And for this reason I dont count them as different possibilities at least for this specific scenario.

>>13068281
I meant == as just a fancy way of showing that MF and FM can be used interchangeably for this case since they are the same thing
>>
>>13067477
>And if both frogs are male you are more likely to hear a croak faster than if only one is male.
Yes, but you are observing UNTILL YOU HEAR A CROAK, what you're saying doesn't make a difference. Lets say you hear a croak every fucking second, If you listen for a second (or half a second if both happen to be male) You will only listen for a second (or half) until you hear 1 croak, and then you consider the original question.

>Not if both frogs are female
we're talking about a situation where he have heard a croak, how could both be female?
>>
>>13067896
>this is not relevant for the given situation. time isnt a factor
This is not an argument. I just explained how it's relevant. We need to know the probability of a male croaking while you were listening. Or equivalently, the amount of time you were listening and the croaking frequency of the male, which would then be used to calculate the Poisson distribution for croaking.
>>
>>13067852
See >>13063281
>>
>>13068276
It's difficult for me to ascertain how I should explaim it to you since you could be exhibiting several different levels of stupidity. Since females never croak they are always silent. Since males do croak they are not always silent. This a silent frog is more likely to be female. If you don't understand this you might need remedial education.
>>
>>13068706
>Yes, but you are observing UNTILL YOU HEAR A CROAK, what you're saying doesn't make a difference.
This has zero affect on what I said.

>Lets say you hear a croak every fucking second, If you listen for a second (or half a second if both happen to be male) You will only listen for a second (or half) until you hear 1 croak, and then you consider the original question.
OK, and? If both are female you would not hear a croak at all. If both are male you would hear a croak faster than if there was 1 male. Using a Poisson distribution for croaking, if you listened for t seconds then the probability of two males is 1/(1+e^t). So your claim that time and rate of croaking are irrelevant is disproven.

>we're talking about a situation where he have heard a croak, how could both be female?
No, we're talking about whether he would have to wait to hear a croak. He would only have to hear a croak if both frogs were not female. Just because something happened doesn't mean it had to happen. I suggest you learn basic probability.
>>
>>13069029
>You hear a croak behind you and turn around to see two frogs of the same species.
And how would we know it one would be the "quiet" frog? We only know that at least one of the frog would be a male, but don't know which. Therefore, the probability of a MF pair would still be double than that of MM. If we would know which frog croaked, then your statement would be correct, but otherwise, it wouldn't be.
>>
>>13069908
>And how would we know it one would be the "quiet" frog?
Because you only heard one croak. It doesn't matter which one.

>We only know that at least one of the frog would be a male
No, we know a frog croaked.

>Therefore, the probability of a MF pair would still be double than that of MM.
No, MF and MM are not equally likely to produce a croak while you were listening, so their probabilities do not remain proportional after getting that information.

>If we would know which frog croaked, then your statement would be correct, but otherwise, it wouldn't be.
It doesn't matter which frog croaked. It's completely symmetric:

Mc M
M Mc
Mc F
F Mc
>>
>>13068607
The order of the frogs does matter. All you know from the given information is that at least one frog is male. MF, FM, and MM are three distinct scenarios.

If I randomly select frog A and frog B from the global pool, assuming 50% male population 50% female population, then the probability that one is male and one is female is 0.5 while the probability that both are male is 0.25.
>>
>>13066092
>A croak came from a specific frog so it's not the same as just giving information about the pair.
It is, since you don't know which one croaked.
>Also, the fact that the other frog didn't croak makes it more likely to be female.
You don't know that. Males don't have to croak. They just sometimes do.
>>
>>13070362
>All you know from the given information is that at least one frog is male
No, you also know a frog croaked.
>>
>>13067852
>Normally it's 50/50. The possibilities are equally split between:
>MF, MM, FM, FF
And how many of those are MM
>>
>>13070594
That just tells you one is male.
>>
>>13070592
>It is, since you don't know which one croaked.
It doesn't matter which one croaked, it's symmetrical:

Mc M
M Mc
Mc F
F Mc

>You don't know that.
I do though. It's simple conditional probability. Let x > 0 be the probability of a male croaking while you were listening.

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

>Males don't have to croak.
I didn't say they did. The fact that they are more likely to croak than females is equivalent to the fact that silent frogs are more likely to be female.
>>
>>13070642
It also makes it more likely both frogs are male since two male frogs are more likely to croak than one.
>>
>>13070658
>Mc M
>M Mc
These are functionally identical.
>silent frogs are more likely to be female
Right, but that's not true, because a silent frog may well be male. It possesses the ability to croak, it just didn't. So they are not equivalent.
>>13070687
Again, doesn't follow. Nothing is said about the likelihood of a male frog ever croaking.
>>
>>13062235
Why are everyone retarded, ordering doesn't matter
P(MM) = 2/3
P(FM) = 1/3
Since a MM pair is twice as likely to croak than an FM pair
>>
>>13070750
>These are functionally identical.
And so are Mc F and F Mc.

>Right, but that's not true, because a silent frog may well be male.
Doesn't follow. I never said they may not be male. I said they are less likely to be male. Are you having trouble reading?

>So they are not equivalent.
What are not equivalent?

>Nothing is said about the likelihood of a male frog ever croaking.
It's said that croaking is random, so two males must be more likely to croak than one. If not, the rate of croaking in one frog would depend on how many frogs are adjacent to it and would not be random.
>>
>>13070941
>I never said they may not be male. I said they are less likely to be male.
And that's the part that's not true.
>>
>>13070956
I already proved it's true and you had no response: >>13070658

Time to stop LARPing and learn basic probability theory.
>>
>>13070933
>P(MM) = 2/3
1/4

>P(FM)
1/2

>Since a MM pair is twice as likely to croak than an FM pair
This is true, but not very helpful. The question is not how likely MM is to croak at least once, it's how likely MM is to produce a single croak. MM is only 2/e^k times more likely to produce a *single* croak than MF, where k is the rate of croaking over the amount of time you were listening. You also failed to take into account that MM is originally half as likely as MF.
>>
>>13070974
So if you see a single frog that hasn't croaked, are you saying it's more likely to be male? Because male and female frogs are equally distributed and male frogs don't have to croak at all. A frog not croaking tells you nothing. A frog croaking tells you it's male. One of two frogs croaking without knowing which tells you at least one is male. That's it.
>>
>>13070974
>>13071003
More likely to be female, sorry
>>
>>13071003
>So if you see a single frog that hasn't croaked, are you saying it's more likely to be [fe]male?
Yes.

>Because male and female frogs are equally distributed and male frogs don't have to croak at all.
Right, but once you start listening for a croak, it's no longer a random frog since listening to it gives you information about its gender.

>A frog not croaking tells you nothing.
Wrong. I already proved it does and you have no response, again.

>One of two frogs croaking without knowing which tells you at least one is male. That's it.
Your incorrect intuition is not a proof.
>>
>>13071043
>Yes.
So, the moment you see a frog, the likelihood of it being female starts to increase the longer it goes without croaking?
>>
>>13071086
Yes. The probability of a female given no croaking over some time t and average rate of male croaking r is 1/(1+e^(-tr)). So at t=0 the probability is 1/2 and as t->inf the probability approaches 1.
>>
>>13071149
Fuck, you're actually right, OP shouldn't have included the detail that croaking is random
>>
>>13071000
>You also failed to take into account that MM is originally half as likely as MF.
No it doesn’t fucking retard since ordering doesn’t matter

Let’s say you are an IDF soldier. Palestinians and jews form pairs uniformly and are in equal proportions. Palestinians throw stones at IDF soldiers at some rate while Jews do not. You get a stone thrown at the back of your skull. It is twice as likely that it came from a pair of Palestinians than a pair consisting of a Jew and a Palestinian and the Jew pair has zero probability. So the pair of Palestinians have probability 2/3 vs the pair consisting of a jew and Palestinian that has the probability 1/3

Fuck this fucking board, only fucking retarded kids here
>>
>>13071254
You forget that the Jew/Palestinian pair is twice as likely to occur.
>>
>>13071254
>No it doesn’t fucking retard since ordering doesn’t matter
LOL, ordering has nothing to do with it. One male and one female are twice as likely as two males regardless of order.

>It is twice as likely that it came from a pair of Palestinians than a pair consisting of a Jew and a Palestinian and the Jew pair has zero probability.
No. Assuming Palestinian rock throwing follows a Poisson distribution with average rate r, the probability of a pair of Palestinians throwing exactly one stone in time t at you is 2tr e^(-2tr). The probability of one Palestinian throwing exactly one stone at you is tr e^(-tr). The former is not twice the latter. You also have to take into account that it's twice as likely a Palestinian and a Jew will pair than two Palestinians, since each member of the group has a 1/2 chance of being Jewish or Palestinian.
>>
1/3.

4 equiprobable events
MM
MF
FM
FF

Given at least one male the outcomes are
MM, MF, FM

so 1/3 that both frogs are male given at least one male.
>>
>>13069058
Scroll up to the top of the thread and nigga read the question.

>No, we're talking about whether he would have to wait to hear a croak.
Read the question nigga with your eyes "You hear a croak behind you" we absolute are not talking about whether he would have to wait to hear a croak, its in present tense.

If the question was "If you where to observe two frogs of unknown gender, for t amount of time, Whats the chances of both frogs being male given you hear 0, 1, 2... 3million croaks, with the the frequency of frogs croaking on average once every x seconds" then yes, what you are proposing is relevant.
>>
>>13072341
Wrong. See >>13063281
>>
>>13072706
consider your cases Mc M and M Mc, what is the probability of each of these cases? p(Mc M) = P(2 males) P(First frog croaks | both male) = 1/4 * 1/2

while P(Mc F) = P(1 male) P(first frog croaks | first frog is male) 1/4 * 1.

I am being loose with the formalisms of the random variables, but the point is, notice that the cases Mc M and M Mc each have half the probability mass that Mc F or F Mc do. The step in which you argue that by "symmetry" the problem reduces to Mc M or Mc F is invalid. You are right in saying that the only cases that matter are

Mc M
M Mc
Mc F
F Mc

but you are wrong in saying all of these scenarios are equiprobable given the current information. Mc M and M Mc have half the probability of either scenario Mc F or F Mc, so we have 1/2(P(Mc M) + P(M Mc) + P(Mc F) + P(F Mc) = 1 and from here we can see that given P(Mc M) = P(M Mc), the answer is 1/3


retard.
>>
>>13072584
>Read the question nigga with your eyes "You hear a croak behind you" we absolute are not talking about whether he would have to wait to hear a croak, its in present tense.
Just because we heard a croak doesn't mean the probability of hearing a croak is 1.

>If the question was "If you where to observe two frogs of unknown gender, for t amount of time, Whats the chances of both frogs being male given you hear 0, 1, 2... 3million croaks, with the the frequency of frogs croaking on average once every x seconds" then yes, what you are proposing is relevant.
It's relevant either way. Your argument basically boils down to "if I didn't think of it then it's not relevant." Your ignorance is not an argument.
>>
>>13069029
>This a silent frog is more likely to be female.
No, it is not. Why the fuck would it be? What general principle did you derive this statement from? Show your work.
>>
>>13070182
>No, MF and MM are not equally likely to produce a croak while you were listening
I wasn't listening. I just heard a croak. If a doctor came in and told you that one frog is male, would that mean the other frog is more likely to be female since the doctor didn't say a thing?
>>
>>13072737
>I am being loose with the formalisms
Yes.
>>
>>13072737
>P(2 males) P(First frog croaks | both male) = 1/4 * 1/2
>while P(Mc F) = P(1 male) P(first frog croaks | first frog is male) 1/4 * 1.
This incorrectly assumes that the probability of some male in the pair croaking is 1. The probability of some male being the first to croak is only 1/2 if it's guaranteed that one or the other will croak. That can only be the case if males croak constantly, in which case there is no male which croaks first. Your reasoning is gibberish. Try again.

>The step in which you argue that by "symmetry" the problem reduces to Mc M or Mc F is invalid.
How so?

>but you are wrong in saying all of these scenarios are equiprobable given the current information
I never said that. If you actually read and unsuspecting my posts you would see I said the opposite.
>>
>>13072769
>Just because we heard a croak doesn't mean the probability of hearing a croak is 1.
No, but it also doesn't mean you can assume whatever the fuck you want about the probability of frogs croaking.
>>
>>13072813
>No, it is not.
Wrong.

>Why the fuck would it be?
I explained it in the post you're replying to. Maybe you should read it again until you understand it.

>I doubt you're competent enough to understand it but the proof is here: >>13070658

Next time put some actual substance into your post instead of just stating your incredulity.
>>
>>13072834
>I wasn't listening. I just heard a croak.
How did you hear it of you weren't listening?

>If a doctor came in and told you that one frog is male, would that mean the other frog is more likely to be female since the doctor didn't say a thing?
What is this gibberish? Did the doctor tell you or not? The probability would depend completely on how the doctor determined that one frog was male.
>>
>>13062235
MF
FM
MM

1/3
>>
>>13072865
"The frog croaked" is an assumes condition. The fact that you calculated the probability that the frog croaked is irrelevant, since it has already happened.

You don't think the probability of flipping tails becomes smaller the more tails in a row you get, don't you? Lmao
>>
>>13072863
I didn't assume anything about the probability of frogs croaking other than what the question says. You ignore that croaking (or lack of croaking) gives more information than just "at least one is male" so you get the wrong answer.
>>
>>13072884
Wrong. See >>13063281 to start.
>>
>>13072876
>What is this gibberish? Did the doctor tell you or not?
He told you a single frog is male. He didn't say a thing about the other frog. Fix your reading comprehension.
>>
>>13072886
>"The frog croaked" is an assumes condition
Right and you used it to calculate probabilities that don't have it as a condition. This is basic stuff.

>The fact that you calculated the probability that the frog croaked is irrelevant
It's highly relevant, since P(MM|C) = P(MM) P(C|MM) / P(C)

You understand nothing about this topic.

>You don't think the probability of flipping tails becomes smaller the more tails in a row you get, don't you?
What does this have to do with anything I said? The probability of a fair coin landing on tails is independent from how many times it landed tails. The probability of two males is not independent from the probability of croaking.
>>
>>13072892
Stfu i'm a statistics major
>>
>>13072893
>He told you a single frog is male.
This is ambiguous. If he looked at only one frog and saw it's male, then it's 1/2. If he looked at both but is only telling you the amount of makes is at least one then it's 1/3. Neither scenario is analogous to a male frog croaking, which is behavior that follows a Poisson distribution.
>>
>>13072876
>The probability would depend completely on how the doctor determined that one frog was male.
OK, I see your problem.
You're trying to calculate the whole world into this simple problem.
You can't calculate the whole world.
If we were to go down this road, we would have to account for the probability that you hear voices, that you misheard a water splash that kinda sounded like a croak, that the frogs don't exist and you are only a brain in a vat... You see where I'm going there?

Do you know why we don't calculate all that shit?
BECAUSE IT'S FUCKING UNKNOWN.
That's how probability/statistics work - you take the known information and you derive the probability from the information YOU KNOW IS TRUE.

The fact that your result >>13063281 contains a variable that wasn't defined in the initial problem only shows that you are incapable of basic statistical thinking.
>>
>>13072932
>>13072876
>>13072887
>>13072916
>>13072865
See: >>13072933
>>
>>13072920
You should probably look for a different major. Does your community college offer a degree in a McWhopper and medium fries?
>>
>>13072769
No, the problem is you are answering the wrong question.
>Just because we heard a croak doesn't mean the probability of hearing a croak is 1.
we're worried about the chance of both frogs being male, Not how frequently they croak, we've already heard a croak How likely that is to happen is irrelevant considering we already know it has.
>but 2 male frogs = increased chance of the situation occurring.
yes, but the situation has occurred, Because the situation has occurred, we then ask the original question OP asked, hence why everyone is talking about the Mc F, Mc M, M Mc, or MF==FM shit, that's the crux of the problem, rather than probability densities and shit.

Even if the chance of a frog croaking is 1/10000000000000000000, by the grace of god we where lucky to have heard a croak (op's question), The fact that we have heard a croak implies at least 1 of the two are male, the chance for a frog to be male is still 50%

If the chance of a frog croaking is 99.99%, you have still only heard one croak. Usually the silence of the second frog would imply it is a female considering how often males croak, but the question starts off with you hearing a croak, with the duration of how long the frogs where in hearing range being a mystery, and instantly considering the probabilities for both being male after just hearing 1 croak (regardless of how long that was).

>It's relevant either way. Your argument basically boils down to "if I didn't think of it then it's not relevant." Your ignorance is not an argument.
funnily enough, that's not an argument.
>>
>>13072933
I believe the most perfect conclusion to this thread and the anon trolling would be:

Take your meds, schizo.
>>
>>13072892
Yeah, it's 1/2
>>
>>13072892
Wrong. See >>13072933 for an explanation of your fallacy.
>>
>>13072933
>If we were to go down this road, we would have to account for the probability that you hear voices
I'm accounting for the probability of croaking, which is not some extraneous thing, it's what the problem is about. You have utterly failed to counter the fact that the probability is dependent on the timing and frequency of the croak.

>The fact that your result >>13063281 (You) # contains a variable that wasn't defined in the initial problem only shows that you are incapable of basic statistical thinking.
I know variables and underdetermined problems must be very scary to a little child like you, but you'll get over it eventually.
>>
>>13072942
>No, the problem is you are answering the wrong question.
I'm answering the question in the OP. The answer is 1/(1+e^(-tr)) where t is the time you were listening and r is the average rate of croaking. Equivalently, 1/(1+1/p) where p is the probability of a male not croaking while you were listening. If your answer is different, it's because you ignored information or made some assumption not given by the problem.

>we're worried about the chance of both frogs being male, Not how frequently they croak
The first is determined by the second.

>How likely that is to happen is irrelevant considering we already know it has.
Completely wrong. The entire field of conditional probability shows the opposite is true.

>yes, but the situation has occurred, Because the situation has occurred, we then ask the original question OP asked
That's nice, how does it respond to anything I said? Why do you waste time repeating this over and over?

>rather than probability densities and shit.
Sample spaces are useless without probabilities attached to them.

>The fact that we have heard a croak implies at least 1 of the two are male, the chance for a frog to be male is still 50%
Nothing you're saying responds to what I said. You're just ignoring what I'm telling you instead of disproving it.

>Usually the silence of the second frog would imply it is a female considering how often males croak
Not usually, always. It's mathematically proven.

You lost, get over it.
>>
What the fuck is going on in here
>>
>>13072961
There is no fallacy explained there, just a child's conception of probability theory.
>>
>>13072954
Wrong. A silent frog is more likely to be female.
>>
>>13073010
Just /sci/ failing miserably at a probability question meant to sort several different kinds of midwits.
>>
>>13072939
All scenarios:
MM
MF
FM
FF

From OP: You hear 1 croak, so there's at least 1 male.

OP is asking us:
P(2M | there's at least 1M)

= P(2M and there's at least 1M) / P(there's at least 1 M)

= P(2M)/P(there's at least 1M)

= (1/4)/(3/4)

= 1/3

Now stfu schizo.
>>
>>13062235
1/3
There are three cases to consider: both are male, both are female, and both are male, and they are all equally likely
>>
>>13073039
Also
>females and males exist in equal proportion, and males and females congregate randomly
That's the reason the prob of each scenario equals 0.25
>>
>>13073039
You missed a few

All scenarios:

M M
Mc M
M Mc
Mc Mc
M F
Mc F
F M
F Mc
F F

OP is asking us:
P(2M | 1c)

= P(2M) P(1c|2M) / P(1c)

= (1/4) 2(1-x)x / ((1/4) 2x(1-x)+(1/2)x) = (1-x)/(2-x) where x is the chance of a male croaking
>>
>>13073044
>There are three cases to consider: both are male, both are female, and both are male
LOL

>and they are all equally likely
Wrong.
>>
>>13062235
>What is the probability both frogs are male?
1/4? how is this hard?
>>
>>13073077
Wrong.
>>
File: WTF.png (195 KB, 467x543)
195 KB
195 KB PNG
You said "croak", not "croaks". The probability is zero that both are male. Only one is male.

Image unrelated.
>>
>>13073092
the probability that each frog is male is 50% right? And they're independent events?
>>
>>13073105
Doesn't follow. A male doesn't have to croak while you're listening.
>>
>>13073106
Right, before listening for croaks. Once you do that, your are getting information about whether they're male, so it's not random.
>>
>>13073117
Croaks doesn't matter, OP is asking what the probability that both frogs are male is
>>
>>13073113
Can we just fucking kill the frogs and eat them?
>>
File: 1610553173103.gif (891 KB, 390x364)
891 KB
891 KB GIF
>>13072976
>
I'm accounting for the probability of croaking, which is not some extraneous thing, it's what the problem is about.
No, it is absolutely NOT what the problem is about. All the information is given as-is. You are supposed to extrapolate the probability from the information you have, not add some magic variable that pulled out of your ass because "muh intuition".

>I know variables and underdetermined problems must be very scary to a little child like you, but you'll get over it eventually.
You're retarded and your "comeback" is pathetic, at best.
OP presented a well-defined problem and asked for the probability of an event - which is a real number between 0 and 1, inclusive.
You provided him an answer that contains a variable that you defined yourself >>13063281 so you have demonstrated that you are incapable of following basic instructions, and don't understand the fundamental principles of probability and statistics.
>>
>>13073142
We can kill shitposters who think they're smart for being contrarian and eat them
>>
>>13062235
Is pic related? If so, what does a baguette have to do with frogs?
>>
>>13073140
>Croaks doesn't matter
Wrong.
>>
>>13073613
>All the information is given as-is.
How would you know?

>You are supposed to extrapolate the probability from the information you have
I did.

>OP presented a well-defined problem and asked for the probability of an event - which is a real number between 0 and 1, inclusive.
That's what my answer is.

>You provided him an answer that contains a variable that you defined yourself >>13063281 (You) # so you have demonstrated that you are incapable of following basic instructions
What "instructions" told you that you are not allowed to use variables?

>don't understand the fundamental principles of probability and statistics.
Nice projection.

There is nothing substantive in your posts. Continue whining about nothing while my solution remains mathematically proven. It's entertaining.
>>
>>13073983
Bump
>>
File: 22.png (27 KB, 300x100)
27 KB
27 KB PNG
>>13073983
>How would you know?
Because that's the way mathematics work - you are given assumptions and you derive conclusions.

>What "instructions" told you that you are not allowed to use variables?
A function is *not* a real number between 0 and 1, inclusive, and is therefore *not* a solution to the OP's problem. Don't you know the difference between functions and numbers?

>Nice projection.
You don't even know what that means, you're just parroting a meme.

>Continue whining about nothing while my solution remains mathematically proven.
Wow, you pulled out a random variable out of your ass, how intelligent. You are really a groundbreaking thinker, and /sci/ is so lucky to have such out-of-the-box thinkers like you! *pats head*
>>
>>13072916
>The probability of two males is not independent from the probability of croaking.
How so? Please describe your line of thought.
>>
File: 1613497362478.jpg (55 KB, 680x467)
55 KB
55 KB JPG
>this thread again
It's 1/3
>>
so before we reach 300 replies... what is the answer according to OP?
I suspect he's wrong tho
>>
>>13062235
Not enough info. We don't know if the females croak with the same frequency around females and males.
>>
File: 1614907461118.png (833 KB, 739x739)
833 KB
833 KB PNG
>>13076264
>>
>>13076276
calling everybody schizo is not a good argument ya know?
>>
>>13076134
>Because that's the way mathematics work - you are given assumptions and you derive conclusions.
That's what I did. You made an incorrect assumption that the probability of croaking is irrelevant to the answer.

>A function is *not* a real number between 0 and 1
(1-x)/(2-x) is not a function, it is the real number value of a function. A function would be f(x) = (1-x)/(2-x)

>You don't even know what that means
Nice projection.

>Wow, you pulled out a random variable out of your ass
There is nothing random about it. The answer to the problem is dependent on it.
>>
>>13076276
I'm treating the problem like my life is on the line, not some ideal no real scenario
>>
>>13070624
FF isn't possible you dolt
>>
>>13076364
>You made an incorrect assumption that the probability of croaking is irrelevant to the answer.
No, you made the incorrect assumption that the probability of croaking is relevant to the answer. The probability of croaking doesn't even make sense in the context of the question as we aren't waiting for the croak.
Now that I think about it, what the fuck even is "the probability of croaking"? The p thingy you pulled out of your ass, what does it signify? Number of croaks per second? Per minute? Some function of it? Be precise.

>(1-x)/(2-x) is not a function
It's literally a function.

>The answer to the problem is dependent on it.
If it's so fundamentally important, then I suppose you will have no problem answering the above questions.
>>
>>13076313
That information isn't necessary to the problem. OP said the male frogs croak randomly and nothing else
>>
>>13076153
I already explained it in the post you're replying to.

P(MM|C) = P(MM) P(C|MM) / P(C)

The only way they could be independent is if P(C|MM) = P(C) but we know this is false since males croak but females don't.
>>
>>13076247
Wrong
>>
>>13076456
>OP said the male frogs croak randomly and nothing else
Well OP should've fucking elaborated what the fuck does he mean by "croak randomly". Croaks are discreet events, what the fuck does "random discreet event" even mean?
The question is ill-posed.
>>
>>13076255
1/(1+e^(tr)) where t is the time you were listening and r is the rate of male croaking.
>>
>>13076264
Females don't croak and croaking is random.
>>
>>13076476
Why did you assume that croaking is uniformly distributed?
>>
this thread
> people who misunderstand basic probability or get tripped up by word problems
> people who get the question right
> schizos who can't arrive at an answer to a word problem without imagining fictitious details bolted on because they are incapable of abstract thought

the third is the worst kind of /sci/ poster
>>
>>13076484
I assumed females since an older version of the thread had females croaking. Randomness isn't uniquely defined in probability. A fair coins result is random. A biased coins result is also random. Their probabilities however aren't the same.
>>
>>13076469
>what the fuck does he mean by "croak randomly"
So I was right about you being a schizo. Obviously you aren't looking at the croaks by themselves but the probability of a frog croaking after x amount of time has elapsed.
>>
>>13076531
>Obviously you aren't looking at the croaks by themselves but the probability of a frog croaking after x amount of time has elapsed.
But how can you be so sure that the probability is uniform?
>>
>>13076516
you forgot
>people who don't understand Bertrand's Paradox
>>
>>13076441
>No, you made the incorrect assumption that the probability of croaking is relevant to the answer.
I didn't assume it, I proved it. But thanks for admitting your answer is based on a fallacy.

>The probability of croaking doesn't even make sense in the context of the question as we aren't waiting for the croak.
What does one have to do with the other?

>Now that I think about it, what the fuck even is "the probability of croaking"?
The probability of a male producing a crisp within a certain interval of time. Since frogs croak randomly, this probability is described by the Poisson distribution.

>The p thingy you pulled out of your ass, what does it signify?
The only time I used p was as the probability of a male croaking 0 times while you were listening.

>Number of croaks per second?
No, that's a rate.

The probability of a male croaking k times is (tr)^k e^(-tr) / k!

>It's literally a function.
No, it's literally an expression of a real number between 0 and 1/2.

>If it's so fundamentally important, then I suppose you will have no problem answering the above questions
I had no problems.
>>
>>13076500
I didn't. Nothing I said implied that.
>>
>>13076544
>But how can you be so sure that the probability is uniform?
Nobody knows if it is uniform, but it is known that the probability is greater than zero.
>>
>>13076516
Let me see your reasoning so I can make fun of you.
>>
>>13062235
>/sci/ finding out probability is only dependant on how precise you want to be and how much you need your model to mimic reality
>>
>>13076523
The only difference between a fair coin and a biased coin is the probability of H and T. Coin flips are still independent from each other, that's what is meant by random.
>>
>>13076652
the answer is 1/3 take your meds
>>
>>13076679
Yeah that's the point of my comment.
>>
>>13076786
Why is it 1/3?
>>
>>13076964
It's clear from context what is meant by random. It means the probability of croaking is independent from any factor besides gender.
>>
>>13076645
>Nobody knows if it is uniform, but it is known that the probability is greater than zero
Retard
>>
>>13076615
>The probability of a male producing a crisp within a certain interval of time.
I said, be precise.
Do you know what "precise" means, you nigger?
What interval of time? Is it independent of the time of day? Time of year? The particular forest you're in? If so, how can you be so sure of all these things?
>>
>>13076615
>Since frogs croak randomly, this probability is described by the Poisson distribution.
I don't see how you've arrived from "frog croak randomly" to "probability is described by the Poisson distribution".
Would you care to elaborate on this part in a more pedantic way?

That is, unless you're an uneducated nigger.
>>
>>13076615
>No, it's literally an expression of a real number between 0 and 1/2.
Lmao ok then please give me (1-x)/(2-x) kilograms of cheese
>>
>>13076615
>I had no problems.
Because you didn't give a real answer, you just parroted all the bullshit you said before without any elaboration. You're not gonna convince anyone that way, autist. Mathematics is indisputable when presented as a well-formed logical argument, which you obviously failed since I'm not convinced.
>>
>>13078383
>What interval of time?
t

>Is it independent of the time of day? Time of year? The particular forest you're in?
Yes, it says croaking is random.
>>
>>13078392
>I don't see how you've arrived from "frog croak randomly" to "probability is described by the Poisson distribution"
If you don't understand the criteria for a Poisson distribution then you should probably learn it before continuing this discussion.

https://en.wikipedia.org/wiki/Poisson_distribution
>>
>>13078403
>Lmao ok then please give me (1-x)/(2-x) kilograms of cheese
No.
>>
>>13079349
>t
Lmao retard
>Yes, it says croaking is random.
Define "random"

>>13079361
So you're incapable of elaborating your argument.
I see, you're just talking shit.

>>13079369
Retard
>>
>>13078416
>Because you didn't give a real answer, you just parroted all the bullshit you said before without any elaboration.
I answered all your questions. If your questions were not specific enough then that's your problem.

>Mathematics is indisputable when presented as a well-formed logical argument, which you obviously failed since I'm not convinced.
Or maybe you're just retarded.

By the way, I gave an explanation without using the Poisson distribution which you have failed to respond to >>13063281

That proof should be simple enough for your feeble brain.
>>
>>13079375
>Lmao retard
Not an argument.

>Define "random"
Independent from the occurence of other events.

>So you're incapable of elaborating your argument.
I just did elaborate my argument. The criteria for a Poisson distribution are right there. What are you whining about? That I won't type it out for you? LOL.

>Retard
Not an argument.
>>
>>13079386
>Or maybe you're just retarded.
no u
>>
>>13079386
>By the way, I gave an explanation without using the Poisson distribution which you have failed to respond to >>13063281
I already explained three times how a function is not a real number between 0 and 1. I'm giving up this discussion since you don't seem to respect basic mathematical formalism.
>>
>>13062235
Just so you guys know, the answer is dependent on the waiting time so if you try to solve it you need to integrate out the waiting time. (if you assume that the croaking follows a poisson process)
[code]
n <- 10000000
filter_time <- 1
crossing(pair=1:n) %>%
mutate(intensity = 1+rbinom(n,1,2/3),
time_to_croak = rexp(n,intensity)) %>%
filter(time_to_croak < filter_time) %>%
group_by(intensity) %>%
summarise(n=n()) %>%
mutate(frequency = n/sum(n))
[/code]

run with filter_time =3 gives 0.677 that its a male pair and with filter_time = 1 gives 0.733 that its a male pair
>>
>>13079568
ah wait just woke up, it should be rbinom(n,1,1/3)
>>
>>13079595
Really when you think about it, waiting time is irrelevant since we have waited until we heard a croak and since the information we can condition on is that one of them is male then the probability that it is a male-male pair or female-male pair are their relative proportions. So P(MM|one male) = 1/3, P(FM|one male)=2/3
>>
>>13079516
>I already explained three times how a function is not a real number between 0 and 1.
What does this have to do with my post?

Since x is a real number between 0 and 1, (1-x)/(2-x) is a real number between 0 and 1/2. You have no actual argument against my proof, so you whine about nothing.
>>
>>13079604
>Really when you think about it, waiting time is irrelevant since we have waited until we heard a croak
How does that make the waiting time irrelevant?

>and since the information we can condition on is that one of them is male
That's not all the information given by the problem. The fact that you ignored information is not an argument for why that information should be ignored.

Two males are more likely to produce a croak than one male, so the condition that you heard a croak does not increase the probably of two males and one male proportionally. Your argument that only the relative proportions of pairs matters is wrong.
>>
>>13079824
>How does that make the waiting time irrelevant?
Because we are waiting until we hear a croak, not recording the time we waited so it is not information we can condition on. If we recorded the time we waited or we checked the pair of frogs behind our back at time T then it would make a difference
>>
>>13079516
You don't even know what mathematical formalism is. Let x = the chance of a male croaking while you were listening.

Mc M - (1/2)(x)(1/2)(1-x)
M Mc - (1/2)(1-x)(1/2)(x)
Mc F - (1/2)(x)(1/2)
F Mc - (1/2)(1/2)(x)

P(Mc M or M Mc | Mc M or M Mc or Mc F or F Mc = 2(1/4)x(1-x) / ( 2(1/4)x(1-x) + 2(1/4)x ) = (1-x)/(2-x)

You lose.
>>
>>13062235
>you hear a croak behind you
Only one frog croaked
>only male of this species croak
Then only one frog is male
The probability of both being male is null
>inb4 they croaked at the same time !
>>
Let's do this again, with Baye's rule. Suppose X is the event that both frogs are male. Suppose Y is the event that at least one frog is male.
Then:
P(X) = 1/4
P(Y) = 3/4
P(Y|X) = 1
P(X|Y) = P(Y|X)P(X)/P(Y) = 1/3
And again, this can easily be verified by a simulation. I've included a simple Python one that even non-programmers should be able to follow.
tinyurl dot com slash ytm39xjm
>>
>>13079951
>Let x = the chance of a male croaking while you were listening.

>Mc M - (1/2)(x)(1/2)(1-x)

You're trolling right? None of what you've written is mathematically formal. You don't write "let x = text." You don't write "text - expression".
>>
>>13080140
>>only male of this species croak
>Then only one frog is male
Doesn't follow. A male frog doesn't have to croak while you were listening.
>>
What type of call was it?
>>
>>13062235
>ignoring the knightian risk that the croaking frog is neither of the two but a different ( out of sight but clearly audible ) frog
>>
>>13080148
>Suppose Y is the event that at least one frog is male.
This is not the condition. The condition is that a frog croaked. P(C|MM) is not 1, it's 2x(1-x).
>>
>>13080384
>You don't write "let x = text."
You have no clue what you're talking about. And yet again you whine about the form of the argument instead of refuting the argument.
>>
if it was a release call they are both male, if it was an advertisement call you have a 50% chance of having two males. IRL croaking isn't random, males and females generally don't aggregate together like this. In the real world you would almost certainly guess both are male. Math fags trying to compensate for their lack of content knowledge with stats is always fun though.
>>
>>13080501
>fails to answer the question
>but at least I know about frog calls
Yeah you really showed them...
>>
>>13080643
>you have a 50% chance of having two males.
I understand you lack any sort of content knowledge on the topic but that shouldn't impair your reading comprehension. Get back to studying for finals lil guy.
>>
>>13080467
You are overcomplicating the question for the sake of pedantry that isn't even correct. May you never have to take an actuarial exam.
>>
>>13080467
>>13080921
I'll elaborate. I am appealing to authority. This is a common question on early actuarial exams. I'd recommend you look into their arguments, but you won't find anything beyond what's in this thread already. You're either right or wrong about this. However, when you run contrary to experts, you're either a genius or wrong. The latter is most likely.
Regarding your actuall response to me: A croak definitively happened. That is a "condition," as you put it, but it would be more correct to call it an assumption. Consequently, it must be the case that at least one frog is male. Conversely, if at least one frog is male and we heard a croak then... we heard a croak. The probability of us hearing a croak and at least one frog being male is exactly the same. It is 1. It happened, by assumption. Your attempt to adopt my argument to yours with some ill-thought substitution is sophomoric.
Additionally, if you continue to disagree, I'd appreciate an explanation of my simulation. It is simple, like the problem statement, and confirms the separate theory I provided while contradicting yours. Why? If you have programming experience, how might you modify it.
>>
>>13080665
>content knowledge
You mean why frogs croak? It's neither impressive nor relevant to the question. This is a math question, sperg.
>>
>>13080921
>You are overcomplicating the question
You are making incorrect assumptions and then crying "overcomplicated" when they get refuted.

>for the sake of pedantry that isn't even correct.
Please explain what isn't correct. I doubt you will.
>>
>>13081001
Your pointless blather makes you sound mentally ill.

>That is a "condition," as you put it, but it would be more correct to call it an assumption.
This is pretty funny considering you made the incorrect assumption that hearing a croak means solely that at least one frog is male. Are you afraid of calculating the probability the question asks for directly?

>Consequently, it must be the case that at least one frog is male.
I never disagreed that at least one frog is male, I disagreed that that is the extent of the information given by a frog croaking.

>The probability of us hearing a croak and at least one frog being male is exactly the same.
Right, under the condition that we heard a croak. But so what? P(at least one male|c) literally has nothing to do with my objection to your argument. Do you even understand it?

>Additionally, if you continue to disagree, I'd appreciate an explanation of my simulation.
I haven't looked at your simulation, but if you can't even correctly describe what the question is asking, then I doubt you can correctly simulate it. You probably just generated random pairs of frogs without a random croaking mechanism. If you do add a croaking mechanism, you'll see the percentage of pairs with wo males among the pairs that generated a croak is dependent on the probability of croaking.
>>
>>13081447
> If you do add a croaking mechanism, you'll see the percentage of pairs with [two] males among the pairs that generated a croak is dependent on the probability of croaking.
You're not assuming that a given MM, MF, or FM selection will necessarily croak. Alternatively, you're making the assumption that we don't continue to sit there until we hear a croak. This is what I meant by pedantry, but perhaps I was rushed in saying it was incorrect.
So let's reconcile these two. I made the assumption that we wait until we hear a croak. In this case, the answer to the question is 1/3 since all arrangements with a male frog will inevitably produce a croak (assuming the frogs don't die first, but let's please not make this more complicated). This is the simplest interpretation, and what we'd consider with the coin-flipping problem, though it certainly is not a reflection of what might happen in reality.
Now let's not make that assumption. Let's fix some period of time to wait, and say that the probability of a croak over this period of time is constant (again, for the sake of simplicity). In such a situation, what you said is correct: the probability of there being two males changes based on the probability of a croak occurring.
I again have code, for those interested.
tinyurl dot com slash 3xmrynj2
>>
I have a question
Am I correct in thinking that for every instant in that the silent frog does not croak, the probability of it being female increases? Even if the observations were made in a single infinitesimal timeframe, the fact that the silent frog did not also croak in that timeframe means that it is infinitesimally less likely that the frog is male, no?
The way I think about it is that male frogs have two states, either croaking or not croaking and females have only one state which is not croaking. After an infinite amount of time passes, the male frog will croak whereas the female frog will never croak. This means that in the entire timeframe that the frog is being observed, it will croak during at least one instant, therefore the longer that the silent frog remains in the state of not croaking, the less likely it is that the frog will ever croak at all.

Imagine two guys, X and Y, come out to shoot hoops every day. X can only score 2 pointer each round whereas it is possible for Y to score a 3 pointer. You have to determine which guy you are observing. One of them comes out to shoot but you don't know which; they score 2 points. If you observe them every day for an entire year and they never once score a 3 pointer, wouldn't you say it's more likely that they can't score a 3 pointer?
How does this affect the base probability of both frogs being male at 1/3?
>>
>>13062235
50%, either they are or they are not.
>>
>>13081839
>You're not assuming that a given MM, MF, or FM selection will necessarily croak.
Right, because that's an incorrect assumption. If a coin lands on heads, does that mean P(Heads) = 1?

>Alternatively, you're making the assumption that we don't continue to sit there until we hear a croak.
I'm not. It makes no difference to my argument.

>I made the assumption that we wait until we hear a croak. In this case, the answer to the question is 1/3 since all arrangements with a male frog will inevitably produce a croak
You're still incorrectly assuming a pair of males is as likely to produce a croak you waited for as a single male. The croak does not increase MM, MF, and FM proportionally. You're failing to understand the problem with your argument.

>Let's fix some period of time to wait
It makes no difference. There is some probability of croaking over any time frame, including the time frame that occurs if you wait for a croak. Once you include this in your calculations, the answer becomes clear.
>>
>>13082353
At this point, you must be trolling.
>>
>>13081980
>Am I correct in thinking that for every instant in that the silent frog does not croak, the probability of it being female increases?
Correct. The probability a silent frog is female over some time t and average croaking rate r is 1/(1+e^(-tr)). So as t->inf the probability approaches 1.

>How does this affect the base probability of both frogs being male at 1/3?
The base probability is 1/2 since the question is simply the probability that the frog which didn't croak is male. As t increases, the probability goes to 0.
>>
>>13082394
I'm not. What don't you understand?
>>
Obviously its 1/3, but I prefer structuring the problem with coinflips, it's easier for people to understand so you can derail threads on other boards. I've been collecting some data (not enough to really draw any conclusions, and since people shitpost it's not really valid) and /mu/ is one of the dumbest boards, even dumber than /b/
>>
>>13082446
>Obviously its 1/3
Wrong.
>>
>>13082413
>The base probability is 1/2 since the question is simply the probability that the frog which didn't croak is male.
Could you elaborate on how this is different from the "at least one coin is heads" version of this problem where the probability is 1/3?
I understood that the sample set was:
MM
MF
FM
>>
>>13063281
How do you know the 9 cases are equiprobable?
>>
>>13084005
>Could you elaborate on how this is different from the "at least one coin is heads" version of this problem where the probability is 1/3?
It's different because a croak has to come from a specific frog. So the question is simply what sex the other frog is. "At least one is heads" is ambiguous but can be interpreted as information not about any specific coin. For example, if you look at both coins and two are heads, then the information "at least one is heads" is not referring to a specific coin. If you only look at one coin and it's heads, then "at least one is heads" refers to a specific coin.
>>
>>13084938
I didn't say they are. They aren't.
>>
>>13062235
Given that you heard the croak and only males croak there's two possible unique permutations
MF
MM
Given the context of the problem and all things being equal that means there's a 0.5 probability of it being a MM outcome
>>
>>13085336
Incorrect, a frog that didn't croak is more likely to be female.
>>
>>13085336
MF is not equally likely to produce a croak as MM, so all things are not equal.
>>
>>13081839
There is no reason to not include the time spent listening/waiting in the condition. If the problem stated we were doing an experiment by where we repeatedly waited to hear a croak, then your method would make sense.
>>
File: 1614304479133.jpg (327 KB, 943x943)
327 KB
327 KB JPG
You must specify the odds that a male frog croaks during observation. It is incorrect to believe that this value does not matter. This is easily demonstrated by assigning a number to the croaking odds and increasing or decreasing that number to extreme values to observe its impact.

Ex: You see 100 frogs, the males have 99% chance of croaking every second, you observe for 1 second and you hear at least one croak. You know there is at least one male frog but you will naturally consider there might be more than one. These frogs are clearly croaking all the time. It is likely that two frogs croaked simultaneously, it is unlikely that several males fail to croak simultaneously, and more importantly the fact you heard a croak seem to have little value for determining the number of females in the group. All it takes is one male and a group will almost always produce a sound. You could theoretically be looking at a group of 99 female frogs and 1 male frog. Your data is not able to argue that it's unlikely.

Now let's take the exact same scenario but modify the odds of croaking from 99% to 1%. This time, the more there are male frogs in the group, the more you are likely to hear a croak. A group of 99 females and one male will rarely produce a croak. Since you heard one, your data now strongly suggests that a group of 99 females is unlikely.

The only thing that changed between both scenario is the croaking odds. It clearly matters. If you do not specify the croaking odds, there is no math problem. For the sake of simplicity, you can make the croaking independent from the length of observation time. Ex: "this frog species is unusual, the male has 50% chance of croaking exactly at noon, and when frogs are close it is impossible to tell whether more than one has croaked, it is currently noon when you hear a croak and you see two frogs". You can now construct a coherent math problem with this scenario.
>>
>>13062235
It's 1/2. Either one are male or both are mail.
>>
>>
>>13090306
Those are not equally likely.
>>
>>13087755
There's no reason not to include this prob as well then >>13080463
>>
>>13062235
I'm noticing a recurring "observation" theme with the probability problems posted to /sci/ lately. It's kind of fascinating and it's definitely a pattern.
>>
>>13091368
There is no reason to include it either. There is no comparison between ignoring information in the condition vs. some made up scenario not implied by the problem.
>>
>>13091510
The crazy thing is consider the problem is the same except you're looking at both frogs when the one on the left croaks. What would the probability be that both are male then?
(MM) MF (50%)
vs.
(MM) MF FM for the original problem (33%)

the observation makes all the difference
>>
>>13062235
not enough information because you do not know for certain the croak originated from the frogs behind you.
>>
>>13091927
Assume it's negligible
>>
>>13092979
then is there any actual difference between this and the boy/girl paradox that i am missing?
"Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?"
where its either 1/2 or 1/3 depending on if you value MF and FM as the same output.
with the given information we have to assume that every outcome is equiprobable unless we take into consideration normal 'real-world' croaking frequency.
>>
>>13093765
>where its either 1/2 or 1/3 depending on if you value MF and FM as the same output.
That has nothing to do with the boy girl paradox. The boy girl paradox is due to it not being clear whether the information given refers to a specific child. A croak on the other hand has to come from a specific frog. Another difference is that a silent frog is more likely to be female, but in the boy girl paradox you are not given information about the other child if it refers to a specific child.
>>
The OP problem specifies that male and female frogs exist in equal proportions but does not specify that they exist in large numbers. The smaller the population, the more it becomes apparent that if you identify one frog as male then the remaining frog is more likely to be female. In the most extreme case, the total population of frogs could be 1 male and 1 female. The croak would 100% identify the remaining frog.
>>
>>13093934
You can assume the population is arbitrarily large



Delete Post: [File Only] Style:
[Disable Mobile View / Use Desktop Site]

[Enable Mobile View / Use Mobile Site]

All trademarks and copyrights on this page are owned by their respective parties. Images uploaded are the responsibility of the Poster. Comments are owned by the Poster.