Can /sci/ do probability yet?

[Anonymous No.11702258]

You are a biologist traveling through a rainforest when you are bitten by a venomous snake. Luckily the antidote to this venom is secreted on the back of the male of a certain species of frog native to this rainforest. You know females and males exist in equal proportion, socialize with each other randomly, and are indistinguishable aside from the fact that females croak randomly but males never croak. Amazingly, you see a frog of this species sitting silently in front of you. You then hear a croak and turn around to see a pair of these frogs where the croak came from. Knowing that you only have enough time to run to the lone frog to lick its back or to the pair to lick both their backs, which direction do you run and what is your chance of survival?

https://www.strawpoll.me/20138536

[Anonymous No.11702294]

>>11702258
1st vote: it doesn't matter, 1/2

Care to explain your answer?

[Anonymous No.11702301]

3 votes, none correct so far.

[Anonymous No.11702302]

>>11702258
This is just anothet version of the Monty hall problem don't fucking @ me or this anime image board ever again you straight nigger faggot
pair 2/3

[Anonymous No.11702304]

>>11702302
>This is just anothet version of the Monty hall problem
Huh? How?

>pair 2/3
Why?

[Anonymous No.11702305]

>>11702258
>Can /sci/ do probability yet?
Yes. The author of this puzzle cannot, though; for the author of this puzzle intended it to have the 2/3 answer that applies to the two-boys paradox, but messed up the scenario so that it doesn't.

Strictly speaking, the probability of the lone frog being a male, or at least one of the pair of frogs being a male, depends on the frequency at which these female frogs croak, relative to the period you have been observing them. If they croak often enough that you would be very likely to have heard a croak for the period you were observing, the probability of the lone frog being male is close to 1. If croaking is rare enough, or you have been listening shortly enough, that you are very unlikely to hear a croak from a female for the listening duration, the probability of the lone frog being male is close to 1/2 -- but never actually 1/2, but rather slightly above it. As this probability of hearing a croak for the listening duration varies, the probability of the lone frog being male can be anything in the interval (1/2, 1), and the probability of the pair containing a male can vary widely as well.

[continued, 1/3]

[Anonymous No.11702308]

>>11702305
[continued 2/3]
If we assume that croaking is rare, or rather than you have been listening for a short enough amount of time to be unlikely to hear a croak even from a female, as I imagine the puzzle is intended to be interpreted, the probability of the lone frog being male is approximately (but not exactly) 1/2. And in this scenario, the probability of the pair of frogs containing a male, is also approximately 1/2. The reason being that a pair of females has a higher probability of croaking in a given interval than a single female; as the probability of a single female croaking for the period approaches zero, the probability of pair of females croaking for the period approaches twice as high a probability. This means that of the initial odds ratio of (1:2:1) for (double male : male and female : double female) gets multiplied by approximately (0:1:2), resulting in an odds ratio of (0:1:1), or a 50% chance of at least one frog of the pair being male.

[Anonymous No.11702310]

Here's an incredibly obtusely verbose problem that makes you imagine some retarded situation that is not even close to a real situation where you would apply this knowledge that includes several bits of misleading obfuscating information so as to make sure that you don't simply understand the problem because clearly deciphering some poorly written obscurantist block of text is what really matters when it comes to calculating probabilities.

[Anonymous No.11702311]

>>11702305
>>11702308
[continued 3/3]
Now, if instead of hearing that croak behind you, you would have learned that one of the frogs was female, while learning no other information (not even probabilistically), then you would get a probability of 2/3 of one frog of the pair being male. This is the result of the two-boys paradox, and this is presumably the situation the author of this puzzle was after.

This is a situation that is very difficult to accomplish in reality though, even as it is common in math puzzles, which is why this puzzle fails in its goals. In almost any real-world scenario in which you learn that a given possibility is impossible, you also probabilistically learn something about which other possibilities are more or less *probable*. In the frogs puzzle, just like almost all other practical scenarios, the croaking information you receive makes certain things impossible, but it also affects the relative probabilities of the things that are still possible, and so doesn't get the 2/3 chance as its answer.

tl;dr The two-boys paradox has a probability of 2/3 in it, but that is only the case because of you receiving a carefully selected set of information that is very rare in realistic situations. If you try to transport that same paradox to a more realistic scenario, you will collapse the probabilities back to approximately 1/2 again unless you are very careful about it, which the author of this puzzle failed to do.

[Anonymous No.11702318]

>>11702310
Please explain what you think is obtuse or misleading.

[Anonymous No.11702326]

>>11702305
>>11702308
>>11702311
Actually the author of this particular puzzle (me) intended familiarity with the Two Child Problem to be a red herring. This is because a youtube video with a very similar problem made that mistake. Also, the Two Child Problem does not have an answer since it's unknown how the information was obtained. This is why it's considered a "paradox" or an ambiguous problem.

[Anonymous No.11702353]

>>11702326
>Actually the author of this particular puzzle (me)
Oh? Are you Derek Abbott, then? Or did you just happen to come up with exactly the same puzzle by coincidence?

>Also, the Two Child Problem does not have an answer since it's unknown how the information was obtained.
Well, there are a lot of versions of that problem, and there is a version that does indeed unambiguously get the 2/3 answer you would mathematically expect. Certainly there are also ambiguous versions, though.

>This is why it's considered a "paradox" or an ambiguous problem.
A paradox is not an ambiguous problem. Usually, a paradox is a completely unambiguous but very counterintuitive problem, and this is no exception.

[Anonymous No.11702375]

>>11702353
>Oh? Are you Derek Abbott, then?
No, that would be the creator of the youtube video.

>Well, there are a lot of versions of that problem
There is an original published version of the problem, which is ambiguous. Other versions illustrate this ambiguity by showing how different interpretations lead to different answers.

>A paradox is not an ambiguous problem.
Yes, that's why I said "paradox" not paradox. Colloquially, ambiguous or intuitive problems like these are called "paradoxes" even though they aren't actually paradoxical.

[Anonymous No.11702397]

>>11702375
>No, that would be the creator of the youtube video.
Oh, you originally authored the puzzle which Derek Abbott then turned into a popular video? Interesting, I fully expected you were bluffing. Correction, then: I meant the maker of that video cannot do probability (because his answer is wrong), no comment on what the original author had in mind.

>intended familiarity with the Two Child Problem to be a red herring
Is it your position that the Derek Abbott video fucked it up, then? Because it seems to follow the line of reasoning of [the unambiguous 2/3 version of] the two child problem to the letter.

>Yes, that's why I said "paradox" not paradox. Colloquially, ambiguous or intuitive problems like these are called "paradoxes" even though they aren't actually paradoxical.
Okay, that's fair enough.

[Anonymous No.11702419]

>>11702397
>Oh, you originally authored the puzzle which Derek Abbott then turned into a popular video?
>Is it your position that the Derek Abbott video fucked it up, then?
Can you read? >>11702326

[Anonymous No.11702451]

>>11702302
in the monty hall problem you know there is only one right answer. there may be two male frogs in this situation.

[Anonymous No.11702491]

All votes are still wrong, LOL.

[Anonymous No.11702500]

>>11702491
its "To the lone frog, not enough information" isnt it, since theoretically it is ever so slightly more likely that the pair are both female?

[Anonymous No.11703219]

>>11702500
No.

[Anonymous No.11703358]

>>11703219
Tell us what answer you think is correct then, faggot.

[Anonymous No.11703381]

>>11702258
The frog in front has a 1/2 chance of being male or female. The two frogs behind me each have a 1/2 chance of being female. I know one of them is female. The other one has a 1/2 chance of also being female. I run to the other two frogs because it is only a 1/4 chance I will die. Another way of looking at this problem is there are three doors. You have to choose an unlocked door.
All doors have a 50% chance of being unlocked. One of the two leftmost doors is unlocked and shuffled with itself and the other leftmost door. If you pick of those two doors, you have a 50% chance of picking the guaranteed unlocked door and a 50% chance of getting lucky with the other door being unlocked anyways. So, basically, there is a 1/2 chance both frogs are female and a 1/2 chance one is male and one is female. Thusly, half the time you cant choose a male frog, half the time you have a possibility to choose a male frog. Thusly, half of those times you choose male and half you choose female. This gives you an overall 75% chance of choosing a female frog.

[Anonymous No.11703385]

>>11703381
Shit I fucked this up i thought we were searching for females. Stay with your frog if you want a male. 50/50.

[Anonymous No.11703392]

>>11703385
Wait fucck we can lick both frogs backs. Both possibilities are exactly equal it doesnt matter what you do. Only one frog in either situation can be male so you can lick both back frogs, licking a female and possibly a male, or you can lick your front frog, possibly licking a male. I suppose id stay with the front frog for the tiny chance both back frogs croaked at the same time in the same way so that i thought only one frog croaked.

[Anonymous No.11703401]

Run to both frogs, apply both.
Double your chances.

[Anonymous No.11703464]

>>11703358
Figuring it out for yourself is the entire point of this thread, my special friend.

[Anonymous No.11703466]

>>11703381
>The frog in front has a 1/2 chance of being male or female.
So a frog which is silent is equally likely to be male or female, even though males are always silent and females are not?

[Anonymous No.11703566]

Well if the populations are exactly equal and we know that there is at least 1 female frog with us there's a slightly higher chance that there is also a male with us.

I guess it doesn't really matter without knowing how many frogs there are total. You definitely can't have them all be males. All females is a possibility, two female pair and a male and you have male/female with a male.

But also I guess you can assume anywhere from 2 males/2 female population up to a million males/million females. I don't know nothing about the probability though.

[Anonymous No.11703624]

Every vote is wrong

[Anonymous No.11705813]

Bump

[Anonymous No.11706362]

Simple answer: pair 2/3

>>11702305
>>11702308
>>11702311
TL:DR, even if you split it up into cases like you did, it is always better to go with the pair. With an infinite amount of time for observation (unlikely, since you have only a finite amount of time for back-licking), then the chance of survival from the single frog will be at best equal to the chance of survival from the pair.
Assume that croaking is a random event that occurs at a constant rate regardless of how long it has been since the last croak. Also assume that during your observation of the frogs, no event that would have changed the probability of a croak has occurred and that a frog by itself is just as likely to croak as a frog in a group. Then the probability to have not heard a croak within time [math]t[/math] is
[eqn]e^\frac{-t}{m}[/eqn]
Where [math]m[/math] is the expected time until a croak is heard. Because, as the problem states, you have been observing the frog for some constant amount of time and the rate of croaking is also constant, then [math]t[/math] and [math]m[/math] are both constants. Let [math]x=\frac{t}{m}[\math]. Then the probability that the first frog is male is

[eqn]\frac{1}{1+e^{-x}}[/eqn]

Using a simplification of Bayes' rule (there was a common factor of [math]\frac{1}{2}[/math] in all terms)
And the probability of there being at least one male in the pair is

[eqn]\frac{\frac{2}{3}e^{-x}}{\frac{2}{3}e^{-x}+\frac{1}{3}e^{-2x}}=\frac{1}{1+\frac{1}{2}e^{-x}}[/eqn]

Notice both are continuous on [math]x \in [0,\infty)[/math]In the case you assumed in your 2/3 post, then either [math]t[/math] is small or [math]m[/math] is large. Hence [math]x=0[/math] and we end up with 1/2 vs 2/3. If both equations are equal, then [math]e^{-x}=0[math]

[Anonymous No.11706812]

>>11706362
>TL:DR, even if you split it up into cases like you did, it is always better to go with the pair. With an infinite amount of time for observation (unlikely, since you have only a finite amount of time for back-licking), then the chance of survival from the single frog will be at best equal to the chance of survival from the pair.
No, where did you get that from anything he said?

Your math is wrong for both. Show your work.

[Anonymous No.11708460]

bump

[Anonymous No.11708480]

>>11706812
>Your math is wrong for both. Show your work.
Both >>11702305 and >>11706362 showed a lot of work. If you think there is something wrong with their reasoning, I suggest YOU show your work.

[Anonymous No.11708943]

>>11708480
Everything >>11702305 said is correct except for some insignificant details, none of it implies your math is correct.

>I suggest YOU show your work.
Why are you deflecting if you just calculated those probabilities? Fine.

Let x = the probability of a female frog croaking while you were listening

Then the probability of the lone frog being male is

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

Probability of the pair containing a male is

P(~FF|C) = (3/4)((1/3)(0)+(2/3)x)/((1/4)(0)+(1/2)x+(1/4)(2x(1-x)) = 1/(2-x)

[Anonymous No.11710140]

Bump