Hi, can someone help me get a bit of an intuition on conditi

>>7933830
>But, suppose we were in the same situation but this time we see which of the two frogs croaked. In this case we lick one of the frogs knowing the other is male and only have 50% chance of living?
the croaking does not tell you the sex of the second frog.

croacking= male = news that you already know

>>7933852
I just meant, if we saw which of the two frogs croaked, say the second one. We'd then lick the first frog only. The chance of this frog being female is then 50% (It can only be male or female and is independent of the other frog?)

But, why would seeing which of the two frogs croaked lower our survival rate from around 67% to 50%.

>Going around and licking frogs
What the hell is wrong with you?

>>7933863
so to recap


also, since you are meant to lick BOTH frogs, if you choose this situation, it does not matter what frog you lick (M first or last).


=>does the knowledge of who is the the masculine sex in the situation B changes something ?

No.

>>7933877
But if we have two frogs and know say, the first is male we just have two cases:
MM and MF
So 50% of selecting a pair with a female?

But if we know one is male but don't know which we have three cases?
MM, FM, MF
2/3 chance of selecting a pair with a female?

Say we have frogs A and B, we have the following possibilities:

A croaked, B is female, A croaked, B is male, B croaked, A is female, B croaked, A is male

from this I get a 50/50 possibility of one female because we essentially have "double dipped" the outcome with two males. I'm not gonna claim any confidence, but I cant seem to make 2/3rd chance of at least one female to add up.

>>7933898

btw this was not a statement, i'm asking if I went wrong here or not.

>>7933903
the point is that you change the experiment if you can identify the sex of at least one the frog.

>>7933903
>>7933898

here again, I gathered that there is a 3/4 chance of getting at least one female, and that one croak only takes out F/F leaving us with 2/3d but just like OP I cant grok the fact that actually seeing which frog croaked should lower the odds.

>>7933830
I'll give you an example that makes more sense.

Let roll a single die. The result can be any of the numbers 1 though 6. Each outcome has a 1/6th probability.

Now suppose I roll the die and tell you the result was even but don't give you any additional information. Can we agree the possibilities are now 2,4,6 and that there is a 1/3 chance of each outcome? Sounds reasonable so far.

Let's do this using Bayes law for conditional probability:

B = probability of getting a 2 = 1/6
A = probability of it being even = 1/2

P(A and B) = 1/6 (2 is even with 100% chance so no real change from P(B))

So P(B given A) = (1/6)/(1/2) = 2/6 = 1/3

All Bays law does is renormalizes a subset so that the probabilities add up to 1 once a subset is selected (via the condition). You can also solve the problem as follows:

2,4,6 all have probability 1/6 before I knew the result was even.

P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 1/2

These do not add to 1. Now that I know that the result is even I need to normalize the probabilities by dividing by 1/2. The above equation then becomes:

P(2) + P(4) + P(6) = 1/3 + 1/3 + 1/3 = 1

I have simply taken the original equation and algebraically divided both sides by 1/2.

Hope this helps.

just lick every goddamn frog you find

>>7933830
elementary questions are always hardest to answer.
>So in the video, it says in the case of the two frogs you have 2/3 odds of licking a female, there are 3 possibilities, MF, FM or MM and two out of three have a female frog, I'm happy with this. But, suppose we were in the same situation but this time we see which of the two frogs croaked. In this case we lick one of the frogs knowing the other is male and only have 50% chance of living?
keep in mind, statistics deals with unknown. If your glass shows which frog is female and which one is male then you have perfect data to know exactly what will happen.
I think the point you are missing is, we know there are two frogs but we don't know which frog is male(croaked).
to visualize it, our sample space
MM
FM
MF
we know, one of them is male so FF is discarded.

you can see from sample space, FM and MF are outcome of this unkown information. if we know which frog is croaked, using some magical glass then sample space becomes smaller. For instance using glass we learned frog number 1 is croaked then our sample space becomes
MM
MF
this time FF and FM are discarded because we know first frog is male. probability becomes 50%

you can play with data if you want. add one more frog etc.

If there were 3 frogs on the clearing and we heard 2 croaks the sample size becomes 8
MMM
MMF
MFM
MFF
FMM
FMF
FFM
FFF

since we know at least 2 frogs are male we are left with MMM, MMF, MFM and FMM. 3 of those result in survival, so the chances are 3/4.

So, the more frogs you add on the other side, the higher your chance of sucess, even if there is only max one female?

>>7933830
The video is incorrect. What it's missing is the chance of a male frog croaking. If you hear only one croak which can only be produced by one male frog, then according to Bayes' theorem the probability of both frogs being male is (1-p)/(2-p) where p is the probability of a male frog croaking in the time you were listening. So the only way to get a 1/3 chance of both frogs being male is if p=1/2, which is a reasonable assumption to make given no other information. BUT then we must ask, what is the probability that the lone frog in front of us is male given it did not croak? This is actually exactly the same, (1-p)/(2-p). So if the probability of a male frog croaking is 1/2 then the probability this lone frog is male is 1/3, NOT 1/2.

Now we go to your variant. If we see which frog croaked, then the probability the other frog is male is again (1-p)/(2-p). So we see that the information of which frog croaked does not in fact raise the probability of us surviving and the paradox is resolved. Also, it is no better to lick the two frogs then the lone frog.

>>7935035
And here is the way to calculate the probabilities if you don't understand, with probability of male frog croaking = x

The probability of the lone frog in front of you being male (M) given it does not croak (-C) when you are listening:

p(M|-C) = p(M) p(-C|M) / ( p(-C|M)p(M)+p(-C|-M)p(-M) )

(1/2)(1-x)/((1-x)(1/2)+1(1/2))

(1-x)/(2-x)

The probability of both of the frogs behind you being male given that only one frog croaked (which must be true since you only heard one croak):

p(MM|C) = p(MM) p(C|MM) / ( p(C|MM)p(MM)+p(C|MF)p(MF)+p(C|FM)p(FM)+p(C|FF)p(FF) )

(1/4)(2(1-x)x) / ( (2(1-x)x)(1/4)+(x)(1/4)+(x)(1/4)+0(1/4) )

(1-x)/(2-x)

The probability of you surviving (aka, licking a female frog:

1 - (1-x)/(2-x) = 1/(2-x)

Hopefully those male frogs croak alot.

>>7933830
Couldn't tell you how to prove this mathematically.

Intuitively, you know one of the frogs is a male in both cases (knowing which one and not knowing which one). The other frog (doesn't matter which) is either male or female.

Assuming you lick both, that means you've covered the male and the unknown regardless. This means you essentially reduce the problem to the second case. Seeing the male that croaked just means you have to lick one less frog.

I think mathematically the problem comes from the fact that we are using MF and FM at the same time? No idea how to prove that using both isn't valid though.

As per the question, though: should you go with A or B? They have equal odds. 50% either way.

If the options were A or C: two frogs behind you and neither croaked, then you choose C (3/4 chance).

>>7935462
>I think mathematically the problem comes from the fact that we are using MF and FM at the same time?
No, the frogs are distinct entities so MF and FM are separate events.

The video's mistake lies in ignoring that a lack of a croak is also information. If the frog in front of us did not croak then this makes it more likely it's female than male, depending on how likely it is for a male to croak. You are correct though that both options give the same chance of survival. The chance of survival is 1/(2-x) where x is the chance of a male frog croaking.

>If the options were A or C: two frogs behind you and neither croaked, then you choose C (3/4 chance).
If neither croaked then the chance of at least one being a female is (3-2x)/(2-x)^2. If this is 3/4 then the x = 0, but since male frogs croak this is impossible. The chance of survival must be greater than 3/4, because no croaking indicates the frogs are more likely to be female.

>>7935035
>>Now we go to your variant. If we see which frog croaked, then the probability the other frog is male is again (1-p)/(2-p).
but if we see one of the frog croaking, say the one on the left, then the cases are : MF, MM. so why is 1/2 not the probability to get a female ?

>>7935501
Because as I explained already, the probability of a frog which did NOT croak being female is greater than 1/2. It's more likely that this was a female that didn't croak than a male that didn't croak, because females never croak.

>>7935505
but what is the point of using priors, here probability of male frog croaking = x, if you have seen the frog croaking ?

>>7935517
...because it affects the problem. Think about it like this: if males ALWAYS croak, then the frog in front of you which did not croak MUST be female, because it didn't croak. And the two frogs behind you MUST contain a female, because one of the frogs didn't croak.

Here, let me try explaining it using sample spaces. Let 1 indicate a croak and 0 indicate no croak. The sample space for the original problem is:

M1 M0
M0 M1
M1 F0
F0 M1

And the variant's sample space is

M1 M0
M1 F0

>>7935523
>And the variant's sample space is
>
>M1 M0
>M1 F0
so you do agree that, in the variant, we have MM or MF, but you think that 1/2 is not the probability of getting a female ?


also, as I understood the OP, only the male croaks, the female never croaks.

>>7935533
>so you do agree that, in the variant, we have MM or MF, but you think that 1/2 is not the probability of getting a female ?
Exactly. Except in the case where the probability of male frogs croaking is 1. Then a male frog not croaking is impossible so M1 M0 cannot occur and M1 F0 is the only option.

>also, as I understood the OP, only the male croaks, the female never croaks.
Yes... I understand. That's why female in the sample space always has a 0 next to it. I am talking about the probability of the *male* croaking over the period of time in which you listened.

>>7933830
Well if you consider the pair of frogs to be either FM, MF or MM, then if you know for a fact that frog A is male, then you only have MF and MM left.
The logical reason for the reduction in survival is that you're licking less frogs.

>>7935669
There is no reduction in survival chance because the video did not calculate probabilities correctly. It attempted to copy the Boy-Girl Problem but by introducing the croaking male it changed the situation. Think about it like this: if the lone frog in front of you didn't croak, then it is more likely to be female than male, because males sometimes croak and females don't. The only way there could be an even chance of it being male or female is if males never croak too. So already we can see the video is wrong.

The chance the lone frog is female turns out to be 1/(2-x) where x is the probability of a male frog croaking once over the time period you were listening. And the chance of the two frogs behind you containing a female also turns out to be 1/(2-x). And this doesn't change when you know which frog croaked. So in this case OP's intuition was correct and.the video is wrong. There is no reason for there to be a reduction in survival chance just because you know which frog is male.

>>7935539
>Exactly
then if the sample space is MM and MF, you do agree that you have 1/2 change to get a female ?

>>7936295
No. I don't think you understand how sample spaces work. The events in a sample space don't necessarily have the same chance of occurring. As I've already explained, the only way the frog which didn't croak can have an equal chance of being male or female is if males never croak. The lack of a croak indicates a higher chance of being female. Get it?

Do you understand what Bayes' theorem is? If not I suggest you learn it and then you can use it to calculate these probabilities given you hear a croak or given you don't hear a croak. See >>7935292

>>7936308
>The events in a sample space don't necessarily have the same chance of occurring
ok I get this.

you discriminate between MM and MF via the probability of croaking (for males, since only the male croak by hypothesis).

so, in the variant,
-if you forget about the general story about croaking (especially for the frog for which you do not know the sex), you would have the story of, say, two usual dices, since, in the general population (in the forest), you have the same chance to find one M as the chance of finding one F (typically because they make the same number of babies of the same sex), you can say that you have equal chance, in the variant experiment, of having MM or MF

-if you introduce the croaking, you modulate the probability for MM and MF, but you still have P[MM or MF] = 1 and the probability to croak, x, modulates as you explained.


can you say that the story with the croaking is like a skewed dice, where one of the face is more likely to appear ?

>>7936344
>-if you forget about the general story about croaking (especially for the frog for which you do not know the sex), you would have the story of, say, two usual dices, since, in the general population (in the forest), you have the same chance to find one M as the chance of finding one F (typically because they make the same number of babies of the same sex), you can say that you have equal chance, in the variant experiment, of having MM or MF
Yes, without conditions, p(MM) = p(MF)

>-if you introduce the croaking, you modulate the probability for MM and MF, but you still have P[MM or MF] = 1 and the probability to croak, x, modulates as you explained.
Yes.

>can you say that the story with the croaking is like a skewed dice, where one of the face is more likely to appear ?
I guess? It's more like the lack croaking gives additional information about the outcome of the dice that favors one outcome over another. The dice itself is not skewed, it's the information you get about the dice that skews the results.

>>7935292
>p(C|MM)
you claim that

p(C|MM) = (2(1-x)x)(1/4)

can you detail your reasoning for this ?

why do you have 2[1-x] before x[1/4] ?

>>7936363
to me x[1/4] is: probability to hear one croak, from a pair of males [which means MM]

which is what we seek...

>>7936363
Actually I claimed that p(C|MM) p(MM) = (2(1-x)x)(1/4)

p(C|MM) = 2(1-x)x because this is asking for the probability that one male croaked (which is x) and the other did not (which is 1-x). There are two ways this can occur :

M1 M0 -> x(1-x)
M0 M1 -> (1-x)x

So the probability of this occuring is x(1-x)+(1-x)x = 2(1-x)x

p(MM) = 1/4 because well you should already understand this. There are four equally likely states without conditions:

MM
MF
FM
FF

>>7936374
>>So the probability of this occuring is x(1-x)+(1-x)x = 2(1-x)x
but then, in the variant, if we see which male croackes, we no longer need to symmetrize, like we do when we *only hear* the croaking without seeing which frog croaked.

>>7936383
Yes, but you still get the same result. In the variant it's exactly the same as analyzing the chance of a lone frog being female if it didn't croak, which I already showed in that post.

>>7936388

if you no longer symmetrize, your line
>p(MM|C) = p(MM) p(C|MM) / ( p(C|MM)p(MM)+p(C|MF)p(MF)+p(C|FM)p(FM)+p(C|FF)p(FF) )

>=(1/4)(2(1-x)x) / ( (2(1-x)x)(1/4)+(x)(1/4)+(x)(1/4)+0(1/4) )

becomes
=[1/4]x (1-x) /[ [1/4]x (1-x +1 +1) ]

so it changes

>>7936398
You have to take into account that we now know the croak came from the first frog and the second frog didn't croak. So p(MM|C) no longer makes sense because it does not describe the condition. It becomes the probability that the second frog is male given it didn't croak:

p(M|-C) = p(M) p(-C|M) / ( p(-C|M)p(M)+p(-C|-M)p(-M) )

(1/2)(1-x)/((1-x)(1/2)+1(1/2))

(1-x)/(2-x)

>>7936398
If you want to change C to mean "the first frog croaked and the second frog didn't" then we get

p(MM|C) = p(MM) p(C|MM) / ( p(C|MM)p(MM)+p(C|MF)p(MF)+p(C|FM)p(FM)+p(C|FF)p(FF) )

(1/4)(x(1-x)) / ( (x)(1-x)(1/4)+(x)(1)(1/4)+(0)(1-x)(1/4)+(0)(1)(1/4) )

x(1-x) / (x(1-x)+x)

(1-x)/(2-x)

So it's the same result, which works intuitively because knowing which frog croaked should not affect your chances of survival.

>>7933830
Here's a more mind-warpy one:

https://en.wikipedia.org/wiki/Monty_Hall_problem

File: Big meme.png (27KB, 699x981px) Image search: [Google]

27KB, 699x981px

>>7933830
Venn Diagrams are always the easiest way to get intuition desu senpai.

>>7936447
yes indeed

>>7934190
>All Bays law does is renormalizes a subset so that the probabilities add up to 1 once a subset is selected (via the condition). You can also solve the problem as follows:
thank you for this. I've read about it in a couple of books, had a course teach it to me and I still couldn't get it. This single post did more than all the resources I mentioned.

>>7938659
It's called the Principle of Proportionality. It's not really the reason why Bayes' Theorem works, it's just a consequence of it.

I emailed the guy who made the video about the mistakes (he's a physicist, cool guy actually). This is his reply:

"Yes, that is a little touch the animators added in there as they wanted
to communicate to the audience how the hero of the story knows
one is male. They didn't like the original version which said that he
is told one is male.

To fix the video, do you have any thoughts on a scenario might
work mathematically but also work visually in a cartoon?

I'm thinking that a possible story "plot" that might work is to
say that in this species of mythical frog, the females are competitive
and fight to the death .......so that would give the clue that the two frogs
sitting next to each other cannot be FF because they are looking
rather placid.

Would you buy that?

Do let me know if you have a solution or a better idea than the one
above!"

Stand back everyone i will explain this. hope OP is still here.

the clue here is that you have 2 frogs on one side and 1 frog on the other side. at the first you get 75% chance in licking a female. if you go for the 2 frogs. but then you know one of them is male. this reduced the probability to 0.67%( think of it as updating your belief state of the world) if you know would see which the frog croaked it would further reduce the belief to 0.5 which is the real probability. you gain information and your probability converges to the real probability which is the distributions of male and female frogs

>>7939087
Except that that is wrong... In this case knowing which frog is male changes nothing.

See >>7935035 and >>7935292

And here is the creator of the video saying that the video's analysis is wrong >>7939039

>>7939108

no that's just about the croaking. its still valid if you know that one of them is male with 100% certainty. no extra P(croaking) involved, that's also what he says in the email . he does't say the whole video is debunked,

>>7939118
No, if you know which frog is croaking you get the exact same result. See >>7936447

>>7933830
OP EVERY SINGLE ANALYSIS PRESENTED SO FAR HAS BEEN WRONG.

This is a variant of a well known problem. Here is the standard solution, assuming the population contains 50% female frogs and making no assumptions about croaking and being near other frogs, except that croaking identifies a frog as male.


The frog you see is female with 50%.
For the two frogs behind you, at most 1 is female. That is, either one of them is female, or none of them are. Both options are equally probable.

You turn around and pick a frog at random. With 50%, it was the one that croaked, and is male. You lose. With 50%, it was the one who didn't croak, so might be female. With 50%, it is indeed female.

Thus, randomly picking among the two gives you the cure 25% of the time.

Taking your chance with the one in front, 50%

No 2/3s here, idiots. We also don't need to slavishly apply Bayes rule and suspend thinking.

>>7939283
>The frog you see is female with 50%.
Nope, wrong. The frog didn't croak so the chance it is female is 1/(2-x) where x is the chance of male frog croaking, so it's always greater than 50%. Think about it for a second and you will understand. How can the frog which didn't croak have an even chance of being male and female when females never croak and males sometimes croak? A female is always going to be more likely than a non-croaking male if males sometimes croak.

>>7939283
>For the two frogs behind you, at most 1 is female. That is, either one of them is female, or none of them are. Both options are equally probable.
Nope, wrong. The probability one of them is female is 1/(2-x). The probability both are male is (1-x)/(2-x). These are only equally probable when male frogs have a 1/2 chance of croaking. The reason is essentially the same as the lone frog. Hearing only one croak indicates there is a male in the group. It also indicates there is a frog which didn't croak in the group, which has a higher chance of being female as I've already shown.

>>7939321

Again. We start not knowing anything about frogs except their population relative frequency.

Then we learn that at least 1 of the 2 frogs behind is male. Nothing else.

If you want to make additional assumptions about croaking or whatever, sure. So, if we assume in some interval of time, a male croaks with probability p and females never do, the chance the one above is female is 1/(2-p).

>>7939376
>Again. We start not knowing anything about frogs except their population relative frequency.
>Then we learn that at least 1 of the 2 frogs behind is male. Nothing else.
That's not what the video presents. It tells us that only males croak. This means that 1 of the 2 frogs behind us is male. It ALSO means that frogs which do not croak are more likely to be female. That's no more "an assumption" then the information that 1 of the 2 frogs behind us is male. They are both directly deduced from the fact that only males croak.