:fa_bar_chart_o: Probability is confusing.



  • Can you solve the frog riddle? - Derek Abbott – 04:31
    — TED-Ed

    It goes like this: you are poisoned and will die, but female frogs of a certain species can cure you. The male and female frogs are identical as far as you can tell except that only the male frogs have a distinctive croak.

    You are between a stump and a clearing. On the stump is a frog of this species. In the clearing are two frogs of this species. You hear the distinctive croak from the clearing. Now is when you have to decide: which way do you go for the best chances of survival?

    The video addresses two wrong ways to solve the problem, and then it says the right way is that the frogs in the clearing can have either MF, FM, or MM, thus giving you 2/3rds chance of survival vs the stump's 1/2 chance. But wait... MF and FM are the same! The order doesn't matter, I just need any female frog! So shouldn't the clearing also be 1/2 chance?

    I don't understand why the order matters. Several other people in the comments have the same concern. The video doesn't address it and nobody in the comments has yet to explain why the order matters. Did they really screw up their logic in the video? TED-ED!? :wtf:

    EDIT: You can lick both frogs in the clearing, you don't have to choose between them.



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  • monty hall



  • In the Monty Hall problem, you know which door is which. In this case you only hear a croak but don't know which frog croaked.


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  • @LB_ said:

    then it says the right way is that the frogs in the clearing can have either MF, FM, or MM, thus giving you 2/3rds chance of survival vs the stump's 1/2 chance.

    It's correct. It's exactly the same as the Monty Hall problem, except that instead of an initial 1/3 chance, you have an initial 1/2 chance: the frog on the stump has a 50% chance of being female. That's the unknown that you initially pick, before anything is known about the other two.

    Then, between the other two frogs, you are told that one of them is male. That's the door opening to reveal a goat. So is the other door more likely to contain the prize? Yes.

    @LB_ said:

    But wait... MF and FM are the same! The order doesn't matter, I just need any female frog! So shouldn't the clearing also be 1/2 chance?

    No, because the scenario where there's one male and one female is twice as likely as the scenario where they're both male.



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  • @Kuro said:

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    @blakeyrat said:

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  • @LB_ said:

    In the Monty Hall problem, you know which door is which. In this case you only hear a croak but don't know which frog croaked.

    Yes, but in the Monty Hall problem you're also only allowed to pick one door.

    In this, one of the doors is opened and you're not allowed to see which, but you get to pick both of the doors. So you don't need to know which of the doors was opened.


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    Now somebody else needs to delete his/her post so we can have three deleted posts to choose from!

    Filed Under: Also: reading is hard, okay?



  • @anotherusername said:

    No, because the scenario where there's one male and one female is twice as likely as the scenario where they're both male.

    Yep. Now, if you were looking at the clearing and saw one frog croaking and could therefore rule it out, then you're back to the 1/2 chance for the remaining frog.



  • @boomzilla said:

    Yep. Now, if you were looking at the clearing and saw one frog croaking and could therefore rule it out, then you're back to the 1/2 chance for the remaining frog.

    It doesn't matter if you see it croaking or not, you can lick both frogs in the clearing.

    @anotherusername I don't see how this is a monty hall problem because we cannot choose to 'switch' to the other frog in the clearing.



  • You still have to pick one of the doors: the one that's completely unknown, or one of the two that you already know something about.

    My gut really wants to say that there's a 50% chance either way, though. So ... I dunno.

    They're independent probabilities; knowing something about one frog shouldn't tell you anything about the other frog...



  • Dude. Monty Hall is a trick question. Trick questions are bullshit. This question is bullshit. Fuck it.

    Make fun of me if you want.



  • @LB_ said:

    It doesn't matter if you see it croaking or not, you can lick both frogs in the clearing.

    It does matter. You have more information in that case than if you simply heard it. In that case, you only have MF / MM (counting the known-male frog first). Knowing which frog is male is additional information that affects the probability calculation.



  • What? No. I was talking about Monty Python in a hallway.


    Filed under: also starting a flame war because it's boring waiting for NodeBB to count the likes on every post ever



  • @blakeyrat said:

    Dude. Monty Hall is a trick question. Trick questions are bullshit. This question is bullshit. Fuck it.

    Make fun of me if you want.

    It's not a trick question. Your gut is just bad at determining probability.



  • @boomzilla said:

    It does matter. You have more information in that case than if you simply heard it. In that case, you only have MF / MM (counting the known-male frog first). Knowing which frog is male is additional information that affects the probability calculation.

    The female frog is the cure to the poison. The male frog makes the croak. Since you can lick both frogs, you can pretend the clearing only has one frog because the male frog eliminates itself. If you could only lick one frog, things would be different.


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    @LB_ said:

    If you could only lick one frog, things would be different.

    I believe that's the point of the question...



  • Yeah whatever. Give me a question with some relevance in a real life situation instead of frog-licking or some game show that hasn't been on the air since like 1960, then maybe I'll pay attention.



  • The video specifically says you can lick both frogs in the clearing. You have to chose between going to the stump and licking one frog with a 50% chance of it being female, or going to the clearing and licking both frogs with one of them being male and other other being 50% chance female. Right?



  • @LB_ said:

    @boomzilla said:
    It does matter. You have more information in that case than if you simply heard it. In that case, you only have MF / MM (counting the known-male frog first). Knowing which frog is male is additional information that affects the probability calculation.

    The female frog is the cure to the poison. The male frog makes the croak. Since you can lick both frogs, you can pretend the clearing only has one frog because the male frog eliminates itself. If you could only lick one frog, things would be different.

    Yes. What you said here is correct and doesn't contradict my also-correct point. One key to understanding probability is that it is always conditional based on what we know. If you saw which frog croaked, the conditioning of the probability is different than if you simply know that one of two frogs croaked.


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    Oh, does it?

    Well, can't watch the video right now, so I missed that little detail :)



  • @boomzilla said:

    If you saw which frog croaked, the conditioning of the probability is different than if you simply know that one of two frogs croaked.

    Why? Is this even a conditional probability problem at all?

    @sloosecannon said:

    Oh, does it?

    I edited the OP to add that detail, sorry.



  • @LB_ said:

    Why? Is this even a conditional probability problem at all?

    #EVERY PROBABILITY PROBLEM IS CONDITIONAL

    Ahem. Sorry for going all fox there.



  • @boomzilla said:

    EVERY PROBABILITY PROBLEM IS CONDITIONAL

    I have made a button that will either light up red or blue when you press it, with equal proportions. What is the chance it will light up red the fifth time I press it?


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  • @LB_ said:

    @boomzilla said:
    EVERY PROBABILITY PROBLEM IS CONDITIONAL

    I have made a button that will either light up red or blue when you press it, with equal proportions. What is the chance it will light up red the fifth time I press it?

    Based on what you said (i.e., the conditions I know about the situation) the probability of red is 50%.

    P( Red | Red or Blue equal proportions) = 0.5



  • Okay, by "conditional" I meant "depends on the previous outcome". Is there a different word for that? Either way, I don't think the chances of the second frog being female depend on whether the first frog is...



  • I would use "dependent." You may have heard the term IID: Independent and Identically Distributed, like your light up button.

    That's not really relevant to the frogs unless you wanted to ask a question like, "You licked the first frog in the clearing, you're still dying, what is the probability that the other frog is female." That's basically equivalent to knowing ahead of time that it was male because you saw it croak. Again: we have additional information than we had originally, so our estimates change.



  • @LB_ said:

    The video specifically says you can lick both frogs in the clearing. You have to chose between going to the stump and licking one frog with a 50% chance of it being female, or going to the clearing and licking both frogs with one of them being male and other other being 50% chance female. Right?

    Think of it this way, maybe it helps you:
    Two frogs at random find themselves in a clearing. We know the probability space is MF,MM,FM,FF. So without hearing anything, you know there's a 75% chance that you'll find at least one female.

    When you hear one croak, without seeing which, all you've eliminated is the chance that they're both female. Or put another way:
    MF -> You hear a croak
    MM -> You hear a croak
    FM -> You hear a croak
    FF -> You DON'T hear a croak

    So, the possibility that there is at least one female and you hear a croak is 2 in 3, or 67%.

    @LB_ said:

    I don't think the chances of the second frog being female depend on whether the first frog is...
    The thing is, you don't know WHICH is the first frog, so you can't discard it and say the second has 50% chance.



  • @blakeyrat said:

    Give me a question with some relevance in a real life situation instead of frog-licking or some game show that hasn't been on the air since like 1960, then maybe I'll pay attention.

    These sorts of calculations definitely come up in gambling situations. Or I guess other games, too where you have incomplete knowledge.



  • @Kian said:

    The thing is, you don't know WHICH is the first frog, so you can't discard it and say the second has 50% chance.

    We are talking about chance of surviving. Since you lick both frogs in the clearing, it doesn't matter if you know which one was male. Licking a male frog won't do anything.



  • Look. When in doubt, simulate.

    There are 3 random values per trial, M or F with equal probability.

    The first random value is the frog on the stump. You know nothing about this frog except that it has a 50% chance of being M or F.

    The second and third random values are the two frogs in the clearing. Beyond that, you know that one of them croaks, so eliminate all trials where both of the frogs are female (approximately 25% of them). Of the remaining trials, in what percent is one of the frogs female?



  • How is Male+Female different from Female+Male? You lick both frogs, as long as one is female you will survive.



  • It's not different. Yes, you lick both frogs. The formula in the "Cure" column is TRUE if either frog is female.



  • @LB_ said:

    How is Male+Female different from Female+Male? You lick both frogs, as long as one is female you will survive.

    Do you agree that without hearing the croak, the probability of one being female is 75% and not 67%? Because that difference also relies on female+male being distinct from male+female.

    The point is, the question is not, "what is the chance that one is female, given that the other is male?" The question is, "what is the chance that having heard a croak, at least one is female?" Hearing a croak only removes the possibility that they were both female. One of them being female is still twice as likely as both being male.



  • Why don't you just lick all three frogs.



  • The stump and clearing are in opposite directions, and presumably you could die before you could reach both of them. In any case, you want the cure as fast as possible; if you're not dead you could always go back and lick the other frog(s).



  • They're both within ear-shot of a frog croak, man-up.



  • @blakeyrat said:

    Why don't you just lick all three frogs.

    Beat me to it. There doesn’t seem to be anything in the question that prevents you from just trying them all, and so automatically getting cured. What it needs is a line that says that licking a male frog will kill you instantly, then the riddle makes sense (in as far as it can, anyway).



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    @Gurth said:

    and so automatically getting cured

    Not exactly. They could all be male...


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  • Okay, let's take a step back. @anotherusername - thank you for suggesting that I simulate it.

    Let's ignore the stump, the controversy is around the field. We know one of the frogs croaked, there can't be two females. We will survive as long as at least one of the frogs there is female.

    #include <iostream>
    #include <random>
     
    int main()
    {
    	std::random_device rd;
    	std::mt19937 gen(rd());
    	std::uniform_int_distribution<> dis(0, 1);
     
    	unsigned const trials = 100000;
    	unsigned survive = 0;
    	for(unsigned i = 0; i < trials; ++i)
    	{
    		int clear1 = dis(gen);
    		int clear2 = dis(gen);
    		if(clear1 && clear2) //both female, one didn't croak
    		{
    			continue;
    		}
    		if(clear1 || clear2) //at least one was female
    		{
    			++survive;
    		}
    	}
    	std::cout << survive << std::endl;
    	std::cout << (survive/1.0/trials)*100.0 << "%" << std::endl;
    }
    

    Do you agree that this program simulates the problem correctly? It seems to be that way to me, but correct me if I am wrong.

    The output I get is this:

    50021 50.021%
    See for yourself: https://ideone.com/46G9yx


  • @LB_ said:

    Do you agree that this program simulates the problem correctly? It seems to be that way to me, but correct me if I am wrong.

    No, because you're counting the scenarios that we know can't be happening, where both frogs in the clearing are female. You have to eliminate those scenarios, and then see how many of the remaining scenarios include 1 female frog.



  • This question actually ignores a very significant detail. What is the probability that a male frog croaks in the time you spent listening? Assuming that they croak randomly, there's a significantly higher chance that you'll hear a croak from the clearing if there are two male frogs in the clearing.



  • @anotherusername said:

    No, because you're counting the scenarios that we know can't be happening, where both frogs in the clearing are female.

    if(clear1 && clear2) //both female, one didn't croak

    Did I miss something else?

    EDIT: Oh, you mean the number of trials...I think I understand now...



  • You're counting it in trials.



  • Fixed for you:
    https://ideone.com/dfi8Cc


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