:fa_bar_chart_o: Probability is confusing.


  • ♿ (Parody)

    @Kian said:

    So to get the right odds, you have to be very clear when asking the question as to whether which frog you saw croak matters or not. Saying you saw the right frog croak is irrelevant if you are not excluding from your chances the people who saw the left frog croak.

    The OP's description of the video (tl;dw;) says that you heard a croak. Not that you saw a frog croak.


  • Banned

    @boomzilla said:

    We don't have any information on the distribution of male and female, so we assume they are equally likely. But you're right, if we had information about the sex ratio the answer would change to reflect that.

    @Kian said:

    The video establishes the distribution for male and female is 50-50. And I'm not assuming they're equally distributed, I'm asserting they are. Yes, if you experimentally take 1000 samples, it's possible you'll run into a weird situation. We could even calculate the odds of getting 960 MM pairs out of 1000 tries. But I'm not doing that.

    You missed the point completely. I wasn't talking about distribution of sexes, I was talking about distribution of alternate universes.


  • I survived the hour long Uno hand

    Which makes it exactly the boy-girl question I linked to, stated in the inverse (what are the odds it's NOT a household with two boys).


  • ♿ (Parody)

    @Gaska said:

    You missed the point completely. I wasn't talking about distribution of sexes, I was talking about distribution of alternate universes.

    Same difference, probability wise.

    @Yamikuronue said:

    Which makes it exactly the boy-girl question I linked to, stated in the inverse (what are the odds it's NOT a household with two boys).

    Yep.


  • Banned

    Let's formulate an equivalent problem. I have two boxes, one has a single black ball, the other has one black ball and one white ball. I pick one ball from each of the boxes at random. If the solution above is correct, then the probability of picking the white ball from the second box depends on whether I know which box is which. It cannot be right.


  • Banned

    @boomzilla said:

    Same difference, probability wise.

    But not in this particular problem. The sexes of frogs are explicitly stated to be distributed equally - and not a word is said about the alternative universe. So, you cannot base your solution on alternative universes, like @Kian did.



  • @Kian said:

    This is what got me mixed up before. Saying "You see one frog croak" is different from saying "You see the left frog croak" or "You see the right frog croak". When you just say "you see one croak", then you are including both scenarios. When you establish "you saw the left frog croak", then you exclude the cases where the right frog croaked and the left one was the female, meaning half of the survival cases, so you're down to 50% odds.

    I think this is the source of all confusion in this discussion.

    @boomzilla said:

    MF and FM being distinct is what "the order matters" means.

    Okay, so I'm still confused then. Probability is hard and I don't have a good intuition for it :p

    My understanding: from a purely mathematical point, the order matters, but in real life, you can just walk around to the other side of the frogs to change the order and make it not matter.


  • kills Dumbledore

    @LB_ said:

    My understanding: from a purely mathematical point, the order matters, but in real life, you can just walk around to the other side of the frogs to change the order and make it not matter

    You arbitrarily label them Frog1 and Frog2. The case where Frog1 = M and Frog2 = F is distinct to Frog1 = F and Frog2 = M. That's what we mean by ordering, and going round to the other side doesn't change Frog1 into Frog2


  • I survived the hour long Uno hand

    @Gaska said:

    the probability of picking the white ball from the second box depends on whether I know which box is which.

    Sure it does. If you want a white ball, you won't select the box that includes only a black ball. So your odds are 50-50 because you're picking from one box with two balls, one of which is white.

    If you don't know which is which, you have to first guess which box contains a white ball at all, and then pick the right ball.


  • I survived the hour long Uno hand

    Or, as I put it in puzzling.se, you label the children Alex and Sam, and no amount of knowing Alex is short for Alexander makes Sam more likely to be short for Samuel, but it does change the odds that both of them are boys because you're already halfway there.

    Knowing Alex is short for Alexandria changes the odds of them both being boys to 0.



  • As a general rule of thumb, if a statistic is intuitive, it's probably wrong.


  • Banned

    @Yamikuronue said:

    If you don't know which is which, you have to first guess which box contains a white ball at all, and then pick the right ball.

    No I don't - because I pick a ball from each box.


  • kills Dumbledore

    @CarrieVS said:

    As a general rule of thumb, if a statistic is intuitive, it's probably wrong.

    74% of all people know that



  • Wait, you hear one frog croak so you know at least one of the two is male, and then the chance of there being two male frogs is bigger if you watched which one croaked and smaller if you closed your eyes? That cannot be right.



  • This post is deleted!

  • ♿ (Parody)

    @Grunnen said:

    Wait, you hear one frog croak so you know at least one of the two is male, and then the chance of there being two male frogs is bigger if you watched which one croaked and smaller if you closed your eyes? That cannot be right.

    Correct: what you said is not right. The frog either is or isn't female. That isn't changing.

    However, increased knowledge regarding one frog improves your estimate about the other frog being female.


  • Winner of the 2016 Presidential Election

    @Jaloopa said:

    83% of all people know that

    FTFY



  • @PJH said:

    @Gurth said:
    There doesn’t seem to be anything in the question that prevents you from just trying them all, and so automatically getting cured.

    p(all three are male) = 1/6.

    ...no.

    p(all three are male) = p(one frog is male)^3 = (1/2)^3 = 1/8



  • @Gaska said:

    The problem with this reasoning is that you assume that the possible universes are distributed equally between the four possible frog sets. What if out of 1000 universes, there were 960 such that both frogs are male?

    You're overcomplicatng this. @Kian was simply using the alternate universes as a way to explain why the probabilities are correct.



  • @LB_ said:

    My understanding: from a purely mathematical point, the order matters, but in real life, you can just walk around to the other side of the frogs to change the order and make it not matter.

    Mathematics often makes its more difficult, because people start talking about distributions and "universes" and other bull, instead of just looking at the possible outcomes. See my example above using coin flips instead of frogs. Compare the differences in possible outcomes based on your knowledge in 3 different states. Probability is just a measurement of knowledge. In addition, not all probabilities will have numbers associated with them.

    When you "walk around" and identify the croaking frog, you are changing your knowledge about the frogs and thus your estimate as to whether you will die or not.



  • @boomzilla said:

    It's not flawed. It's simulating the case where you saw a particular frog croak.

    Yes, but by simulating the situation where you see one frog croak, you're implicitly assuming that there's one other frog in the simulation that can't croak. When in reality, as soon as a frog (either frog) croaks, it becomes "the frog that croaked" and the other frog is "the frog that didn't".


  • Discourse touched me in a no-no place

    @anotherusername said:

    @PJH said:
    @Gurth said:
    There doesn’t seem to be anything in the question that prevents you from just trying them all, and so automatically getting cured.

    p(all three are male) = 1/6.

    ...no.

    p(all three are male) = p(one frog is male)^3 = (1/2)^3 = 1/8

    Wrong. You already know one of the frogs in the clearing is male. It's the other two, the other clearing frog and the one on the stump, that are unknown.

    There are 6 variations of this:

    clearing - stump
    MF - M
    FM - M
    MM - M
    MF - F
    FM - F
    MM - F
    

    The two impossible ones (accounting for your 'out of 8' solution) being:

    FF - M
    FF - F
    

    And only one of the 6 allowable combitions has all three as male.



  • This post is deleted!


  • @Gaska said:

    Let's formulate an equivalent problem. I have two boxes, one has a single black ball, the other has one black ball and one white ball.

    This is not a truly equivalent problem because the distribution of black and white is set and known. In the frog problem, the M/F distribution of the frogs is unknown, other than that at least one in the clearing is male.

    We've already discussed that the probability of a frog in the clearing being F is ~67%. In your ball problem picking a ball from each box, there are two outcomes: BW and BB. There is a 50% chance of getting the white ball.



  • Oh, I thought you were talking about the general case.

    Yeah, in the situation where you heard one of the frogs in the clearing croak (or saw it, I think), then the probability increases to 1/6.



  • @Gaska said:

    Let's formulate an equivalent problem. I have two boxes, one has a single black ball, the other has one black ball and one white ball. I pick one ball from each of the boxes at random. If the solution above is correct, then the probability of picking the white ball from the second box depends on whether I know which box is which. It cannot be right.

    A truly equivalent problem would be:

    I have two boxes with balls that can be white or black. Box 1 has one ball. Box 2 has two balls. I can chose to take the ball from box one, or both from box 2. I win money if I end up with a white ball. Before I choose, someone tells me that one of the balls in Box 2 is black.

    What is the chance i win money if I select Box 1? 1/2. W or B
    What is the chance i win money if I select Box 2? 2/3. BB, WB, BW.

    Adding information to the problem changes my chances.

    I have two boxes with balls that can be white or black. Box 1 has one ball. Box 2 has two balls. I can chose to take the ball from box one, or both from box 2. I win money if I end up with a white ball. Before I choose, someone removes a ball in Box 2 and gives it to me, and it is black.
    What is the chance i win money if I select Box 1? 1/2. W or B What is the chance i win money if I select Box 2? 1/2. BB, BW

  • ♿ (Parody)

    @Gaska said:

    If the solution above is correct, then the probability of picking the white ball from the second box depends on whether I know which box is which. It cannot be right.

    We don't know anything about the stump frog, so this isn't equivalent.

    Additionally, you're confusing the prior probability with different estimates based on different knowledge.


  • Banned

    @abarker said:

    You're overcomplicatng this. @Kian was simply using the alternate universes as a way to explain why the probabilities are correct.

    If his explanation is using wrong assumption, then his explanation is wrong, so it's still unexplained to me.

    @abarker said:

    This is not a truly equivalent problem because the distribution of black and white is set and known. In the frog problem, the M/F distribution of the frogs is unknown, other than that at least one in the clearing is male.

    The video explicitly states that M/F distribution is 50/50. If I get any individual frog, its 50% likely to be male and 50% likely to be female - except for the one I heard croaking, for which it's 100% male and 0% female. Just like the black-or-white box and the only-black box.

    @abarker said:

    In your ball problem picking a ball from each box, there are two outcomes: BW and BB.

    Or WB, if the only-black box is second.


  • ♿ (Parody)

    @anotherusername said:

    Yes, but by simulating the situation where you see one frog croak, you're implicitly assuming that there's one other frog in the simulation that can't croak.

    No I'm not. I'm just assuming that it didn't croak. Otherwise the result of the simulation would be 1 or 0.

    @anotherusername said:

    When in reality, as soon as a frog (either frog) croaks, it becomes "the frog that croaked" and the other frog is "the frog that didn't".

    That's what we're simulating. We don't know anything about the other frog except that it has a 0.5 chance of being male and a 0.5 chance of being female.


  • ♿ (Parody)

    @Gaska said:

    The video explicitly states that M/F distribution is 50/50. If I get any individual frog, its 50% likely to be male and 50% likely to be female - except for the one I heard croaking, for which it's 100% male and 0% female. Just like the black-or-white box and the only-black box.

    But that's assuming you know which one croaked. You don't know that in the video. That was an additional wrinkle introduced in subsequent discussion.



  • Hmm. I think I see the 'problem'. First I thought that if something is true, there should be multiple ways to arrive to the same result.

      P(frog A is female or frog B is female | frog A croaks or frog B croaks)
    = P((frog A is female or frog B is female) and (frog A croaks or frog B croaks)) / P(frog A croaks or frog B croaks)
    = P((frog A is female or frog B is female) and (frog A croaks or frog B croaks)) / 0,75
    
        (frog A is female or frog B is female) (frog A croaks or frog B croaks)
    MM  false                                  true
    MF  true                                   true
    FM  true                                   true
    FF      true                                   false
    
    = 0,5 / 0,75 = 2/3
    

    This assumes two things:

    • If there is a male frog, there is a 100% chance that it will croak
    • If there are two male frogs, only one croaks at the same time and you cannot distinguish their 'voices'.
      Are these assumptions logical?


  • @boomzilla said:

    We don't know anything about the other frog except that it has a 0.5 chance of being male and a 0.5 chance of being female.

    We don't even need to know the "chance" of it being a specific gender - we just need to know that its gender can take 1 of 2 states, "male" or "female". But that truth might just confuse folks more.


  • Banned

    @boomzilla said:

    But that's assuming you know which one croaked.

    No matter if any frog croaked at all, if I go to the two, pick one and examine it individually, there's 50% chance for either gender. Only if I examine the two frogs together things get interesting.


  • ♿ (Parody)

    @NTW said:

    We don't even need to know the "chance" of it being a specific gender - we just need to know that its gender can take 1 of 2 states, "male" or "female". But that truth might just confuse folks more.

    Well, I think it was stated in the video that M/F had equal distribution. But yes, if we didn't know anything, that's what we'd assume.


  • ♿ (Parody)

    @Gaska said:

    @boomzilla said:
    But that's assuming you know which one croaked.

    No matter if any frog croaked at all, if I go to the two, pick one and examine it individually, there's 50% chance for either gender. Only if I examine the two frogs together things get interesting.

    False. Let's look at the possibilities:

    MM
    MF
    FM

    You know that you have one of these three cases. Therefore you have a 1/3 chance of female and 2/3 chance of male if you go and try one of the frogs in the clearing, assuming one croaked but you don't know which one.


  • Discourse touched me in a no-no place

    @Grunnen said:

    If there are two male frogs, only one croaks at the same time and you cannot distinguish their 'voices'.

    Same time as... what?



  • Probability has to align with reality, though. It wouldn't be useful if it didn't align with reality.

    If you repeat an experiment 1000 times with you blindfolded, so that you only hear a frog croak (completely omitting the experimental tries where neither of the frogs croaked), you'll find that one of the two frogs is female around 2/3 of the time. We agree on that point.

    Now, if there was another observer who wasn't blindfolded, so he was able to see which frog croaked, the results would still be that one of the two frogs was female around 2/3 of the time. That can't change. His reality can't be different than yours. So neither can the probability, because it must match what actually happened. If it doesn't, there's no use to calculating it.


  • Banned

    @boomzilla said:

    False. Let's look at the possibilities:

    M
    F


    I did. Still looks like a single frog examined in isolation has 50% chance of being either.


  • Banned

    @PJH said:

    Same time as... what?

    As the time you're observing the croak.


  • ♿ (Parody)

    @anotherusername said:

    Now, if there was another observer who wasn't blindfolded, so he was able to see which frog croaked, the results would still be that one of the two frogs was female around 2/3 of the time. That can't change. His reality can't be different than yours. So neither can the probability, because it must match what actually happened. If it doesn't, there's no use to calculating it.

    Let's all say it again: all probability is conditional.

    So when we're talking about "probability" we're not talking about reality. That's either 0 or 1. We're talking about our estimates of what the reality is based upon what we know. If two people know different things, they'll likely have different estimates.


  • Banned

    @anotherusername said:

    His reality can't be different than yours.

    It can, actually - you just can't talk to each other 🚎


  • ♿ (Parody)

    @Gaska said:

    @boomzilla said:
    False. Let's look at the possibilities:

    M
    F


    I did. Still looks like a single frog examined in isolation has 50% chance of being either.

    That makes sense if you change the situation to having no knowledge about the frogs' sexes. It's just not the situation described in the video.


  • Banned

    @boomzilla said:

    So when we're talking about "probability" we're not talking about reality. That's either 0 or 1.

    Or something in between 🚎


  • ♿ (Parody)

    @Gaska said:

    @boomzilla said:
    So when we're talking about "probability" we're not talking about reality. That's either 0 or 1.

    Or something in between 🚎

    Quantum is too weird to be confused with reality. 🍹



  • To reply to myself:

    Let's change the assumption of the 100% chance that a male frog will croak. Let's change it to just 1%. Then the calculation becomes:

      P(frog A is female or frog B is female | frog A croaks or frog B croaks)
    = P((frog A is female or frog B is female) and (frog A croaks or frog B croaks)) / P(frog A croaks or frog B croaks)
    = P((frog A is female or frog B is female) and (frog A croaks or frog B croaks)) / 0.0199
    
        (frog A is female or frog B is female) (frog A croaks or frog B croaks)
    MM  0                                      0.0199
    MF  1                                      0.01
    FM  1                                      0.01
    FF  1                                      0
    
    = 0.009975 / 0.0199 =~ 1/2
    

    Which confirms the simulation result of someone else earlier in this topic.

    Thus, I would say:

    • If you hear just only one single croak and then nothing, the male frogs apparently don't croak much and the probability of a female is close to 1/2.
    • If the male frogs croak a lot (and hence the probability of a female would be 2/3), you'd probably hear them croak simultaneously if there were two male frogs, so if you don't hear that, the probability of a female rather approaches 1.


  • @boomzilla said:

    So when we're talking about "probability" we're not talking about reality. That's either 0 or 1. We're talking about our estimates of what the reality is based upon what we know.

    Reality is either 1 or 0 in one trial of that experiment. But if you repeat it an infinite number of times, the average of reality must converge to your probability. That's what probability is. If it wasn't, it wouldn't be useful.



  • Yes, it's the 'Law of Large Numbers'.


  • ♿ (Parody)

    @anotherusername said:

    @boomzilla said:
    So when we're talking about "probability" we're not talking about reality. That's either 0 or 1. We're talking about our estimates of what the reality is based upon what we know.

    Reality is either 1 or 0 in one trial of that experiment. But if you repeat the experiment an infinite number of times, the average of reality must converge to your probability. That's what probability is. If it wasn't, it wouldn't be useful.

    That's one ([frequentist][1]) definition.

    Nevertheless, your intuition about the different observers is misleading you.
    [1]: https://en.wikipedia.org/wiki/Frequentist_probability


  • ♿ (Parody)

    @Grunnen said:

    Let's change the assumption of the 100% chance that a male frog will croak.

    There was no assumption about how frequent they croak. It was simply that one did croak. You could alternatively talk about how frequently they sit on stumps vs hang out in clearings. It wouldn't change the probabilities at all, because we came upon a situation where one was on the stump and two were in the clearing.

    You could say something about the a priori chances of that happening, but it wouldn't change the probabilities in the problem as presented.



  • @anotherusername said:

    But if you repeat it an infinite number of times, the average of reality must converge to your probability. That's what probability is. If it wasn't, it wouldn't be useful.

    So what's the probability of Donald Trump being elected? The 2016 election cannot be repeated an infinite number of times.

    Probability is not relative frequency.


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