When intuition and math probably look wrong
sciencenews.orgI know Gary, the person who presented this, and I was there when it happened. He fully intended this to ignite the argument it has.
Firstly, as presented it is clearly ambiguous. It is intended to be ambiguous, but in such a way that people who are familiar with the original version will get suckered into believing that it's well formed.
Secondly, if presented precisely, the answer usually given is either 13/27 or 1/2, depending on which version.
Finally, this is like the Monty Hall problem all over again. There are people arguing vehemently and without listening at all, demonstrating clearly that they are excellent at missing the point.
In case you're wondering, here's one statement and answer.
Suppose on knock on people's doors and ask - Do you have exactly two children? If they answer no, I move on. If they answer yes I then ask - Is at least one of them a boy born on a Tuesday? If they say no, I move on.
If they look surprised and say "Yes," what is the probability that they have two boys?
Answer: 13/27.
Yes, it really is.
If you replace the second question with "Is at least one of them a boy with red hair, left-handed, plays piano, was born on Tuesday, and has a cracked left upper incisor" then if the answer is "Yes" then the probability of both children being boys is almost exactly 50%.
If, instead, you replace the second question with "Is at least one a boy" then the probability of two boys is 1/3.
Finally, suppose you see a parent that you know has two children in the park with a boy. Now the probability of two boys is 50%, because, assuming uniform probabilities, having two boys makes it more likely you see them with a boy.
tl;dr: It's hard, and depends precisely on the assumptions you make.
Another riddle where the math is easy, but translating the riddle into math is hard. Which is to say, it's not really a math riddle at all. In fact, the riddle is complicated by the fact that it's posed as a riddle -- if you had to answer the question "in real life," you would probably know how you came about the information, so the problem would be straightforward.
I had a math teacher that got mad when I used the word "intuitively". Probably because most people don't understand that when your intuition doesn't match the mathematical result you have to hit your intuition with a hammer until it does.
Ok, I was about to rage about yet another article going on about the Two Children problem and getting it wrong by leaving out a whole host of children (i.e. children in families of more or less than two children). Then it surprised me by not only acknowledging it, but acknowledging that the the 50% answer is correct when we are selecting from an arbitrary family (as the original problem is usually presented).
It then goes on to acknowledge that the 33% answer would be correct if we specifically choose a two child family that fits the parameters of the problem beforehand.
It's all about why the information was selected.
> But since the boy could be either the younger or the older child, the analysis is more subtle. Devlin started by listing the children’s sexes in the order of their birth
Personally, when I read that I could spot the error. For lazy people (tl;dr types) ... order doesn't matter as you're not given any info about that order.
So given 2 children, there are only 3 possibilities ...
And that's it ... a 50% probability that both children are boys.boy, boy girl, girl boy, girl So if you're told one of them is a boy ... boy, girl boy, boyActually, I'm going to stop now: http://xkcd.com/386/
It's pretty clear I won't convince you.
Let me leave you with these questions:
If I toss two coins until at least one shows a Head, what's the probability that both are Heads?
If I roll two dice until at least one shows a 6, what's the probability that both are 6's?
If I spin two roulette wheels until at least one shows a Red-23, what's the probability that both are red?
Are you sure?
It's been fun :-)
> If I toss two coins until at least one shows a Head, what's the probability that both are Heads?
I answered in my last reply ...
The problem is that head_a and head_b are independent events, and that P(head_a) = 1 (you made sure of that).P(head_a AND head_b | head_b) = 1/3
QED :)P(head_a AND head_b | head_a) = P(head_a AND head_b) / P(head_a) = P(head_a AND head_b) = P(head_a) = 1/2I have no idea what problem you think you're solving, but it's not the one I posed. I never said a specific coin was heads, I just said at least one of them was heads.
Given that I make money on these sorts of questions (doing it for real in a semi-military context where it's important to be right, and you get tested against reality) I feel pretty confident that I know what I'm talking about. You're clearly talking about something completely different, and I really don't understand what you're saying.
And that's the last I'll say.
Look, don't take this the wrong way ... probabilities are easy to get wrong ... I'm only having this conversation with you because my math skills are rusty.
> I never said a specific coin was heads
OK, so let's make it mathematically correct (let's say we're painting them) ... making an effort here :)
Making the problem ...P( blue_head AND red_head | blue_head OR red_head) P(blue_head OR red_head) = 1 you said that you're retrying until this happens normally this would be 3/4
But ...P( blue AND red | blue OR red) = P( blue OR red | blue AND red ) * P( blue AND red ) / P(blue OR red) (applied bayes) = 1 * P(blue) * P(red) / 1
Show me the error.P(blue OR red) = P(blue) + P(red) - P(blue) * P(red) = 1 => P(blue) * P(red) = 1 - P(blue) - P(red) = 1 - 1/4 - 1/4 = 1/2[EDITED] ... modified the stuff as I've totally fucked up the previous version :) ... as I said, I'm rusty
Your first error is here:
This is supposed to be the probability without conditionals, in which case P(BH or RH) = 3/4. Using the Bayesian formula is what is taking into account the fact that at least one head has shown.P(BH or RH) = 1Your second error is here:
That's probably a typo, because you probably know that P(BH)=1/2.P( blue_head ) * P( red_head ) (independent events) = 1/4 * 1/4Your third error is here:
1/4 * 1/4 = 1/161/4 * 1/4 = 1/2Even if you correct this to the 1/2 * 1/2 that you intended, you now get that P(BH and RH)=1/4. Surely you realise that this must be wrong. I've already told you that there's at least one head, so you can't get the same answer as when I've given you no information.
Here's the correct Bayesian sum:
The fact that I'm using the conditional probability is what uses the information I've given you, so you can't use that again and claim P(RH or BH)=1.P(RH and BH | RH or BH) = P(RH and BH and (RH or BH)) / P(RH or BH) = P(RH and BH) / (3/4) = P(RH) * P(BH) / (3/4) = 1/2 * 1/2 / (3/4) = 1/4 / (3/4) = 1/3Show me the error.
Here, do an experiment. Toss two coins of different denominations. Consider those cases where there's at least one head. How often are they both heads? Are you a programmer? Run the simulation - show me the code.
Here, let me help:
#!/usr/bin/python import random def toss(): if random.random() <= 0.5: return 'Head' return 'tail' trials = 0 two_heads = 0 for trial in xrange(10000): coin0 = toss() coin1 = toss() if coin0 != 'Head' and coin1 != 'Head': continue trials += 1 if coin0 == 'Head' and coin1 == 'Head': two_heads += 1 print trials,float(two_heads)/trialsNow replying to the modified version. My response to your first version is here: http://news.ycombinator.com/item?id=1473088
Anyway ...
Don't take this the wrong way ... probabilities are easy to get wrong ... I'm only having the conversation with you because you're getting it completely wrong, and yet you appear to want to learn.
You say:
This is an incorrect application of Bayes' Theorem. The whole point of taking P(X|Y) isn't that P(Y) can be then taken as being 1. P(Y) remains the probability of Y. The point is that when we then only consider those events where Y occurs, then we have conditioned our probabilities, and the formula gives us what we want. Specifically, relative to Y, the probability of X changes.P(blue_head OR red_head) = 1 you said that you're retrying until this happens normally this would be 3/4Using "BH" for "Blue coin shows Head" and "RH" for "Red coin shows Head" we have:
You then go on to say:P(BH and RH | BH or RH) = P(BH and RH and (BH or RH)) / P(BH or RH) = P(BH and RH) / P(BH or RH) = 0.25 / 0.75 = 1/3
Taking just the second part of this:P(BH OR RH) = P(BH) + P(RH) - P(BH) * P(RH) = 1
and adding P(BH) * P(RH) to both sides we get:P(BH) + P(RH) - P(BH) * P(RH) = 1
Since by symmetry P(RH)=P(BH), and letting x=P(BH)=P(RH) this simplifies toP(BH) + P(RH) = 1 + P(BH) * P(RH)
The only solution to that is x=1, so P(BH)=P(RH)=1x + x = 1 + x^2Your math is clearly screwed at this point.
So I've given you the correct interpretation, I've shown you where your calculations are wrong, and I've explained your incorrect use of Bayes' Formula.
Twice.
multiplication and addition are not the same thing 1/4 * 1/4 = 1/16
Oh shit ... I'm tired :)
did you just prove that 1 == 1/2?
You're making the mistake he mentions in the article.
You're only selecting two child families.
Flip one million coins. Select two at random. If one of them is heads, what is the probability that the second coin is heads?
You can't mean that. There are four possibilities:
Of those only one is precluded by saying (at least) one child is a boy.Older child boy, younger child boy Older child boy, younger child girl Older child girl, younger child boy Older child girl, younger child girlNo there aren't ... notions of order like older/younger don't enter the equation (as the problem was stated).
Oh well, I guess this is what the article is talking about :)
Older/younger doesn't exactly enter into it, but some ordering is necessary to properly describe the set of probabilities. This can be arbitrary, but birth order makes the most intuitive sense. The set of probabilities of sex distribution in two-child families is [(B, B), (B, G), (G, B), (G, G)], for any arbitrary ordering, this cannot be reduced to [(B,B), (B,G), (G,G)] as you have done without adding a variable to double the probability of (B,G).
Assuming a coin-flip probability for boy/girl distribution, you get the 1/3 answer if we select for two-child families where at least one child is a boy: [(B,B), (B,G), (G,B)]. If we don't pre-select for having at least one boy, (i.e. if we select the family because we just met the father socially), the probability rises to 1/2, because we have two cases to consider, each with a 1/2 probability: [(B,B), (B,G)], and [(B,B), (G,B)].
Yes, but this is wrong ... the sex of the second child is in no way connected to the sex of the first child.
He didn't ask ... what's the sex of the second child? No ... he asked ...
Choosing a distribution that involves both children in this case is wrong, hence my answer that ordering doesn't matter. 33% is just wrong.p(boys = 2 | boy >= 1) == p(child = boy)What I wrote is not wrong unless, when you wrote: "So given 2 children, there are only 3 possibilities ... boy, boy girl, girl boy, girl" you are asserting that in the absence of any information about the sex of either child, 2 boys is a 1/3 probability, instead of (1/2)*(1/2)= 1/4.
Let's toss two coins until at least one shows a head. By your reasoning the odds of them both being heads is 1/2. It's not. Try it.
Suppose I roll two dice until at least one of them shows a 6. What's the odds of both being 6's? I've said nothing about the red die versus the blue die, but the underlying truth requires that the situations are kept separate. It's only - as far as we know - in quantum mechanics where you deliberately lose the distinction.
I've done these as real world experiments as I explore them with kids, and I have a lot of direct experience. If you disagree then I'd be delighted to gamble with you.
> Let's toss two coins until at least one shows a head. By your reasoning the odds of them both being heads is 1/2. It's not. Try it.
You haven't read the article then ... the problem as stated in the article is that you know one coin is going to be a head, so what's the probability of the other one also being a head?
Of course ... the events aren't connected ... the second coin toss doesn't depend in any way on the first coin.
That's why I think there's something wrong about the article ... saying that the probability is 33% fails both intuition and elementary probabilistic.
Here's the problem as Gary originally stated it, and as the article quotes it:
You say:> I have two children, one of whom is a son born > on a Tuesday. What is the probability that I > have two boys?the problem as stated in the article is that you know one coin is going to be a head,
No. The point of the article is that you don't know how or why you are given this information.
Suppose I toss two coins until I get one that's a head, then I tell you that I have two coins, and one is a head. I've complied with the problem as described. The probability that they are both heads is 1/3.
I was there when this problem was posed. I was in the room when the questions were asked, and Gary clarified. I had lunch with Gary afterwards, and he said it was deliberate that it was ambiguous.
It seems to me that you're missing the point. Perhaps you should explain clearly exactly how you think the situation arises where the information given is as described.
> The probability that they are both heads is 1/3.
The probability they are both heads is 1/2.
You're trying to formulate this as ....
But this is wrong ... you know that you have a head ... which makes ...P(head AND head | one is a head) = P(head AND head) / P(one is a head) = 1/4 * 4/3 = 4/12 = 1/3
1/3 would be the probability only on the first try (instead of stopping when you've got a head).P(one is a head) = 1 && P(head_a AND head_b) = P(head_a) = P(head_b)
isn't this the same kind of "intuitive" reasoning that fails at simpson's paradox? http://en.wikipedia.org/wiki/Simpsons_paradox
conditioning upon more events can lead to a higher probability. p(boy = 2 | boy >= 1) < p(boy = 2 | boy >= 1, tuesday) (or more precisely, conditioning upon more events can yield a distribution with less entropy)
You know, the problem with that conditional probability is that the sex of the second child is in no way conditioned by the sex of the first child, so ...
And this was the original problem that led them to the 33% probability.p(boy = 2 | boy >= 1) = p(any child = boy)in the sample space i was intending, that would not be the case. i was imagining boy as a random variable that counts instances in an order tuple of genders (the underlying sample space).
you're right, without this explicit construction, it's problematic.
This is why I'm glad I was a mathematics major :)
The answer given is not even wrong.
The statement "I have two children, one of whom is a son born on a Tuesday" is semantically ambiguous. It can mean (1) "I have two children, and the quantity of them who are males born on a Tuesday is exactly one", (2) "I have two children, at least one of whom is a male born on a Tuesday", or even (3) "I have two children, and the maximum quantity of males born on the same Tuesday is one".
I agree with you that it's not even wrong, but I disagree with your reasoning. I think (2) is the natural intended meaning of the question.
The problem I have is attempting to assign a probability to something that isn't reasonably known to be based on randomness. Did the questioner arrive at this statement by picking an arbitrary child and then declaring his gender and day of birth? Or did he pick his favorite child and declare his gender and day of birth? Would he have used this same question if they had both been born on Tuesday, or would have have picked a different distinguishing feature?