# Brain-aching list of counterintuitive probability riddles

Originally published at: https://boingboing.net/2018/06/06/sleeping-beauty-epistemology.html

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It’s threads like that that remind me why I am no good at mathematics.

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It is practice more than anything. The more you work at this kind of problem the more you can just go aha bam!

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Yes, I know.

But things like this:

A quick intuitive explanation of the rough order of the answer to the birthday problem is to notice that for birthdays to be all distinct each pair has to be distinct, and there are Θ(n2) pairs for n people. Thus roughly you expect that, if n∈Θ(k−−√) as k→∞, where k is the number of possible birthdays, then there are Θ(k) pairs each of which has probability 1k of being equal, and so the total probability of distinct birthdays is roughly (1−1k)Θ(k) which tends to exp(−c) for some positive constant c. – user21820 Feb 13 '17 at 4:17

make the motivation to practice tend towards zero.

That’s the ‘intuitive’ explanation?

People who are good at mathematics tend to be really crap at explaining it to people who aren’t.

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I’m sure you’ve heard the old joke about the Math professor. During a lecture, about to complete a derivation he concludes, “so it is clearly obvious that…” He stops mid-sentence. He looks at the board for a minute, then retreats off stage and starts scribbling. After fifteen minutes he comes back on stage and continues, “okay, it IS obvious that…”

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Maybe if I could master bayesian thinking, these kinds of problems wouldn’t just piss me off. But it’s easier just to avoid this sort of word problem in the first place.

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That’s my approach and I was a math major in college.

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I think the confusion here is that “intuitive” is used not as “intuitive for people that don’t have a similar education” but instead as “assuming you have a similar level of education in math but have not yet attacked a problem like this…”.

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I haven’t been able to make it intuitive to myself, but I know how to figure out these sorts of questions - count the possibilities.

1. A family has two children, one of which is a girl. The following are the combinations gg gb bg bb
bb is not allowed since both cannot be boys. 1/3 remaining have two girls
2. A family has two children, the elder is a girl. The combinations are gg gb bg bb bg and bb are not allowed since the eldest most be a girl. 1/2 remaining have two girls.

So knowing the elder child is a girl improves the odds that are both girls from 1/3 to 1/2. I cannot make that intuitive to myself.

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See, now, that’s what bugs me about problems like this. The same logic would apply for the girl to be the younger sibling in the first case, with the same factors of chance. I.e. the girl could be either younger or older, but not both at the same time, (even in the case of fraternal twins, they do still come out one at a time!).

So “logically” the odds for the older case should apply to the unknown case!

Obviously (intuitively? lol) they don’t because it is that very unknown factor which expands the range of possibilities and skews the probability. But my poor brain collapses all these subtle nuances into the simplest possible perspective and comes up with 50/50 for both. (And I am fairly good at math, but I have to sit down like you did and walk through it step by step, the logical leaps don’t work for me either.)

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Wait wait wait… The probability of returning to the origin is strictly between 0 and 1? Strictly? Well since the probability of anything is between 0 and 1 (or 100%), yeah, that makes sense, Professor Tautology.

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“Strictly” is the key word.

In this context, “strictly between 0 and 1” means the probability is neither 0 nor 1.

http://mathworld.wolfram.com/StrictlyBetween.html

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AAAAAAAAAAAAAAAAAAAAAAAGH!!

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@roomwithaview that’s why the Monty Hall question causes so much debate! The answer of 1/3 and 1/2 is correct, but counterintuitive.

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So I think the answer is 1/2. Actually the link to the wikipedia page in the quoted comments from the original post explains a different way of looking at it pretty well.

If we start with a universe that has four possibilities: BB (child 1 is boy, child 2 is boy), BG, GB, GG and restrict it to those possibilities where we could correctly say that one child is a boy, then the answer appears to be 1/3. The question you have to ask is, what information would you have if we were in the GG universe?

If the answer is “they would have told me one child is a girl” then they might also have said that in the BG or GB universes. So now we have three (not independent) unknowns to consider. First child’s sex, second child’s sex, and what information we are given about the situation. That gives us six possibilities:

BBB (child 1 is boy, child 2 is boy, we are told there’s a boy), BGB, BGG, GBB, GBG, GGG. But these aren’t equally likely. The tree looks like this:

So there is a .25 chance of getting BBB, but only a .125 chance of BGB and GBB - the other two ways we get told there’s a boy. That’s an equal chance of each.

To get one in three, you’d have to assume the alternative to being told that one child is a boy is being told nothing at all.

So it all comes down to which of these things you think a parent is interested in communicating to you:

• The sex of one of their children (probability is 1/2)
• That they have a boy or not (probability is 1/3)

The probabilities are the same. SMASH THE PATRIARCHY!

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My take:

(1) 0%.
(2) ~1/2

If the family has two children, one of which is a girl, then the probability the other child is a girl is zero, because if the other child was also girl, the original premise (one is a girl) is violated (because two are girls instead.)

If the family has two children, and the older is a girl, then the probability the younger is also a girl is ~1/2. The probability of events are independent, so the question can be condensed to “What is the probability that a child is a girl?”

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Wait, except that the GG case might be two district possibilities.

I guess it depends on what is meant by the family saying one of the children is a girl.

Suppose you know that a child’s name is “Jessie”, but you don’t know whether it’s a boy or a girl, or whether it’s older or younger.

Then they tell you Jessie is a girl. Surely there are now four possibilities:

1. Jessie is the first G in GG (she’s the elder sister to a girl)
2. She’s the second G in GG
3. She’s the G in GB
4. She’s the G in BG

So there would be 50% change the other sibling is a girl.

So if “You learn that they have two children, one of which is a girl” means “You learn that there are two children, and happen to ask about one of the children and are told she is a girl,” then I think the answer is 50%.

If it means “you ask them point-blank: ‘do you have a girl in the family?’, and they answer ‘yes’” then I think the 33% answer is correct.

In real life there if a parent says they have one girl, there is a 90% chance that their other child is a boy (or neither), and there is a 10% chance they are going to get their hand cut off:

If you change the problem so that you are the one who is inquiring (rather than them volunteering the information) then it gets rid of the ambiguity, because you know what is being measured. Otherwise you don’t know whether “we have a girl” is an answer to the question “Do you have a girl in the family?” or an answer to the question “tell me something about the gender of one of your children.”

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I wish that I had Jessie’s Girl! (duh nuh na nuh) / I wish that I had Jessie’s Girl! (duh nuh na nuh) / Why can’t I find a woman like that!

The odds of this are less than 1/3, but are much more intuitive.

I’m going to suggest that people not beat themselves up if they have trouble with the BG puzzle, as that one has some extremely subtle assumption issues that even trip up those of us in the business.

Incidentally, as a pedagogical issue when I teach probability I sometimes wonder how much emphasis I should give in class to unusual examples. I worry that while they’re fun, they tend to distract attention from the core applications of the subject and make it look more ad hoc than it actually is.

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