March 21, 2006 9:52 PM Subscribe

Besides the Monty Hall paradox, what are some other counterintuitive math problems that require a bit of a mind warp?

My partner is doing a stats class in which the Monty Hall paradox was brought up. After a bit of time educating ourselves on the internet as to why you should always switch doors (http://en.wikipedia.org/wiki/Monty_Hall_problem) we need more! What are some other strange math/probability questions that go against all gut instinct?
posted by arcticwoman to Education (89 answers total) 12 users marked this as a favorite

My partner is doing a stats class in which the Monty Hall paradox was brought up. After a bit of time educating ourselves on the internet as to why you should always switch doors (http://en.wikipedia.org/wiki/Monty_Hall_problem) we need more! What are some other strange math/probability questions that go against all gut instinct?

e^(pi * i) + 1 = 0

The most amazingly improbably nonsensical thing I have ever come across that is absolutely true. Makes the Monty Hall paradox look obvious. Not only is it true, but the proof is almost as beautiful as the equation itself.

Why should the ratio of the circumpherence of a circle to its diameter have*anything* to do with the constant for continuously calculated rate of return have *anything* to do with the square root of -1, a number that, contrary to the other two, *doesn't even exist*, have *anything* to do with 1 or 0, two otherwise *normal* numbers? The really odd thing? Once you have a good solid year or so of calculus under your belt, this equation is almost so *obvious* as to be self-evident, which for me was final and conclusive proof that math makes you insane.

The best thing about this is that equation is how it neatly ties up all of mathematics from kindergarten up through about freshman or sophomore year of college. You have the most important numbers in: statistics, geometry and trigonometry, calculus, and arithmatic (the last two, repressenting the multiplicative and additive identities.)

posted by ChasFile at 10:08 PM on March 21, 2006

The most amazingly improbably nonsensical thing I have ever come across that is absolutely true. Makes the Monty Hall paradox look obvious. Not only is it true, but the proof is almost as beautiful as the equation itself.

Why should the ratio of the circumpherence of a circle to its diameter have

The best thing about this is that equation is how it neatly ties up all of mathematics from kindergarten up through about freshman or sophomore year of college. You have the most important numbers in: statistics, geometry and trigonometry, calculus, and arithmatic (the last two, repressenting the multiplicative and additive identities.)

posted by ChasFile at 10:08 PM on March 21, 2006

Oh, man. Statistics and probability can be, and often are, totally counterintuitive.

Bayesian statistics can be especially rich in this kind of thing. Consider, for instance, the rate of false positives in what seems, intuitively, like an accurate test.

posted by mr_roboto at 10:11 PM on March 21, 2006

Bayesian statistics can be especially rich in this kind of thing. Consider, for instance, the rate of false positives in what seems, intuitively, like an accurate test.

posted by mr_roboto at 10:11 PM on March 21, 2006

Me: I hear you have two children.

Him: Yes, the oldest is named Tim.

(the chances are 50-50 that his other child is a girl)

Me: I hear you have two children.

Him: Yes. i think you've met Tim before.

(The chances that his other child is a girl is 2/3)

posted by vacapinta at 10:14 PM on March 21, 2006

Him: Yes, the oldest is named Tim.

(the chances are 50-50 that his other child is a girl)

Me: I hear you have two children.

Him: Yes. i think you've met Tim before.

(The chances that his other child is a girl is 2/3)

posted by vacapinta at 10:14 PM on March 21, 2006

The obvious one is the number of people who must be collected together in a room before it is more likely than not that two of them share the same birthday.

The number is much smaller than many people think.

The solution is arrived at as follows. If there are N people, the probability that *none* of them have the same birthday is

1 if N=1

1 x 364/365 for N=2 (since the second person has 364 days to choose from)

1 x 364/365 x 363x365 for N=3

and so on.

So the critical number is N where

365/365 x (365-1)/365 x (365-2)/365... (365-(N-1))/365 < 0.5br>

You can do the math but it turns out to be about 30. A favorite trick of stats teachers is to ask people in a classroom to guess whether it's more likely or not that two of them share the same birthday, and then go through the class and find two people who share one.

posted by unSane at 10:17 PM on March 21, 2006

The number is much smaller than many people think.

The solution is arrived at as follows. If there are N people, the probability that *none* of them have the same birthday is

1 if N=1

1 x 364/365 for N=2 (since the second person has 364 days to choose from)

1 x 364/365 x 363x365 for N=3

and so on.

So the critical number is N where

365/365 x (365-1)/365 x (365-2)/365... (365-(N-1))/365 < 0.5br>

You can do the math but it turns out to be about 30. A favorite trick of stats teachers is to ask people in a classroom to guess whether it's more likely or not that two of them share the same birthday, and then go through the class and find two people who share one.

posted by unSane at 10:17 PM on March 21, 2006

Then there's this (not really a paradox, but thinking about it still makes my brain hurt):

When mathematicians refer to the square root of negative 1, they speak of**i**; engineers generally use **j**.

It's my considered opinion that**i** is not the same as **j**; actually **i** = -**j**.

I have yet to figure out which of them is actually negative :)

posted by flabdablet at 10:22 PM on March 21, 2006

When mathematicians refer to the square root of negative 1, they speak of

It's my considered opinion that

I have yet to figure out which of them is actually negative :)

posted by flabdablet at 10:22 PM on March 21, 2006

Oh, and of course there's the most famous logical intuition-breaker of them all:

Godel's incompleteness theorum.

Perhaps my greatest all-time holy-wow-it-makes-ALOT-of-sense-now-that-I-get-it-fuck-thats-beautiful moment, similar but greater in magnitude to the Monty Hall problem, was when I was shown Cantor's diagonal proof. So amazingly stupidly brilliant that it makes you feel amazingly stupid for taking so long to understand how stupidly effortless yet effortlessly brilliant it is.

posted by ChasFile at 10:25 PM on March 21, 2006

Godel's incompleteness theorum.

Perhaps my greatest all-time holy-wow-it-makes-ALOT-of-sense-now-that-I-get-it-fuck-thats-beautiful moment, similar but greater in magnitude to the Monty Hall problem, was when I was shown Cantor's diagonal proof. So amazingly stupidly brilliant that it makes you feel amazingly stupid for taking so long to understand how stupidly effortless yet effortlessly brilliant it is.

posted by ChasFile at 10:25 PM on March 21, 2006

vacapinta: huh? Is it because Tim can be considered a gender-neutral name?

posted by divabat at 10:27 PM on March 21, 2006

posted by divabat at 10:27 PM on March 21, 2006

diva; no, its for the same reasons as the Monty Hall problem; hence no further explanation is really nescessary in the context of this question.

Remember: probability is the mathematics of knowledge.

posted by ChasFile at 10:38 PM on March 21, 2006

Remember: probability is the mathematics of knowledge.

posted by ChasFile at 10:38 PM on March 21, 2006

Banach-Tarski paradox.

You can take a solid ball, cut it into a finite number of pieces (five, to be precise) and reassemble those pieces into two identical (in every respect) copies of the original ball.

posted by vernondalhart at 11:09 PM on March 21, 2006

You can take a solid ball, cut it into a finite number of pieces (five, to be precise) and reassemble those pieces into two identical (in every respect) copies of the original ball.

posted by vernondalhart at 11:09 PM on March 21, 2006

Related: What's a catchy anagram of Banach-Tarski?

Banach-Tarski Banach-Tarski

posted by vernondalhart at 11:09 PM on March 21, 2006

Banach-Tarski Banach-Tarski

posted by vernondalhart at 11:09 PM on March 21, 2006

This whole system of declaring probability by defining equally possible scenarios, eliminating knowns at a partial point of completion of the overall game, and then recalculating the remaining scenarios as having a total probability of 1 seems faulty to me.

posted by scarabic at 11:15 PM on March 21, 2006

posted by scarabic at 11:15 PM on March 21, 2006

A wonderfully misleading stats question to throw in on a test or just mix in with a lesson:

You are getting dressed in a dark room, and you know that your drawer has M black socks and N brown socks in it. How many socks do you need to pull out to make sure you have a matching pair?

posted by tkolar at 11:24 PM on March 21, 2006

You are getting dressed in a dark room, and you know that your drawer has M black socks and N brown socks in it. How many socks do you need to pull out to make sure you have a matching pair?

posted by tkolar at 11:24 PM on March 21, 2006

tkolar: three, but there's no guarantee you'll actually end up wearing the matching pair unless you put them on where you can see what you're doing.

Is this another birthday parasox?

posted by flabdablet at 11:31 PM on March 21, 2006

Is this another birthday parasox?

posted by flabdablet at 11:31 PM on March 21, 2006

divabat, there are four possibilities for two children:

Option 1: older brother, younger brother

Option 2: older brother, younger sister

Option 3: older sister, younger brother

Option 4: older sister, younger sister

if the older child is tim, then options 3 and 4 can't be true. and the probability that the other child is a girl is one half.

if ONE child is tim, then only option 4 can't be true, and in two of the the three remaining options, the other child is a girl.

posted by clarahamster at 11:31 PM on March 21, 2006

Option 1: older brother, younger brother

Option 2: older brother, younger sister

Option 3: older sister, younger brother

Option 4: older sister, younger sister

if the older child is tim, then options 3 and 4 can't be true. and the probability that the other child is a girl is one half.

if ONE child is tim, then only option 4 can't be true, and in two of the the three remaining options, the other child is a girl.

posted by clarahamster at 11:31 PM on March 21, 2006

There is Gabriel's horn. The area under the curve y=1/x for x>1 is infinite. Rotate the curve around the x axis, and you create something of a cone shape that only needs pi units of paint to fill.

There is the Brouwer fixed point theorem. "take two equal size sheets of graph paper with coordinate systems on them, lay one flat on the table and crumple up (but don't rip) the other one and place it any way you like on top of the first. Then there will be at least one point of the crumpled sheet that lies exactly on top of the corresponding point (i.e. the point with the same coordinates) of the flat sheet."

Also, it is not a math problem, but I like:

Why do mirrors only reverse left and right while leaving top and bottom alone?

Even if you lie on your side, the mirror still does its job and switches your left and right hands.

posted by gearspring at 11:41 PM on March 21, 2006

There is the Brouwer fixed point theorem. "take two equal size sheets of graph paper with coordinate systems on them, lay one flat on the table and crumple up (but don't rip) the other one and place it any way you like on top of the first. Then there will be at least one point of the crumpled sheet that lies exactly on top of the corresponding point (i.e. the point with the same coordinates) of the flat sheet."

Also, it is not a math problem, but I like:

Why do mirrors only reverse left and right while leaving top and bottom alone?

Even if you lie on your side, the mirror still does its job and switches your left and right hands.

posted by gearspring at 11:41 PM on March 21, 2006

Next time you're staying at Hilbert's Hotel, check out the library.

posted by flabdablet at 11:56 PM on March 21, 2006

posted by flabdablet at 11:56 PM on March 21, 2006

I can't resist taking the bait here on divabat's problem. Clarabat gives the following rationale:

Option 1: older brother, younger brother

Option 2: older brother, younger sister

Option 3: older sister, younger brother

Option 4: older sister, younger sister

If one child is male then we rule out option 4 which apparently leaves us with three equally likely possibilities, in two of which the other child is a girl.*However*! We need to count the first case twice, because Tim can take either place in it. The chance is still actually 50%. To put it in a clearer way: we have a child called Tim, who has another sibling, possibly older, possibly younger. Also the other sibling is possibly male, possibly female. So we have four possibilities:

Tim has an older brother

Tim has an older sister

Tim has a younger brother

Tim has a younger sister

These are obviously all equally likely.

posted by teleskiving at 1:48 AM on March 22, 2006

Option 1: older brother, younger brother

Option 2: older brother, younger sister

Option 3: older sister, younger brother

Option 4: older sister, younger sister

If one child is male then we rule out option 4 which apparently leaves us with three equally likely possibilities, in two of which the other child is a girl.

Tim has an older brother

Tim has an older sister

Tim has a younger brother

Tim has a younger sister

These are obviously all equally likely.

posted by teleskiving at 1:48 AM on March 22, 2006

Reminiscent of (but completely different from) the Monty Hall problem:

There is a paradox (I thought it was called "Newman's Paradox" but googling that term doesn't turn it up) which goes something like this:

There are two boxes. You may choose to take one or both. One of the boxes contains $5,000 in cash. The other contains either nothing or $1,000,000 in cash. The boxes are unmarked - if you choose one, you don't know which one you'll get.

Here's the catch: the "variable" box isn't random. It's loaded by someone who will try to guess whether you'll choose one box or two. If he thinks you'll take one box, he'll load the variable box with $1,000,000. If he thinks you'll take both, he'll leave the variable box empty.

So ... you're faced with two boxes. The decision is made, the boxes are loaded. Nothing you can do now to change that. Obviously you should take both boxes, right? Except that if it's "obvious" then the box-loader would make sure the variable box is empty and all you'll get is $5,000. Better to take just one and have a 50/50 shot at $1,000,000. Right? Except ... if that $1,000,000 is there, better to take both boxes and have a 100% shot at it ... except if you take both boxes, one'll be empty ...

If anyone else can remember the actual name of this paradox and point to a better explanation, that'd be great.

posted by zanni at 2:07 AM on March 22, 2006

There is a paradox (I thought it was called "Newman's Paradox" but googling that term doesn't turn it up) which goes something like this:

There are two boxes. You may choose to take one or both. One of the boxes contains $5,000 in cash. The other contains either nothing or $1,000,000 in cash. The boxes are unmarked - if you choose one, you don't know which one you'll get.

Here's the catch: the "variable" box isn't random. It's loaded by someone who will try to guess whether you'll choose one box or two. If he thinks you'll take one box, he'll load the variable box with $1,000,000. If he thinks you'll take both, he'll leave the variable box empty.

So ... you're faced with two boxes. The decision is made, the boxes are loaded. Nothing you can do now to change that. Obviously you should take both boxes, right? Except that if it's "obvious" then the box-loader would make sure the variable box is empty and all you'll get is $5,000. Better to take just one and have a 50/50 shot at $1,000,000. Right? Except ... if that $1,000,000 is there, better to take both boxes and have a 100% shot at it ... except if you take both boxes, one'll be empty ...

If anyone else can remember the actual name of this paradox and point to a better explanation, that'd be great.

posted by zanni at 2:07 AM on March 22, 2006

In a similar vein to the birthday paradox, and showing that humans really have a poor handle on probability:

The government decrees that every family can continue to have children until they have a boy. At that point, they must stop having children.

What does this do to the male/female ratio of the population?

posted by Leon at 2:12 AM on March 22, 2006

The government decrees that every family can continue to have children until they have a boy. At that point, they must stop having children.

What does this do to the male/female ratio of the population?

posted by Leon at 2:12 AM on March 22, 2006

Both times you meet Tim's father, he mentions Tim only. Clearly, he's a misogynist and the other child is a girl.

(that's what i thought the answer was.)

posted by b33j at 2:26 AM on March 22, 2006

(that's what i thought the answer was.)

posted by b33j at 2:26 AM on March 22, 2006

You ought to go look at the archive of puzzlers on the Car Talk site. They are very often statistical in nature and sometimes quite intricate. I seem to recall a "prisoner lightswitch" one that was diabolical.

posted by Rhomboid at 2:40 AM on March 22, 2006

posted by Rhomboid at 2:40 AM on March 22, 2006

But teleskiving your logic does not hold.

If you knew absolutely nothing of the situation except that there were two siblings, then there can only be four possible configurations, as outlined. If you then learn that one of the two is male then that doesn't change the fact that there can still only be four possible combinations of children, one of which is now impossible. Learning that one is male does not suddenly cause another outcome to suddenly become possible -- case 1 is case 1 regardless of whether Tim is the older or younger brother, it's still the same thing. So counting it twice is nonsense.

Put another way, your two outcomes "Tim has an older brother" and "Tim has a younger brother" are both describing the same outcome as Case 1: there are two boys. It doesn't matter which is which, because that is not part of the question. The question is only "what are the chances that the other sibling is female", and therefore the "two brothers" case is really the same outcome regardless of who is who.

posted by Rhomboid at 2:52 AM on March 22, 2006

If you knew absolutely nothing of the situation except that there were two siblings, then there can only be four possible configurations, as outlined. If you then learn that one of the two is male then that doesn't change the fact that there can still only be four possible combinations of children, one of which is now impossible. Learning that one is male does not suddenly cause another outcome to suddenly become possible -- case 1 is case 1 regardless of whether Tim is the older or younger brother, it's still the same thing. So counting it twice is nonsense.

Put another way, your two outcomes "Tim has an older brother" and "Tim has a younger brother" are both describing the same outcome as Case 1: there are two boys. It doesn't matter which is which, because that is not part of the question. The question is only "what are the chances that the other sibling is female", and therefore the "two brothers" case is really the same outcome regardless of who is who.

posted by Rhomboid at 2:52 AM on March 22, 2006

Since I need to apologise to clara*hamster* for getting her name wrong (sorry!), I can't resist another post. Essentially the argument I made can be boiled down to: the chance that the one child named by the father will be male is higher in the case of two males than in the case of one male and two females, and this compensates for the fact that there are two ways of having a girl and a boy and only one way of having two boys.

This is a very Monty Hall like problem in that it is really dependent on the rules behind the information given to you - in order for MH to be really fair it should be stated that*Monty always opens a door*. Likewise, vacapinta's problem needs to be stated as "A man has two children. You ask if he has at least one male child, and he answers 'Yes' ". At that point it is safe to assume that he is more likely than not that he also has a female child.

posted by teleskiving at 3:01 AM on March 22, 2006

This is a very Monty Hall like problem in that it is really dependent on the rules behind the information given to you - in order for MH to be really fair it should be stated that

posted by teleskiving at 3:01 AM on March 22, 2006

Rhomboid, let me put it another way. A man has two children. We start by observing one of the children at random, and note that it is male. This is obviously more likely in the case where both children are male. I've stated above what I think is a more reasonable condition to get vacapinta's answer, what do you think?

posted by teleskiving at 3:05 AM on March 22, 2006

posted by teleskiving at 3:05 AM on March 22, 2006

Just thought of another statistical one that's ridiculous at first glance - 30% of numbers begin with 1.

posted by Leon at 3:24 AM on March 22, 2006

posted by Leon at 3:24 AM on March 22, 2006

Leon: just as a guess, very little?

My assumptions: assume that every family has children, and continues to have children until they have a boy. Birth probabilities are exactly 50/50.

In the first batch of kids, half the parents have boys and stop breeding; the remainder have girls. Gender balance: matched.

In the next batch of kids, the half still in the gene pool have 50/50 male and female, so half stop. Sex ratio: still balanced.

This continues ad-infinitum, until the random pool gets small enough that all boys are born at once. The boys of that final generation will be the only 'unmatched' ones, and will represent the net imbalance from the 'breed until you have a boy' program.

(I'm discounting really freakish events, like everyone in the country simultaneously having a boy.)

It would appear that, given a large starting population, it wouldn't matter that much. if the population were small, random fluctutations would probably result in some extra males. If it were very small, it could wipe out the population.

Unless, of course, I missed something, which is entirely possible, what with these being trick questions and all. :)

posted by Malor at 3:54 AM on March 22, 2006

My assumptions: assume that every family has children, and continues to have children until they have a boy. Birth probabilities are exactly 50/50.

In the first batch of kids, half the parents have boys and stop breeding; the remainder have girls. Gender balance: matched.

In the next batch of kids, the half still in the gene pool have 50/50 male and female, so half stop. Sex ratio: still balanced.

This continues ad-infinitum, until the random pool gets small enough that all boys are born at once. The boys of that final generation will be the only 'unmatched' ones, and will represent the net imbalance from the 'breed until you have a boy' program.

(I'm discounting really freakish events, like everyone in the country simultaneously having a boy.)

It would appear that, given a large starting population, it wouldn't matter that much. if the population were small, random fluctutations would probably result in some extra males. If it were very small, it could wipe out the population.

Unless, of course, I missed something, which is entirely possible, what with these being trick questions and all. :)

posted by Malor at 3:54 AM on March 22, 2006

Okay, two children. Each has a 50/50 chance of being male or female. This makes four possible combinations, each equally likely to happen - so each has an equal 25% chance of happening.Rhomboid, let me put it another way. A man has two children.

Okay, since we know that one of the children is male, this eliminates one of the four possible outcomes. But this is all the the knowledge tells us. It does not change anything else, there are still three possible outcomes each with a 1 in 3 chance of happening.We start by observing one of the children at random, and note that it is male.

And this is where you stop making logical sense. The "male/male" state is still just one of the three remaining possibilities. It does not become any more likely to happen than it was before, so the chances of the other child being female are still 2 out of 3, corresponding to the 3 equally possible outcomes.This is obviously more likely in the case where both children are male.

posted by Rhomboid at 3:57 AM on March 22, 2006

Leon: I had to do a similar question for my Probability course recently, find the expected number of boys in a family if you stop after the first girl if the first child is a boy, or the second girl if the first is a girl. Answer is 5/4, and the probability of getting more than one boy is only 1/4.

Anyway, my contribution: There are more real numbers between 0 and 1 than there are integers (...-2, -1, 0, 1, 2...). Or between 0 and 0.1, or any interval you care to choose. Also, 1 + 2 + 3...= -1/12. Sort of.

posted by Orange Goblin at 4:07 AM on March 22, 2006

Anyway, my contribution: There are more real numbers between 0 and 1 than there are integers (...-2, -1, 0, 1, 2...). Or between 0 and 0.1, or any interval you care to choose. Also, 1 + 2 + 3...= -1/12. Sort of.

posted by Orange Goblin at 4:07 AM on March 22, 2006

Malor: that's pretty much it, though the answer is "no change", rather than "almost no change". A (hopefully) simple explanation:

Think of each birth as a coin toss. Whatever rules you impose on a sequence of coin tosses, the odds of each individual coin coming up heads or tails remain unchanged. So, the odds of each individual baby being male or female remain the same (105M/100F at birth, ISTR).

If you simulate the births with a coin, moving from family to family every time you get a boy, it becomes intuitively obvious.

posted by Leon at 4:12 AM on March 22, 2006

Think of each birth as a coin toss. Whatever rules you impose on a sequence of coin tosses, the odds of each individual coin coming up heads or tails remain unchanged. So, the odds of each individual baby being male or female remain the same (105M/100F at birth, ISTR).

If you simulate the births with a coin, moving from family to family every time you get a boy, it becomes intuitively obvious.

posted by Leon at 4:12 AM on March 22, 2006

Some great ones at the rec.puzzles archive particularly in decision and probability. This is a good one:

posted by edd at 4:19 AM on March 22, 2006

Someone has prepared two envelopes containing money. One contains twice as much money as the other. You have decided to pick one envelope, but then the following argument occurs to you: Suppose my chosen envelope contains $X, then the other envelope either contains $X/2 or $2X. Both cases are equally likely, so my expectation if I take the other envelope is .5 * $X/2 + .5 * $2X = $1.25X, which is higher than my current $X, so I should change my mind and take the other envelope. But then I can apply the argument all over again. Something is wrong here! Where did I go wrong?One from the probability section:

Suppose that it is equally likely for a pregnancy to deliver a baby boy as it is to deliver a baby girl. Suppose that for a large society of people, every family continues to have children until they have a boy, then they stop having children. After 1,000 generations of families, what is the ratio of males to females?Most of the problems already mentioned, if not all, are contained within, along with solutions.

posted by edd at 4:19 AM on March 22, 2006

(and then I spot that Leon posted that second one earlier... see - I *said* they were all in the archive!)

posted by edd at 4:20 AM on March 22, 2006

posted by edd at 4:20 AM on March 22, 2006

OK, Rhomboid, I'll try again (I hope you are having fun here too!).

This is in fact exactly the opposite of the way the Monty Hall problem works. In MH the way people get duped is exactly because they ignore the influence that the observation has on the likelihood of the possible outcomes. Your assumption is that the only possible influence of this observation is that it rules out possible outcomes but it cannot influence the relative likelihoods of outcomes that are still possible.

For example, I have two boxes of socks, box A and box B. Box A has 9 red socks and 1 blue sock and box B has 9 blue socks and 1 red sock. My friend Bob picks a box - with a 50/50 chance - picks a red sock out of it and gives it to me.

Now I need to guess which box it came from. Are A and B equally likely? No, because I'm more likely to get a red sock from box A than from box B. Similarly, I'm more likely to get a boy from a pair of two boys than I am from a pair of a boy and a girl.

posted by teleskiving at 4:43 AM on March 22, 2006

This is in fact exactly the opposite of the way the Monty Hall problem works. In MH the way people get duped is exactly because they ignore the influence that the observation has on the likelihood of the possible outcomes. Your assumption is that the only possible influence of this observation is that it rules out possible outcomes but it cannot influence the relative likelihoods of outcomes that are still possible.

For example, I have two boxes of socks, box A and box B. Box A has 9 red socks and 1 blue sock and box B has 9 blue socks and 1 red sock. My friend Bob picks a box - with a 50/50 chance - picks a red sock out of it and gives it to me.

Now I need to guess which box it came from. Are A and B equally likely? No, because I'm more likely to get a red sock from box A than from box B. Similarly, I'm more likely to get a boy from a pair of two boys than I am from a pair of a boy and a girl.

posted by teleskiving at 4:43 AM on March 22, 2006

Feynman, among others, thought so as well.

It's a degenerate case of the following formula:

e^(in) = cos(x) + i(Sin(x))

Which makes a kind of sense once you consider that the line of imaginary numbers can be mapped orthogonally to the line of real numbers to make a plane of complex numbers. Trigonometric operations on this plane can be performed using Euclidean gemetries. From these you can derive the standard trigonometric ratios.

The fundamental relationship is:

z = x + iy = r*(cos(n )+ i sin(n)) = r*e^(in)

posted by meehawl at 4:52 AM on March 22, 2006

I think I've already posted so much in this thread that one more will make no difference. In order to fix vacapinta's problem you need to provide the information "There is at least one boy" without providing any other information that skews the outcome e.g. as soon as you make it implicit that "I observed one of the two siblings at random and determined that they were male", you've lost the outcome that you want.

I would be interested to see if anyone is capable of phrasing the fixed problem in a way that doesn't seem contrived or give the game away, I've thought about it a bit but am drawing a blank.

It's about time I got some work done...

posted by teleskiving at 5:08 AM on March 22, 2006

I would be interested to see if anyone is capable of phrasing the fixed problem in a way that doesn't seem contrived or give the game away, I've thought about it a bit but am drawing a blank.

It's about time I got some work done...

posted by teleskiving at 5:08 AM on March 22, 2006

I thought mr_roboto's example was striking enough I'd put the essence of it (deleting the actual math) in the thread:

posted by languagehat at 6:59 AM on March 22, 2006

Suppose that a test for a particular disease has a very high success rate:I'm completely confused by the boy/girl problem. Thanks, guys!

* if a tested patient has the disease, the test accurately reports this, a 'positive', 99% of the time (or, with probability 0.99), and

* if a tested patient does not have the disease, the test accurately reports that, a 'negative', 95% of the time (i.e. with probability 0.95).

Suppose also, however, that only 0.1% of the population have that disease (i.e. with probability 0.001). We now have all the information required to use Bayes' theorem to calculate the probability that, given the test was positive, that it is a false positive...

Despite the apparent high accuracy of the test, the incidence of the disease is so low (one in a thousand) that the vast majority of patients who test positive (98 in a hundred) do not have the disease.

posted by languagehat at 6:59 AM on March 22, 2006

About the birthday paradox, there is something that doesn't add up to me. In it you are assuming that birthday dates are equiprobable and as far as my common sense can guide me, they are not. For some reason, at least in Argentina, birthdays are a little bit more common in Spring (9 months after the summer?).

Does this happen anywhere else?

Wouldn't this explain why I know only 2 people who share my birthday?

posted by Manouk at 8:20 AM on March 22, 2006

Does this happen anywhere else?

Wouldn't this explain why I know only 2 people who share my birthday?

posted by Manouk at 8:20 AM on March 22, 2006

teleskiving writes...

*...Now I need to guess which box it came from. Are A and B equally likely? No, because I'm more likely to get a red sock from box A than from box B. Similarly, I'm more likely to get a boy from a pair of two boys than I am from a pair of a boy and a girl.*

This is a very nice description.

Although I enjoy these sort of counter-intuitive questions, as time goes by I've come to really started to enjoy accessible explanations of them more than the questions themselves.

Well done.

posted by tkolar at 8:22 AM on March 22, 2006

This is a very nice description.

Although I enjoy these sort of counter-intuitive questions, as time goes by I've come to really started to enjoy accessible explanations of them more than the questions themselves.

Well done.

posted by tkolar at 8:22 AM on March 22, 2006

I believe that both Einstein and Erdos came down on the wrong side of that one.

posted by StickyCarpet at 8:22 AM on March 22, 2006

The unintuitive nature of the birthday paradox is probably not strongly tied to the uniform distribution of birthdays it assumes. If you were to take the extreme case of all birthdays being on the same day then obviously it only takes two people to get over that 50% probability of a match within the group. If you start gradually spreading the distribution from there I'm guessing that the number taken will rise until you reach the a maximum case for the uniform distribution.

Since you only know 2 people who share your birthday, chances are there are significant numbers of other people who share each others birthday.

Remember that the birthday paradox only applies to any coincident birthdays from a group. By talking about*your* birthday specifically, you're changing the problem, and you'd need nearly two hundred people assuming a uniform distribution of birthdays to find a match for your own.

posted by edd at 8:26 AM on March 22, 2006

Since you only know 2 people who share your birthday, chances are there are significant numbers of other people who share each others birthday.

Remember that the birthday paradox only applies to any coincident birthdays from a group. By talking about

posted by edd at 8:26 AM on March 22, 2006

This isn't strictly what you are looking for, but...

It was pointed out to me a few years ago that you can encode all of human knowledge onto a stick.

1) Mark a zero and a one on the stick

2) Encode the sum of human knowledge into one very long integer

3) Place a decimal point in front of it, and mark that spot between 0 and 1 on the stick.

posted by tkolar at 8:27 AM on March 22, 2006

It was pointed out to me a few years ago that you can encode all of human knowledge onto a stick.

1) Mark a zero and a one on the stick

2) Encode the sum of human knowledge into one very long integer

3) Place a decimal point in front of it, and mark that spot between 0 and 1 on the stick.

posted by tkolar at 8:27 AM on March 22, 2006

I was going to post this one as my own question, but now's as good a time as any.

Back as a Boy Scout in the eighties, our Scoutmaster dropped one on us at a campout.*It was about several people staying in several rooms at a hotel and splitting the bill. Somewhere a dollar ends up missing.* Certainly blew a lot of 12-year-old minds. Anyone know this one?

If this puzzle could be found, it totally fits what arcticwoman is looking for.

posted by sourwookie at 8:46 AM on March 22, 2006

Back as a Boy Scout in the eighties, our Scoutmaster dropped one on us at a campout.

If this puzzle could be found, it totally fits what arcticwoman is looking for.

posted by sourwookie at 8:46 AM on March 22, 2006

The case of Sleeping Beauty.

I have never spent so much time being confused by such a seemingly simple story. For many paradoxes, I can settle on one side or another, but with this one, I can convince myself of one answer, and seconds later, convince myself that the other answer is true.

posted by Maxwell_Smart at 9:00 AM on March 22, 2006

I have never spent so much time being confused by such a seemingly simple story. For many paradoxes, I can settle on one side or another, but with this one, I can convince myself of one answer, and seconds later, convince myself that the other answer is true.

posted by Maxwell_Smart at 9:00 AM on March 22, 2006

Sourwookie: Here you go:

posted by grahamwell at 9:06 AM on March 22, 2006

Three men stay at a hotel for the night. The innkeeper charges thirty dollars per room per night. The men rent one room; each pays ten dollars. The bellhop leads the men to their room. Later, the innkeeper discovers he has overcharged the men and asks the bellhop to return five dollars to them. On the way upstairs, the bellhop realizes that five dollars can't be evenly split among three men, so he decides to keep two dollars for himself and return one dollar to each man.The answer here. Many more such puzzles at Brainfood (a wonderful site for puzzles).

At this point, the men have paid nine dollars each, totalling 27. The bellhop has two, which adds up to 29. Where did the thirtieth dollar go?

posted by grahamwell at 9:06 AM on March 22, 2006

Manouk: Yes, birth rates in the US, for example, are up to 12% higher in summer than winter.

Also, the birthday paradox doesn't mean that 1 in 23 people will share*your* birthday. In a group of 23 people, it's likely that two will share a birthday... but it's also highly likely that most of the other 21 will not. Only about 1 in 365 people you know should share your birthday; slightly less if you were born in the winter, I suppose.

posted by mbrubeck at 9:11 AM on March 22, 2006

Also, the birthday paradox doesn't mean that 1 in 23 people will share

posted by mbrubeck at 9:11 AM on March 22, 2006

Count me in as a halfer on the Sleeping Beauty problem!

posted by Justinian at 9:27 AM on March 22, 2006

posted by Justinian at 9:27 AM on March 22, 2006

Grahamwell: Thanks! That only took 20 years to find!

posted by sourwookie at 9:42 AM on March 22, 2006

posted by sourwookie at 9:42 AM on March 22, 2006

I never understood what was so vexing about the prisoner's dilemma. It's not a dilemma at all. You have persistent, unilateral incentive to screw the other guy over. Doesn't matter what he does.

posted by TunnelArmr at 10:00 AM on March 22, 2006

A simpler version of the men-in-a-hotel problem is "I have eleven fingers." (counting on left hand:) ten, nine, eight, seven, six... (showing fingers on right hand:) plus five is eleven.

posted by ook at 10:58 AM on March 22, 2006

posted by ook at 10:58 AM on March 22, 2006

Yes, that's why it's a dilemma. If you both cooperated, you would have come out far ahead.

posted by kindall at 11:21 AM on March 22, 2006

A side question: Would a CGI video of the Banach-Tarski paradox be completely disappointing to watch? (And why or why not?)

posted by skryche at 11:23 AM on March 22, 2006

posted by skryche at 11:23 AM on March 22, 2006

grahamwell,

That hotel paradox reminds me of the wise man and the camel inheritance puzzle.

A certain man died leaving three sons and 17 camels. In his will he left 1/2 his inheritance to his first son, 1/3 to the second and 1/9 to the last. Unfortunately 17 is prime and a fraction of a camel isn't much use, except perhaps as a meal.

As a wise man was riding by on his camel he noticed the confusion and offered his aid. First, he added his own camel to the herd, making 18 camels. He gave 9 (1/2) to the first son, 6 (1/3) to the second son and 2 (1/9) the youngest. Then he mounted his own camel and rode away.

posted by Octaviuz at 11:31 AM on March 22, 2006

That hotel paradox reminds me of the wise man and the camel inheritance puzzle.

A certain man died leaving three sons and 17 camels. In his will he left 1/2 his inheritance to his first son, 1/3 to the second and 1/9 to the last. Unfortunately 17 is prime and a fraction of a camel isn't much use, except perhaps as a meal.

As a wise man was riding by on his camel he noticed the confusion and offered his aid. First, he added his own camel to the herd, making 18 camels. He gave 9 (1/2) to the first son, 6 (1/3) to the second son and 2 (1/9) the youngest. Then he mounted his own camel and rode away.

posted by Octaviuz at 11:31 AM on March 22, 2006

You tell me :)

posted by mr_roboto at 11:33 AM on March 22, 2006

I'd argue that it also has profound policy implications. Think about the rate of false positives in racial profiling, for instance.

posted by mr_roboto at 11:36 AM on March 22, 2006

skryche: Regarding the Banach-Tarski video: you don't actually cut the ball into pieces, rather you "construct" (loaded term) those pieces using the axiom of choice. So they might look kinda wacky in a video, hard to say. The point of the BT paradox is that there is no "nice" volume function (one that doesn't change under rigid motions) that works on all subsets of R^3 (since these pieces in particular would change volume en route from one ball to two). And the existence of these wacky pieces is yet another terrible consequence of the axiom of choice!

Still, nice video, mr_roboto!

I loved that anagram joke, vernondalhart. I can only contribute a Banach-Tarski haiku:

Cut up a green pea,

Reassemble it so it's

As big as the sun!

posted by blue grama at 1:21 PM on March 22, 2006

Still, nice video, mr_roboto!

I loved that anagram joke, vernondalhart. I can only contribute a Banach-Tarski haiku:

Cut up a green pea,

Reassemble it so it's

As big as the sun!

posted by blue grama at 1:21 PM on March 22, 2006

Furthering Tims dilemma... if you add another feature to the family, for instance length, then there are 10 possibilities. Suddenly, older Tim possibility of a younger of one in three, whereas some Tim will have equal chances.

*puts away papers full of younger-taller-bigger Tims*

posted by Psychnic at 2:01 PM on March 22, 2006

*puts away papers full of younger-taller-bigger Tims*

posted by Psychnic at 2:01 PM on March 22, 2006

OK, I'm still puzzling over Vacapinta's conundrum

*Me: I hear you have two children.*

Him: Yes. i think you've met Tim before.

(The chances that his other child is a girl is 2/3)

Seems to me, there are only three combinations:

boyboy

boygirl

girlgirl

Statistically, each is equally possible. If Tim is a boy, then the only two combinations remaining are:

boyboy and boygirl. Girlgirl is no longer possible.

Therefor the likelihood of child two being female is 50/50.

What am I missing?

posted by johngumbo at 7:18 PM on March 22, 2006

Him: Yes. i think you've met Tim before.

(The chances that his other child is a girl is 2/3)

Seems to me, there are only three combinations:

boyboy

boygirl

girlgirl

Statistically, each is equally possible. If Tim is a boy, then the only two combinations remaining are:

boyboy and boygirl. Girlgirl is no longer possible.

Therefor the likelihood of child two being female is 50/50.

What am I missing?

posted by johngumbo at 7:18 PM on March 22, 2006

Thanks, blue grama. I can't claim to have come up with it, but I do love to pass it on.

posted by vernondalhart at 8:11 PM on March 22, 2006

posted by vernondalhart at 8:11 PM on March 22, 2006

johngumbo, you're missing girlboy.

posted by youarenothere at 8:21 PM on March 22, 2006

posted by youarenothere at 8:21 PM on March 22, 2006

Ok, just to throw a wrench into the tim/boy/girl problem, let's assume Vacapinta's original probabilities are correct:

If the guy says "the older one is named Tim" then it's 50/50 boy/girl.

If the guy says "one of them is named Tim" then it's 2/3 likely to be female.

So... The guy says "one of them is named Tim."

At this point we can say there is a 50% chance Tim is the older sibling - in which case the male/female probability of the other sibling is 50/50.

There is also a 50% chance that Tim is the younger sibling -in which case the male/female probability of the other sibling is 50/50.

Add it all up and the probability is still 50/50.

It would seem that one of the two probabilities stated in the original problem must not be correct, as they contradict each other. All in all it doesn't make sense to me, but I know well enough that you shouldn't be able to get different answers by calculating a problem in different ways, and I don't see how I've introduced any new assumptions here beyond what are implied by the original problem.

posted by Wingy at 7:52 AM on March 23, 2006

If the guy says "the older one is named Tim" then it's 50/50 boy/girl.

If the guy says "one of them is named Tim" then it's 2/3 likely to be female.

So... The guy says "one of them is named Tim."

At this point we can say there is a 50% chance Tim is the older sibling - in which case the male/female probability of the other sibling is 50/50.

There is also a 50% chance that Tim is the younger sibling -in which case the male/female probability of the other sibling is 50/50.

Add it all up and the probability is still 50/50.

It would seem that one of the two probabilities stated in the original problem must not be correct, as they contradict each other. All in all it doesn't make sense to me, but I know well enough that you shouldn't be able to get different answers by calculating a problem in different ways, and I don't see how I've introduced any new assumptions here beyond what are implied by the original problem.

posted by Wingy at 7:52 AM on March 23, 2006

First, a key fact that I think we can all agree on: **There are twice as many families with one boy and one girl as there are with two boys.** Select 100 two-child families at random. There will be approximately 25 with two boys, 25 with two girls, and 50 with one of each.

Now select one of those families at random, and ask the father if he has at least one boy. If the answer is yes, then it must be one of the 25 families with two boys or one of the 50 families with a boy and a girl.

This means that a boy from a randomly selected family is*more likely* to have a sister than a brother. Yes, this is counter-intuitive.

The number of boys and girls is equal, but you are more likely to choose a split family (because there are more of them). Yet a family with two boys and a family with one boy both give you the same answer; thus the boys from the two-boy family carry half as much individual "weight" in this random trial.

However, if you select a random**child** instead of a random **family**, then the odds are different (because the random selection process is different).

Choose a family at random, then ask the father to flip a coin. If it's heads, he will tell you only the sex of his oldest child; tails, he will tell you only the sex of the younger child. Now if he says he has a son, the odds are 1/2 that the son has a sister. The odds of choosing a two-boy family are still only 25%, but you now are twice as likely to hear about a son from a two-boy family as from a one-boy family.

The original description of the puzzle, where the father volunteers the information rather than being asked, does not make it clear what random trial is being performed. Would the father only mention a boy child (as in my first experiment)? Or is he picking a child at random and mentioning it regardless of sex (as in my second experiment)? I think this ambiguity is the source of the disagreement.

posted by mbrubeck at 8:59 AM on March 23, 2006 [1 favorite]

Now select one of those families at random, and ask the father if he has at least one boy. If the answer is yes, then it must be one of the 25 families with two boys or one of the 50 families with a boy and a girl.

This means that a boy from a randomly selected family is

The number of boys and girls is equal, but you are more likely to choose a split family (because there are more of them). Yet a family with two boys and a family with one boy both give you the same answer; thus the boys from the two-boy family carry half as much individual "weight" in this random trial.

However, if you select a random

Choose a family at random, then ask the father to flip a coin. If it's heads, he will tell you only the sex of his oldest child; tails, he will tell you only the sex of the younger child. Now if he says he has a son, the odds are 1/2 that the son has a sister. The odds of choosing a two-boy family are still only 25%, but you now are twice as likely to hear about a son from a two-boy family as from a one-boy family.

The original description of the puzzle, where the father volunteers the information rather than being asked, does not make it clear what random trial is being performed. Would the father only mention a boy child (as in my first experiment)? Or is he picking a child at random and mentioning it regardless of sex (as in my second experiment)? I think this ambiguity is the source of the disagreement.

posted by mbrubeck at 8:59 AM on March 23, 2006 [1 favorite]

You're right that this is the only real ambiguity in the way the problem is posed. But to be honest, I can't think of any justification for the assertion that the father would only mention a boy child. Who would assume that, really?

posted by teleskiving at 10:01 AM on March 23, 2006

Curiously, I performed an experiment with my 5th grade math whiz class day before yesterday after reading the beginnings of this thread. We rolled a dice to determine the gender of twenty pairs of siblings.

5 of the pairs proved to be boy/boy, 5 were girl/girl, 10 were boy/girl (and girl/boy).

So working with these actual numbers, the chances of a randomly chosen boy from the group having a sister was 2/3.

It was very cool, and somewhat fortunate, to watch the numbers play out so perfectly to my prediction!

posted by carterk at 9:00 AM on March 24, 2006

5 of the pairs proved to be boy/boy, 5 were girl/girl, 10 were boy/girl (and girl/boy).

So working with these actual numbers, the chances of a randomly chosen boy from the group having a sister was 2/3.

It was very cool, and somewhat fortunate, to watch the numbers play out so perfectly to my prediction!

posted by carterk at 9:00 AM on March 24, 2006

Dude, you have ten boys in boy/boy pairs and ten boys in boy/girl pairs. Pick a boy at random, there is a 50/50 chance he is in a boy/boy pair and has no sister. How did you get 2/3?

posted by teleskiving at 3:31 PM on March 24, 2006

Me: I hear you have two children.

Him: Yes, the oldest is named Tim.

(the chances are 50-50 that his other child is a girl)

Me: I hear you have two children.

Him: Yes. i think you've met Tim before.

(The chances that his other child is a girl is 2/3)

This is incorrect. The probability in both cases is 0.5 because the gender of the two children is independent. It's not like the Monty-hall problem at all, because the parent isn't guessing the one he 'knows' is a boy. He's just stating (randomly) that one of his children is a boy, as if in the Monty hall problem Monty might open the door with the car.

If, in the Monty hall problem, the announcer could open

posted by delmoi at 7:35 PM on March 27, 2006 [1 favorite]

No, they're not describing the same case.

Look at it this way

case 1) Older brother, younger brother

case 2) Older brother, younger sister

case 3) Older sister younger sister

case 4) older sister younger sister.

Once the father tells you the that one of his children is a boy, the probability changes to this:

case 1) 50%

case 2) 25%

case 3) 25%

case 4) 0%

So you have a 50/50 chance of having the second sibling be a boy.

posted by delmoi at 7:39 PM on March 27, 2006

another way to think about it, there are eight 'spots' that Tim can fit into, but four of them are closed off. Tim has two chances to land in case one, and one chance in the other two cases.

posted by delmoi at 7:39 PM on March 27, 2006

posted by delmoi at 7:39 PM on March 27, 2006

No, this is where you're wrong. As I've said before, the first case now has a 50% chance of happening.

posted by delmoi at 7:46 PM on March 27, 2006

Now select one of those families at random, and ask the father if he has at least one boy. If the answer is yes, then it must be one of the 25 families with two boys or one of the 50 families with a boy and a girl.

Let's think about that a bit more. While there are 100 families in your example, there are only 75 families with at least one boy. 25 with two boys, and 50 with one boy and one girl.

But at the same time are 150 children. 75 of them are boys. Of the boys 50 of them have a brother and 25 of them have a sister. If you were to chose a boy at random, you'd have a 50/25, or 2/3 chance of chosing a boy

In other words the odds are reversed now.

The point is there are lots of diffrent ways to

I belive the origional construction vacapinta gives us 50% chance of tim's sibling being a girl.

back after the daily show.

posted by delmoi at 8:01 PM on March 27, 2006

Actually, I think vacapinta just mixed up the two examples. If you know the oldest one is a boy, then you can say that there's a 2/3rds chance that the other is a girl. I think.

posted by delmoi at 8:19 PM on March 27, 2006

posted by delmoi at 8:19 PM on March 27, 2006

Sorry, delmoi, but vacapinta has it exactly right and you have it exactly backwards.

Assumptions, upon which I presume we will agree:

1. There are two children in this family (as agreed to by father)

2. Each child is either a boy or a girl (intersex children swept under the rug as usual)

3. Boys and girls are equally likely to be born (close enough to true for our purposes)

4. Tim is a boy's name

Assumptions 2 and 3 together lead directly to an even distribution of the following four cases across the population of two-child families:

a) boy born first, boy born second

b) boy born first, girl born second

c) girl born first, boy born second

d) girl born first, girl born second

That is to say: there will be, across any sufficiently large population of two-child families,**equal numbers of families** in **each of those four categories**.

In Vacapinta's first variation, we find out that the eldest child is Tim. This means that the family cannot be of type (c) or (d); it must be one of type (a) or (b). Because there are equal numbers of**families** in those **two** categories, and we have no further information, the chances of Tim having a sister - that is, the chance that his family is type (b) - are one in two.

In the second variation, we find out that one of the children (we don't know which) is Tim. This means that the family cannot be of type (d); it must be one of type (a), (b) or (c). Because there are equal numbers of**families** in those **three** categories, and we have no further information, the chances of Tim having a sister - that is, the chance that this family is type (b) or (c) - are two in three.

The absolute number of boys in**all families combined** is a total red herring. It's a very effective red herring. Lots of people here can't seem to smell anything else :)

posted by flabdablet at 11:08 PM on March 27, 2006

Assumptions, upon which I presume we will agree:

1. There are two children in this family (as agreed to by father)

2. Each child is either a boy or a girl (intersex children swept under the rug as usual)

3. Boys and girls are equally likely to be born (close enough to true for our purposes)

4. Tim is a boy's name

Assumptions 2 and 3 together lead directly to an even distribution of the following four cases across the population of two-child families:

a) boy born first, boy born second

b) boy born first, girl born second

c) girl born first, boy born second

d) girl born first, girl born second

That is to say: there will be, across any sufficiently large population of two-child families,

In Vacapinta's first variation, we find out that the eldest child is Tim. This means that the family cannot be of type (c) or (d); it must be one of type (a) or (b). Because there are equal numbers of

In the second variation, we find out that one of the children (we don't know which) is Tim. This means that the family cannot be of type (d); it must be one of type (a), (b) or (c). Because there are equal numbers of

The absolute number of boys in

posted by flabdablet at 11:08 PM on March 27, 2006

You guys are still disagreeing about whether a father with two boys is twice as likely to mention a boy. As I said above, the random trial is not well-defined in the original question. I provided two ways to clarify the original question, one that leads to delmoi's result and one that leads to flabdablbet's.

In short: If the father is choosing a child at random and then mentioning its name/sex (as delmoi assumes), then if he mentions a boy there is a 50% chance the boy has a sister. If the father is volunteering "At least one of my children (Tim) is a boy," and would not have said anything about whether he has any girls, then you are more likely to hear this from a father with a boy and a girl; then there is a 2/3 chance that Tim has a sister.

posted by mbrubeck at 6:24 AM on March 28, 2006

In short: If the father is choosing a child at random and then mentioning its name/sex (as delmoi assumes), then if he mentions a boy there is a 50% chance the boy has a sister. If the father is volunteering "At least one of my children (Tim) is a boy," and would not have said anything about whether he has any girls, then you are more likely to hear this from a father with a boy and a girl; then there is a 2/3 chance that Tim has a sister.

posted by mbrubeck at 6:24 AM on March 28, 2006

flabdablet: Like I said, there are many different ways to construct the problem, and depending on how you construct it, you get different answers. I now think both of the original formulations give you 50%, I just threw that last comment out there without thinking it all the way though (I thought that would explain why vacapinta messed up his original formulation)

In order to get**two thirds** you have to have this conversation:

"Me: I hear you have two children; I've met your son"

"Him: Yes. i think you've met Tim before."

It has to be asked this way, such that the fact he has a boy is known**before** the question is asked, not after.

It all comes down to what you know**before** the question, and what you know **after** the question.

And like I said above, if you took all the boys from sibling pairs with at least one boy, and asked if they had a brother or sister, you'd only get a**1/3** chance of having a sister.

So depending on how you ask the question you can have 2/3rds, 1/2, or 1/3rd.

posted by delmoi at 7:13 AM on March 28, 2006

In order to get

"Me: I hear you have two children; I've met your son"

"Him: Yes. i think you've met Tim before."

It has to be asked this way, such that the fact he has a boy is known

It all comes down to what you know

And like I said above, if you took all the boys from sibling pairs with at least one boy, and asked if they had a brother or sister, you'd only get a

So depending on how you ask the question you can have 2/3rds, 1/2, or 1/3rd.

posted by delmoi at 7:13 AM on March 28, 2006

2. Each child is either a boy or a girl (intersex children swept under the rug as usual)

3. Boys and girls are equally likely to be born (close enough to true for our purposes)

4. Tim is a boy's name

Assumptions 2 and 3 together lead directly to an even distribution of the following four cases across the population of two-child families:

a) boy born first, boy born second

b) boy born first, girl born second

c) girl born first, boy born second

d) girl born first, girl born second

You're actually implicitly adding a 5th assumption: "

That's what I don't agree with.

posted by delmoi at 7:28 AM on March 28, 2006

I'd just like to mention that I have a long history with the MHP and put up one of the first, possibly *the* first, web sites devoted to it. For the longest time it was the #1 Google for it, but I changed ISPs and, well, you know. But I still am cited at Mathworld. Which makes me absurdly happy, especially considering that I don't think that highly of Wolfram.

Anyway, I hope this isn't obnoxious self-linking, but if you're curious about the MHP, you might want to take a look at my page. I gave it its own domain a year ago or so. And if you'd like to discuss it, I'd be happy to. I now have more than a decade of discussing the MHP with correspondents.

posted by Ethereal Bligh at 4:47 AM on March 29, 2006

Anyway, I hope this isn't obnoxious self-linking, but if you're curious about the MHP, you might want to take a look at my page. I gave it its own domain a year ago or so. And if you'd like to discuss it, I'd be happy to. I now have more than a decade of discussing the MHP with correspondents.

posted by Ethereal Bligh at 4:47 AM on March 29, 2006

No, I'm not doing that at all.

The process by which the father chose what to say is irrelevant; a red herring.

The simple fact is that the father has told us that one of his children is called Tim. Therefore, we know he doesn't have two girls.

There is no "distribution of a probability" between "the other probabilities"; the effect of the information we now possess is simply to

The probabilities attached to the remaining family types have

The fact that they are also equal to the number of two-girl families - that is, to the number of families in the category we've just eliminated - is also irrelevant.

It's the cunningly concealed red herring in this problem that makes it reminiscent of the Monty Hall problem. Both problems require attention to the effect of new information on the situation that presents itself. The MHP, though,

When the contestant first picks a door, the chances of having picked the winning door are clearly 1 in 3.

Monty then opens a door without a prize behind it. Monty always has at least one empty door available for this, regardless of which door the contestant originally picked; therefore, Monty's door-opening action has

The probability of the prize being behind the remaining door is 1 (the prize is certain to be

There are scenarios where Monty has to choose which of two prizeless doors to open, and there are scenarios where Father has to choose which sex of child to mention. But in neither case do these choices, or the processes by which they are made, have any effect at all on the hard information available to us - which is what probability calculations must be based on.

posted by flabdablet at 7:36 AM on March 29, 2006

OK, back to the socks. Drawer A has 9 red socks, one blue sock. Drawer B has 9 blue socks, one red sock. Drawer C also has 9 blue socks and one red sock. Drawer D has 10 green socks. Someone picks a sock from one of the drawers at random (either A,B, C, or D with equal probabilities) and gives it to me. It is red.

Which drawer did the sock come from? Obviously not drawer D. Would you guess A, B, or C? It is pretty obvious that you should guess drawer A in this case. Altogether we have 11 red socks, 9 of which are from drawer A, and all socks are equally likely, so 9 out of 11 times we get a red sock, it will be from drawer A. That this is a probabilistically expressed fact does not mean that we can just say "it's not hard information, throw it away". If we throw it away, we will get the wrong answer. If we go on to ask what the probability of getting another red sock from the same drawer is (without having yet categorically determined which drawer was selected), we*have *to take this information into account.

This is exactly the same as the boy/girl problem, I have only changed the numbers of each type of element in each category to make the principle clearer and also been more explicit about the sampling process to get rid of the "more likely to mention a boy than a girl" aspect.

posted by teleskiving at 9:36 PM on March 29, 2006

Which drawer did the sock come from? Obviously not drawer D. Would you guess A, B, or C? It is pretty obvious that you should guess drawer A in this case. Altogether we have 11 red socks, 9 of which are from drawer A, and all socks are equally likely, so 9 out of 11 times we get a red sock, it will be from drawer A. That this is a probabilistically expressed fact does not mean that we can just say "it's not hard information, throw it away". If we throw it away, we will get the wrong answer. If we go on to ask what the probability of getting another red sock from the same drawer is (without having yet categorically determined which drawer was selected), we

This is exactly the same as the boy/girl problem, I have only changed the numbers of each type of element in each category to make the principle clearer and also been more explicit about the sampling process to get rid of the "more likely to mention a boy than a girl" aspect.

posted by teleskiving at 9:36 PM on March 29, 2006

Picking a sock at random is not analogous to meeting Tim's father. Tim's father is more like a drawer than a sock.

Vacapinta's second problem, translated into the terms of your analogy, goes like this:

Me: I hear you hold ten socks.

Drawer: Yes. Here's a red one.

What is the probability that another sock from the same drawer will be red?

Solution: there are four drawers. We pick one. We can't tell the drawers apart, so our chances of having got any particular drawer are equal.

Having made a choice of one of four equally likely drawers, we now pull out a sock. It's red. That means that we can't possibly have picked drawer D, which contains only green socks.

There is no bias built into the way we chose our drawer; the only thing we know for sure now is that it isn't drawer D. A, B and C are equally likely choices. *

If we are in fact working with drawer A, we know it must now contain eight red socks and one blue sock, so the probability of drawing a second red sock will be 8/9.

If we are in fact working with drawer B, we know it must now contain only blue socks, so the probability of drawing a second red sock will be 0.

If we are in fact working with drawer C, we know it must now contain only blue socks, so the probability of drawing a second red sock will be 0.

So the overall probability we'll get another red sock is

(1/3 * 8/9) + (1/3 * 0) + (1/3 * 0) = 8/27.

If you have an alternative method for working this out, what does it predict the result should be?

We can find out which method is actually correct by coding the thing up as a Monte Carlo simulation:

0. Load drawers with socks; clear trial count and match count.

1. Pick any drawer at random.

2. Pull out any sock at random from that drawer.

3. If it's red, go to step 5.

4. Replace the sock and go back to step 1.

We're now at the "Yes. Here's a red one" point, analogous to having been told "Yes. I think you've met Tim".

5. Increment trial count.

6. Pull out a second sock from the same drawer at random.

7. If it's red, increment match count.

8. Replace both socks.

9. If trial count is under 1 million, go back to step 1.

Now divide the match count by the trial count. I'll bet you a dollar the result is very close to 8/27.

* If the scent of red herring is overwhelming at this point, imagine rolling a die that's got a red face on 1, partly-red and partly-blue faces on 2 through 5, and a green face on 6. If you can't see red on the face you've just rolled, roll again. Now, what are the chances you've rolled a three? Are you any more likely to have rolled a one?

Keep in mind that the question "how likely are we to find ourselves working with drawer A, given that our chosen drawer has just yielded a red sock?", while interesting in itself, is**not analogous to the problem as given**.

posted by flabdablet at 2:52 AM on March 30, 2006

Vacapinta's second problem, translated into the terms of your analogy, goes like this:

Me: I hear you hold ten socks.

Drawer: Yes. Here's a red one.

What is the probability that another sock from the same drawer will be red?

Solution: there are four drawers. We pick one. We can't tell the drawers apart, so our chances of having got any particular drawer are equal.

Having made a choice of one of four equally likely drawers, we now pull out a sock. It's red. That means that we can't possibly have picked drawer D, which contains only green socks.

There is no bias built into the way we chose our drawer; the only thing we know for sure now is that it isn't drawer D. A, B and C are equally likely choices. *

If we are in fact working with drawer A, we know it must now contain eight red socks and one blue sock, so the probability of drawing a second red sock will be 8/9.

If we are in fact working with drawer B, we know it must now contain only blue socks, so the probability of drawing a second red sock will be 0.

If we are in fact working with drawer C, we know it must now contain only blue socks, so the probability of drawing a second red sock will be 0.

So the overall probability we'll get another red sock is

(1/3 * 8/9) + (1/3 * 0) + (1/3 * 0) = 8/27.

If you have an alternative method for working this out, what does it predict the result should be?

We can find out which method is actually correct by coding the thing up as a Monte Carlo simulation:

0. Load drawers with socks; clear trial count and match count.

1. Pick any drawer at random.

2. Pull out any sock at random from that drawer.

3. If it's red, go to step 5.

4. Replace the sock and go back to step 1.

We're now at the "Yes. Here's a red one" point, analogous to having been told "Yes. I think you've met Tim".

5. Increment trial count.

6. Pull out a second sock from the same drawer at random.

7. If it's red, increment match count.

8. Replace both socks.

9. If trial count is under 1 million, go back to step 1.

Now divide the match count by the trial count. I'll bet you a dollar the result is very close to 8/27.

* If the scent of red herring is overwhelming at this point, imagine rolling a die that's got a red face on 1, partly-red and partly-blue faces on 2 through 5, and a green face on 6. If you can't see red on the face you've just rolled, roll again. Now, what are the chances you've rolled a three? Are you any more likely to have rolled a one?

Keep in mind that the question "how likely are we to find ourselves working with drawer A, given that our chosen drawer has just yielded a red sock?", while interesting in itself, is

posted by flabdablet at 2:52 AM on March 30, 2006

flabdablet, thanks for casting this in a testable way. Looking at your proposed model, I accept your bet and contend that the result will be closer to 72/90 - a big difference! I should have a few minutes spare to code this this evening if someone hasn't already done it by then.

The problem in your die-rolling example is that it is the equivalent of observing both children at the same time - if there is a red part, then we always see it. This is just like asking directly "Do you have at least one boy?" or "Does this drawer have at least one red sock?" which are different to "Is the one sock that you have pulled out red?". The answers to those questions carry less information.

posted by teleskiving at 5:35 AM on March 30, 2006

The problem in your die-rolling example is that it is the equivalent of observing both children at the same time - if there is a red part, then we always see it. This is just like asking directly "Do you have at least one boy?" or "Does this drawer have at least one red sock?" which are different to "Is the one sock that you have pulled out red?". The answers to those questions carry less information.

posted by teleskiving at 5:35 AM on March 30, 2006

My code for the drawer game is here. I get near enough 800000 positive trials every time, which is equal to (72/90) * 1000000 trials. So I think I'm right. What do you think?

Bu the way, the 72/90 comes from (9/10) chance of an observed red sock coming from drawer A multiplied by (8/9) chance of getting a second red sock from drawer A.

posted by teleskiving at 9:19 AM on March 30, 2006

Bu the way, the 72/90 comes from (9/10) chance of an observed red sock coming from drawer A multiplied by (8/9) chance of getting a second red sock from drawer A.

posted by teleskiving at 9:19 AM on March 30, 2006

I think I owe you a dollar.

I now see that the final probability hinges very sensitively on the process by which Tim's father chooses which child to mention, which is what you've been saying since very early on.

If we recast the simulation so that steps 2 and 3 are replaced by one that reads

2. If there is a red sock in the drawer, go to step 5.

which is analogous to Tim's father saying "yes" when asked directly whether he has at least one male child, it ought to come up with 8/27.

Thanks for helping me change my mind! My email's in my profile if you want to tell me where to send your dollar.

posted by flabdablet at 9:27 PM on March 30, 2006

I now see that the final probability hinges very sensitively on the process by which Tim's father chooses which child to mention, which is what you've been saying since very early on.

If we recast the simulation so that steps 2 and 3 are replaced by one that reads

2. If there is a red sock in the drawer, go to step 5.

which is analogous to Tim's father saying "yes" when asked directly whether he has at least one male child, it ought to come up with 8/27.

Thanks for helping me change my mind! My email's in my profile if you want to tell me where to send your dollar.

posted by flabdablet at 9:27 PM on March 30, 2006

Good for you, flabdablet. My experience over the years corresponding with people who disagree with the correct answer to the MHP has been that few of them have the grace to admit they were wrong. I am only guessing about the people I didn't hear back from at all (after several rounds of emails), but there is a good chunk of them who will "admit" they were wrong by way of not admitting they were wrong because their misunderstanding of the problem was someone else's fault.

I think the single most frustrating correspondence I've had over the last 12 years on the MHP was a science writer from the Houston Chronicle (I thinkāis that the big daily there?) who came upon my site and was considering doing a short article on the MHP. Except he thought I was wrong and that it was really 50-50. So after trying the various strategies I've learned to explain the problem, I finally asked him to call up a math prof at one of the Houston universities. How hard would that be? And then he would at least have an authoritative answer, instead of relying on his intuition. But he said he didn't want to call anyone. "Why not?" I asked him. He replied that he didn't care that much. Well, he cared enough to argue about it with me through several emails. And the man was a damn journalist for crying out loud. That's what really pissed me off.

To be fair, I've had some really successful conversations where the correspondent finally saw the light and was very happy about it and thanked me for the trouble of discussing it with them. And I get occasional emails from people who say that they had not understood the answer even after discussing it with others and reading some other explanations, but my page made it clear to them and they thanked me for it.

I think the MHP is very interesting in itself. But I'm more interested in why people have trouble with it, and in discovering new approaches to help explain it and increasing my comprehension of how people think about it. Secondarily, I'm also very interested in the emotional psychology of being very wrong but very certain, what people will and will not do to discover the answer, and how they react when they discover they were wrong. One thing that has always baffled me is why people won't simply empirically test it. That's what I did: my intuition insisted it was 50-50, but I knew that supposedly that wasn't correct, so I wrote a simple program to see. Once I had that evidence, then the "only" thing left was figuring out where I had gone wrong.

Incidentally, I recall one extremely frustrating exchange where I had suggested writing a little program to test it and the person insisted that doing so wouldn't prove anything because the program would use pseudorandom, not true random, numbers in its calculations. I said, well, okay, that's not correct but even if it were you could make the numbers truly random by connecting your pseudorandom number generator to some real world seed that is truly random. Nope, he said, that wouldn't be random and so it wouldn't prove anything. And anyway, he knew it was 50-50, so why go to the trouble?

posted by Ethereal Bligh at 11:11 PM on March 30, 2006

I think the single most frustrating correspondence I've had over the last 12 years on the MHP was a science writer from the Houston Chronicle (I thinkāis that the big daily there?) who came upon my site and was considering doing a short article on the MHP. Except he thought I was wrong and that it was really 50-50. So after trying the various strategies I've learned to explain the problem, I finally asked him to call up a math prof at one of the Houston universities. How hard would that be? And then he would at least have an authoritative answer, instead of relying on his intuition. But he said he didn't want to call anyone. "Why not?" I asked him. He replied that he didn't care that much. Well, he cared enough to argue about it with me through several emails. And the man was a damn journalist for crying out loud. That's what really pissed me off.

To be fair, I've had some really successful conversations where the correspondent finally saw the light and was very happy about it and thanked me for the trouble of discussing it with them. And I get occasional emails from people who say that they had not understood the answer even after discussing it with others and reading some other explanations, but my page made it clear to them and they thanked me for it.

I think the MHP is very interesting in itself. But I'm more interested in why people have trouble with it, and in discovering new approaches to help explain it and increasing my comprehension of how people think about it. Secondarily, I'm also very interested in the emotional psychology of being very wrong but very certain, what people will and will not do to discover the answer, and how they react when they discover they were wrong. One thing that has always baffled me is why people won't simply empirically test it. That's what I did: my intuition insisted it was 50-50, but I knew that supposedly that wasn't correct, so I wrote a simple program to see. Once I had that evidence, then the "only" thing left was figuring out where I had gone wrong.

Incidentally, I recall one extremely frustrating exchange where I had suggested writing a little program to test it and the person insisted that doing so wouldn't prove anything because the program would use pseudorandom, not true random, numbers in its calculations. I said, well, okay, that's not correct but even if it were you could make the numbers truly random by connecting your pseudorandom number generator to some real world seed that is truly random. Nope, he said, that wouldn't be random and so it wouldn't prove anything. And anyway, he knew it was 50-50, so why go to the trouble?

posted by Ethereal Bligh at 11:11 PM on March 30, 2006

Pretty much, except that under these new rules we're only taking out one sock, so the figure is slightly different. We agree that under the "If there is a red sock in the drawer" condition, all socks are equally likely, and 10 out of the 30 socks are red so we just get a 1/3 chance of a red sock. If we wanted to get 8/27 we should replace step two of your original description with "If there is at one least red sock in the drawer, take a red sock, otherwise take any other sock at random". I only mention this for the sake of tidiness - it seems that we agree on the principles now.

Thanks for a fun exchange! Please drop the dollar in any charity collection box that takes your fancy.

posted by teleskiving at 11:17 PM on March 30, 2006

This thread is closed to new comments.

posted by flabdablet at 10:00 PM on March 21, 2006