# Eliminating the irrelevant

by on 10 June 2010 · 23 comments

After going to all the trouble of explaining the “classical” analysis of the Tuesday’s Child problem, here I will explain why I think that analysis ends up with the wrong answer.

This will get into the nuts and bolts of the mathematics of probability again, but I will start with some of the questions which led me to rethink this problem.

In her Math Blog, Tanya Khovanova approaches the problem in a very interesting way. Paraphrasing Tanya, if Mr Smith tells you that one of his two children is a boy born on Tuesday, you will (if you accept the classical reasoning) revise the probability of two boys from 1/4 to 13/27. But, you would do the same if he said the boy was born on Wednesday, or indeed any day of the week. So if he said he had a boy born on **@day (his voice was muffled and you couldn’t quite catch which day he said), wouldn’t you then revise your probability to 13/27 too? After all, you’d make that revision regardless of what day he actually said! If so, couldn’t you then revise the probability to 13/27 even if he doesn’t mention a day of the week and just said he had a boy? This clearly makes no sense and suggests there is a problem with the classical analysis.

The crux of the problem is that if Mr Smith is volunteering information, he could just as easily have spoken of a day other than Tuesday. Indeed, he may have said ‘I have two girls’, ‘Anyone for tennis?’ or ‘The UK doesn’t have a solvency problem the way Greece does’. The situation is very different if you are asking the questions of Mr Smith. If you ask him whether he has a boy born on Tuesday, he is restricted to saying ‘yes’ or ‘no’. At least, that’s the case if you allow the one big assumption I am making here, namely that whatever Mr Smith says, he is telling the truth (even about macroeconomics).

So, exactly how the information is revealed makes a difference. That in itself is well understood, but at this point many people will simply give up saying that there are too many options open to Mr Smith so it is impossible to come to any sensible conclusions. So, they will either give up on the problem or rephrase it to one that can be answered, such as where you ask the questions of Mr Smith, as I did in the Tuesday’s Child post. But, if you rephrase the problem, you are solving a different problem and giving up on the original one. What I will argue here is that all those options open to Mr Smith do not matter at all. Mathematically we can eliminate “irrelevant” options and come up with a perfectly defensible solution to the problem.

To ease into the mathematics, I will go back to Gardner’s simpler two boys problem:

Mr Smith says, ‘I have two children and at least one of them is a boy.’ What is the probability that the other child is a boy?

Before trying to answer that, I will instead ask if Mr Smith had two boys, what is the probability that he would say ‘I have two children and at least one of them is a boy’. It is impossible to say: there are so many other things he could have said instead. But what if I asked about the probability that Mr Smith would say ‘I have two children and at least one of them is a boy’. Again it’s impossible to say, so instead I will ask how does that probability compare to the probability of him saying ‘I have two children and at least one of them is a girl’? I would argue that, by symmetry, these two probabilities have to be the same (whatever they may be). As soon as you accept that, the classical solution falls apart.

Now for the mathematics. To save a bit of space I’ll use X to denote the event “Mr Smith says, ‘I have two children and at least one of them is a boy’.” and I will write Y for “Mr Smith says, ‘I have two children and at least one of them is a girl’.” In mathematical terms, my symmetry argument is that $P(X | BG) = P(Y | BG)$, even though we don’t know what either of these probabilities are. This is the sort of reasoning I will use to calculate the conditional probability:

$P(BB | X) = P(BB)\frac{P(X | BB)}{P(X)}$

Even though I will not be able to calculate either the top or the bottom term of this fraction, I will be able to calculate the ratio. To do this, I will use an equation I call “eliminating the irrelevant”. For any events A, B and Z, if A is a subset of Z and Z is independent of B then a bit of algebra will show that.

$\frac{P(A | B)}{P(A)} = \frac{P(A | B\cap Z)}{P(A|Z)}.$

I can apply that to the current problem by setting $Z = X\cup Y$ as long as Z is independent of BB. While we have not been told that is the case, it would seem to be an extremely reasonable assumption. If you think of all the other things Mr Smith might have said, is there any reason to assume that he would be more likely to say something other than X or Y simply because he had two boys rather than two girls? I don’t think so. It’s certainly possible that there could be a link, in much the same way that a coin could be biased, but in the absence of any other information we would start by assuming heads and tails are equally likely.

Note that Y by itself is not independent of BB, nor is X. What I am arguing is that the fact that Mr Smith has two boys (BB) does not affect the likelihood that he will say either X or Y as opposed to something else entirely.

The formula above thus allows us to calculate the conditional probability we are after as follows:

$P(BB | X) = P(BB)\frac{P(X | BB\cap Z)}{P(X | Z)}.$

In one fell swoop, we have eliminated the irrelevant alternatives for Mr Smith and now have something we can work with, even though we will never know what the probability of X actually is!

The top line of the fraction is straightforward. If Mr Smith is restricted to X or Y and has two boys, only X is possible, so $P(X | BB\cap Z) = 1$. The bottom line requires a bit more work. As in the last post, we can break it up into disjoint alternatives:

$P(X | Z) = P(X | Z\cap BB) P(BB) + P(X | Z\cap BG) P(BG) +P(X | Z\cap GB) P(GB).$

Note that I have taken the liberty of dropping one term here as I know that $P(X | Z\cap GG)=0$. By symmetry, since Mr Smith is restricted to X and Y, $P(X | Z\cap BG)=\frac{1}{2}$ and the same applies for the GB term. So, we now have

$P(X | Z) = 1\times\frac{1}{4}+ \frac{1}{2}\times\frac{1}{4} + \frac{1}{2}\times\frac{1}{4} = \frac{1}{2}.$

Now we’ve got everything we need and so

$P(BB | X) = \frac{1}{4}\times\frac{1}{\frac{1}{2}} = \frac{1}{2}.$

So instead of the classical conclusion that the probability that Mr Smith has two boys is 1/3, we have arrived at a figure of 1/2. The key to this result is that when we ask Mr Smith whether he has at least one boy, the probability that he says ‘yes’ is the same whether he has BB, BG or GB. When he makes an utterance of his own accord, although we don’t know what the probability values $P(X | BB)$, $P(X | BG)$ or $P(X | GB)$ are, I am contending that $P(X | BB)$ is double the other two (because in those cases he could have volunteered information about a girl). So, when Mr Smith volunteers that he has at least one boy, he is giving more information to us than if he simply answers ‘yes’ to our question.

The same reasoning will also give a probability of 1/2 for two boys in the Tuesday’s child case (there you would set Z to cover the 14 options for Mr Smith of saying one of boy/girl and one of Monday/Tuesday/Wednesday, etc).

To sum up, if Mr Smith volunteers ‘I have two children and at least one is a boy,’ then the probability that he has two boys is 1/2, whereas if you ask him ‘do you have two children and at least one boy’ and he answers ‘yes’, the probability that he has two boys is 1/3. Likewise, if he volunteers that he has a boy born on a Tuesday, the probability that he has two boys is 1/2 but if you ask him whether he does and he says yes, the probability of two boys is 13/27.

This gives the satisfying (and, I think, intuitive) result that the day of the week does not matter and Tanya’s paradox of a muffled word changing the probabilities is resolved.

### Possibly Related Posts (automatically generated):

1 Ramanan June 11, 2010 at 2:27 am

Once a rigorous proof is given, things become easy to see.

Imagine millions of Mr Smiths appearing and asking you the question. Half of the ensemble of Mr Smith will have a girl as the other child and a half of them a boy. The trick is that none of the Mr Smith in the ensemble is asking “I have daughter … ”

Can’t one simply the proof? We need to evaluate P(BB|BX) which is 1/2.

2 Stubborn Mule June 11, 2010 at 3:35 pm

Ramanan: I’m not sure what you mean by P(BB|BX) there.

3 Mark L June 11, 2010 at 7:18 pm

Another interesting post from the Mule, but alas this time I have to disagree with the conclusion. What you’ve done here is to speculate on what Mr Smith might have, could have, or would have said. True, your speculations do sound reasonable, seductive even: perhaps he was indeed equally likely to say “at least one of them is a girl” as to say “at least one of them is a boy” when he had a boy and a girl. But it is still speculation. And by speculating in this way, we can make the answer come out to almost any probability we like.

For example, I could suggest that Mr Smith was equally likely to say “at least one of them is a material girl”, “at least one of them is a girl who just wants to have fun” or “at least one of them is a boy” (since he’s not a Rolling Stone baby-boomer or a Triple J gen-Yer, and who ever heard of a “material boy”?). Adjusting the definition of Z for this, and following your reasoning about assuming coins aren’t biased, makes the answer 60%.

Or perhaps Mr Smith comes from a society where it is taboo to mention having two sons, which is why he leaves it so vague, and thus he would simply say so if he had exactly one boy. Then the answer is 100%.

For (almost) any answer I’d like to have, I can invent a speculation about Mr Smith’s counterfactual behaviour that yields that answer (so long as it is no smaller then the classical answer of 1/3). The fact is that we observed Mr Smith to say the original sentence, and we have no idea about what he might have said otherwise. The classical approach is to assume that truthful agents always make either the statement they did make (when it’s true) or otherwise say nothing, unless the wording indicates that we should consider a more random behaviour.

So your speculation about Mr Smith has introduced a source of randomness that isn’t in the original puzzle statement, and by introducing it you’ve changed the answer. It’s a neat trick, and a particularly attractive one in this case since it means that “irrelevant options” get cancelled out. However, as you point out, rewording the problem (so that you ask the questions of Mr Smith) brings all the paradoxes of irrelevant options straight back again. So you haven’t resolved those paradoxes in general — you’ve just banished them from this particular wording of this particular puzzle. And you’ve done it by introducing an extraneous source of randomness, a technique that can give almost any answer you choose.

So thanks for tempting solution, but no thanks. I think I’ll stick to the classical answer of 1/3.

Cheers,
Mark

4 Stubborn Mule June 11, 2010 at 7:26 pm

Mark: I could (almost) understand if your critique would lead you to say that there is no solution, but why accept the classical one? If your concern is all the possible counterfactuals, why is restricting to ‘I have at least one boy’ or ‘I do not have at least one boy’ make sense? And how do you deal with Tanya’s muffled day-of-the-week paradox?

5 Stubborn Mule June 11, 2010 at 7:30 pm

Mark: I would also say that your material girl example is not convincing. While I would happily say that symmetry gives reasonable grounds to a prior for rolling a dice of a probability 1/6 for each side, I would not say there is any natural prior to assign probabilities to whether a man in the street would prefer apples to oranges. My choice of Z strikes me as in the spirit of the dice, yours in the spirit of the fruit.

6 Mark L June 11, 2010 at 7:35 pm

Mule: The classical solution does not introduce any new sources of randomness that were not clearly part of the puzzle description. That seems like a fair approach, given that introducing new random sources changes the answer.

I don’t “deal with” Tanya’s muffled day-of-the-week paradox — I recognise it, and other phenomena of its kind, as one deficiency of the Bayesian probability framework. How do you deal with it, in the case that questions are posed to Mr Smith?

7 Stubborn Mule June 11, 2010 at 7:45 pm

Mark: does it really arise when you ask the question? The mother of all assumptions made here is that all statements are truthful. If Mr Smith cannot hear you, he cannot answer so must ask you to repeat it. If you allow for untruthful (or mistaken answers), you’re in a whole other world of pain!

8 Mark L June 11, 2010 at 7:56 pm

Mule: Agreed. Unless the puzzle wording indicates otherwise, we assume that people are honest (and accurate and informed). Questioning this is clearly out of scope.

To make the muffled word version speculation-proof we could simply ask the question through an interpreter, who tells us afterwards that he thinks the days of the week were probably translated incorrectly and that Mr Smith could equally well have understood the question to be any day of the week.

9 Stubborn Mule June 11, 2010 at 8:39 pm

Mark: in which case I don’t see how Tanya’s paradox arises in the case where you ask the question.

10 Danny Yee June 11, 2010 at 8:50 pm

This debate is not over probability but over the interpretation of puzzle questions.

If presented with this in an exam or a quiz, I would go with Mark’s approach. However if presented with this in most “ordinary” contexts I can imagine, I think I’d go with Sean’s interpretation. Let’s say it comes out in passing in a conversation that someone has two children and has a son whose birthday has some property. I think in “most” conversations, at least in English – can we generate a measure over a corpus here? – the context would be that our speaker would be just as likely to have mentioned a daughter, or other days of the week, if they were the case.

11 Ramanan June 11, 2010 at 9:07 pm

Sean,

By P(BB|BX), I meant: probability that both are boys, given one is a boy and the other unknown. I used X as “unknown”, its not the same X you used.

So what I am saying is that the question is just reduced to asking, what is the probability that the other is a boy and its half. Such solutions (mine) are tricky and even if correct, they appear correct only after a proper solution is given.

12 Stubborn Mule June 11, 2010 at 9:08 pm

Danny: what if Mr Smith made the statement to you and someone (who couldn’t have rigged things) offered you a bet on Boy-Boy. What odds would you take?

13 firefly June 11, 2010 at 9:50 pm

Hello Mule:)

I have spent my spare time trying to independently reason why in the simplified two child probability puzzle in the classical solution we eliminated the GG and not the GB and BG false combinations to arrive at 1/3 as the answer with classical reasoning? Instead of 1/4? I was wondering why it was considered to be good reasoning to separately eliminate just one known false combination from the probability calculation in the classical solution?

If we don’t eliminate any of the false combinations we once again arrive at 1/4 or if we eliminate all the extraneous variables even eliminating the known sibling himself from the start and we only try to calculate the probability for the unknown child we arrive at 1/2.

Not considering myself a good mathematician this has been somewhat of a challenge but very interesting as an exercise in reasoning. When I came back to ask this question I noticed that you have already gone a long way to answering it for me he he :)

You have great timing :)

Cheers

14 Mark L June 11, 2010 at 11:46 pm

Mark: in which case I don’t see how Tanya’s paradox arises in the case where you ask the question.

Mule: You aren’t called S. Mule for nothing!

To make a puzzle that, even under your interpretations, generates Tanya’s paradox, we proceed as follows: You pose the question “Do you have two children, one of whom is a son born on a Tuesday?” to Mr Smith through an interpreter. He responds and the interpreter translates his response as “yes”. Mr Smith leaves and at this point you think the probability of two sons is 13/27. But then the interpreter says ‘Hmm, there may have been a translation problem there. I definitely used the word for one day of the week, and Mr Smith understood me. But I’m not sure it was Tuesday. Maybe it was Wednesday.’ You consider this and realise that if Mr Smith understood Wednesday then everything is essentially the same. The probability is still 13/27. But then the interpreter says ‘Well actually, I really don’t know which day of the week I said. I always get them confused. We don’t have these stupid weekdays in my native language’. You ask the interpreter what his best guess is as to which day he said and he responds that ‘It could equally well have been any day of the week’.

Now no matter what day you think Mr Smith understood, you will always get 13/27. But all you really know is that he has two children and one is a boy, which gives a probability of 1/3. The paradox is alive and well, living happily in the sunny exile of another rewording.

15 Stubborn Mule June 12, 2010 at 1:55 pm

Firefly: I’m glad my timing is good! I think I will have to take a break from probability for a post or two before getting back to Monty Hall and the envelope though.

Mark: yes, I pride myself on my stubborn-ness! I don’t think that the scenario you pose does actually create a paradox. It’s just that the questioner has reasoned incorrectly. The key point is that asking questions limits the options for Mr Smith. If we don’t know what has been asked, then options have not been limited. Assuming that the translator has no difficulty with the word ‘boy’, but doesn’t know which day was used, you are back to the classical version of the simple Gardner problem and the probability of two boys is 1/3 (it’s 1/2 if there’s also confusion about the words ‘boy’ and ‘girl’). If the uncertainty was just whether the day used by the translator was ‘Tuesday’ or ‘Wednesday’ then we’ve restricted Mr Smith’s options to these two days. This is not the same as restricting to one or other of these days. In fact, it’s not too hard to determine that the probability of two boys is 6/13. When we are asking questions, if we just ask whether Mr Smith has one boy, the probability of two boys is 1/3. The more restrictive the rest of our question is (i.e. the closer it gets to identifying one of the children), the closer the boy-boy probability gets to 1/2. An extreme example is if we ask ‘Is your elder child a boy?’ and Mr Smith says ‘yes’. Since restricting to only Tuesday or Wednesday is less restrictive than restricting to Tuesday, the result we get, 6/13, is closer to 1/3 than 13/27 is (the probability for the classical Tuesday’s child puzzle).

So, I still see no paradoxes arising with a Tuesday’s child probability of 13/27 when we ask questions and 1/2 if Mr Smith volunteers the information.

16 evo June 15, 2010 at 12:11 pm

Not sure if this has been previously raised, but an application of a similar probabilistic question arises in bridge, and the so-called “principle of restricted choice”. For this and other “bridge paradoxes”, have a look at http://bit.ly/aLoWUD.

17 Tommo1 June 16, 2010 at 5:29 am

I think that we are making the first part too complicated!
‘I have two children and at least one of them is a boy.’
From our choices BB, BG, GB, GG, if he is referring to the first boy then the choices left are BB or BG. This leaves it as probability of 1/2 fortwo boys.
However, if the one he is referring to is the second boy then we are left with BB or GB. Probability of two boys still = 1/2.
It doesn’t matter if we ask him or he volunteers the information the probability is still = 1/2. He cannot refer to the one boy and simultaneously mean both the first and second individuals as one situation excludes the other.

18 Tommo1 June 17, 2010 at 7:43 am

I have to stop thinking about this constantly, my wife is starting to notice I answer, “Yes dear.” to everything and anything she says. I must pay attention apparently and stop day dreaming.

19 Stubborn Mule June 18, 2010 at 11:53 am

Tommo1: that Wikipedia article came up on the Stable. I hadn’t seen it before writing these posts, but it certainly gives further illustration as to the level of controversy the puzzle has generated!

20 carrpin December 23, 2011 at 3:10 pm

The answer is 1/2 in all cases. Never 1/3. The “classical” explanation is flawed.

21 Stubborn Mule December 25, 2011 at 7:44 am

@carrpin: do you have an argument to go with your assertion?

22 carrpin December 25, 2011 at 1:52 pm

In the end, it has all the makings of a trick question, leading many to think that the obvious answer must be incorrect. In simplest terms, there exists a child; that child is either a boy (50%) or a girl (50%), regardless of the gender of the “known” child, the birth order of the two children, and, most definitely, the day of week that the “known” child was born.

When trying to break down the question into possible “outcomes”, so many (vos Savant et al) included either too few or too many possible combinations. So Mr. Smith has 2 children and one is a boy. What is the probability that the other child is a boy. The possible combinations are:

B/B
B/G

And that is all. Order does not matter.

Now, if one is so inclined to analyze it further, the same answer holds true. Most have presented the following combinations:

B/B
B/G
G/B
G/G

Eliminating G/G, leaves 3 possibilities, with only one including a second boy. 1/3. However, if one is going to include both B/G and G/B, implying birth order, then the possibilities must be:

B/b
b/B
B/G
G/B
G/G

Where B = the “known” boy. And eliminating G/G leaves 4 possibilities, with 2 including a second boy. 1/2.