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Probabability Paradox* — part 2

August 27, 2013

You may see this title and immediately think that this is the second part of a two part post; you’d be wrong. This title is an example where intuition can be wrong — or an example of sneaky buggery. Or something? This is to prepare you for a little probability problem. At first, probability sounds like it must be the most intuitive thing ever; if something happens 50% of the time then there’s a good chance that it will happen in around 50 of 100 trials. But it turns out that probability can actually be quite the sneaky bugger. We’ve already seen the Monty Hall problem here, which seemed like quite a simple little probability problem but turned out to be quite counter-intuitive. Today we’ll look at another little probability paradox* that I was recently reminded of.

*Not actually a paradox.

*Not actually a paradox.

First let’s look at a simple problem.

Mrs Robinson — here’s to her — has two children; at least one of whom is a son. What’s the probability that she has two sons?

How stupidly simple does this problem sound? It’s about 50%, right… We know that one of them is a son, so the other one has a 50-50 chance?
Of course not! Probability is far too sneaky for that.

Since we know that she’s got at least one son, there are three possibilities:

  • She has two sons,
  • The older sibling is a son while the younger is a daughter,
  • Or the older is a daughter while the younger is a son.

We’re assuming here that they weren’t at the the exact same time, but this is only so that I can easily talk about the different scenarios. You can convince yourself that these outcomes are equally likely by considering the 4th possibility; if we didn’t know about one of her sons then she could have two daughters. These 4 outcomes are clearly equally likely before we have any knowledge of a son. By saying that at least one is indeed a son, all we’ve done is rule out the fourth option.

Since there are three equally likely outcomes, there is a one in three chance that she has two sons! Take that, intuition!

So now let’s change the problem ever so slightly.

Mrs Robinson has two children; at least one of whom is a son who was born on a Tuesday. What’s the probability that she has two sons?

Oh get stuffed! What does Tuesday have to do with anything? These are probably things that you’re thinking to yourself. The answer is obviously the same as before… right… isn’t it?

Ha! Probability and it’s swift kick to the brainbits strikes again.

Let’s write down all of the (equally likely) possible cases again:

  • The older is a son born on a Tuesday and the younger is a daughter.
  • The older is a son born on a Tuesday and the younger is a son born on a Wednesday.
  • The older is a son born on a Thursday and the younger is a son born on a Tuesday.
  • The older is a son born on Tuesday and the younger is a son born on a Thursday.
  • … There appears to be a whole lot of other possibilities in this scenario.

The difference here, is that we are being more specific about the boy now. If we had said “This is Johnny, Mrs Robinson’s son; he’s got one sibling.” then it would be obvious that there was a 50-50 chance of his sibling being a brother. The more specific we are about her son, the closer the probability gets to a 50-50. If we add up all of the cases in the Tuesday example, we’d actually find that there is a 13/27 chance of Mrs Robinson having two sons.

Convince yourself! Write up a table of the possibilities and check it out.

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From → Math

6 Comments
  1. PalmerEldritch permalink

    “Mrs Robinson — here’s to her — has two children; at least one of whom is a son. What’s the probability that she has two sons?”

    The question is ambiguous. If Mrs R was randomly selected from all 2 children families that had at least 1 boy then the answer is 1/3. If Mrs R was selected from all 2 children families and one of her children was randomly chosen then the answer is 1/2.

    • I had the same thought when I first heard this problem, but there actually isn’t a problem with the question. She’s randomly selected from the entire population and then as you are given more information, the probabilities change.

      If I flip two coins, what’s the probability that both are heads? Then what if I peak, and tell you that at least one of them is a head? The physical situation hasn’t changed, but when you have different information then your expectations must change.

      Thanks for the comment though 🙂

      • PalmerEldritch permalink

        The answer to your coin toss problem is 1/2.

        As to Mrs R, if she’s randomly chosen from all 2 children families and the gender of 1 her children is given, then the answer to that problem is 1/2 too.

        It can only be 1/3 if it’s been predetermined that if Mrs R has a boy then that is the information that is given (and if she has 2 girls then what?). Since the problem doesn’t mention that precondition it can’t be assumed.

        Thanks for your reply 🙂

  2. The answer to the coin toss problem is again 1/3. There are 4 possibilities, all of which are equally likely: HH, HT, TH, TT. By telling you that at least one of them is a head, all I’ve done is said that it’s not TT. So it must be one of the three remaining (equally likely) options.

    The Mrs R problem is identical. The key is in the fact that we aren’t specifying which of the children is known to be male, just that one of them is. This is like in the coin problem, saying that the first coin is a head, rather than at least one of them being a head. You can see that the answer does become 1/2 in this case.

    Have a read over the post again and if it still doesn’t click, you should be able to find somebody else’s explanation on Google easy enough. It’s a well-known problem 🙂

    • PalmerEldritch permalink

      Perhaps you didn’t understand my previous comment (or maybe didn’t read it), so let’s try another approach.

      I’ll flip 2 fair coins:,
      1) What’s the probability both coins land the same side up?
      2) What’s the probability both coins land the same side up if I truthfully tell you 1 landed heads?
      3) What’s the probability both coins land the same side up if I truthfully tell you 1 landed tails?

      If your answer to 2) and 3) is different to your answer to 1) then I’ve got a betting proposition that you might be interested in. 🙂

      You fail to recognise that there is a difference between the set of cases where the facts, as presented, are true; and the set of cases where those facts would be what gets presented. If the coins land heads and tails there are two possible “truths” I could tell you, but I only tell you one of them.

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