A probability question!

After writing yesterday’s post about probabilities and statistics I remembered a particular question that I would often ask people to illustrate how easy it is to get probabilities wrong. Since I’ve never done a poll before, I thought I would try one and post it here. Very few people are reading this, so I may well get no responses. This question also happens to actually be true for me, but that isn’t really relevant.

Advertisements

14 thoughts on “A probability question!

  1. the question is not that well posed; it’s not clear whether it means ‘at least one is a girl’ or ‘only one is a girl’.

  2. Okay, I guess one could argue that it wasn’t clear. What I meant was at least one is a girl. The other could be a girl or a boy and the question is, “what is the chance that the other is a boy?”.

  3. As expected, not many read this post or even attempted the poll. For those who may be interested, the answer to the question is two-thirds. If you have two children then (if we assume for any particular birth there is an equal chance of having a boy or a girl) there are 4 possible combinations – boy-boy, girl-girl, boy-girl, girl-boy. Since the question said “one of them is a girl” this means that it can’t be boy-boy. Of the 3 remaining possibilities, 2 have boys and hence the chance that the other child is a boy is two-thirds.

    On the other hand, if the question had said “I have two children, the eldest of which is a girl. What is the chance that the other is a boy?”, then the answer would be 50%.

    • No I’m not. That’s the point. If all you know is that I have two children, at least one of which is a girl then the chance that the other is a boy is 2/3. As I pointed out above, if you knew that the oldest child was a girl, then it would indeed be 1/2.

      Another way to look at it is the following. Imagine you took all the families in the country that had 2 children. From that group you take all that have at least one girl. You would find that 2/3 of the siblings in that group would be boys.

  4. “I have two children, one of whom is a girl. What is the chance that the other is a boy?” is a statement made by one of the parents. The assumption you are making is that a parent of BG/GB family makes this statement a 100% of the time, whereas a more reasonable assumption is it is equally likely they would say “I have two children, one of whom is a BOY…….”. So although there are twice as many BG families as GG families, only 50% of BG parents would make this statement, making it a 50/50 proposition.

    I agree entirely with what you say in your last paragraph, but that is not the question you posed.

    • You seem to be criticising the setup of my question. Fine, but this isn’t a post about pedantry, it’s a post about probability. However, I still think your counterclaim is wrong. The whole point of my question is to illustrate the point you make at the end. Assuming that the parents are not lying, then I am only sampling 3/4 of the population of families who have two children. That is the fundamental point. Parents who have 2 boys cannot say “one of my children is a girl”, because that is not true. I agree that those who have a boy and a girl could have said “I have two children, one of whom is a boy” but that isn’t what I chose for them to ask. I didn’t say “You meet someone who has two children, at least one of which is a girl. They will randomly tell you the gender of one of their children, what is the chance that the other is of the opposite gender?” .

      Therefore, if you were to meet someone and they said to you ” I have two children, one of my children is a girl, what is the gender of my other child” what would you answer? Probability says that you should answer boy because 2/3 of those who have two children and who can say “one of my children is a girl” also have a boy. That is essentially what I was trying to illustrate, which you actually seem to understand.

  5. I do understand what you were trying to illustrate, but I think you posed the question imperfectly to get an unambiguous answer of 2/3.

    If you accept the proposition that the parent of a BG family is equally likely to volunteer the information “I have two children, one of whom is a girl..” as “I have two children, one of whom is a boy..” , then the answer to the question posed is 1/2. And, in the absence of any other information, that is the most reasonable assumption to make.

    If you presuppose a bias towards girls such that the answer is 2/3, then the answer to the question “I have two children, one of whom is a boy. What is the chance that the other is a girl?” would have to be zero, as none of the parents of BG families would make this statement.

    This isn’t pedantry. The reason many people get the Monty Hall Problem wrong is because it is frequently poorly stated (the constraints “has to open a goat door”, “must always offer the option to switch” are often omitted entirely).

    “My neighbour has 2 children. I asked him if (at least) one was a girl. He said ‘Yes’. What is the chance the other is a boy?” is unambiguous. Just saying, not criticising.

    • I was thinking a little more about the ambiguity of the question I posed. You did have me convinced, a while ago, that maybe there was an ambiguity, but I now think that isn’t the case. Imagine someone were to walk up to you and say “I have two children, at least one of which is a girl.” With this information, the chance that the other is a boy is 2/3. I don’t think you need to consider the possibility that they could have said something different. As soon as they’ve said “at least one is a girl” 1/4 of the possibilities (i.e., boy-boy) have been removed and hence 2/3 of the remaining possibilities have the other child being a boy.

      I’m not trying to extend the debate, but I think the ambiguity you claim exists actually doesn’t (or at least I think this is the case).

      • Well you think the answer is 2/3, I think it’s 1/2 for reasons I’ve already given ( I think you have to remove 50% of the boy-girl possibilities as well).
        I’m aware a lot of people think the answer is 1/2 for completely the wrong reasons, but that in itself doesn’t make the answer wrong. It’s easy to rephrase the question so that 2/3 is the only correct answer.

      • I understand what you’re suggesting, but I’m now disagreeing. I do agree that it could be rephrased, but I I don’t think – even given the current form of the problem – that it’s correct to discount half of the boy-girl possibilities. If somebody walked up to me and said “I have two children, at least one is a girl. I will give you a prize if you correctly guess the gender of my other child.” Probability says, that you should guess boy because 2/3 of people who have 2 children, at least one a girl, also have a boy. That’s – in my view at least – all that matters. The fact that 2/3 of the people who could have made the above claim could have made a different claim is not relevant.

  6. I realise that you are pointing out a subtlety to the way in which I have posed the question and I do understand your subtlety. I also agree that maybe it could be seen as ambiguous if someone was really going to stop and think “I have to include the possibility that someone with two children, one boy and one girl, may have posed the question differently”. I did consider asking it the way I posed at the end of my last comment (i.e., more like the Monty Hall problem) but thought it wouldn’t illustrate what I was trying to illustrate (I wanted to ask for the probability rather than the options being boy or girl). Also (as I mentioned in the post) it applies to me, hence I asked it the way I did (although I guess I could have said “my neighbour asked me if one of my children is a girl…”).

    Here’s my question to you. Most of those who did the poll (not many addmitedly) selected “half”. Do you think that is because they factored in the possibility that BG/GB parents could have said “at least one is a boy” instead of “at least one is a girl” or because they understood what I was asking, but misunderstood the basic way in which to determine the answer (i.e., they fundamentally misunderstand the Monty Hall problem). Every time I’ve discussed the Monty Hall problem with people their confusion does not seem to stem from a misunderstanding of the process (as you seem to claim above), but a genuine misunderstanding of how to calculate probabilities, which – in a sense – is the point of this post.

  7. Oh I agree with you. I’m sure most of those polled who answered 1/2 did so because they didn’t understand the relative probablities involved (I answered 1/2 BTW, but I guess you suspected that !). Same with the MHP, most people who get it wrong do so because of “2 doors, 1 car, it must be 50/50” syndrome. Still, I have read threads where some haven’t understood the problem correctly because it’s been poorly defined, which leads to pages of pointless arguments

    I wasn’t trying to be pedantic, I was just pointing out a potential for misunderstanding in the way the question was posed.

    Have you considered the Sleeping Beauty Paradox? I’ve read convincing arguments for the 1/3 solution but still don’t fully “get it”.

    Palmer

    • That’s interesting. I had never heard of that before. My gut feeling is that if one considered it in the same way as the Monty Hall problem, the chance of the throw being heads is 1/3. One could argue that 2 times out of 3 when Sleeping Beauty is woken, the throw will have been tails. If she then said “tails” (assuming the question is “what side did the coin land on?”) she would be correct 2 times out of 3.

      I gather that there is debate about this and it may well be an example of where the question is not well posed (possibly intentionally so). It seems that if, for example, there was a prize associated with this. The coin is only tossed once (on the Monday) and either lands heads or tails. Sleeping Beauty gets a prize if she correctly states which side the coin landed on. One could argue that if Sleeping Beauty says “tails” she will win 2 times out of 3. However, she only gets one attempt when heads lands, but 2 attempts when tails lands. I think in this case it is 50:50. She has an equal chance of winning whether she says heads or tails (assuming that she can’t win 2 prizes per experiment).

      Maybe, it’s poorly posed question. Imagine we repeat the experiment 1000 times. The coin will land heads up 500 times and tails up 500 times. Sleeping Beauty will therefore be woken 1500 times (500 heads, 1000 tails). If she’s asked the question every time, what side did the coin land on and she replied “tails” she would be correct 2/3 of the time. If, however, she can only win a prize once per experiment then she would only win 500 prizes (so in a sense the probability is 50:50).

      I guess I don’t really get it either.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s