A tale of two probabilities

Introduction

Moderator. Two members in our Statistics Department – Tom and Jerry – have been scrapping for the last few weeks. At the centre of their controversy is this simple problem:

Tom says the answer is ½, and Jerry insists it's ⅓. To put an end to their squabbling, they've agreed to this open debate. May the better one win! We'll tape and transcribe everything. I'll add some relevant references in places for the sake of our students. Tom, you won the coin flip, so you go first.

The debate

Tom. Thank you. Actually, this shouldn't be much of a debate, because Jerry and I have already agreed that having kids is like flipping fair coins. That is, we are assuming that the chance of getting a boy is the same as getting a girl – exactly ½ (this is approximately true in reality). And, as with coin-flip outcomes, we assume the genders of different children are independent. So who cares whether we notice that one is a girl? The other kid is also a coin flip – no disrespect intended – so obviously the chance that it's also a girl is ½. It seems a little hard to convince my esteemed colleague of this down-to-earth logic! (Carlton and Stansfield 2005)

Jerry. Your logic is more simplistic than down-to-earth. To solve this problem you need to go back to the sample space. The sample space for the original experiment of picking a random two-child family and observing the sexes of the children is {BB, BG, GB, GG}, where the first letter refers to the first-born, the second to the second-born. We stipulate that before getting any additional information, the probability of each point of this sample space is ¼, but as soon as we learn that one is a girl, the problem becomes one of conditional probability. The conditioning event is {BG, GB, GG}. Within this event there are three possible outcomes of equal probability, one of which is GG. This leads to a conditional probability of ⅓ for GG.

Tom. But I can play the sample-space game too! We both agree that having a child is like flipping a coin, so let's just state the problem in terms of heads and tails:

Seeing heads first tells us that both TT and TH are eliminated. The condition is the event {HH, HT}, within which there are two equally likely possible outcomes, one of which is HH. This leads to a conditional probability of ½ for HH.

Jerry. Whoa! That's not a fair translation. In each of the sample points HH, HT, TH, TT, the first letter stands for the outcome of the first flip, and the second for that of the second. Similarly for BB, BG, GB, GG. You solved your coin problem perfectly, but that doesn't touch the problem we're debating about! The big thing is, we never knew whether the girl we noticed was the younger child or the older. In your coin problem, we did – we knew that the heads we saw was the “older” flip, which makes the problem straightforward. If we'd been told that the girl we saw was the older one, then I'd agree with the answer ½, and we wouldn't be having this debate. (Bar-Hillel and Falk 1982; Gardner 1959, p. 51)

Tom. I see your point, but I see a potentially fatal weakness in your solution too. In it, you conveniently assume that all three possibilities BG, GB, GG are equally likely. But is that justified? Seeing a girl indeed removes BB, leaving three points, but I can't believe that assigning them equal probabilities models this situation. Maybe observing a girl imparts different probabilities to the three options. You have to ask exactly how did this observation come about?

Jerry. Hah! You see that I've already won the debate, so now you're kicking up dust to confuse the issue. What, exactly, is your point?

Tom. We're both fans of Feller's classic text, aren't we? He explains at the very outset that probabilities refer to events which are outcomes of well-defined real or conceptual statistical experiments. So when you assign a probability to GG, you need to consider the actual experimental mechanism that gave rise to the event a girl was noticed. You never bothered to tell us what the mechanism is; how was the existence of a girl ascertained? This could totally sink your boat. (Feller 1957, pp. 7–9, 22; Fisher 1934; Nickerson 1996)

Jerry (not expecting that from Tom, and now slightly defensive). Well, of course Feller and Fisher are right. I suppose one reasonable assumption could be that the girl we saw was an entirely random choice out of the couple's two children. We could have met the girl in town with her mum, who always decides at random which of her two children will accompany her. Note also that the chances that this meeting would take place are immaterial. To play it safe, let's agree to define the conditioning event precisely as “One child, chosen at random from the two-child family, was observed and it was a girl.” (Glickman 1982)

Tom. Where does that get you?

Jerry (turning to the blackboard). OK, let's denote that conditioning event by g. The complementary event, denoted b, is the observation that the randomly chosen child was a boy. Our sample space now consists of eight ordered triplets in which each of the previous four points is intersected with either g or b. These eight sample points are not equiprobable anymore (e.g. the probability of GGb is zero, whereas that of GGg is ¼). However, for practical reasons I'll skip working with that detailed sample space, and employ directly the resultant Bayes' formula for calculating the desired conditional probability. This problem is paradigmatic for applying Bayes' theorem. The prior equal probabilities of BB, BG, GB, GG are all ¼, and, Bayes' formula gives us their posterior conditional probabilities, given g:

Similarly, we find that P(BG| g) =P(GB|g) =¼ and P(BB|g) = 0. Egad! That would make the problem's answer ½, wouldn't it?! (Falk 1993, Problem 2.4.11)

Tom (smiling triumphantly). And what about your “convenient” assumption that GG, BG and GB are all equally likely? Instead, the probabilities are ½, ¼ and ¼.

Jerry. Well, you were the one who said we need to consider the mechanism giving rise to the event a girl was noticed, so I just quickly chose something to fill the bill. My mechanism of the mum always randomly choosing which child to take was a bit rash.

Tom. You got a better one?

Jerry. Look, there are probably dozens of background scenarios that could potentially affect the answer. (Nickerson 1996)

Tom (tauntingly). Choose one for our audience.

Jerry (thinking, thinking). Wait a minute. What about this? Suppose I learn that a two-child family has at least one daughter by meeting the parents in a girl-scout parents' meeting. The a priori uniform sample space is again {BB, BG, GB, GG}. My conditioning event is g'= {BG, GB, GG}. Our question is P(GG|g')=? Clearly,equals the probability that both children are girls, namely ¼, and the probability of at least one daughter is ¾. Therefore

(Loyer 1983)

Tom (puzzled silence).

Conclusions

Moderator. I never thought the debate would go like this! It looks like you guys have just shown that the right answer should be “Not enough information to decide” since the background mechanism isn't specified. Once the scenario that yielded the observation is spelled out, it determines the answer, and different underlying processes may end up in different answers. Both of you are apparently right (applause from the audience).

A graduate student from the department. Gee! Leafing through Feller's book while the debate was going on, I was delighted to find a problem concerning families with two children that is pertinent to this discussion. Paraphrasing Feller's words to fit our case, he in fact maintained that ½ answers the following question: A girl is chosen at random and found to come from a family with two children; what is the probability that the other child is a girl? The answer ⅓ might refer to a card file of families that have at least one girl, from which one card is chosen at random, and the question is about the probability of two girls; whereas the previous question refers to a file of girls. (Feller 1957, p. 107; Glickman 1982; Wikipedia on-line 2009/2011)

I wonder, however, which of the scenarios – that of meeting the girl and her mum in town, or that of meeting the parents in a girl-scout parents' meeting – makes more sense in real life. (Falk 1993, Problem 2.4.11)

Professor of Statistics in the audience. This is a nonmathematical question, which is out of the scope of this mathematical debate. I see the stories' main contribution in outlining possible directions of occurrences which may lead to probabilities of either ½ or ⅓ for GG. It is important to note that the two scenarios are uncertain events. Yet the question of the real-life chances of these meetings is beside the point. No matter how unlikely they are, they reduce all the relevant probabilities by the same factor, and the relative probabilities of BG, GB, and GG, which determine the sought conditional probability of GG, remain unchanged.

Such scenarios can only approach the mathematical model, or better said, allude to it. A real mother would not flip a coin to decide which child will go out with her, even if she tries to be fair; and parents who have two daughters might be more likely to attend a girl-scout parents' meeting than parents of one daughter. It is barely possible to tell a real story in a way that makes the conditional probability of GG exactly½ or ⅓. The attempts to invent scenarios that match the models perfectly are commendable, but it is a lost cause. Feller solved it the easy way by suggesting randomly drawing a card from a file of cards of either girls or families with at least one girl. He was right, but his procedures fit the models because they are mathematical to start with. Still, I believe the two scenarios serve the didactic purpose of highlighting, in principle, the difference between procedures that lead to the two answers.

Department member in the audience. This problem has been intensively discussed in statistical and psychological writings. I've taught probability and statistics for a long time, and my students get awfully familiar with sample spaces. But every time I get to the above problem, the conditioning event {BG, GB, GG} turns into a red herring. Given that the family has at least one daughter, and even after being told that the one girl had been observed at random, they usually continue to assign equal probabilities to BG, GB, GG and get ⅓. They are not alone in this; recent research by Falk and Lann found that students widely assume equal probabilities of all outcomes in diverse contexts, including cases in which this assumption is definitely not appropriate. This tendency is not easy to fix! (Falk and Lann 2008)

Extensions

Another audience member. If there's not enough information to decide a two-child problem, then for a family with three children, things would probably be just as bad.

Moderator. That seems easy enough to believe. However, I think it would be helpful to look at an example. How about something like this?

Any reactions?

Jerry. We could begin by eliminating all cases with two or three boys from the sample space of the eight possible birth-ordered triplets. That would leave {GGG, GGB, GBG, BGG} as the conditioning event. If we can justifiably consider these four possibilities equally likely, then the probability of GGG is obviously ¼. But if you know that at least two children are girls as a result of seeing two randomly chosen children, then GGG is the most likely to yield this observation. Using Bayes' theorem again: let G₃ denote the event that all three of them are girls, and let g₂ denote the event of seeing that two randomly chosen children are girls. Then

Tom. Why not go even further, replacing 3 by n? Extending the above, Bayes' formula gives

Wow (getting excited)! Given that n-1 random children are girls, the probability that the nth child is a girl is ½. But this is just what I told you at the beginning! This is the essence of the independence between the outcomes. Who cares whether we notice that n-1 random children are girls, the gender of the nth child is also a coin-flip outcome. In fact, the answer ½ does little more than stating the obvious. We have come full circle in finding what should have been self-evident in the first place.

My lesson from all our deliberations is that one's assumptions should be transparent, and the statistical mechanism behind the scenes unequivocally explicated.

The curtain falls

Moderator. This has been a wild ride. Thank you Tom and Jerry, and all other participants for a lively debate and instructive conclusions. (I suggest Falk 2011 as extra reading).