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Heads Or Tails? Sugar Or Snails?

After several weeks of heavier items, here is a puzzle just for the fun of it. I was challenged by it, but instead of deriving the answer from scratch, I recognized it as being related to a probability puzzle we have featured before in several modes. Here is the puzzle, followed by the related one that helped me to quickly solve it, and finally the “from scratch” solution. No peeking.

Assume you have a friend you have not seen in years, but you know that in the intervening time he has grown a family with three children. You do not know their names or genders. Your friend invites you to visit. When you arrive, you are met at the door by a boy who says you must be his father’s friend. He is the only child at home at that time. He says that his parents are expecting you and that his siblings will arrive later. He is polite and takes you through the house to the back patio where his parents are enjoying the summer afternoon outside. As you pass through the kitchen, you notice a picture on the refrigerator. It is a girl in school clothes. You realize from her family resemblance that she must be the boy’s sister — one of your friend’s three children.

As you greet your friend and meet his wife for the first time, you wonder whether the third child is a boy or girl. Since you saw the boy and a picture of a girl, he has either two boys and one girl or two girls and one boy. What is the probability of either combination? In the absence of extraneous factors such as a genetic disposition favoring boys or girls, it seems obvious that the probabilities are equal. Is that right?

This is a puzzle that seems more complicated than it is. When you strip away some of the narrative, you might recognize it as a variation of the puzzle of two children (or two coins, whatever) where the probabilities of some feature (boy or girl: head or tails) is independent for each event.

In the more common version (and more easily understood), a man says he has two children. At least one is a boy. What is the probability the other is a boy? Naively one might answer that the probability of either gender is 0.5, but by the conditions of the puzzle, the children could be BG, BG, or BB with the GG combination ruled out. Or the three possible combinations, in two of them the second child is a girl. So the probability of the second child being a girl is twice that of being a boy.

This solution seems counter-intuitive. However, if we change the puzzle slightly and say the oldest child is a boy, then the probability changes. Why?

With that simpler puzzle in mind, we see that the same analysis holds for the case of three children. The possible combinations are BBG, BGB, or BGG with the possibility of GGG being excluded by the condition of having seen at least on boy child. Therefore the probability of the third child being a girl is again twice that of being a boy.

How would this analysis change if your friend greeted you and said, “I see you have met my youngest child” as the boy escorted you to through the kitchen?

Suppose he then said that his oldest child does not play sports, but his daughter likes soccer. What are the probabilities then?

The clues in all these variations depend on whether events are independent or linked in some way. Sometimes a subtle change in wording can significantly change the probability of an event. Clever gamblers can make a lot of money on this.

In response to the interest my original tutorial generated, I have completely rewritten and expanded it. Check out the tutorial availability through Lockergnome. The new version is over 100 pages long with chapters that alternate between discussion of the theoretical aspects and puzzles just for the fun of it. Puzzle lovers will be glad to know that I included an answers section that includes discussions as to why the answer is correct and how it was obtained. Most of the material has appeared in these columns, but some is new. Most of the discussions are expanded compared to what they were in the original column format.

[tags]statistics, probability, chance, puzzle[/tags]

4 Comments

Sherman, I believe you’re mistaken in your analysis. Given that you know there is one boy (whom you’ve met) and one girl (whose picture you’ve seen), the combinations you defined are repetitive. You said the possible combinations are BBG, BGB, or BGG. You know one boy and one girl, so the first two options here (BBG and BGB) are the same - two boys and one girl. I suggest that the possible combinations are really just BGB and BGG since the first two children are known.

From a pure probability standpoint, and assuming that birth rates are truly random between boys and girls (50% each), the probability of the third child being a boy or a girl is one in two, or 50%.

Or, put another way, you need to look at all of the possible combinations (BBB, BBG, BGB, BGG, GBB, GBG, GGB, and GGG). Take away the two that you cannot be true (BBB and GGG) and you’re left with three permutations that have two boys and one girl (BBG, BGB, GBB) and three that have two girls and one boy (BGG, GBG, GGB). Three out of six or, again, 50%.

I have to take exception. In the first example listed, because age is not a factor, order is irrelevant. Hence BGB and BBG are identical (otherwise, you must also consider GBB), hence the remaining available options are [Some combination of one girl and two boys] and [Some combination of two girls and one boy]. The probablilty is equal.

Taken from a different angle, accept that order is relevant. As the order is not age, then it must be the order in which you became aware of them. Thus, as your set already contains “BG” in that order, the remaining two options are “BGB” and “BGG”. Equal probability.

So, the given answer to the “common” question, the posible answers are [A boy or girl]B and BB. Even odds.

Taking on your final two questions, age (and thus order) seems to be becoming a factor now. Going from oldest to youngest, for the first question the options seem to be BGB, GBB, and GGB. But once again, age of the older two children between them is not relevant, so the options are [A girl and a boy in any order]B or GGB. Equal probability.

Last question. Age of the older two children has now become relevant. We must discard one of the two groups that could be combined above [A girl and a boy in any order], and much to our luck, the uncombined group as well. So our options are GBB (discarded), BGB, and GGB (discarded). Even if we didn’t have the verbal cue that our friend has a single girl, we now know that the girl on the ‘fridge is a soccer fan, and is sibling to two boys.

The problem with all questions of this type is they try to consider the question *for all possible couples simultaneously*. But that is not what the questions are considering. The questions are being asked about a single, specific, couple. And for any single couple, the odds governing the gender of the next child to be revealed to you are even. It realy is the coin flip question. If I flip 99 heads in a row, the odds of flipping a hundredth head are even, even though the odds against flipping 100 heads in a row are astronomical.

Simply put, when considering a single couple, once the gender of only a single child is unknown, *and no other significant information about that child is available*, then the odds really are 50/50.

I hereby recind everything I said above, with the exception of the answer to the last problem (which is deterministic). I still don’t quite make the connection yet, but I performed a brute-force analysis of the two-child problem for 4.5 million randomly-generated couples, and it clearly is a 33 percent chance. A co-worker with a math professor father is helping me work through the rest.

BC, I believe you have got to be a geek, in the best possible sense of the word! :-) Maybe more specifically a math geek or statistics geek. 4.5 million simulations! I love it!

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