Wild About Math! Making Math fun and accessible

26Sep/117

Two new videos by James Tanton

After a hiatus of several months Dr. Tanton is making videos again. Here are two new ones.

Lulu has two children. You are told that at least one of her children is a boy who was born on a Tuesday. What is the probability that her other child is also a boy?

The answer will surprise you!

Here is a cute geometry puzzle: Imagine you are an archeologist and have come across just a small section of a rim of an ancient wheel. What size wheel did it come from?

This is a great puzzle to give to geometry students too. Hand out a picture of an arc of a circle and ask if is possible to find the measure of that arc using only basic tools - and them have students actually do it.

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  1. This is exactly the kind of math you shouldn’t try to learn. The “math” of the ambiguous and prosperous tricky problems. It is more interesting to learn the ambiguities of the language that leads to those confusions. So, you go to the supermarket, meet a mother, who turns out to be with a son, she mentions that his son (the one that is with her) has one sibling (that is, she has two children), then you comment: “Most likely (2/3) your other child is a girl and less likely (1/3) your other child is a boy”. You will look like an idiot *and* is wrong. The answer (1/3 vs 1/2) depends on how you find out that the at least one of the children is a boy; as explained here: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox. The second question is even more ambiguous, it depends on how you find out that there at least one boy and how you find out that it was born on a Tuesday.

  2. The Tuesday Boy problem bugged me for a long time, and I’ve written a refutation of it.

    Basically, my problem starts with the question about Mike, and I propose a game. I’ll flip two coins, announce one, and you try to guess the other. In Mike’s problem, we see that given that Mike has at least one boy, there’s a 2/3 chance he has a girl. So the same should hold true for coins. I flip two coins, and I say one of them is heads. So you would then guess tails, and you’d be right 2/3 of the time, right?

    Wrong. Quite easily demonstrable that it’s wrong, too. Given the strategy, you will win if I flip one of each, and lose if I flip doubles. Thus you’ll win exactly 1/2 the time. So what gives?

    It took me a few days to figure out what was going on there. If you’re interested in my solution, I wrote it up here: http://blog.asmor.com/2010/07/12/the-surprisingly-paradoxical-gender-problem/

  3. I really like the videos! ! I will send them to all my friends!

  4. on the Boy-Boy Paradox:
    This is very good representation of the known paradox presented by M Gardner back in 1959 (with a bit of an extra twist at the end, I loved the addition of the Tusesday condition for the full effect).

    The whole thing though it is not true. Please visit Wikipedia for a full explanation (http://en.wikipedia.org/wiki/Boy_or_Girl_paradox) but in brief:
    - The trick is located on the method that is used to solve the second and the third questions. This is done firstly by showing the first question where the model applies.

    Let me explain: In the first question the order is important, as we are choosing the first son to be a boy, therefore it make sense to diferenciate the couple GB from BG, which correctly leads as to a possible outcome of {BG, BB} and so to the 1/2 probability.
    Issue arises in the second question when the order it is not important, but the author still uses the same model as in the first question… so the group of possible outcomes becomes {BG, GB, BB} instead of the correct one which only holds {BG, BB} as the option GB is equivalent to BG if order is not important. Out of this 2 possible outcomes only one is correct so –> 1/2!!! (that makes more sense!)

    Obviulsy the same approach is used with much more spectacular efect in the thrid question…but basically is the same..model is not correct.

    Just as an addition, another way of seeing the same: think of the second problem as composed by 2 problems similar to the question number 1, where the model is correct. In this way you get 2 sets of possible outcomes {Bg, Bb} and {gB, bB}. Please note that Bb and bB are not the same because the separation. when you put them toghether again you get {Bg, Bb, gB, bB} and only 2 of those are good Bb and bB…so 2/4 –> 1/2!!! (again..that makes sense)
    Congratualtions on a great post…it is very enjoyable!

  5. Dr. Tanton, are you familiar with the Monty Hall problem? On a game show, you are offered the choice of three doors. One conceals a new car, but the other two conceal goats. You choose Door #1, but before opening it the host opens Door #3 to reveal a goat. He then offers to let you switch to Door #2. Should you?

    If you use the same tree diagram you used for Mike, you get just one level with three branches in it. Ordering the prizes by door number, the branches represent the cases CGG, GCG, and GGC. The last one is eliminated since we know Door #3 has a goat, so the remaining two doors appear to each have a 1/2 chance to have the car. There is no benefit to switching. But if you look up the literature on this problem, you will find that this is not the right answer.

    To get the right answer, you must add another level to your tree, discriminating which door the host chooses to open. Theoretically, there are now nine options. But if we assume the host never opens either your door, or the door with the car, there are actually only four: CGG2, CGG3, GCG3, and GGC2. Their initial probabilities are 1/6, 1/6, 1/3, and 1/3, respectively. The difference is that only in the CGG branch does the host have a choice of doors to open, and we can only assume he chooses randomly. Once we know that the actual case can only be CGG3 or GCG3, the modified probabilities become 1/3 and 2/3, respectively. Switching to Door #2 will double your chances.

    My point, like alfC’s, is that for Mike you have to might have to add another level to your tree (I say “might” based on your wording; it is definitely true for Lulu). That level discriminates the cases where you know, or are told, that Mike has at least one girl. If you gained that information in an unbiased manner, half of the BG or GB cases will end up with you knowing about a girl. The answer is 1/(1+2P), where P is the chance you will learn about a boy in those cases. It is 1/3 only if P=1, and 1/2 if P=1/2. But just like in the Monty Hall problem, we can only assume it is determined randomly, so P=1/2. The same reasoning leads to Lulu’s answer being (1+12P)/(1+26P).

    What you calculated is the proportion of at-least-one-boy families that have two boys, or the proportion of at-least-one-Tuesday-boy families that have two boys. That isn’t the same thing as probability, unless you learned “at least one” whatever in a biased manner. And, if you disallow repeated names in a family, it turns out that the proportion of Poindexter families that have two boys is slightly *greater* than 1/2, not slightly less as you might think the answer is. This is because a younger-child Poindexter is more likely to have an older brother than an older sister, since that case removes one name from the pool of possibilities for the second son, increasing the likelihood the parents will choose “Poindexter.”

  6. I’m replying here, to Dr. Tranton’s video response to these criticisms on YouTube, because there is no character limit.

    Sorry Dr. Tanton, but “specificity” is a sufficient, but not necessary, condition for the answer to be 1/2. In general, the answer to the “Mike” problem is 1/(1+2P), where P is the probability you would learn “one is a boy” when only one is. The answer to the Lulu problem is (1+12Q)/(1+26Q) where Q is the probability you would learn “one is a boy born on Tuesday” when only one is. Your answers require P=Q=1, which is only true if you sought a family to fit the fact. It is mere coincidence that if only one child fits, that child is “specified.” If both fit, neither is “specified” by the information.

    You might better see this if you change the problem to be about dice: If I tell you that one of two dice landed on a three, are the chances I rolled doubles 1/11 (the answer based on the solution method you used to get 1/3 for Mike) or 1/6? Or something else? If it is 1/11, is it 1/11 no matter what I tell you about one die? And if you say yes, please apply the Law of Total Probability to the results – you should get that the chances are 1/11 even if I don’t tell you anything.

    In fact, these two problems (Mike, and my dice) are variations of Bertrand’s Box Paradox. Usually it gets compared to the Monty Hall Problem because the numbers are the same, but logically these two are better matches. A set of two variables, where each is equally likely to have any one of N discrete values, is chosen at random. You find out what one (unspecified, as far as you know) value is. Arguments can be made that the probability that both (not “the other” as you phrased the Mike question, which implies a specification) have that value is 1/N, or 1/(2N-1). But the 1/(2N-1) answer has to be true no matter what value you learn about, and the law of Total Probability then creates the paradox Bertrand observed.

    The conclusion is that the answer to your Mike question has to be 1/2, UNLESS you are told how the information “is one a boy” was specifically sought.

  7. I don’t disagree with you and I agree with your point. Like Bertrand’s paradox, the question requires a statement of what probability measure comes with the problem. What I don’t think you like is my translation of this into a video to be accessible to folk thinking about this sort of thing for the first time – and maybe I am wrong to have taken the simplistic approach I have taken.

    BUT … I will say that some of the responses I’ve received to the addendum have been good and have people thinking on good lines! For example, one person realises that in the absence of all information about under what circumstances I would be told that at least one is a boy, all bets are completely off. (For example, suppose I am told that I will be told at least one is a boy only by people that have two boys?)

    It sounds like I need to add a third video directly taking the lines you would like me to take. I will happily do that.

    Thanks for being so adamant about this! And I apologise for giving the impression I don’t get it. (Actually, maybe I still have missed your point, and I still don’t get it. Please feel free to let me know.)

    Cheers,

    J.


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