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Probability Fallacies

January 14, 2008

So, I was teaching probability to my middle school group, using the Challenge Math chapter, and following up with some MATHCOUNTS problems I found here and here and here. (The latter, I discussed previously here.)

We discussed the usual rules about how AND means you multiply probabilities, and OR means you add them.

But then, when I tried to extend some of the MATHCOUNTS problems, I ran into some problems with those simple rules.

Consider this problem:

A football stadium has 7 gates for fans to enter or exit. The gates are numbered consecutively from 1 through 7. The probability of entering or exiting through each gate is equal. What is the probability Sylvia enters the stadium through an even numbered gate and leaves the stadium through an odd numbered gate? Express your answer as a common fraction.

Easy enough. 3/7 probability of entering through an even gate times 4/7 probability of exiting through an odd numbered gate = 12/49.

Now change one word:

A football stadium has 7 gates for fans to enter or exit. The gates are numbered consecutively from 1 through 7. The probability of entering or exiting through each gate is equal. What is the probability Sylvia enters the stadium through an even numbered gate OR leaves the stadium through an odd numbered gate? Express your answer as a common fraction.

The naive answer is 3/7 probability of entering through an even gate plus 4/7 probability of exiting through an odd numbered gate = 7/7 = 1.

I know why that is wrong. And I know that the problem is ambiguously worded (is that an inclusive or exclusive or?)  And I know what problem it is a correct answer to:

A football stadium has 7 gates for fans to enter or exit. The gates are numbered consecutively from 1 through 7. The probability of entering or exiting through each gate is equal. What is the probability Sylvia enters the stadium through an even numbered gate OR enters the stadium through an odd numbered gate? Express your answer as a common fraction.

The question is how to explain to middle schoolers what’s wrong with the fallacious reasoning above. The best I’ve been able to come up with is that we add when the question gives alternatives (OR) for different acceptable outcomes of the same event, but we multiply when we want to combine (AND) outcomes of different events.

Any better suggestions?

And another thing that bothered my middle schoolers was figuring out the probabilities of different outcomes of rolling two dice. Why do we count 1+2 and 2+1 separately, but only count 1+1 once? I tried to show this one with a tree diagram, showing that 1+1 only occupies one of the 36 leaves, whereas 1+2 and 2+1 each occupy their own leaf. I also thought about saying that 1 one a red die and 2 on a blue die is distinguishable from 1 on the blue die and 2 on the red die, but 1 on a red die and 1 on a blue die isn’t distinguishable from 1 on the blue die and 1 on the red die.

Again, I would appreciate further insights into making this more clear and intuitive to the kids.

Thanks in advance!

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15 Comments leave one →
  1. Efrique permalink
    January 15, 2008 12:02 am

    I also thought about saying that 1 one a red die and 2 on a blue die is distinguishable from 1 on the blue die and 2 on the red die, but 1 on a red die and 1 on a blue die isn’t distinguishable from 1 on the blue die and 1 on the red die.

    Take a single die and roll it twice, tabulating “first roll” vs “second roll”. Note that the outcomes form ordered pairs – (1,2) is not the same as (2,1), but (1,1) is a single event.

    Point out that if you roll two dice, even two apparently indistinguishable dice, it’s the same experiment as rolling one die twice (if it helps, make the intermediate step of considering which die hit the table first – even infinitesimally earlier).

    If time permits, demonstrate it by actually rolling dice and seeing that a sum of 3 is actually twice as likely as a sum of 2 (combining rolls across a whole class). [If that requires too many rolls, you could try it it with two-sided dice … that is, coins (count heads as 1 and tails as 2). ]
    You will probably want more than 30 outcomes of those two totals, so if you’re doing it with dice, a total number of rolls around the 400 mark (a round number a bit bigger than 30 x 12) should give enough rolls for the relative difference to show up pretty well.

  2. January 15, 2008 8:36 am

    The first puzzle is very like another one MathCounts likes: “With two spinners (numbered in various ways from one problem to the next), what is the probability that the product of the numbers spun will be even?”

    It is both an AND and an OR problem, so it makes a good review. What is the probability that the first spinner was even, OR that the first was odd AND the second one even? With the gates, assuming inclusive OR, you would look for the probability that the first gate was even, OR that both gates were odd.

    The main thing that students need to realize is that you can’t just add the probabilities as in your fallacy, because there is overlap. Some of the second-gate-odd possibilities are already counted in the first-gate-even set.

  3. January 15, 2008 9:09 am

    The main thing that students need to realize is that you can’t just add the probabilities as in your fallacy, because there is overlap. Some of the second-gate-odd possibilities are already counted in the first-gate-even set.

    Thanks, that’s a helpful way of explaining it.

  4. January 15, 2008 9:11 am

    Efrique — I have tried using the order of results as well, but it is still not really sinking in to them. I would have thought the tree of possible results would be convincing, but I’m afraid that at the moment most of them just count doubles once “because I said to”. Sigh…

    I could do the experiment with them, but I want them to also understand why the results come out as they do.

    Thanks!

  5. garbo permalink
    January 15, 2008 7:43 pm

    Sometimes it’s easier to figure out the probability that the outcome WON’T happen, then subtract that from one. In the gate problem, for the situation described not to occur, she would need to enter through an odd gate AND leave through an even gate: 4/7*3/7=12/49. So the probability for the situation in the problem to occur is 37/49. I don’t know if this would be any more convincing to your students though. I also may be missing something.

  6. January 15, 2008 7:57 pm

    Garbo, yes, if it’s an inclusive or, finding the complementary probability is a great approach. I have shown them this in the past, but I think it’s worth showing again. Thanks for the reminder to keep bringing this up. :)

  7. January 17, 2008 7:24 am

    Garbo’s on the money here. Try a Venn Diagram, with one circle for “enters even” and one circle for “exits odd”, and then you 2 main options.

    1. Since there are only 49 pairs, you could place them in the diagram. Really, not so bad, though a bit primititive.

    2. Identify which region you want. Point to the part of “enters even” that does not overlap. Does this meet the condition? Do that for all four. And then draw Garbo’s result as a conclusion.

    Note, Garbo’s result would be the conclusion of #1 as well.

    Great stuff.

  8. January 17, 2008 9:14 am

    Thanks, JD. The Venn Diagram is a nice idea for the enter/exit scenario. It gets to Denise’s point as well, that you can’t just add because of the overlap.

  9. wolfdog permalink
    January 31, 2008 8:18 am

    There’s something interesting going on here, which can be expressed in terms of sets or in terms of probabilities, pretty much to the same end.

    With sets, you usually learn that |A union B| = |A| + |B| as long as A and B are disjoint. If they aren’t disjoint, then the righthand side is an overcount of the number of elements in the union; you need the little more sophisticated
    |A union B| = |A| + |B| – |A intersect B|
    formula, a basic form of the inclusion-exclusion principle.

    The same same thing’s true of probabilities – if E and F are mutually exclusive events, then
    Prob(E OR F) = Prob(E) + Prob(F)
    is legit. But if it’s possible for both E and F to occur at the then you need the analagous
    Prob(E OR F) = Prob(E) + Prob(F) – Prob(E AND F)
    to take care of the overcount.

    I think that’s exactly the way I’d want to explain it – in terms of overcount, where you’ve double-counted the outcomes where she enters even and exits odd. They can definitely get the idea of “mutually exclusive” and it’s always an important thing to have in mind when thinking about probabilities involving more than one event.

  10. January 31, 2008 10:33 am

    Thanks Wolfdog. That is of course closely related to the Venn Diagram that JD suggested, but having more than one way to see it is always helpful.

  11. February 1, 2008 1:29 am

    I would stick with Woldog’s more general formulas:
    |A union B| = |A| + |B| – |A intersect B| and
    Prob(E OR F) = Prob(E) + Prob(F) – Prob(E AND F)

    All that “disjoint” or “mutually exclusive” really mean is that the term to be subtracted equals 0. That’s just a special case (of course we are glad for the easier calculation, when it’s available).

    Jonathan

  12. February 1, 2008 8:09 am

    It’s important to recognize that the inclusion/exclusion formula here is for an INCLUSIVE OR (which I think is a little counter-intuitive, since you are subtracting off P(both)):

    Prob(E OR F OR Both) = Prob(E) + Prob(F) – Prob(E AND F)

    For an EXCLUSIVE OR you need to subtract the union twice, because it’s counted in both of the individual probabilities and you don’t want it either time:

    Prob(E OR F BUT NOT Both) = Prob(E) + Prob(F) – 2xProb(E AND F)

    This is easiest to see if you draw the Venn Diagram, and/or write out the individual paired probabilities.

    The usual inclusion/exclusion formula has to subtract off P(both) not because we don’t want P(both) but because it is double counted in the P(E) +P(F) formula. A formula for an exclusive OR needs to subtract it off twice, because it is double counted, and we don’t want even one count of it.

  13. Matt permalink
    November 9, 2011 9:35 pm

    Hi – I know this is 3 years late but I stumbled across your blog while looking for something else.

    The best way I can think of to explain this to your kids is using the two different colored dice, and drawing a 6×6 chart on the chalkboard of all 36 possibilities, and noting that each box has a 1/36 probability outcome. Then it is easy to see that the odds of rolling a 1 and a 2 is equal to 1/36+1/36=1/18 but the odds of rolling a 1 and a 1 is 1/36. Now you can turn back and apply this reasoning to the original stadium problem, and any other problem whose events are not mutually exclusive.

  14. April 16, 2014 5:07 am

    The gates problem: the events (leaving and entertaining) are independent. If students have a full understanding of what that means they should be able to comprehend what’s going on.

    Re the dice. The reason you count 1 and 1 once is that it only happens once. There’s only one way to get a one on each dice. There are 2 ways to get a 2 and a 1!

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