I saw this one in class last week:

Probability of flipping 10 heads when flipping 10 coins: (1/2)^10

Probability of flipping 9 heads when flipping 10 coins: (1/2)^9

  • Well, yes, I can see that. Even if your mind is telling you, ‘the tail must figure in somewhere’, that would just lead back to (1/2)^10, which can’t be right either, as the probability should be different. Might be more interesting to see where a die roll question would go… would (5/6) factor in then, as the fail case is different?

    IMO, visualizing tree diagrams or constructing sets becomes rather tricky beyond about 3 cases, so unless you’re aware (consciously or otherwise) of the binomial distribution, even an adult can miss the “order matters” issue as soon as you don’t have a simple event.

  • An answer I would expect to see from a more sophisticated, but more confused, student would be 10 * (1/2)^9 where the 10 represents the number of places that the 1 tail could reside. I think the mistake above can be resolved by asking the (relatively obvious to us) questions
    1 – What is the probability of ten heads in a row? Good, you got that right.
    2 – If I have tossed a coin 9 times and I saw heads each time, what is the probability of getting a head on the next toss?
    If the student answers question 2 correctly, then proceed to build up the correct answer to the question posed in the post. If not, go back and talk about conditional probability at the appropriate level.