The Shooting Stars puzzle
Probability questions can be very slippery
19 October 2018
Jen Rogers, the VP of the Royal Statistical Society, set an interesting probability question on the radio this morning. The puzzle was this (I've slightly changed the wording):
Two friends are stargazing. They know that if they stare at the heavens for an hour, the chance of seeing a shooting star is 90%. But it's a cold night so they only want to spend ten minutes outside. What is the chance that they will see a shooting star?
[Spoiler alert - solution coming]
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Ten minutes is one sixth of an hour, so a first hunch might say that the chance is one sixth of 90%, so 15%.
But is that the way that shooting stars work? I don't know much about astronomy, but I believe that shooting stars are pretty much an unpredictable, random phenomenon, in which case their appearance is more like rolling a dice or tossing a coin.
So perhaps a more realistic way to think about shooting stars would be to say that there is a chance 'p' that we will see a shooting star in any ten minute interval, and a chance of '1-p' that we won't see one. And each ten minute interval is independent of the others, in the same way that every roll of a dice is independent of every other one.
In the shooting star problem, we know that the chance of seeing (at least one) shooting star in an hour is 90%. This means the chance of NOT seeing a shooting star during that hour is 10%.
But one hour of stargazing is six lots of ten minutes, so the question is really: if there's a 10% chance shooting star not appearing in six consecutive periods of ten minutes, what's the chance of it not appearing in one ten minute period? Or, mathematically: What is 'p' if (1-p)^6 = 10%.
A few presses on the calculator tell me that p=32% - in other words, there is roughly a 1 in 3 chance of seeing a shooting star if I wait for ten minutes. That's twice as likely as my original hunch of 15% - which I find quite surprising. It's a deceptively tricky question which would defeat many A Level maths students.
However, this solution DOES depend on my assumption about how shooting stars behave.
Suppose the question had been about Bill Smith, who every day leaves the office randomly sometime between 5pm and 6:07pm. A surveillance officer keeps watch outside the office building between 5pm and 6pm each day, and after several weeks' surveillance he reports that there is a 90% chance that he will see Bill Smith at some point during the hour. The officer hands over the job to a lazy colleague who couldn't be bothered to wait around for an hour, and decides he'll just keep watch for ten minutes. What is the chance that he'll see Bill Smith? The numbers in this problem are the same, yet the answer this time really is 15%.
So: if shooting stars behave like dice then the answer is 32%, but if they behave like Bill Smith, the answer is 15%.
It shows just how slippery probability questions can be*. Indeed, these questions are impossible to answer if you don't know the behaviour (or distribution) of the thing you are observing.
* In previous blogs I discussed two famous probability puzzles, Monty Hall and the Tuesday Boy problem, both of them notoriously slippery.