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Let's say you have a digital clock that only shows you the hour, not the minutes. If you wanted to guess the correct time, down to the minute, you'd have a 1 in 60 chance of getting the minute correct. However, let's say you noted the time, came back 10 minutes later, rechecked the hour and then guessed. If the hour has changed, you know that the minutes have to be < 10, and if it didn't, the minutes have to be $\geq$ 10. There's a $\frac{1}{6}$ chance that the hour changed, which gives you a $\frac{1}{10}$ chance of guessing correctly, and a $\frac{5}{6}$ chance that it doesn't change, giving your guess a $\frac{1}{50}$ chance of getting the minute right, so your overall chance of guessing correctly is $\frac{1}{6}\cdot\frac{1}{10}+\frac{5}{6}\cdot\frac{1}{50}$, or $\frac{1}{30}$. Replace the 10 minute wait with an $n$ minute wait, however, everything still cancels, leaving you with $\frac{1}{30}$ again (as long as $n$ is not a multiple of 60).

Something seems wrong with the fact that any wait at all doubles your chances of guessing right. I wrote a small python script to test and it thinks empirically the chance with a 10 minute wait is $\frac{1}{60}$. I trust the maths more than I trust my ability to spot errors in the code though.

Is the maths right? Does waiting even a single minute double your chances? Or am I missing something?

Elliott Price
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2 Answers2

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Let's walk through the probabilities carefully. If it is currently $M$ minutes past the hour, then $n$ minutes later it is $M+n$ minutes past. So the strategy is:

  1. If the hour has changed, then we know $M+n \geq 60$, so we guess $M$ uniformly from $[60 - n, 60)$.

  2. If the hour has not changed, we guess $M$ uniformly from $[0, 60 - n)$.

In the case of 1, our probability of being correct is $\frac{1}{n}$, and in the case of 2, our probability of being correct is $\frac{1}{60 - n}$.

Then, assuming that a priori our chance of looking at the clock at a particular value of $M$ is uniformly distributed across the hour, then there's an $\frac{n}{60}$ chance that $M \geq 60 - n$, and a $\frac{60 - n}{60}$ chance that $M < 60 - n$. Then the overall probability you guess correctly is $\frac{1}{n} \frac{n}{60} + \frac{1}{60 - n} \frac{60 - n}{60} = \frac{1}{30}$.

It feels unintuitive because of the sudden jump from $n = 0$, where this strategy would include a $\frac{0}{0}$ term that in reality contributes 0 (because there's no chance we'll see the hour change, so we just ignore that term). It's a weird discontinuity that confuses us when the rest of the distribution is constant.

ConMan
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Super interesting question! I think your math is completely correct: waiting for any amount $n$ of minutes doubles your chances. You implicitly use the law of total probability (see https://en.wikipedia.org/wiki/Law_of_total_probability), but your calculations are right.

I do think the result is somewhat counter-intuitive, so I am not surprised you were confused. As for your Python script, could you maybe tell us what it does exactly? There must be some kind of mistake in it. Also notice that it does not make sense intuitively that the probability would remain $1/60$ for $n=10$. You suddenly have much more information, so it would be weird if your chances stay the same.

I do not see a quick, better intuitive explanation for this phenomenon, except for the calculation itself. Perhaps I will think of one later.

naiht
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