I've tried searching for this kind of question, but the closest I've found is this, but the math is a bit over my head, and it's also tagged with "normal distribution", which I don't think applies to my question.

I found a few others, but they seem to be the inverse of what I'm trying to figure out, and it doesn't appear to be straight forward to reverse their calculations.

My Question:

Basically, I'm trying to understand how you calculate the probability of an event, when time is the only deciding factor.

Let's say that you're looking at a digital clock on the side of a building which only displays hours & minutes (i.e. no seconds), but the clock is partially obstructed by a tree or utility pole or something. So, at any given moment, you can only see either the minutes or the hours, but never both at the same time.

You start out standing in a location such that you can see the minutes, which read "00". Then, you walk to a new position where you can see the hours, which read "12". If it took you "x" seconds to walk to that new location, what is the probability that the minutes have since changed to "01".

My gut reaction is simply to say x/60. But is that actually true? or is there a better way of understanding this kind of thing.


P.S. Obviously, if this where a riddle, then you could just look at the hours first, and then look at the minutes, and know for sure. But I'm just using this example to provide a foundation for understanding the math.


1 Answer 1


If you assume that (a) the clock is "correct", in that it changes display to a new minute at the instant that the minute starts (to save having to distinguish clock-time from actual time); and (b) the instant in time that you saw the minute digits was uniformly distributed in $[0,60)$ seconds (which is probably a reasonable model), then $x$ seconds later, the time is uniformly distributed on $[x,60+x)$ seconds after $12:00$.

The probability that the full time display at that point is $12:01$ or later is $\min(x/60,1)$ and the chance that it's still showing $12:00$ - i.e. that the correct time is still in $[12:00, 12:01)$ - is $\max(0,1-\frac{x}{60})$.

  • $\begingroup$ That makes sense. Thank you, sir. $\endgroup$
    – Yurelle
    Aug 8, 2021 at 2:13

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