What are the chances on watching the clock? [closed]

Question

This is a relatively straightforward little puzzle, but the answers are somewhat unintuitive or can prompt a new perspective. It might be an interesting little puzzle for a probability teaching example or job interview question.
I hope it helps get you thinking, in an useful way!

For the purpose of this puzzle, you have a digital clock that displays hours and minutes.
The clock [display] changes instantaneously after 60 seconds, every 60 seconds (which you can approximate to 59.9999999999... if that helps below). For periods relevant to this puzzle, the clock is guaranteed to work perfectly and without ceasing.

1. If you stare at the clock for 6 seconds beginning at a random starting time, what is the probability you will observe it changing?
2. If you do what's described in #1 ten times independently of each other, what is the probability you will observe the clock changing? (Guidance question under spoiler tag).

   Should this be higher than, lower than, or the same as the answer to #1?

3. If you wanted the probability of observing a change to be 50%*, how many times would you have to do what's described in #1?
4. If the ten times described in #2 are sequential, so that you're watching for one continuous period, then what is the probability you will observe the clock changing?

Have fun!

(*): If you can't get 50% exactly, get as close as you can.

Answer 1

1.

You will observe the clock change time if you look at it any time from (instantaneously) after 6 seconds until right before the clock would normally change (aka xx:xx:00), so the probability is $\frac{6}{60}=\frac{1}{10}.$

2.

The probability that you don't see the clock change in one glance is $1-\frac{1}{10}=\frac{9}{10}.$ So the probability that you don't observe the clock change in any of those 10 instances is $(\frac{9}{10})^{10}$ and so the probability that you do see the clock change is $\boxed{1-\left(\frac{9}{10}\right)^{10}}$ which is around $0.651.$

3.

We're basically trying to solve $1-(0.9)^n=0.5$, so $\boxed{n=7}$ (actually, n comes out to $\log_{0.9}0.5,$ which is around $6.579$).

4.

This is basically equivalent to watching the clock for a minute. That means that no matter what, at some point in the interval you'll observe the clock changing (unless you want to count the period from $0.000\dots 1$ seconds past the minute to $59.999\dots$ seconds past the minute as a valid interval), so the probability is $\boxed{1}.$

This definitely feels more like math than puzzles, tbh.

Answer 2

1.

1 in 10, so 10%

2.

This depends on whether you can time the intervals or not. If you take 10 consecutive intervals of 6 seconds (or choose them $60 \cdot n$ seconds apart, where $n$ is a natural number), you're guaranteed to see a change, so it's 100%. If the intervals are random and independent, the chance that you don't see a change is $(9/10)^{10} \approx 0.35$, so the chance that you do see at least one change is 1 minus that, so $\approx 0.65$.

3.

Assuming the intervals are chosen independently, the solution is $\frac{\log(50%)}{\log(9/10)} \approx 6.58$. So 6 is too few times, and 7 too many, but the answer for 7 (52.2%) is closer than the one for 6 (46.9%).

4.

See the answer to 2.