1st assumption that I made: I thought that you don't need to do a calculation for a whole day, but finding out the probability during one hour would be enough since each hour would have same proportion of satisfying the condition.
A. So let's set one clock to 12:00, and other to 1:00, which makes 1 hour difference.
- For 12:00, the time period that make minute/hour hands acute angle should be $0 $ ~ $180/11$ and $540/11$ ~ $0$
(if we set $x$ as a single minute, it should be clear. $5.5x < 90$ for the first hand, $360-6x + 0.5x <90$ for the second hand.
- For 1:00, the time period that make minute/hour hands acute angle should be $0 $ ~ $240/11$ and $600/11$ ~ $0$
So the overlapping time should be $0 $ ~ $180/11$ and $600/11$ ~ $0$ during one hour, making it $2/33$
B. for $x=2,3,4,5,6$
Should divide $x$ to $2$ cases
a. when time difference is within $3$ hours
b. more than $3$ hours
I did this because the proportion of each time gets quite different after $3$ hours
for $x = 2$, its quite same as part A, just to move $60/11$ from each hands, making the answer $1/22$
When it comes to $x = 3$, however, the total time that forms acute angle between 3:00 ~ 4:00 becomes $360/11$ (starts from 0, and ends at $x$ satisfying $6x – (90 + 0.5x) < 90$ which makes $x = 360/11$) when $x = 1$ or $2$ the time period was $300/11$
So the answer for each hour should be, $1/22$ for $3$ hours, $1/33$ for $4$ hours, $1/66$ for $5$ hours, and $0$ for $6$ hours.
I intentionally skipped other hours since all of them could be dealt with using symmetry.
I solved this problem with my 1st assumption, but not sure whether this assumption would be valid.
Do you think there is a better way, or legitimate way to approach this problem?
Best Answer
Each clock individually alternates between $p$ hours in an acute state followed by $p$ hours in an obtuse state, where $p=\frac{6}{11}$. If the offset between them is $x=0$ hours, then they are perfectly in-sync, and so both clocks are acute with probability $\frac{1}{2}$. If the offset between them is $x=p$ hours, then they are perfectly out-of-sync, and so both clocks are acute with probability $0$.
Clearly, as we vary $x$ from $0$ to $p$, the probability shifts linearly, and so in the range $x\in\left[0, p\right]$, the probability must be $\frac{p-x}{2p}$.
As we vary $x$ from $p$ to $2p$, the probability must linearly go back down to zero, and then this pattern must recur with period $2p$, forming a triangular wave pattern. Using the standard formula for triangular waves, we find that the desired probability is
$$f(x) = \left|\frac{x-p}{2p} - \left\lfloor\frac{x-p}{2p} + \frac{1}{2}\right\rfloor \right|$$