How do I determine the time difference in seconds between two clocks that show only minutes? So, after a power outage, I found myself having to reset a bunch of clocks in my apartment. The most handy clock I had was my cell phone, which only shows time down to the minute, on the home screen. That got me thinking, how do I tell how far off my oven, microwave, etc. are off from my cell phone without using another clock?
To put it more concretely, I know HH:MM from Clock A. I know HH:MM from Clock B. Is there a way to determine how far off Clock A is from Clock B down to the second, without using a third clock? 
I've been thinking in terms of random sampling. I know that if Clock A is 5 sec. ahead of Clock B, then if I take a random sample, I have a 55/60 chance of finding that Clock A and Clock B read the same time, and a 5/60 chance of finding that Clock A reads a minute ahead of Clock B. But how do I exploit that to find an unknown offset?
Is what I'm trying to do even possible? Am I going about it in completely the wrong way?
 A: Your method will work, with some uncertainty, assuming that you can sample randomly and can read both clocks at the same time. You can apply a Bayesian technique.
Assuming your sample size is big enough to see a distinction between the two clocks, you could have started with the difference being uniform between $0$ and $1$ minute.  This is a $Beta(1,1)$ distribution and you can take it a prior. If you then observe $55$ identical clock readings and $5$ clock A ahead readings, your posterior distribution will be $Beta(6,56)$ which has a mean of $\frac{6}{62}$ minutes, a mode of $\frac{5}{60}$ minutes and a standard deviation of $\sqrt{\frac{6\times 56}{62^2 \times 63}}$ minutes, corresponding to  about $5.81$, $5$, and $2.34$ seconds respectively.
It gets slightly more complicated if all your observations are identical and so you do not know which clock is ahead of the other, but your central estimate will presumably be $0$ and the standard deviation of the posterior distribution perhaps of the order of $\frac{\sqrt{2}}{n+2}$ minutes if you have $n$ identical observations.   
