# A perpetually lagging watch and its time accuracy

A 24-hour, date-less, digital wrist-watch loses accuracy everyday, falling behind atomic time by an additional 15 minutes (every day). On the date of its manufacture, 1.1.2000 00h:00m:00s, it was perfectly synchronized to atomic time, and the malfunctioning applied every 24 hours thereafter.

A customer buys the watch on 1.1.2017 00h:00m:00s and assumes, rather stupidly, that the time displayed is accurate. After purchase, when mentioning the time to any one who asked, what is the probability that the time our buyer mentions is within +/- 1 minute of atomic time?

(Assume everlasting battery, consider leap years if needed, post purchase the watch still malfunctions everyday as usual, watch displays ONLY hh:mm, and not seconds).

Notes:

• Beginning probability student here, solving some harder problems (from the competitive math space)
• Not looking for a number answer or an outright solution, but hints, possible approaches.
• Welcome to StackExchange. Please can you add your thoughts to the question - the site is not here to just blindly answer homework style questions. Adding your thoughts means that people can give you an answer at your level of knowledge and can spot if/where you have gone wrong in what you have done already – lioness99a Jul 6 '17 at 8:59
• @lioness99a Added. Not looking for a straight solution at all. – Jim T Jul 6 '17 at 9:02

A beginning:

A lot of the given information is irrelevant: You can forget about dates and leap years.

Since the watch looses $15$ minutes, or ${1\over4}$ hour, on every $24$ true hours it has lost $24$ hours after exactly $96\cdot24$ true hours. It follows that everything repeats after $96$ days, so that you can take this as your basic interval for probability considerations. You now have to compute the length of the time interval centered at an exact hit during which the watch is off less than $1$ minute from true time.

You could calculate the time difference at the time of purchase, and then calculate how long it takes for the watch to be back in sync with the atomic clock for the first time. You could then calculate the duration of a complete cycle, the sum of the time it takes for the clock to run out of sync with the atomic clock, and the time it takes for the clock to run back in sync. You could take into account daylight saving time, leap years and seconds, but... The above steps are in fact not necessary.

Since the difference between the watch and the atomic clock is uniformly distributed, the probability of the watch being no more than one minute ahead or behind, is simply:

$$\frac{2}{24 \cdot 60} = \frac{1}{720}$$