Will I see you?

The following graph shows Dev's arrival time on one axis and Akshatha's on another axis, with a green area containing coordinates where both lunchtimes overlap, and red areas containing coordinates where the lunchtimes do not overlap:

Exactly $\frac{3}{4}$ of the graph is green, so it is more likely that both are there at the same time .

The chance of their missing each other is infinitesimally small. The only way they could miss each other is if one of them arrives at 1:00 and leaves at 1:30, and the other arrives at 1:30. ANY other arrival times give overlap.

Let $s$ be the time the first arrival finishes, $t$ be the time the last arrival comes. Notice that $s$ lies in $[1:30, 2:30]$ and $t$ lies in $[(s- 0:30), 2:00]$ . If $s > t$ they will be there together at some time. Which is clearly more likely to happen according to the intervals $s, t$ lie in.

You could graph the probability that person 2 sees person 1 as a function of the arrival time of person 1 $P(x)$ . This can be done by thinking about what fraction of the total possible arrival times person 2 could arrive and see person 1. notice that as long as person 2's arrival time is anywhere within 30 minutes of person 1's arrival time, that they will see each other . Using the variable x as person 1's arrival time, in hours, it is easy to see that $P(1)=0.5$ , $P(1.5)=1$ , and $P(2)=0.5$ . Notice that if you change x, $\frac{dP(x)}{dx}$ =(constant) as you could think of Person 2's arrival times that result in them seeing person 1 for some amount of time as being on a bar of length 1 that is centered at the arrival time of person 1, to change the arrival time of person 1 is to change how much of the total arrival time is contained within the bar centered at person 1's arrival time. Logically then, $P(x)$ will increase linearly for x between 1 and 1.5, while it will decrease linearly for x between 1.5 and 2.
Now to solve the problem, take the ratio of the area under the graph of $P(x)$ to the total area of the box that contains all possible probability values and all possible arrival times. You could do two separate integrals (as the function is not differentiable at x =1.5) but it is less time consuming to use geometry. using x in hours, The area under the graph would be 0.75, the total area of the relevant graphing space is 1. and thus the ratio of area under graph to area of graphing space is $\frac{3}{4}$ , This fraction represents the overall probability that they end up in the canteen at the same time, for any length of time. and thus it is most likely that they do see each other or are mutually there for some period of time.

If someone arrives at 1:01 (the earliest time they can), there is >50% chance of them being there at the same time. The same is true for 1:30. There is exactly 29 minutes from 1:01 - 1:30 and 1:30 - 1:59. For the times between the number, the chances of them being there at the same time is 50%. Therefore, the average chance of them being there at the same time is >50%, meaning that the answer is yes.

I figure if lunch is over at 2:00 at the latest, it only leaves from 1-1:30 as "random-times-of-arrival", anytime after that renders u late, with 1:00 being the only way to exhaust 30mins without them seeing one another. Every "rtoa" from 1:01 forward is additional time spent with one another.

An hour is divided into 2 half-hours. Anywhere between 7.5 and 52.5 is higlighted, so 45 is divided into the hour equally and takes up 45 mins, which is ¾ an hour, which is more than the remaining fifteen minutes of the hour. ¾ is more than ¼, so it is more likely.

Let's name the guys Mutt and Jeff. If Mutt arrives exactly at 1:00, there's a 50-50 chance Jeff arrives within the next half-hour.

If Mutt arrives exactly at 2:00, there's a 50-50 chance Jeff arrived within the previous half-hour.

If Mutt arrives at 1:00 $+ t$ minutes, with $0 < t < 30$ , then the probability Jeff arrives either before Mutt or within the half hour after Mutt is $\frac{t + 30}{60} > \frac{1}{2}$ .

Similarly, if Mutt arrives at 2:00 $- t$ minutes, with $0 < t < 30$ , then the probability Jeff arrives either before Mutt or within the half hour after Mutt is $\frac{t + 30}{60} > \frac{1}{2}$ .

Finally, if Mutt arrrives at exactly 2:30, Jeff arrives within a half-hour of Mutt's arrival with probability 1.

So whenever Mutt arrives, Jeff will arrive within a half-hour with probability at least 1/2. With with a positive probability, this conditional probability will be greater than 1/2. Thus, as we sum (or integrate) over the possible times of Mutt's arrival, we see that the plain probability Jeff will arrive within a half-hour of Mutt is greater than 1/2.

They will meet so long as one arrives within half an hour of the other, either within 30mins before or 30mins after. For all starting times for one person, the other has a window that is longer than or equal to 30 minutes to arrive. This is more than half of the total 1 hour limit.