This week the #puzzle is: Can You Meet Me at the Mall? #ExpectedValue #average #maximum #probabilities

You and two friends have arranged to meet at a popular downtown mall between 3 p.m. and 4 p.m. one afternoon. However, you neglected to specify a time within that one-hour window. Therefore, the three of you will be arriving at randomly selected times between 3 p.m. and 4 p.m. Once each of you arrives at the mall, you will be there for exactly 15 minutes. When the 15 minutes are up, you leave.

At some point (or points) during the hour, there will be a maximum number of friends at the mall. This maximum could be one (sad!), two, or three. On average, what would you expect this maximum number of friends to be?

And for extra credit:

Instead of three total friends, now suppose there are four total friends (yourself included). As before, you all arrive at random times during the hour and each stay for 15 minutes.

On average, what would you expect the maximum number of friends meeting up to be?

Can You Meet Me at the Mall?

Highlight to reveal (possibly incorrect) solution:

Program 1 Program 2

• My 1st shot at this was a monte carlo program. It says: 2.03123 friends.
• My 2nd shot was to try to estimate the value by systematically going through a lot of options. Just like small rectangles estimate an integral. This program says: 2.03222 friends.
• I therefore say a good guess would be 2.03.

And for extra credit:

• Using the same programs I get 2.47646 and 2.47802.
• A good guess seems to be 2.48.

Desmos 1, 2 and 3. Probability of two events occurring at overlapping times. Desmos 4.

I actually tried to get something out of playing with desmos, but didn’t quite get there. I also tried modeling a version of the puzzle with 2 friends. Especially the no. 4 above is nice.

Using the article, I get a general expression. T is the length of the interval. The probability that 2 out of 2 friends are there at the same time is 2T – T². For T = 1/4, this is 7/16. This fits with my programs and with Desmos.