Care for a test of mathematical intuition?

They will both be around 30.

That was my original thought. Think of it this way. Choosing one number divides 60 into two equal parts. Suppose we choose 100 numbers each time. Surely the minimums and maximums will diverge from 30. There will be divergence even when choosing two numbers.

Bacon.

I would imagine the low numbers would average 15 and the high numbers would average 45.

You are diving the numbers into two ranges 1–30 and 31— 60, You select a number from each range. The numbers chosen would tend to average out to the middle of their respective ranges.

The low/high numbers that you pick could also be, for example, 59/60.
Maybe the average for each would shift by a fraction, because the low can never be 60, and the high can never be 1, but that is it. The two picks are not otherwise related. But that is only if you make it a rule that you must discard a roll, if both picks come up 1, or both come up 60. I think of this as just another variation of the gambler’s fallacy.

@zenvelo has the right idea, but the answer is even simpler. In the case of just one number, it divided 0 to 60 into two equal parts. In the case of two numbers, they divide 0 to 60 into 3 equal parts, giving 20 for the average minimum and 40 for the average maximum. Another way of looking at it is to say say that the average minimum is half way between 0 and the average maximum and the average maximum is half way between the average minimum and 60.

This problem is a special case of what is called order statistics.

It seems you are right.
I just tested this in excel.
Populated 2 columns, A and B with a random number between 0 to 60 for 485 rows :”=RANDBETWEEN(0,60)”.
Then populated rows D and E with the maximums/minimums from the A/B columns ”=MIN(A1:B1)”, and ”=MAX(A1:B1)”.
Then calculated the average of columns D and E. ”=AVERAGE(D1:D485)”, and ”=AVERAGE(E1:E485)
The averages for D and E do indeed come out to 19.587 for the lows, and 39.882 for the highs.

However, the averages for columns A and B are at 29.858 and 29.612, so it is the sorting that is the deciding factor.

My brain just blew up. More coffee please.

@LostInParadise ”...In the case of two numbers, they divide 0 to 60 into 3 equal parts, giving 20 for the average minimum and 40 for the average maximum.”

That does not follow from your description of the problem at all. The two numbers fit into two pre-selected ranges. The division is at 30, and the averages must follow as 15 and 45.

Maybe I did not explain the problem correctly. There are no two pre-selected ranges. Each time two numbers are selected and each number is chosen randomly from 0 to 60 independently.

@ragingloli, I also ran a simulation and got similar results. I also came up with a quirky derivation without calculus. I hope that you can follow it.

Here is how it works for the minimum value.

Let 60x be the average minimum. We want to show that x=⅓

From the single number case, half the time a number chosen from 0 to 60 will be under 30 and half the time it will be between 30 and 60. That means that for two numbers, there are four cases:
A: ¼ of the time both numbers are below 30
B: ¼ of the time both numbers are above 30
C: ¼ of the time the first number is below 30 and the second is above 30
D: ¼ of the time the first number is above 30 and the second number is below 30

Let’s compute the average from the 4 cases
Case A is the same as the original problem with 30 instead of 60, so that gives
(¼)30x
Case B is equivalent to choosing a number from 0 to 30 (as in case A) and then adding 30, so the average is
¼(30x+30)
Cases C and D can be combined so that ½ the time the minimum will come from a single number drawn form 0 to 30, with average value 15
giving (½)15

Putting everything together, we get:
60x = ¼(30x) + ¼(30x+30) + ½(15)
60x=15x + 15
45x=15
x=⅓

Yeah, I found your original description to be imprecise, so I didn’t know what you meant exactly.

That’s often the pitfall of many story problems. The author tends to have some specific math in mind, and then tries to translate it into non-mathematical language that could be interpreted in other ways (and/or that sounds like a unreal situation).

I don’t consider it reasonable that the chosen numbers between 0 and 60 will be somewhere “in the middle” if the “randomly” instruction is followed. I didn’t go beyond that.

My intuition said 15 and 45. But that is wrong. Your answer is correct – and @ragingloli proved it! Nice!