Help me please ! Maths?

Let's say the arrivals are all exactly on some minute,

and they can be at 1:00 through 1:59. (60 arrival times).

Let's say I am one of the two people.

Let x = my arrival time.

Then the other guy must arrive no earlier than x - 5 (if other guy arrives first)

and no later than x + 5 (if other guy arrives second).

So we have roughly a 10/60 chance = 1/6 of a successful meeting.

Roughly because, if one of the arrivals is in the first 5 minute of the hour

(1:00 to 1:04), then there isn't 5 minutes of buffer before that, but rather

0, 1, 2, 3, or 4.

And if one of them arrives at 1:56 or later, there is only

4, 3, 2, 1 or 0 minutes of buffer after that.

So we have the following summation:

If "my" arrival time is:

1:00 ... 5/60

1:01 ... 6/60

1:02 ... 7/60

1:03 ... 8/60

1:04 ... 9/60

1:05 through 1:54 ... 10/60

1:55 ... 9/60

1:56 ... 8/60

1:57 ... 7/60

1:58 ... 6/60

1:59 ... 5/60

So we have

50 * 10/60 + 2 * (5+6+7+8+9) / 60 all divided by 60 =

570 / 3600 = 0.158333 ← Answer

Confimed by simulation using 1 second resolution:

1 million trials, 158842 successes.