Re: arrival times; was: Geiger (a challenge)

[brian whatcott <inet@INTELLISYS.NET>]

At 14:18 7/8/00 -0400, John D wrote:
> ...
> Here's an even simpler arrival-interval puzzle, which doesn't require a
> Geiger counter or dice or anything else.
>
> Part 1:  We have a bus stop.  Busses arrive every ten minutes like
> clockwork, all day and all night.  If the last one was at 8:15 the next one
> will be at 8:25, guaranteed.

>   Question 1a: If you show up at the bus stop at a random time, how long
> do you have to wait for the arrival of the next bus, on average?

>   Question 1b: If you show up just as the previous bus is leaving, how
> long do you have to wait for the arrival of the next bus, on average?

NIT PICKER'S SPECIAL!
Answer 1a: 0 to 5 minutes on average, not enough data provided to decide
within the range.

Answer 1b: 0 to 5 minutes on average, not enough data provided to decide
within the range.

These legitimate answers are offered in the spirit of
encouraging puzzlement.

Aw shucks! John just closed off the nit, like this:
***********************************************************
Part 1:  Suppose busses drive past your window every ten minutes like
clockwork, all day and all night.  If the last one was at 8:15 the next one
will be at 8:25, guaranteed.
    Question 1a: If you start looking at a random time, how long
do you have to wait for the passage of the next bus, on average?
    Question 1b: If you start looking just as one bus is passing, how
long do you have to wait for the passage of the next bus, on average?

(Busses can hold on station for an inderterminate time.
It is not unknown for a bus to depart when the succeeding bus arrives)
*************************************************************


brian whatcott <inet@intellisys.net>
Altus OK