Friday Puzzle -- A five-minute meeting

None

Lee Schierer McKean, PA

My guess is no probability, because Jerry won't get his driveway shoveled out in time from the latest snowstorm.

Lee

Re: Friday Puzzle -- A five-minute meeting

David Weaver

Off the top of my head, easiest done as a conditional probability.

So, take ben's arrival time as a given. Jerry's will determine then if they have a five minute overlap.

Taking the endpoints away, assuming ben comes in between 2:10 and 3:50, Jerry can come in either ten minutes before ben does or up to 10 minutes after and and they still have a 5 minute meeting.

That's a 20 minute span where the universe of possibilities is the full hour.

That is also 2/3rds of the times ben could arrive. So the total cumulative probability for that part is 2/3 * 1/3 = 2/9 (we have to add the other probability to that later)

Now, the other 1/3rd of the time has to be a density function of some sort, because if ben comes in at 2:00, jerry must come in within 10 minutes. If Ben comes in at 2:59, Jerry must've been there 10 minutes before or 1 minute after, etc.

So the first density function would have a probability that starts at 1/6th and approaches 1/3rd right before it gets to ben coming in at 3:10 (this is hard stuff for a guy who hasn't done calculus in a while, but I think I can get around it because it's linear)

Same thing at 3:50, it's 1/3rd probability at that point grading down to 1/6th at 3:00.

wait...i think i have a math error...need to redo the last part.

Ok..here goes, had to draw a picture with triangles on it, how sad is that?!

for the two endpoint probabilities since they're linear, for each I get (1.5)*((1/6)^(2))

That works out to be 1/12th added to the 2/9ths, which is a sum total of 3/36 + 8/36 = 11/36

I'll go with that until I find something else wrong with the math!

Close

Alex Y

Very close, in fact. But reread the question and look for a small error in your setup.

BTW, don't apologize for drawing a picture; I think that is the clearest way to see this one.

Re: Close

David Weaver

Cruddo - got immersed in the detail and totally missed the last condition!!

BRB

Re: Close

David Weaver

I'll just cut my picture back 5 minutes on the last side.

That'll give me (for the equal probability middle part of the picture) (35/60)*(1/3) = 35/180 = 7/36

Then mash the two ends on again at 2*1.5*((1/6)^(2)) = 1/12 or 3/36

so 10/36

or 5/18

Alex Y

Yes, the meeting has to be completed by 3:00, a condition I missed on my first try as well.

To illustrate the solution David is talking about, here is the diagram. Think of Ben's arrival time on the x axis and Jerry's on the y axis. The inner shape is the solution space, diagramming the parts David mentioned.

The area of possible arrivals is 3600 minutes^2. I've put a 5-minute grid on this diagram, so the possible arrivals are 144 squares, while the area inside the solution space is 40 squares.

Re: 

David Weaver

in my mind was a probability density function over time from 0 to 1 hour, but same area! It would've been a PDF that at time zero was the area under a line going from 1/6th in height to 1/3rd at time = 10 minutes, then staying at 1/3rd for another 35, and then 1/3rd to 1/6th at 45-55 minutes (none after that obviously because it is impossible to get a 5 minute meeting in with less than 5 minutes remaining).

I did eventually do it with integrals of three PDFs, but I couldn't break out the cob webs at first, i haven't done a PDF in probably 11 years. Or even an integral for that matter.

I haven't been checking in a lot lately, but I checked in today hoping there'd be a dandy little brain stretcher like this one.

Thanks for taking the time!

Coming up

Alex Y

Glad you liked it. Come back next week for a tricky magic square, then another probability toughie the following week.