Walk in the Park
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Alice and Bob agree to meet in the park. They agree to each show up some random time between 12:00 PM and 1:00 PM. Wait 20 minutes for the other person, or until one o'clock, whichever comes first.
Assuming they stick to their word, what is the probability that they will meet?
HintThey are equally likely to show up at any time.
However their probability of meeting will change depending on the time.
Try looking at it through only one person's perspective.
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Answer
Their probability of meeting is 5/9
There are four events that can happen.
1. Bob arrives first and Alice meets him while he is waiting.
2. Bob arrives first, leaves and then Alice shows up.
3. Alice arrives first and Bob meets her while she is waiting.
4. Alice arrives first, leaves and then Bob shows up.
Also there are 3 main time segments that different things can happen during. from 12:00 12:00-12:20, 12:20-12:40, 12:40-1:00.
There is a 1/3 chance of Alice showing up in any of these times.
If Alice shows up from 12:20-12:40
Then there are 20 minutes before and after Alice's arrival that Bob could arrive and they would meet. So 40 minutes out of the hour that Bob could arrive and they will meet. (40/60)
If Alice shows up between 12:00 and 12:20 there are 20 minutes after her arrival, and all the time between 12:00 and when she arrived. This time averages out to be 10 minutes (remember we are assuming she showed up between 12:00 and 12:20) So that's 30 minutes out of the hour that Bob could show up and they would meet. (30/60)
From 12:40-1:00 you have a similar situation. There's twenty minutes before Alice's arrival that Bob could have shown up. And all the time till 1:00, which again averages out to 10 minutes. So again 30 minutes out of the hour.
(30/60)
Each of those three are equally likely with 1/3 probability so
1/3(30/60 + 40/60 + 30/60)
1/3(5/3)
5/9
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Comments
katiebeth123 
Mar 21, 2005
| OMG SHUT UP |
krishnan  
Mar 23, 2005
| Nice teaser. |
mschup
Mar 24, 2005
| i have a headache. |
psychedelictoad 
Mar 28, 2005
| That teaser hurt my soul. |
runscapekiller
Mar 31, 2005
| is an anser soposed 2 b that long  |
Dedrik
Apr 06, 2005
| I know, but it was the only way I could think to explain it. Believeme I am open to a shorter explination  |
ANgelStAR  
Apr 10, 2005
| ok, i did not get that!   |
monkeygirl
Apr 11, 2005
| confusing... |
cehnehdeh
Apr 17, 2005
| WAY to complicated and purely math no fun!   |
pi202
Apr 19, 2005
| Really nice teaser. A good bit of hard thought never hurt anyone. Well done! |
duluoz_jack
Apr 22, 2005
| If they have both agreed to wait until 1 0 then the probability is 100% that they will meet. Or what am I missing? |
jinzcarmela
Apr 29, 2005
| What the heck was that? I almost got my brains lost |
brianz
Apr 30, 2005
| Nice teaser. |
BrownEyes
Jun 16, 2005
| hard, HARD, HARD!!!
 |
babygrl329
Jun 24, 2005
| confusing |
bjoel4evr
Jul 25, 2005
| Nice one. I think your answer explained the teaser very well. |
cazjackrabbit
Aug 24, 2005
| But couldn't the waiting time have been like 12.37-12.57? Ok... well even if I completely missed that it was still HARD!!! But it was good in a way. If you have hours to do it in- or you could just do it till one o'clock, whichever comes first. Ok that was a lame attempt at a joke. |
shoulder_angel
Sep 22, 2005
| I didn't get your explanation... I guess that I should just stop reading probability teasers 'cause they make me feel dumb . Oh, well |
shoulder_angel
Sep 22, 2005
| Oh... and don't be such a know-it-all bjoel4evr! |
MissleMan27
Jan 01, 2006
| I didn't understand any of that! |
mr_brainiac
Jan 04, 2006
| I agree with the Missileman. Deceptively confusing puzzle looks easy until you jump in. |
kayleeskitties
Jan 21, 2006
| Good teaser, boggled my brain there for a moment but I got it!  |
paul726
Feb 11, 2006
| Excellent job. Well thought out! |
(user deleted)
Apr 27, 2006
| This riddle has a flaw. You can't assume that they will show up at 120. 12:20 or 12:40 unless you put that in the problem itself... any random time means any random time. It is impossible to solve otherwise. |
Dedrik
Aug 09, 2006
| I didn't assume she showed up at any of those times. I suggested breaking the hour up into different pieces as the probability of meeting up is different during each interval. |
javaguru
Dec 09, 2008
| I approached the problem slightly differently. I broke the time period into the three 20 minutes sections. There are then three relevant possibilities:
1) They arrive in the same period. They always meet in these cases.
2) They arrive in non-adjacent periods. They never meet in these cases.
3) The arrive in adjacent periods. As explained above, there is a 50% chance they meet in these cases.
There are 3 x 3 = 9 combinations of periods they can arrive in.
There are 3 combinations where they both arrive in the same period: {(1,1), (2,2), (3,3)}
There are two combinations where they arrive in non-adjacent periods: {(1,3), (3,1)}
That leaves four combinations where they arrive in adjacent periods {(1,2), (2,1), (2,3), (3,2)}.
Therefore there is a:
3/9 x 1 + 2/9 x 0 + 4/9 * .5 = 5/9
probability of meeting. |
opqpop
Sep 28, 2010
| This is a standard geometric probability problem. Let the x axis be Alice's arrival time and y axis be Bob's arrival time. We're interested in the region inside a 60minute * 60minute square that corresponds to the event that they meet.
Notice this region is bounded by the line y=x shifted 20minutes up and 20 minutes down. This is because whatever time Alice arrives, Bob needs to be within 20 minutes of her time for them to meet.
The area of this region is 60 * 60 - 40 * 40. 40 * 40 is the area of the sum of the two triangles that aren't in this region. You'll see it clearly if you draw the graph.
The total area is 60 * 60. Hence (60-40)(60+40) = 2000 / 3600, so the answer is 5/9. |
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