Uniform Probabilities on a Square

# Uniform Probabilities on a Square

In this problem, we will be
helping Romeo and Juliet meet up for a date. And in the process, also we’ll
review some concepts in basic probability theory, including
sample spaces and probability laws. This problem, the basic setup
is that Romeo and Juliet are trying to meet up for a date. And let’s say they’re trying
to meet up for lunch tomorrow at noon. But they’re not necessarily
punctual. So they may arrive on time with
a delay of 0, or they may actually be up to 1 hour late
and arrive at 1:00 PM. So the other thing that we
assume in this problem is that all pairs of arrival times– so
the time that Romeo arrives paired with the time they
Juliet arrives– all of these pairs are
equally likely. And I’ve put this in quotes,
because we haven’t really specify exactly what
this means. And we’ll come back to
that in a little bit. The last important thing is that
each person will wait for 15 minutes for the other
person to arrive. If within that 15-minute
window the other person doesn’t arrive, then they’ll
give up and they’ll end up not meeting up for lunch. So to solve this problem, let’s
first try to set up a sample space and come up with a
probability law to describe this scenario. And let’s actually start
with a simpler version of this problem. And instead of assuming that
they can arrive at any delay between 0 and 1 hour, let’s
pretend instead that Romeo and Juliet can only arrive in
15-minute increments. So Romeo can arrive on time
with a delay 0, or be 15 minutes late, 30 minutes
late, 45 minutes late, or one hour late. But none of the other
times are possible. And the same thing for Juliet. Let’s start out with just the
simple case first, because it helps us get the intuition for
the problem, and it’s an easier case to analyze. So it’s actually easy
to visualize this. It’s a nice visual tool
to group this sample space into a grid. So the horizontal axis here
represents the arrival time of Romeo, and the vertical
axis represents the arrival time of Juliet. And so, for example, this point
here would represent Romeo arriving 15 minutes
late and Juliet arriving 30 minutes late. So this is our sample
space now. This is our omega. And now let’s try to assign
a probability law. And we’ll continue to assume
that all pairs of arrival times are equally likely. And now we can actually
specifically specify what this term means. And in particular, we’ll be
invoking the discrete uniform law, which basically says that
all of these points, which are just outcomes in our
probabilistic experiment– all of these outcomes
are equally likely. And so since there are 25 of
them, each one of these outcomes has a probability
of 1 over 25. So now we’ve specified our
sample space and our probability law. So now let’s try to answer
the question, what is the probability that Romeo
and Juliet will meet up for their date? So all that amounts to now is
just identifying which of these 25 outcomes results in
Romeo and Juliet arriving within 15 minutes
that I’ve picked out. If Romeo arrives 15 minutes
late and Juliet arrives 30 minutes late, then they will
arrive within 15 minutes of each other. So this outcome does result in
the two of them meeting. And so we can actually highlight
all of these. And it turns out that these
outcomes that I’m highlighting result in the two them arriving
within 15 minutes of each other. So because each one has a
probability of 1 over 25, all we really need to do now
is just count how many outcomes there are. So there’s 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 13. So the probability in the end
is for the discrete case. The discrete case– I’m referring to the case where
we simplified it and considered only arrival times
with increments of 15 minutes. In this case, the probability
is 13 over 25. So now we have an idea of how
to solve this problem. It amounts to basically coming
up with a sample space, a probability law, and then
identifying the events of interest in calculating the
probability of that event. So now let’s actually solve the
problem that we really are interested in, which is that
instead of confining Romeo and Juliet to arrive in only
15-minute minute increments, really, time is continuous,
and Romeo and Juliet can arrive at any time. So they don’t necessarily have
to arrive 15 minutes late. Romeo could arrive 15 minutes
and 37 seconds late if he wanted to. So now our new sample space is
actually just, instead of only these 25 points in the grid,
it’s this entire square. So any point within the square
could be a possible pair of meeting times between
Romeo and Juliet. So that is our new sample
space, our new omega. And now let’s assign a
new probability law. And now, instead of being in the
discrete world, we’re in the continuous world. And the analogy here
is to consider probabilities as areas. So the area of this entire
square is one. And that also corresponds to the
probability of omega, the sample space. And imagine just spreading
probability evenly across this square so that the probability
of any event– which in this case
would just be any shape within this square– is exactly equal to the
area of that shape. So now that is our new
sample space and our new probability law. So what we have to do now is
just to identify the event of interest, which is still the
event that Romeo and Juliet arrive within 15 minutes
of each other. So let’s do that. If Romeo and Juliet arrive both
on time, then obviously they’ll meet. And if Romeo’s on time and
Juliet is 15 minutes late, then they will still meet. And in fact, any pairs of
meeting times between these would still work, because now
Romeo can be on time, and Juliet can arrive at any time
between 0 and 15 minutes late. But you notice that if Juliet is
even a tiny bit later than 15 minutes, then they won’t
end up meeting. So this segment here is part
of the event of interest. And similarly, this
segment here is also part of the event. And if you take this exercise
and extend it, you can actually verify that the event
of interest is this strip shape in the middle
of the square. Which, if you think about it,
makes sense, because you want the arrival times between Romeo
and Juliet to be close to each other, so you would be
expect it to be somewhere close to a diagonal
in this square. So now we have our event
of interest. We have our sample space and
our probability law. So all we have to do now is
just calculate what this probability is. And we’ve already said that
the probability in this probability law is just areas. So now it actually just boils
down to not a probability problem, but a problem
in geometry. So to calculate this area, you
can do it in lots of ways. One way is to calculate the area
of the square, which is 1, and subtract the areas
of these two triangles. So let’s do that. So in the continuous case, the
probability of meeting is going to be 1 minus the
area of this triangle. The base here is 3/4 and
3/4, so it’s 1/2 times 3/4 times 3/4. That’s the area of one
of these triangles. There’s two of them, so
we’ll multiply by two. And we end up with 1
minus 9/16, or 7/16 as our final answer. So in this problem, we’ve
reviewed some basic concepts of probability, and that’s
also helped us solve this problem of helping Romeo and
Juliet meet up for a date. And if you wanted to, you could
even extend this problem even further and turn
that they arrive within 15 minutes of each other, what
is the probability that they’ll meet, let’s say that
Romeo really wants to meet up with Juliet, and he wants to
assure himself a least, say, a 90% chance of meeting Juliet. Then you can ask, if he wants to
have at least a 90% chance of meeting her, how long should
he be willing to wait? And so that’s the flip
side of the problem. And you can see that with just
some basic concepts of probability, you can answer
some already pretty interesting problems. So I hope this problem was
interesting, and we’ll see you next time.

## 10 Replies to “Uniform Probabilities on a Square”

1. Rajagopalan P says:

you asked in the end one question that how long he's willing to wait so that there is a 90% probability of meeting her? i think even he waits for one hour and she waits for 15 minutes den also there is no 90% probability of meeting. (provided that the meeting time does not extend 60 minutes)… could u please answer the same question?

2. Randy Diaz says:

awesome content!!! thsnk you for this!

I feel like you guys are nervous though. dont be afraid to over explain !

again thank you!

3. ROB LAW says:

Did not make it clear what the initial question is! Jumping on tangents without explanation! You did not explain why you chose 15 mins increments. Just because the problem states the wait time is 15 mins does not justify the lack of explanation. . .

4. Borut Levart says:

I'm not sure why the discrete approximation is more than 50 % and the continuous answer is less than 50 %. Is that because some of the points of the discrete case fall on the edge of the continuous region?

5. Alex Zhang says:

27.75?

6. Ashish Patel says:

Nice explaination.
But I have one doubt, what is the condition:
"All pairs of arrival times are EQUALLY LIKELY" is not considered.
How will the solution change?

7. Omar Reyes Ortiz says:

So discrete vs continuous don’t necessarily need to have the same answer ?

8. ilaniam_rakup says:

given that Juliet only wait for 15 min, romeo cannot have 90% chance of meeting. the max chance if romeo is willing to wait 1 hr and juliet can only wait 15 min is 71.8%, to have 90% chance of him meeting her is that he must be willing to wait 1hr and tell juliet to wait upto 35minutes, in that case the probablity is 90.5%.

9. Bishshoy Das says:

But if one of them arrive exactly 1 hour late, they should wait for another 15 minutes right? Otherwise, there is no point in coming.

10. Eric Fay says:

About 41 min for a probability of 90%. p=1-(1-x)^2 => x=1-√(1-p) where x is in hours