# 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

of each other. So let’s start with this one

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

it on its head. And instead of calculating given

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”

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?

awesome content!!! thsnk you for this!

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

again thank you!

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. . .

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?

27.75?

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?

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

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%.

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.

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