Probability that 3 people meet at the same place

Question

$3$ people arrive at a place separately at a random time between $9$ am and $10$ am. Each person remains there for $15$ minutes. What is the probability that all $3$ meet at the same time?

I understand how this works if it's only $2$ people. If $9:00$ is $0$ and $10:00$ is $1$, then it can be graphed $2d$. There would be a $x$ and a $y$ for each person. Since one person has to be within $\dfrac14$ of an hour to another person, then it can be said that $x- \dfrac14 < y < x + \dfrac14$.

Here is a visual representation of that work:

So then the probability would be $\dfrac7{16}$ for two people case

But then we turn to $3$ people. If I were to use a similar approach then I could say that

$$y - \dfrac14 < x < y + \dfrac14$$

$$z - \dfrac14 < x < z + \dfrac14$$

But then I don't know what this graph shape would look like? Or do I need more inequalities (two with $y$ in the middle, two with $z$ in the middle?) Bit confused from here.

Answer 1

Suppose I wanted the probability that they all meet at 9:15. What is the probability? Suppose I represent the probability that they meet at time $t$ as $f(t)$. Can you find a formula for $f(t)$? How do you go from $f(t)$ to the total probability?

Answer 2

My method of working it out was different from yours. It gives the same answer, of course, but I think you'll find it easier to generalize. Suppose we know when $x$ arrives. What is the probability that $y$ arrives in time to meet him? If $x$ arrives between $9:15$ and $9:45$ there is a half-hour period in which $y$ can arrive so that they meet. The probability that both these events occur is $1/4$. If $x$ arrives at $9:10$ though, there are only $25$ minutes when $y$ could arrive. Similar remarks apply if $x$ arrives after $9:45$. We can compute the sum of these probabilities as $$ 2\int_{3/4}^1{(.25+(1-x))dx} = 2\frac{5}{4}\frac{1}{4}-\int_{3/4}^1{2xdx}=\frac{5}{8}-(1-\frac{9}{16})=\frac{3}{16}. $$ Together with the 1/4 we had to begin with, this gives $7/16$.

You can easily extend this to three people meeting, for $y$ and $z$ have to arrive in the same interval, if they are both to meet $x$.

Answer 3

Using Geometry.

The combined grey areas cover the two person case.

and the area is $1 - (\frac {3}{4})^2$

Now we need to exclude the cases when z is early.

If $x,y$ meet up at some time in the light grey area, then it is impossible for z to be early. If $x,y$ meet up at some time in the dark grey area, then we have this pair of triangular prisms to exclude.

And there is also a pair of triangular prisms to exclude when z is late, and these excluded region are symmetrical.

Exclude 4 triangular prisms each $\frac 14 \times 34 \times 34$

$\frac 7{16} - (\frac 34)(\frac 14)(\frac 34)\cdot 2 = \frac 5{32}$

Using calculus:

WLOG assume $x\ge y\ge z$

$\frac {\int_9^{9.25}\int_{9}^x\int_{x}^z \ dy\ dz\ dx+\int_{9.25}^{10}\int_{x-0.25}^x\int_{z}^x \ dy\ dz\ dx}{\int_8^9\int_{8}^x\int_{z}^x \ dy\ dz\ dx}$