Geometric probability and checking by simulation

What you will be able to do

• find a probability as a ratio of lengths on a line, or of areas in a square;
• set up the unit square for two numbers chosen at random;
• check an answer with a simulation, and say how large the usual error is.

The bus that comes every fifteen minutes

A bus arrives at a stop every $15$ minutes. You reach the stop at a random moment. What is the chance that you wait at most $5$ minutes? The moments in a gap of $15$ minutes are too many to list. But they are spread evenly. So the chance of an event is the length of the time that makes it happen, divided by the whole length. You wait at most $5$ minutes if you arrive in the last $5$ of the $15$ minutes, so the chance is $\dfrac{5}{15} = \dfrac{1}{3}$.

The rule

If a point is chosen at random on a line segment, with all positions equally likely, then

In a region of the plane, replace length with area. This is called geometric probability. Three facts are worth remembering.

• The words "at random" mean that every position is equally likely. Say it, because in geometry another meaning would give another answer.
• The chance that the point lands on one exact position is $0$, and yet the point lands somewhere. So an event of chance $0$ is not always impossible.
• The chance of a part does not depend on where the part is, only on its size.

A rod. A rod of length $12\ \text{cm}$ is cut at a random point. For both pieces to be at least $3\ \text{cm}$, the cut must lie between $3\ \text{cm}$ and $9\ \text{cm}$ from one end. That is a length of $6\ \text{cm}$, so the chance is $\dfrac{6}{12} = \dfrac{1}{2}$.

Two numbers at random: the square

Two friends each arrive at a random time in one hour, and each waits $15$ minutes. The pair of arrival times is a point $(x, y)$ in a square with sides from $0$ to $60$ minutes. Every point of the square is equally likely. They meet when $|x - y| \le 15$. This is a band along the diagonal. The part of the square where they do not meet is two triangles, each with sides $45$ minutes. So

Checking an answer by simulation

Sometimes a probability is hard to be sure about. A simulation lets a computer repeat the experiment many times with random numbers, and counts how often the event happens. The fraction of successes is an estimate of the probability.

How good is the estimate? With $N$ trials and a true probability $p$, the usual size of the error is about $\sqrt{\dfrac{p(1 - p)}{N}}$. For $N = 10\,000$ and $p = 0.5$ that is $0.005$. Ten times more trials make the error only about $3$ times smaller.

Three rules for a simulation.

1. It estimates. It does not prove. It can show that a formula is off by a lot, or that it is plausible.
2. The error shrinks slowly. Quote the answer as close to the estimate, never as equal.
3. The seed. A seed makes the random numbers the same every run, so you can repeat a result. A different seed gives a slightly different estimate, and both are fine.

A worked example

Estimate the chance that the two friends meet, when each waits $15$ minutes, by simulation. Then compare with $\dfrac{7}{16}$.

1. The experiment. Choose $x$ and $y$ at random between $0$ and $60$. The event is $|x - y| \le 15$.
2. The estimate. In $20\,000$ trials, a run of the program below gives a number near $0.44$. The exact value is $\dfrac{7}{16} = 0.4375$.
3. The error. The usual error is $\sqrt{\dfrac{0.4375 \times 0.5625}{20\,000}} \approx 0.0035$. A result between $0.43$ and $0.445$ is expected. A result of $0.40$ would be a warning that something is wrong.

Practise with new numbers

Try it in Python

The program simulates two experiments from this chapter: a shared birthday in a group of $23$, and two friends who meet. It prints each estimate next to the exact value, and the usual size of the error.

Common mistakes

• Not saying what "at random" means. Equally likely positions give lengths and areas. Another rule gives another answer.
• Using a length in a two-dimensional problem. Two random numbers need the square, and an area.
• Trusting one short simulation. With $200$ trials the error can be about $0.035$.
• Calling an estimate exact. A simulation gives an estimate.

Remember these

• Geometric probability: $\dfrac{\text{length}}{\text{total length}}$ on a line, $\dfrac{\text{area}}{\text{total area}}$ in a region.
• Two numbers in $[0, a]$: the sample space is a square of side $a$. Two friends who wait $w$ meet with probability $1 - \left(\dfrac{a - w}{a}\right)^2$.
• A simulation has a usual error of about $\sqrt{\dfrac{p(1 - p)}{N}}$.
• A simulation checks an answer. It does not prove it.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A point is chosen at random on a segment of length $30\ \text{cm}$. Find the probability that it lies within $4\ \text{cm}$ of the midpoint.

2. A bus comes every $20$ minutes. You arrive at random. Find the probability that you wait more than $5$ minutes.

3. Two friends arrive at random in one hour and each waits $10$ minutes. Find the probability that they meet.

4. A simulation of $10\,000$ trials estimates a probability near $0.2$. About how large is the usual error?

5. Why is the chance of hitting one exact point on a line equal to $0$, although the point must land somewhere?

References

• OpenStax, Introductory Statistics, Chapter 5, sections on continuous probability functions and the uniform distribution.