Simulating the birthday problem

What is it about?

What is the probability that in a class of $n$ people at least two share a birthday? Many people guess only a few percent for $25$ people – in fact it is almost $57\,\%$ .

Model: Each of the $365$ days is equally likely (no leap days, no twins), and the birthdays are independent.

Calculation via the complement (view “Calculation”): Instead of “at least two equal” we compute “all different”. The people arrive one by one: the first may have any day, the second must hit one of the $364$ free days, the third one of the remaining $363$ … By the path rule we multiply:

P(\text{all different}) = \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \cdot \ldots \cdot \frac{365 - n + 1}{365}, \qquad P(\text{at least two equal}) = 1 - P(\text{all different}).

$n$	$10$	$23$	$30$	$50$
$P(\text{at least two equal})$	$11.7\,\%$	$50.7\,\%$	$70.6\,\%$	$97.0\,\%$

Why so high? – Counting pairs (view “Pairs”): It is not about one particular day but about any pair. $n$ people form $\frac{n \cdot (n-1)}{2}$ pairs, already $253$ for $23$ people. Each pair shares a birthday with probability $\frac{1}{365}$ . If we pretend the pairs are independent, we get the good approximation

P \approx 1 - \left(\frac{364}{365}\right)^{\frac{n(n-1)}{2}}, \qquad n = 23: \; 1 - \left(\tfrac{364}{365}\right)^{253} \approx 50.0\,\%.

Compare: “on my birthday”. Here only the $n$ pairs “me – other person” count:

P(\text{someone on my day}) = 1 - \left(\frac{364}{365}\right)^{n}, \qquad n = 25: \approx 6.6\,\%.

Only from $253$ other people on is this value above $50\,\%$ .

Simulation (view “Many classes”): Each tile is a randomly drawn class: filled with, light with a border without a shared birthday. The relative frequency $h = \frac{\text{hits}}{\text{classes}}$ varies a lot at first and then approaches the probability $P$ (empirical law of large numbers). The light funnel shows where $h$ lies after $N$ classes in about $95\,\%$ of all cases: $P \pm 1.96 \cdot \sqrt{\frac{P(1-P)}{N}}$ .

Same month, same weekday: The same calculation works with $12$ months or $7$ weekdays. With only $5$ people, two share a birth month with probability $61.8\,\%$ . From $13$ people on it is certain: there are only $12$ months, so two must share one (pigeonhole principle). The product then contains the factor $\frac{0}{12}$ .

Try it

Press New class several times. How often do you find a shared day? One class
Tap your birthday in the calendar and compare the two questions. My birthday
Start 1000 classes and watch the curve of the relative frequency settle into the funnel. Simulation
Watch the free days shrink. From which person on is $P(\text{all different})$ smaller than $0.5$ ? Calculation
Tap a person in the pairs view: how many pairs does it form? How many pairs are there in total? Pairs
Switch the pairs view to “on my birthday”. Why are there suddenly far fewer lines? Only my pairs
Choose “same month” and increase $n$ from $5$ to $13$ . What happens at $13$ ? Same month · 13 people

Tasks

Task 1: Three people

Find the probability that at least two of three people share a birthday. Load task

Show solution

$P = 1 - \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365} \approx 1 - 0.9918 = 0.0082$ , so about $0.82\,\%$ . Show in the simulation

Task 2: Your class

Find the probability for a class size of $28$ . About how many of $1000$ classes of this size have a shared birthday? Check with the simulation. Load task

Show solution

For $n = 28$ , $P \approx 65.4\,\%$ , so about $654$ of $1000$ classes. The simulation gives values around this; after $1000$ classes $h$ almost always lies between about $0.62$ and $0.68$ . Show in the simulation

Task 3: Pairs and approximation

How many pairs do $30$ people form? Use this to compute the approximation and compare it with the exact value $70.6\,\%$ . Load task

Show solution

$\frac{30 \cdot 29}{2} = 435$ pairs. Approximation: $1 - \left(\frac{364}{365}\right)^{435} \approx 69.7\,\%$ – just below the exact value $70.6\,\%$ . Show in the simulation

Task 4: My birthday

What is the probability that one of $25$ other people has a birthday on exactly your birthday? Why is this so much less than $56.9\,\%$ ? Load task

Show solution

$1 - \left(\frac{364}{365}\right)^{25} \approx 6.6\,\%$ . Here someone has to hit one particular day, and there are only $25$ pairs “me – other”. In the birthday problem any of $300$ pairs is enough. Show in the simulation

Task 5: Same weekday

What is the probability that at least two of $4$ people were born on the same weekday? From how many people on is it certain? Load task

Show solution

$P = 1 - \frac{7 \cdot 6 \cdot 5 \cdot 4}{7^4} = 1 - \frac{840}{2401} \approx 65.0\,\%$ . From $8$ people on it is certain (pigeonhole principle: only $7$ weekdays). Show in the simulation

Task 6: Model check

In reality, birthdays are not spread quite evenly over the year. Does this make a shared birthday more or less likely?

Show solution

Rather more likely: if some days are more frequent, birthdays cluster there and matches become more likely. The even distribution is the “least favourable” case for shared birthdays.

Ideas for the classroom

Use: Lower secondary, multi-stage random experiments and stochastic simulations. Let the class estimate before calculating – and perhaps take a look at the birthdays in your own class.
Simulation and calculation: The view “Many classes” makes the surprising size plausible, the view “Calculation” explains it. The funnel shows how much relative frequencies still vary after $100$ classes.
Same month works well as a game in class: even in small groups a shared month almost always turns up.
Compare groups: Each device draws its own classes. Collect the numbers of hits after $100$ classes each on the board – the spread between the groups fits the funnel well.
Common stumbling blocks:
- The birthday problem is confused with “someone shares my birthday”.
- Computing $\frac{n}{365}$ – which is even greater than $1$ for large $n$ .
- Adding the factors $\frac{364}{365}, \frac{363}{365}, \ldots$ instead of multiplying them.