Probability Foundations II
Agenda
- Lecture, Combinatorics and Computing Probabilities
- WS Part 1
- Lecture, Simulating Code in R
- WS Part 2
- Fun Live Experiment: The Birthday Problem!
Lecture Part 1
Combinatorics and Computing Probabilities
60:00
Break
05:00
Worksheet Part 1
25:00
Lecture Part 2
Simulation Code in R
25:00
Break
05:00
Worksheet Part 2
20:00
Fun Experiment, The Birthday Problem!
- Here is a question: How many people do you need in a room before there is a 50% chance at least 2 of them share a birthday. Guesses?
-Let’s see if we can work it out with probability, and then see if anyone in the room shares a birthday (don’t check yet!)
Setting Up the Problem
Say we have \(K\) people in a room
Define the event \(A = \{\text{At least two people in a room share a birthday}\}\)
Consider the complement event, \(A^{C} = \{\text{no one in the room shares a birthday}\}\)
In this case, computing \(P(A^{C})\) seems like the easier route, so we will do that
Two People In a Room
- Consider two people in a room. Total number of birthdays possible is \(365^2\).
- For them NOT to share a birthday, \(365 \times 364\). Person 1 can have any day of the year, person 2 has to have not that day.
\[P(\text{2 people don't share birthday out of 2}) = \frac{365 \times 364}{365 \times 365}\]
\[P(\text{At least 2 people share a birthday out of 2}) = 1- \frac{365 \times 364}{365 \times 365}\]
Three People In a Room
- Now consider a third person, total birthdays possible \(365^3\)
- Yet third person cannot share first or second’s birthday, so only 363 days left for them
\[P(\text{2 people don't share birthday out of 3}) = \frac{365 \times 364 \times 363}{365 \times 365 \times 365}\]
- Notice numbers on top are getting smaller, on bottom staying same size.
\[P(\text{At least 2 people share a birthday out of 3}) = 1-\frac{365 \times 364 \times 363}{365 \times 365 \times 365}\]
K People In a Room
- Recognizing the pattern here, we see that with \(K\) people in a room, the probability no one shares a birthday is
\[P(\text{K people don't share birthday}) = \frac{365}{365} \times \frac{364}{364}\times \frac{363}{365} \times \cdots \frac{(365-K+1)}{365}\] \[P(\text{At least 2 people share a birthday out of K}) = 1-\frac{365}{365} \times \frac{364}{364}\times \frac{363}{365} \times \cdots \frac{(365-K+1)}{365}\]
It turns out, this equals 0.5 at 23 people! Not really that many at all!
Let’s test it out, make (hopefully 2) groups of 23 and lets see if anyone in those groups shares a birthday!
