Help with probability (the birthday problem)
- Thread starter Jaskaran
- Start date
What is the probability that 2 or more people in a group of 20 have the same birthday?.
You're correct. The 'trick' is to count the probability that no one has the same birthday and subtract from 1.
The denominator, all possibilities of birthdays for 20 different people is \(\displaystyle 365^{20}\)
The numerator, the possibilities where evryone has a different birthday is \(\displaystyle P(365,20)\)
Thus, we have \(\displaystyle 1-\frac{P(365,20)}{365^{20}}\)
You're correct. The 'trick' is to count the probability that no one has the same birthday and subtract from 1.
The denominator, all possibilities of birthdays for 20 different people is \(\displaystyle 365^{20}\)
The numerator, the possibilities where evryone has a different birthday is \(\displaystyle P(365,20)\)
Thus, we have \(\displaystyle 1-\frac{P(365,20)}{365^{20}}\)
Hello, Jaskaran!
This is a classic (old) problem with a surprising result.
Consider the case with 30 people in the classroom.
We will calculate the probability that no one has the same birthday.
. . That is, that everyone has a different birthday.
The first person can have any birthday. .He/she can be ignored.
The second person can have any of the other 364 birthdays.
. . This probability is \(\displaystyle \frac{364}{365\)
The third person can have any of the other 363 birthdays.
. . This probability is \(\displaystyle \frac{363}{365}\)
The fourth person can have any of the other 362 birthdays.
. . This probability is \(\displaystyle \frac{362}{365}\)
. . . . . \(\displaystyle \vdots\)
The 29th person can have any of the other 337 birthdays.
. . This probability is \(\displaystyle \frac{337}{365}\)
The 30th person can have any of the other 336 birthdays.
. . This probability is \(\displaystyle \frac{336}{365}\)
The probability that no one has the same birthday is the product of these probabilities.
. . \(\displaystyle P(\text{no matches}) \:=\:\frac{364}{365}\cdot\frac{363}{365}\cdot\frac{362}{365}\,\cdots\,\frac{337}{365}\cdot\frac{336}{365} \:=\:0.299387334\)
That is, the probability of no matching birthdays is about 30%.
Therefore, in a group of 30 people, the probability of some matching birthdays
. . (two or more) is a surprising 70%.
This is a classic (old) problem with a surprising result.
Consider the case with 30 people in the classroom.
We will calculate the probability that no one has the same birthday.
. . That is, that everyone has a different birthday.
The first person can have any birthday. .He/she can be ignored.
The second person can have any of the other 364 birthdays.
. . This probability is \(\displaystyle \frac{364}{365\)
The third person can have any of the other 363 birthdays.
. . This probability is \(\displaystyle \frac{363}{365}\)
The fourth person can have any of the other 362 birthdays.
. . This probability is \(\displaystyle \frac{362}{365}\)
. . . . . \(\displaystyle \vdots\)
The 29th person can have any of the other 337 birthdays.
. . This probability is \(\displaystyle \frac{337}{365}\)
The 30th person can have any of the other 336 birthdays.
. . This probability is \(\displaystyle \frac{336}{365}\)
The probability that no one has the same birthday is the product of these probabilities.
. . \(\displaystyle P(\text{no matches}) \:=\:\frac{364}{365}\cdot\frac{363}{365}\cdot\frac{362}{365}\,\cdots\,\frac{337}{365}\cdot\frac{336}{365} \:=\:0.299387334\)
That is, the probability of no matching birthdays is about 30%.
Therefore, in a group of 30 people, the probability of some matching birthdays
. . (two or more) is a surprising 70%.
