Find the probability that among 25 randomly selected people, at least 2 have the same birthday

Question

For the above classic birthday problem, a simulation begins by representing birthdays by integers from 1- 365, where 1 represents January 1st and so on. Randomly generate 20 different groups of 25 birthdays. Use the result to estimate the probability that among 25 randomly selected people, at least 2 have the same birthday.

Answer 1

The Classic Birthday Problem:

 

The givens and the perspective:

(1) In the room, there are 25 people (no twins) who have independent birthdays.

(2) Assume the 365 days in each year have an equal (uniform) probability that each day could be the birthdate for a person (p=1/365 for each day).

(3) Now, restate the problem from the perspective of the calendar dates. As you select people from the room, how likely is it that the remaining people in the room have unique birthdates (note: the probability that at least two of them share a birthdate is 100% minus this value).

Solution:

For each person selected (and date eliminated), the number of open dates is one less. So, for only two people in a room, the probability that they share a birthdate is (1 – probability of unique birthdays)
P₁ = 1 - (365/365)*(364/365).

For just three people in a room, the probability that at least two share a birthdate is
P₂ = 1 - (365/365)*(364/365)*(363/365).
. . .
For 25 people in a room, the probability that at least two of them share a birthdate is
P₂₅ = 1 - (365/365)*(364/365)*(363/365)*(362/365)* … *(341/365) [note: 25 ratios]
P₂₅ = 0.5687

The 50% probability is first exceeded with only 23 people in the room (P=0.5073). Perhaps surprisingly, 99% probability is achieved with only 57 people in the room. When the room holds 366 people, there must be a duplicate birthday!

 

Simulation:

 

  Note that a PSEUDO-RANDOM NUMBER generator is used.

 

  "Randomly generate 20 different groups of 25 birthdays.  Use the result ..."   means

      Do this 20 times and store resulting counter

         Generate a group of 25 birthdays (that is, 25 random integers)

         Determine whether any two or more of them is the same

             Set a counter to 0

             For each birthday

                For each remaining birthday   [note: previous are already considered]

                   If this birthday = remaining birthday THEN increment counter

                Next remaining birthday

              Next birthday

         Store resulting counter in an array

       End Do

 

"Use the result to estimate the probability"    means

   Count the non-zero counters 

             [in the array of counters;  these groups had multiple birthdays]

       Set counter to 0

       For 20 array elements

           If array element <> 0 Then increment counter

       End For

   Output this counter

 

Now, note the difference between (1) Theoretical Probability (the mathematical value obtained in the "Solution" above), the Simulated Probability (from running this computer program), and (3) the Experimental Probability (from actually asking "people in a room").  The Simulated Probability and the Experimental Probability change each time.  But -- with many, many of these simulations or experiments, we expect then to approach the Theoretical Probability.