What is in Birth Day…

In a class of 50, I had this bet.. There will be atleast 2 people in class who share same birth date(date and month only and not year).

To clarify, person A with DOB 1st January 1999 shares birth day with person  X with DOB 1st January 1965. Years can be different but dates and months are same.

What is probability that such an event might or might not occur?

Before arriving solution to above problem, brief basics of probability..

P(Event) = Count of events that meet condition / Total number Events.

  • First person:
    • As he / she is first person, DOB can be any of 365 days of total 365 days.
  • Second person:
    • First person has already chosen a DOB, assuming, in class all of them have different DOB, then there are only 364 days remaining. So probability of second person is 364/365
  • Third person:
    • First and second already shared DOB,
      • Probability: 363/365
  • Fifty’th person:
    • All prior 49 have already taken 49 days and so there are only 365- 49 days = 316 days. Probability = 316/365

As each person DOB is independent,

Combined Probability: 1×364/365…………316/365 = 0.029626

That implies there is 0.029626 = 0.03 or 3% chances that 2 persons do not share same DOB. Conversely 1-0.29626 = .97034 = 97.034% that there will be atleast 2 persons sharing same DOB.

Find surprising.. try it out..

Until next time..



My Phone Number revealed….

During recent session in Intro slide I put my entire phone number except last three digits. It is stupid to do that.. may be or may be not..

Phone number is a 10 digit number with each digit ranging from 0-9. For example if my phone number is 3021067894 then

  1. Each position of phone number can vary from 0-9
  2. And every position is independent of other position. Example 3 at first position is not dependent upon 0 at second position.

Now coming to solve “Guessing Phone Number Problem”, for each position, from total 10 numbers only 1 can be selected. So probability of selecting a number from total ten number is 1/10. And since each position is independent of every other position we have (1/10)*(1/10)…

As I revealed 7 digits of my 10 digit phone number only 3 positions have to be guessed, simple right??

Total number of combinations: (10)*(10)*(10) = 1000 and

In that only 1 number is correct = 1.

So, likely chances of getting correct number (IN FIRST TRY) is 1/1000 = 0.001 or conversely 1-0.001 = .999.

That implies 99.9% times people still will not be able to find my number Smile.

Me being stupid:

Assumption here is only 1 try what if someone writes a loop and puts all numbers together, then probability he has my number is 1 (Certain).

May be it not that stupid after all…

  • I am not a celebrity for people to search for my number spending their valuable time
  • Even if they write a loop to get my correct numbers, what is feedback mechanism to check if they have right number or if I have given first 7 digits right Smile. May be call all numbers or may be use Machine Learning to guess my numbers…

Until next guess….