Sunday, May 21, 2006

Birthday matching

At a movie theater, the manager announces that he would give a free ticket to the first person in line whose birthday is the same as someone who has already bought a ticket. You have the option of getting in line at any time. Assuming that you don't know anyone else's birthday and that birthdays are distributed randomly throughout the year what position in line gives you the greatest chance of being the first duplicate birthday so you can get a free ticket?


Anonymous Anonymous said...

Bobach writes:
I would ask to be 23rd in line.
First figure out the chances all n people all have different birthdays. The first person could be any day. The second person has 364 out of 365 chances of not having
that same birthday. The third person has a 363 out of 365 chance of not having either of the 2 birthdays of those who came before. Leading to:
pn(n) = 1 times (365-1)/365 times
(365-2)/365 ... times
where pn(n) is defined as the probability that n people all have different birthdays.
The probability of at least 2 having the same birthday is 1-pn(n).
The key is to find at what n, 1-pn(n) > 0.5. This happens to be 23, where the probablility of 1-pn(n) is approximately 0.507.
While it could happen earlier if you wait that long, and you certainly could be in line too soon, this represents your best odds to be the first.

5:53 PM, May 21, 2006  
Blogger Karthik Nagaraj said...

You are right. Its surprising how many people cannot believe that in such a low number of people your birthday could match with them with a probability of a little over 50%.

Lets add a twist to the problem: What happens if you do not know how many people are already inside and you just arrived and this was told to you by somebody. Does it change the position you decide to wager on? Think about it!

2:17 PM, May 22, 2006  
Blogger Karthik Nagaraj said...

This puzzle is SOLVED!

9:53 AM, June 03, 2006  

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