Birthday matching
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As the name of this site is, 'PuzzleGuru', it is infact not run by a Puzzle Guru at all. It is run by all of us for all of us to become puzzle gurus and increase our technical know-how and improve our knowledge in problem solving. So let's join our hands and try our heads together on this.
3 Comments:
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
(365-n)/365
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.
Bobach,
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!
This puzzle is SOLVED!
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