Source: wikipedia
Problem: There is a big line of people waiting outside a theater for buying tickets. The theater owner comes out and announces that the first person to have a birthday same as someone standing before him in the line gets a free ticket. Where will you stand to maximize your chance.
Problem: There is a big line of people waiting outside a theater for buying tickets. The theater owner comes out and announces that the first person to have a birthday same as someone standing before him in the line gets a free ticket. Where will you stand to maximize your chance.
When I talk that what is the minimum number of people required so that at least two people share their b’days ,its 23. Here we look for the cases where probability touches 50% for the ist time and it occurs at 23rd position. In this case we have 23C2 number of pairs to compare their respective b’days. If you put p(x)= 1-e^(-x^2)/2*365 > 0.5
ReplyDeleteYou will get approx 23.
While ,in case there is a contest I chose P(x)-p(x-1) to be maximized.its nothing but mathematical form of the statement that kth person has k choices of matching bdays and also nobody has won yet who are standing in front of him.
Here no 50% thing comes,its just a simple case of finding the maxima in the curve of p(x)-p(x-1) .
In simple words for the 2nd person to win ,the conditions to be satisfied are:
1.Nobody in front of him won yet
2.he has 1 choice of dates out of 365 to match his bdays(because only one person is standing in front of him)
For kth person in queue,his winning is dependent on:
1.no body in k-1 persons standing before him won yet.
2. kth person has k-1 out of 365 to chose from ,to match his b’day.
Find the probability ,p(k)= p(no one won yet) *[(k-1)/365]
This gets maximized at 20.