At a movie theater, the manager announces that they will 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, that birthdays are distributed randomly throughout the year, etc., what position in line gives you the greatest chance of being the first duplicate birthday?

from http://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml

I've really enjoyed working to solve this one.  Though others have used good approximating formulae to find the best position, I was not completely convinced the given equations were accurate enough, and hence was not sure their conclusions were accurate.  So, I delved into the mathematics of probabilities and binomial coefficients....

Given that each person in a line is born on a non-leap day with each birthday as likely as the other, the probability of having the same birthday as one of the others prior in line and no others prior in line have matching birthdays can be given in terms of p, such that p is the position in line and 0 <= (p – 1) <= 365. The expression is:

           P(p) = [ (p - 1) * 365! ] / [ 365^p * (366 - p)! ]

• p =  1; P( 1) =  0
• p =  2; P( 2) =~ 0.00273972602739726027397260273973
• p =  3; P( 3) =~ 0.00546443985738412460123850628636
• p =  4; P( 4) =~ 0.00815174658176892012431333335047
• p =  5; P( 5) =~ 0.01077966123324328432877233487806
• p =  6; P( 6) =~ 0.01332690994931789603659867428418
• p =  7; P( 7) =~ 0.01577321944686392078852226655004
• p =  8; P( 8) =~ 0.01809958925569362234774267389966
• p =  9; P( 9) =~ 0.02028854153749766512874169003861
• p = 10; P(10) =~ 0.02232434382191095823926268838837
• p = 11; P(11) =~ 0.02419320061065540679810507478314
• p = 12; P(12) =~ 0.02588341051633133247852063480223
• p = 13; P(13) =~ 0.02738548639436500631849455956161
• p = 14; P(14) =~ 0.02869223677254361050127203968681
• p = 15; P(15) =~ 0.02979880775871336407803342709728
• p = 16; P(16) =~ 0.03070268548916357375163326686247
• p = 17; P(17) =~ 0.03140366004371068730760662911960
• p = 18; P(18) =~ 0.03190375257522868284178598126145
• p = 19; P(19) =~ 0.03220710815974737619999152080926
• p = 20; P(20) =~ 0.03231985754904329547740397209976 <<<
• p = 21; P(21) =~ 0.03224995158462575410290345882197
• p = 22; P(22) =~ 0.03200697249734432718842952865276
• ...

The maximum value, before the probability begins decreasing, occurs when p is 20. Therefore, being the 20th person in line gives one the greatest probability of winning the free ticket.

Posted by Dej Mar on 2008-03-18 04:45:15