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 Birthday Line (Posted on 2004-03-29)
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

 See The Solution Submitted by Victor Zapana Rating: 3.5556 (9 votes)

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 re(3): Function [with observations] | Comment 14 of 21 |
(In reply to re(2): Function [with observations] by Richard)

What I did was simpilify the problem.

I supposed there was a 6 day 'year' and stated making a tree diagram.

P(1) = 0/6 (Six days, none are winners)
P(2) = 6/36 (36 pairs of days, six are winners)
p(3) = 60/216 (216 trios of days, 6 ways to match 1st person times 5 ways to match second times 2 winning days)
p(4) = 3x4x5x6/6^4
p(5) = 4x3x4x5x6/6^5
p(6) = 5x2x3x4x5x6/6^6
p(7) = 6x1x2x3x4x5x6/7^6

I figured this was right because these added up to 1

For 6 days in a year maximum probability p(3)=p(4)
Once I had a formula I made a TI-83 calculator program to make a list of probabilitites for each number of days to see what the max. prob. was for each.

The numbers with ties were 2, 6, 12, 20, 30, 42, 56, overflow (57^58 is too big)
with ties for 2&3, 3&4, 4&5, 5&6, 6&7, 7&8, 8&9 respectively.
This fits a polonomial of degree two. I can't see how to prove it does, so it will have to do.

Hope this makes a shred of sense.

-Jer

 Posted by Jer on 2004-04-02 14:32:18

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