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Topic: HARD: BIRTHDAY LINE (Read 1453 times) |
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S. Owen
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HARD: BIRTHDAY LINE
« on: Aug 14th, 2002, 7:15am » |
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Is there a more elegant way to analyze this?
Your probability of getting a free ticket when you are the nth person is line is:
(probability that none of the first n-1 people share a birthday) * (probability that you share a birthday with one of the first n-1 people)
(1 * 364/365 * 363/365 * ... * (365-(n-2))/365) * ((n-1)/365)
Fire up that computer and you'll find that this is maximized for n = 15.
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Tingx
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Hey Owen!
Did what you did, and arrived at the same expression for P(n)...
Continued like this, we want least n such that P(n) > P(n+1), so get (after some simplifying)
n2+n-365 > 0
(n-18.6)(n+19.6) > 0
Inequality is satisfied in the regions
n > 18.6 or n < -19.6 (rejected)
So, n = 19.
Did not have time to check the working, but idea seems right to me, but answer different from yours. 
Cheers, Tingx
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Tingx
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oops...should be like this....
n2-n-365 > 0
(n+18.6)(n-19.6) > 0
Inequality is satisfied in the regions
n > 19.6 or n < -18.6 (rejected)
So, n = 20.
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S. Owen
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Re: HARD: BIRTHDAY LINE
« Reply #3 on: Sep 2nd, 2002, 8:07pm » |
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You are right - I ran the numbers again and got 20! I'm not sure what I did wrong the first time. Thanks for pointing out an easier way to solve it.
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