Re: Birthday Problem
From: Buster (noone_at_nowhere.com)
Date: 04/12/04
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Date: Mon, 12 Apr 2004 20:09:31 +0100
John Carson wrote:
> "Buster" wrote
>
>>John Carson wrote:
>>
>>>"osmium" wrote
>>>
>>>>Given a bazillion people in a room.
>>>>The fist person will not match anyone. The next person will have a
>>>>*different* birthday with probability 364/365. The next 363/365.
>>>>And so on.
>>>
>>>Right answer to the wrong question.
>>
>>It's Step One of a correct solution.
>
> Perhaps. Provided you are prepared to go around the world a couple of times
> in order to get there.
Let p (n) be the probability that two or more people in a room have the
same birthday when n people are in the room.
"The first person will not match anyone."
Therefore the probability that all the birthdays are different is 1.
Therefore p (0) = 0;
"The next person will have a different birthday with probability
364/365."
Therefore the probability that all the birthdays are different is
1 * (364/365).
Therefore p (1) = 1 - (1 * (364/365)) = (1/365) < 95%.
"The next 363/365.".
Therefore, the probability that all the birthdays are different is
1 * (364/365) * (363*365).
Therefore, p (2) = 1 - (1 * (364/365) * (363 / 365)) < 95%.
In general, if there are n people in a room, the probability that
all the birthdays are different is /* */.
Therefore, p (n) = 1 - /* */.
This is the solution I was taught as an undergraduate.
-- Regards, Buster.
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