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What is the smallest number of people in a room to assure that the probability that at least two were born on the same day of the week is at least 40%?

I understand when approaching this type of problem, you simplify it so there's only 365 days. Also, I thought you go about the question by finding the probability that no one is born on the same day of the week. Then you subtract by 1 to get the solution:

Therefore, if the first person can have a birthday on any of 365 days, and the second is (365-8) because 1 week has to be removed (since question asks at least two born on the same day). I thought the answer is:


The solution is 4 people but when I enter r=4, I get 5.6% which is obviously wrong. Any help is appreciated. Thank you.

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With "day of the week" you should work with $7$ instead of $365$. – Hagen von Eitzen May 2 '13 at 22:36
Thanks. Misunderstood the question! – user1527227 May 2 '13 at 22:49
To @HagenvonEitzen's point, your wording is for "born on same day of the week" but your work appears to be pointing towards "born on the same week". Both are interesting questions. Which are you after? – CommonerG May 2 '13 at 22:52
up vote 3 down vote accepted

Here are the first several results for the same day of the week: $$ \begin{align} 1-\frac77&=0&\text{$1$ person}&(0\%)\\ 1-\frac77\frac67&=\frac17&\text{$2$ people}&(14.29\%)\\ 1-\frac77\frac67\frac57&=\frac{19}{49}&\text{$3$ people}&(38.78\%)\\ 1-\frac77\frac67\frac57\frac47&=\frac{223}{343}&\text{$4$ people}&(65.01\%)\\ \end{align} $$

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Apparently I misunderstood the question. Thanks to Hagen's comment, this is how I approached the question:

First, find the probability that $r$ people will each have been born on different days of the week, and then subtract this from 1 to get probability that at least two from $r$ people will have been born on the same day of the week:

$Pr(\text{At least 2 of r were born on the same day of the week.)}=1-\cfrac{1-7(7-1)...(7-r+1)}{7^r}$

Substituting: r=3, Pr=39%. r=4, Pr=65%!

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