\problemdata      
        {probability-conditional-7} % latex-friendly label for the prob.
        {conditional probability, false proof} % The topic(s) of the problem content.
        {F01 Tut11} % Source (if known)
        {S02 cp9F; F01Tut11} % Usage (list ones you are aware of).
        {Albert Meyer, S02} % Last revision info (author, date).

\begin{problem} \textbf{(Birthdays Again)}
Suppose there are 100 people in a room. Assume that their birthdays
are independent and uniformly distributed. As stated in lecture notes, 
with probability $> 99\%$ there will be two that have the same
birthday. 

Now suppose you find out the birthdays of all the people in the room 
except one---call her ``Jane''---and find all 99 dates to be different.

\begin{problemparts}

\ppart 
What's wrong with the following argument:
\begin{quotation}\noindent
With probability greater than 99\%, some pair of people in the room
have the same birthday.  Since the 99 people we asked all had
different birthdays, it follows that with probability greater
than~99\% Jane has the same birthday as some other person in the
room. 
\end{quotation}

\solution{
Here's the problem with the argument: Let $A$ be the event that some two
people in the room have the same birthday.  Let $B$ be the event that the
99 people we asked all have different birthdays.

It is true that $\pr{A} > 0.99$ (that is indeed the probability
spoken about in the lecture notes). However, that is the \emph{a
priori} probability, i.e., assuming all the birthdays are
uniform and independent, with no other constraints. The argument
above makes the erroneous assumption that event $A$ has
probability of at least $99\%$ even once we know that event $B$
holds. But once we know that event $B$ holds, the birthdays are
no longer independent. Thus $\pr{A \mid B}$ is not necessarily
equal to $\pr{A}$ (in our case, they are actually quite
different, as will be computed in part~(b)).
}

\ppart 
What is the actual probability that Jane has the same birthday
as some other person in the room?

\solution{
Let $S$ be the set of birthdays of the 99 people in the room
other than Jane. By assumption, $|S| = 99$. Let $b$ be the date
of Jane's birthday. Since $b$ is uniformly distributed over a
set of size 365, and $b$ is independent of all the birthdays in
$S$, we have
\[
\pr{A\mid B} = \pr{b\in S} = 
\frac{\card{S}}{365} = 
\frac{99}{365} \approx 27.1\%,
\]
where $B$ is the event that the 99 people we asked all have
different birthdays.
}

\end{problemparts}
\end{problem}
