Probability of winning sweepstakes when some information is not given I wanted to give my students an interesting probability problem, and found myself perplexed by it.
Below is the probability problem I came up with. 
To win some prize a participant in a sweepstakes has to guess a four digit number. The designers of the sweepstakes created it in such a way that for every digit, any number from 0 to 9 is possible.
The creators of the sweepstakes did not inform well the participants about how the digits were selected. What is the probability that the first participant wins the prize if she thinks that no number can be repeated in the sequence?
Intuition: The probability of the first participant winning when understanding the rules is 1/10000 (1 out of 10^4 possibilities). By thinking that no number can be repeated, the participant may not consider numbers that may very well be the winning one. Thus, there probability of winning then should be lower than 1/10000. 
Let W = win (chose winning number), C = winning number has no repeated digits. Then 
P(W) = P(W|C)P(C) + P(W|~C)P(~C)
Naturally, if the winning number had repeated digits, then the participant can't win due to the misunderstanding. Thus P(W|~C)=0 and
P(W) = P(W|C)P(C)
Now P(W|C) = 1/5040 (1 out of 10*9*8*7 possibilities) and P(C) = 5040/10000 (from 10*9*8*7/10000). Therefore P(W) = 1/5040 (5040/10000)=1/10000, same as if the participant did not misunderstand the rules.
Where did I go wrong?

It makes no difference how you come up with your guess, the probability of guessing correctly will always be $\frac 1{10^4}$.
To write this out in your case:  Let $p(n)$ be the probability that $n$ is the correct answer (so $p(n)=\frac 1{10^4}$ as we are assuming that the true distribution is uniform).  Let $\psi(n)$ be the probability that you guess $n$.
Now, under the false rule you have introduced, you are choosing uniformly at random from the $5040$ possible four digit numbers with distinct digits.  Thus:  $$  \psi(n) =
\begin{cases}
\frac {1}{5040},  & \text{if $n$ has distinct digits} \\
0, & \text{if $n$ does not have distinct digits}
\end{cases}$$
The probability of guessing correctly is given by $$P=\sum_{n=0}^{9999} p(n)\psi(n)=\frac 1{10^4}\times \sum_{n=0}^{9999} \psi(n)$$
But that last sum is $\frac 1{5040}$ times the number of $n$ with distinct digits, hence the sum is $1$, making $P=\frac 1{10^4}$ as claimed.
The problem with your calculation comes when you write $P(C)=\frac {5040}{10000}$.  What does that mean?  $P(C)$ is, I believe,  the probability that, following the false rule, I choose a number consistent with the false rule...so $P(C)=1$.  If, instead, you meant that $P(C)$ was the probability that a uniformly random number in the range happened to satisfy the false rule, then $P(C)$ is what you say but in that case of course $P(\sim C)$ would not be $0$.  Indeed we'd have $P(\sim C)=\frac {4960}{10000}$ in which case your calculation would give the correct answer of $\frac 1{10^4}$.  Note that whatever you might have meant by $P(C)$ we should certainly have $P(C)+P(\sim C)=1$.
