Calculating the probabilty of drawing a specific number A friend and I got into a small argument the other day, but neither of us are good enough with probability to get a definitive answer.
Say there is game with 100 participants. Each person draws a random number from a bag, the numbers being 1-100. Every person draws one number, and the number is not put back into the bag after it has been drawn. Drawing #1 wins you the lottery, and I have been given the choise of where in the line of 100 people drawing numbers I want to stand.
Question is then, where should I stand to have biggest chance of drawing #1?
We ended up concluding the probability is the same regardless of where in the line you stand, you'll always have a 1/100 chance. We arrived at this answer by using a smaller example of 4 people and 4 numbers. Intuitively though, neither of us could really accept that answer, but we can't really see any other way to look at it.
 A: You got the right idea. One way to see why is to calculate the odds of winning for each participant. Let's call the participants $P_1,P_2,\ldots,P_{100}$.
For $P_1$, the probability of winning is $\frac{1}{100}$.
For $P_2$, the probability of winning is $\frac{1}{99}$, if $P_2$ gets to play. $P_2$ only gets to play if $P_1$ loses, and the probability that $P_1$ loses is $\frac{99}{100}$. Hence the probability that $P_2$ wins is $\frac{99}{100}\cdot\frac{1}{99}=\frac{1}{100}$.
For $P_3$ to win, $P_1$ and $P_2$ must lose. Hence the probability of $P_3$ winning is $\frac{99}{100}\cdot\frac{98}{99}\cdot\frac{1}{98}=\frac{1}{100}$.
The probability that $P_4$ wins is $\frac{99}{100}\cdot\frac{98}{99}\cdot\frac{97}{98}\cdot\frac{1}{97}=\frac{1}{100}$.
The probability that $P_5$ wins is $\frac{99}{100}\cdot\frac{98}{99}\cdot\frac{97}{98}\cdot\frac{96}{97}\cdot\frac{1}{96}=\frac{1}{100}$.
For each $P_n$, the probability that $P_n$ wins will be a similar telescoping product that is equal to $\frac{1}{100}$.
A: This point is often a source of confusion.
If you like, you can prove it inductively.  Small $n$ pose no difficulty, so suppose we have shown it up to collections of size $n-1$.  
Now, you have a collection with $n$ tickets (and one winning one). Draw one from the lot.  The probability that it is the winning one is  $\frac 1{n}$.  If it is not the first choice then you have a collection of $n-1$ tickets with one winning one.  By the inductive hypothesis, the probability that the winning one is in any of the slots numbered $2$ to $n-1$ is now $\frac 1{n-1}$.  Thus, prior to drawing the first ticket, the probability that the winning one was in a specifed slot between $2$ and $n-1$ was $$\frac {n-1}n\times \frac 1{n-1}=\frac 1{n}$$ as desired.
