Which person has the highest probability of picking the correct ball? 100 balls in one urn, each with different number from 1-100 written on. 
100 people line up to randomly pick one ball for each of themselves. 
Which person (by the order of picking) has the highest probability of picking ball with #99 on it?
 A: Ms Nobody. Or to put it another way, Ms Everybody.  All sequences of numbers are equally likely. So for every $n$ with $1\le n\le 100$, the probability that Ball $99$ is in the $n$-th ball to be chosen is $\frac{1}{100}$. 
A: The order should not make a difference; everybody has an equal chance of selecting the right ball. Here's a proof by the naive approach:
Suppose you're the $n^{th}$ person to pick a ball.  Then the probability that you pick the correct ball is the probability that the following two events occur:


*

*those before you don't pick ball #99

*after that happens, you pick #99


The probability that no one before you picks the correct ball is
$$
\frac{99}{100}\cdot\frac{98}{99}\cdot \cdots\cdot\frac{100-n}{101-n}=\frac{100-n}{100}
$$
The probability that you pick the correct ball (given that those before you didn't pick the correct ball) is
$$
\frac{1}{100-n}
$$
Since there are only $100-n$ balls left. Thus, the probability that both events occur is
$$
\frac{100-n}{100}\cdot \frac{1}{100-n}=\frac{1}{100}
$$
Which is to say that each person has a $\frac{1}{100}$ chance of picking the correct ball, regardless of his/her position in the lineup.
A: Imagine if there were only $3$ balls. They can be taken out in the following order
$$
123\\
132\\
213\\
231\\
312\\
321\\
$$
You see that out of the $6$ possibilities, the ball, say $2$, comes out twice at first, twice second, and twice last. This implies that wherever you line up, you are as likely to pick it than someone else.
Note that in those kind of problem, you can often reduce the complexity to deduce the solution. The next step would be to explain carefully why this remains true with $100$ balls. You can have a look at Omnomnomnom's answer to see the argument.
A: Let $k\in\{1,\ldots,100\}$.
\begin{align*}
P_k\equiv&\,\mathbb{P}(\text{person $k$ picks the right ball})\\=&\,\mathbb{P}(\text{person $k$ picks the right ball }\textit{and}\text{ no person $\{1,\ldots,k-1\}$ has picked it yet})\\=&\,\mathbb{P}\left(\text{person $k$ picks the right ball }\big|\text{ no person $\{1,\ldots,k-1\}$ has picked it yet}\right)\\\times&\,\mathbb{P}(\text{no person $\{1,\ldots,k-1\}$ has picked it yet})\\=&\,\frac{1}{101-k}\times\mathbb{P}(\text{person $\{k,\ldots,100\}$ will pick it})=\frac{1}{101-k}\sum_{\ell=k}^{100} P_{\ell}\\=&\,\frac{1}{101-k}\left(1-\sum_{\ell=1}^{k-1}P_{\ell}\right),
\end{align*}
because the events defining $\{P_k\}_{k=1}^{100}$ are mutually exclusive and one of them must happen. Thus, we can compute $P_k$ inductively. Of course, we know that $P_1=1/100$. Then, $$P_2=\frac{1}{99}\left(1-\frac{1}{100}\right)=\frac{1}{100}.$$ We can prove that $P_k=1/100$ for all $k\in\{1,\ldots,100\}$ inductively: if $P_{\ell}=1/100$ for all $\ell\in\{1,\ldots,k\}$ some $k\in\{1,\ldots99\}$ , then $$P_{k+1}=\frac{1}{101-(k+1)}\left(1-\frac{(k+1)-1}{100}\right)=\frac{1}{100}.$$
