Calculate average/probability of an action happening in a shrinking pool I'm not sure how to title this actually, so I'll try to explain it better.
Let's say I have a bag of 50 items. In there is one particular item that I want. Obviously, the probably of getting that item on the first try is 1/50. After the first try, the item I pick is removed from the pool, so my chance of picking it the second time is 1/49, this continues until I pick the item, all the way up to a probably of 1.
However, once I pick the item from that bag, it is thrown away and a new one is brought out where I then repeat the process.
So my question would be, how do I calculate, on average, how many tries it would take me to pick the same item 10 times (not in a row)?
Thanks!
 A: Simplify the question. In mathematical terms, you want the expected number of tries to obtain the same item over 10 instances of playing the game. This is equivalent to 10 times of the expected number to obtain the same item over a single game (since I believe you're assuming the games are independent). 
It would help by introducing a new random variable. 
Consider the random variable given by:
$$
V_n = \begin{cases}
1 & \text{if a ball has not been picked before the n-th draw} \\
0 & \text{if a ball has been picked before the n-th draw} \\
\end{cases}
$$
In other words, this is the random variable that specifies if I will (or will not) make the n-th draw.
Then we would like to calculate the expected value of:
$$
\mathbb{E}\Big[ \sum_{n=1}^{50} V_n \Big] = \sum_{n=1}^{50} \mathbb{E}[V_n ]
$$
(i.e. What is the expected value of the sum of all my indicator variables $V_n$)
Now, what is $\mathbb{E}[V_n ]$? 
Well, for n = 1, $\mathbb{E}[V_1]$ is just $1$. For $n >2$
$$
\begin{align*}
\mathbb{E}[V_n ] &= Pr(V_n = 1) \times 1 + Pr(V_n = 0) \times 0 \\
&= \frac{49}{50}\frac{48}{49}\frac{47}{48} ... \frac{50 - n + 1}{50 - n + 2} \times 1 \\
&= \frac{50 - n + 1}{50} \\
&= 1 - \frac{n-1}{50}
\end{align*}
$$
Note that I have sacrificed a little bit of mathematical integrity to illustrate the cancellation(s) in line 2. And that the final expression also holds for $n = 1$. 
Thus, we get:
$$
\begin{align*}
\sum_{n=1}^{50} \mathbb{E}[V_n ] &= \sum_{n=1}^{50}  1 - \frac{n-1}{50} \\
&= 51 - \frac{1}{50}\sum_{n=1}^{50} n \\
&= 51 - \frac{1275}{50} \\
&= 25.5
\end{align*}
$$
Hence, multiplying this quantity by 10 (since we play the game 10 times) gives 255, which is the answer you want. 
