how many times in average do I have to roll a dice to get a 1? I've read that the answer to such a problem is the inverse probability. So here getting a one has probably 1/6, so the number of tries you would be expected to run in order to get a 1 is 6.
I'm not sure I understand this.
 A: Possible sequences you might roll are:
6 with probability $1/6$
16, 26, 36, 46, 56 with collective probability $5/6^2$
116, 126, ..., 556 with collective probability $5^2/6^3$
The pattern is that the probability it takes $n$ rolls is $\frac{1}{6}\left(\frac{5}{6}\right)^{n-1}$. To average over all possible values of $n$, you multiply each $n$ by its probability and sum:
$$\sum_{n=1}^{\infty}n\cdot\frac{1}{6}\left(\frac{5}{6}\right)^{n-1}$$
There are several ways to calculate the value of an infinite sum like this one. But the result is $6$. If you repeated this approach using $p$ in place of $1/6$, you would find the expected valus is $1/p$.
A: You are thinking correct. The answer is indeed 6.
You can read here further if you want deeper understanding. I hope this helps.
A: You can roll a 1 in your first throw with a chance of $\frac{1}{6}$
You can roll a 1 in your second throw (while not on your first) with a chance of $\frac{5}{6}*\frac{1}{6}$
etc.
Now, the expected number of throws until you get a 1 is the sum of all possible numbers of throws (which goes from 1 to infinity) multiplied by the probability of that event happening. So, where $P(i)$ is the probability of taking $i$ throws to roll a 1:
$$E = P(1)*1 + P(2)*2 + P(3) * 3 + ... = $$
$$\frac{1}{6}*1 + \frac{5}{6}*\frac{1}{6} * 2 + ... =$$
$$\frac{1}{6}*\sum^\infty_{i=1}{(\frac{5}{6})^{i-1}}*i =$$
$$\frac{1}{6}*\frac{1}{(1-\frac{5}{6})^2}= \frac{1}{6}*\frac{1}{\frac{1}{6}^2}= \frac{1}{6}*6^2= 6$$
Here I used:
$$\sum_{i=1}^{\infty}k^{i-1}*i=\left(\sum_{i=0}^{\infty}k^{i}\right)'=\frac{1}{(1-k)^2} $$
A: Here is one way to think about the problem. It also gives insight into finding the solution to more complex expectations. First, let us define $X_i$ to be the value of the $i$th roll, $i \in \mathbb{N}$. Next, define $N$ to be the first roll when we observe a 1. Note that you are asking for $\mathbb{E}(N)$. We can condition on the first roll to get an expression in terms of conditional expectations, i.e., $$\mathbb{E}(N) = \mathbb{E}(N|X_1=1)\mathbb{P}(X_1 = 1) + \mathbb{E}(N|X_1\neq1)\mathbb{P}(X_1 \neq 1).$$ Now, it is obvious that if we roll a 1 on the first go, that $N = 1$, hence the conditional expectation $\mathbb{E}(N|X_1=1) = 1$. The conditional probabilities are trivial, that is $\mathbb{P}(X_1 = 1) = \frac{1}{6}, \mathbb{P}(X_1 \neq 1) = \frac{5}{6}$. The interesting step is noticing that $$\mathbb{E}(N|X_1 \neq 1) = \mathbb{E}(N) + 1.$$ This relationship occurs because when we fail to roll a 1 on the first try, we have that the process probabilistically renews (because these rolls are independent) and so our new expectation is just one more than it was when we first started rolling. Hence we simply solve $$\mathbb{E}(N) = \frac{1}{6} + \left(\mathbb{E}(N) + 1\right)\cdot \frac{5}{6} = 6.$$
