Probability after repetition and series So I was wondering about how often you would have to try something for a certain probability if you had a 15% winrate. Then I found something weird:
The average loss rate l with n tries is obviously $0.85^n$. The win rate would be $1-0.85^n$.
Now, I am pretty bad with probabilities but the win chance should also be $\sum_{i=0}^{n-1} {0.15 * 0.85^i}$, right?
Now to my problem:
How would you show this?$$1-0.85^n = \sum_{i=0}^{n-1} {0.15 * 0.85^i}$$
Also, that $\sum_{i=0}^\infty {0.15 * 0.85^i}=1$
 A: 
Now to my problem:
  How would you show this?$$1-0.85^n = \sum_{i=0}^{n-1} {0.15 * 0.85^i}$$

More generally, let us show that $$1-x^n = \sum_{i=0}^{n-1} (1-x)x^i.$$
To show this, start from the RHS and expand each product as $$(1-x)x^i=x^i-x^{i+1},$$ thus the RHS is $$\sum_{i=0}^{n-1}x^i-\sum_{i=0}^{n-1}x^{i+1}=\sum_{i=0}^{n-1}x^i-\sum_{i=1}^{n}x^i=x^0-x^{n}.$$ Finally, note that $x^0=1$, hence our general claim holds, and, to solve your problem, use the claim for $x=0.85$. As a bonus, note that, if $|x|\lt1$ then $x^n\to0$ when $n\to\infty$ hence $$\sum_{i=0}^{\infty} (1-x)x^i=1,$$ a fact at the basis of every geometric distribution.
A: Actually $1-0.85^n$ is the probability that someone wins at least once.
This could be obtained otherwise usiong the expression:
$$\zeta=\sum_{k\ge1}^n \binom{n}{k}(0.15)^k(0.85)^{n-k}$$
which is actually the binomial expression without the first term, actually:
$$(0.15+0.85)^n=\sum_{k=0}^n \binom{n}{k}(0.15)^k(0.85)^{n-k}=\binom n0(0.15)^0(0.85)^n+\zeta=0.85^n+\zeta\\\zeta=1-0.85^n$$

Actually using the fact involving geometric progression $$\sum_{k=0}^n\alpha=\frac{\alpha^{n+1}-1}{\alpha-1}$$
$$\sum_{i=0}^{n-1} {0.15 * 0.85^i}=0.15\sum_{k=0}^{n-1}(0.85)^k=0.15\frac{1-(0.85)^n}{1-(0.85)}1-(0.85)^n$$ 
