Calculating probability of success with changing odds I use binomial probability to calculate a static probability percentage over x number of trials to see how often I would get at least 1 success. But, how would I calculate it when after each attempt the probability of success changes? 
Say in 5 attempts, the probability starts out as 10% on the first attempt and grows by an additional 10% after each loss. Whats the probability of winning at least once in those 5 tries? And how would I incorporate it for very large amount of trials, say 100 attempts (starting off with 10% plus 0.1% after each loss). 
Thanks in advance. 
 A: If the change is exactly like "probability of success increases after failure", then it's simple to calculate probability of having no successes.
Probability of loss at $n$-th trial given first $n - 1$ were lost is $p - xn$ (where $p$ is probability of loss at first attempt, and $x$ is increment of winning probability). Then probability of loss at first $n$ attempts is $\prod\limits_{k=0}^{n-1} (p - k\cdot x)$, and so probability ow winning is $1 - \prod\limits_{k=0}^{n - 1} (p - k\cdot x)$.
For your first example, it's $1 - 0.9 \cdot 0.8 \cdot 0.7 \cdot 0.6 \cdot 0.5 = 0.8488$.
I doubt there is closed-form expression for it, but if $\frac{kx}{p}$ is small (ie total probability increase is much smaller then initial loss probability), we can rewrite this product as $\prod\limits_{k=0}^{n - 1}(p - kx) = p^n \cdot \prod\limits_{k=0}^{n - 1}\left(1 - \frac{kx}{p}\right)$ and then use $\ln \prod\limits_{k=0}^{n - 1}\left(1 - \frac{kx}{p}\right) = \sum\limits_{k=0}^{n-1}\ln (1 - \frac{kx}{p}) \approx -\sum\limits_{k=0}^{n-1} \frac{kx}{p} = -\frac{n \cdot (n - 1) \cdot x}{2p}$, so probability of winning is approximately $1 - \exp(-\frac{n \cdot (n - 1) \cdot x}{2p})$.
A: Here is an idea. For the case of $5$ attempts we can write the $S_5=X_1+\dotsb+X_5$ where $X_i$ are bernoulli trials with $X_1\sim \text{Ber}(0.1)$ (the probability of success is $0.1$), $X_2\mid X_1=0\sim \text{Ber}(0.2)$ while $X_2\mid X_1=1\sim \text{Ber}(0.1)$ and $X_3\mid X_1=x_1, X_2=x_2$ being analogously defined for $x_i\in \{0, 1\}$ based on the scheme described in your problem formulation (e.g. $X_3\mid X_1=1, X_2=0\sim \text{Ber}(0.2)$, $X_3\mid X_1=0, X_2=0\sim \text{Ber}(0.3)$ and so on) We define $X_j\mid X1,\dotsc ,X_{j-1}$ similarly.
So the probability of not winning in the five trials $P(S_5= 0)=P(X_1=0,\dotsc, X_5=0)$ is given by
$$
\begin{align}
P(X_1=0,\dotsc, X_5=0)&=P(X_1=0)P(X_2=0\mid X_1=0)\dotsb P(X_5=0\mid X_1=0,\dotsc, X_4=0)\\
&=0.9(0.8)(0.7)(0.6)(0.5)
\end{align}
\\
$$
and hence the probability of no wins is given by
$$
P(S_5\geq 1)=1-0.9(0.8)(0.7)(0.6)(0.5)
$$
