Probability with Level up system I'm been wondering about this level up system (in gaming) for quite a while. Every time a character levels up, he has a chance to gain +1 damage.
Let $b$ be the base percentage gain, $0<b<100$.
$p_1 = b$
If the character rolls and succeeds, ie. $r_1 \leq p_1$, then he gets +1 damage and $p_2 = b$.
If the character rolls and fails, ie. $r_1 > p_1$, then he gets +0 damage and $p_2 = p_1 + b$.
If $p_i > 100$, then you have guaranteed +1 damage, and we actually roll $p_{i} = p_i - 100$, for a chance of another +1.
Example:
Let $b = 60$. If the first level up fail, then $p_2 = 120$. This results in a guaranteed +1 damage and we roll $p_2 = 20$. If the roll fails, $p_3 = 80$. If the second roll gives us another +1 damage, then $p_3 = 60$.
So my question is, this is equivalent to what fixed percentage chance per level up?
EDIT: I'm so sorry to the two that answered below me, I typed in something wrong the first time around.
Simulation: http://i.imgur.com/Ou6bDqp.png (each simulation is 1000 rolls)
Code: http://i.imgur.com/bTYXYYP.png
EDIT2: Based on the graph of the simulation, there is a disjoint at $b = 50$, almost like it's two different curves. Why is this?
EDIT3: Added Brian Tung's estimation.
 A: I did some calculations for a few rolls, assuming $b = 1/n$.
For $n=2$, the probabilities correspond to (for 1,2... rolls)  $p = 1/2, 5/8, 5/8, 41/64$.  Looks like the limit is approaching $1/2 + 1/7 \approx 65\%$.
For $n=3$, I got $1/3, 4/9, 13/27 \approx 48\%$.  Heading toward $50\%$, but things get more complicated after 3 rolls, which is the first time you have a guaranteed success.
I'd recommend doing a simulation to find the actual probabilities.
Note that the answer is not a continuous function of $b$.  For example, for $b = 50\%$, you never have the case where $p_i > 1$, so all $p_i$ are either $50\%$ or $100\%$.  While if $b$ is slightly greater than $50\%$,
$$ \lim_{\varepsilon\to 0^+} f(50\% + \varepsilon) = 50\%$$
This is true because if $p_i = 50\% + \varepsilon$ and you fail, then $p_{i+1} = 100\% + 2\varepsilon$ and $p_{i+2} = 2 \varepsilon$.  As $\varepsilon$ vanishes, this is equivalent to taking a string of equally-distributed successes and failures, and inserting "success, fail" after each failure.  The original string has $50\%$ success, the final string added the same number of successes and failures, so must also be $50\%$ success.  This is clearly less than $f(50\%) \approx 65\%$.
Graph the result as a function of $b$, and you'll see a sharp dip at $b = 50\%$, other dips at every $1/n$, and smaller dips at other rational numbers with small denominators (for example, $b = 2/3$).
If remove the $p_i > 1$ rule, and just set the probability back to $b$ each time you succeed, then the function will be more smooth.
A: I haven't the time to sketch this out in any attractive way, but for the case considered by user3294068 (where $b = 1/n$ for some integer $n$), we can model this as a discrete-time Markov chain, with state $k$, $1 \leq k \leq n$, representing a probability $k/n$ of increased damage.  This chain has the transition probabilities
$$
p_{k, k+1} = \frac{n-k}{n} \qquad 1 \leq k < n
$$
$$
p_{k, 1} = \frac{k}{n} \qquad 1 \leq k \leq n
$$
and $p_{k, j} = 0$ otherwise.  Since we go (or return immediately) to state $1$ whenever we get $+1$ damage on the level-up, the proportion of level-ups that result in $+1$ damage is equal to the proportion of time the chain spends in state $1$.  We can determine this by tabulating the probability flux balance equations
$$
\pi_k = \frac{n-(k-1)}{n}\pi_{k-1} \qquad 1 < k \leq n
$$
This leads to the expression
$$
\pi_k = \frac{(n-1)!}{(n-k)!n^{k-1}}\pi_1 \qquad 1 \leq k \leq n
$$
and therefore
$$
\pi_1 = \left( \sum_{k=1}^n \frac{(n-1)!}{(n-k)!n^{k-1}} \right)^{-1}
$$
Wolfram Alpha says this comes out to
$$
\pi_1 = \left[\left(\frac{e}{n}\right)^n \Gamma(n+1, n)-1\right]^{-1}
$$
where $\Gamma(\cdot, \cdot)$ is the incomplete gamma function.  I had previously written that $\Gamma(n+1, n) \to n!$ as $n \to \infty$, but that turns out not to be true; it's something like $n!/2$, with convergence being rather slow.  This would yield an asymptotic value of
$$
\pi_1 \to \frac{2}{\sqrt{2\pi n}-2}
$$
However, since $\frac{\Gamma(n+1, n)}{n!}$ settles only slowly to somewhere in the vicinity of $1/2$, this asymptotic expression isn't very good until $n$ is quite large (say, maybe about $100$?).
Interesting question!  I'm investigating further.
ETA: I have, essentially empirically, obtained an approximate expression
$$
\pi_1 \doteq \frac{4.6}{5.6\sqrt{n}-1} = \frac{4.6\sqrt{b}}{56-\sqrt{b}}
$$
That should be fairly close for a decent range of $n$ or $b$.
