Probability with a chance to try again If I play a game where the chance to win is 40%, and every time I lose I get a 25% chance to try again, what's my actual chance to win?
Every time you try you have an $0.4$ chance of succeeding and an $0.6$ chance of failing. At each failure you have an $0.25$ chance of trying again. 

The chance you win after one try is $p_1 = 0.4$. The chance you win at your second try is the chance of first failing, then being allowed to try again, and then actually succeeding, so $p_2 = 0.6 \cdot 0.25 \cdot 0.4$. 
Now in general in order to try an $n-th$ time, you first have to fail $n-1$ times and have been allowed to retry $n-1$ times. Since these events are independent, we have that $p_n = 0.4(0.6 \cdot 0.25)^n$, and the chance that you succeed at any time is simply the sum of all these $p_n$'s. 

Or, $$p_{win} = \sum_{n=0}^{\infty}p_n=\sum_{n=0}^{\infty}0.4(0.6 \cdot 0.25)^n=0.4\sum_{n=0}^{\infty}(0.6 \cdot 0.25)^n = \frac{0.4}{1-0.6\cdot 0.25}\approx0.47.$$ Where in the second to last step I have used the formula for a geometric sequence.
Hint: $p = 0.4 + 0.6 \times 0.25 p$