In an online game the chance of succeeding at an action starts at 5% and goes up 5% every time the action fails.
Upon success the chance resets back to 5%.
(So you know how they say a die has no memory of prior rolls? Here it does.)
The question: How many attempts do I need to have a 50/50 chance of success?
Alternate question: What fixed percent chance would duplicate the overall success rate of this system? (Over many rolls.)
This probability tree represents the game and guides the calculations:
The blue node at the left represents the start. At this point there is a 5% chance of success (leading to the up and left). If success is achieved now, only one attempt is made, as indicated in the orange circle.
Lacking success, we progress down and to the right to the next blue node. This time the chance of success is 10% and the chance of failure 90%. And so on, up to the 20th blue node where the chance of success (after 20 tries) equals 100% (and so there is no possibility of failure).
By axiomatic laws of probability, the chance of arriving at any terminal orange node is the product of the chances along the edges leading to that node. For instance, the chance of success in exactly two attempts is 95% times 10%, equal to $19/200,$ and the chance of success in exactly three attempts equals 95% times 90% times 15%, equal to $513/4000.$
By carrying out all the multiplications we compute the chances of success after exactly $k$ attempts for $k=1, 2, \ldots, 20$:
$$\frac{1}{20},\frac{19}{200},\frac{513}{4000},\frac{2907}{20000},\frac{2907}{20000},\frac{26163}{200000},\ldots,\frac{14849255421}{640000000000000000}.$$
From these values all quantities of interest may be calculated. For instance, the chance of success after four attempts is $\frac{1}{20}+\frac{19}{200}+\frac{513}{4000}+\frac{2907}{20000} = \frac{2093}{5000} = 0.4186$ and the chance of success after five attempts is greater by $\frac{2907}{20000}$ again, giving $\frac{11279}{20000} = 0.56395.$ Therefore the median number of attempts lies between four and five.
The expected (mean) number of attempts is (by definition) obtained by multiplying the number of attempts by its chance and summing over all possible numbers. The value is $\frac{3387894135040576041}{640000000000000000}$, approximately equal to $5.29358$. Its reciprocal, approximately $18.8908$%, answers the last question: the fixed chance of success with the same expected number of attempts.
A similar probability tree can help answer any similar questions where the value of $5$% may differ and can even change from one attempt to the next: just write in the appropriate probabilities and do comparable calculations.
Make a 19-state Markov chain and solve for its stationary distribution, then look at the probability of the first state. This will be the long-term expected number of successes per mouse click.
