Let me preface this by saying that this is similar to the following question: Calculating Probability with Changing Odds. However I do have some differences in the problem that I am having trouble figuring out.

The problem is:

Imagine you have the chance at succeeding at some task with a base rate of $R$. Everytime you fail to succeed, the possible success rate increases by $10\%$. That is the new success rate for the next try will be $R + (R * 0.10)$ This relation holds true until a maximum success rate of $2R$. How do you calculate the average amount of tries to succeed? And how do you calculate the probability that you will fail $N$ number of times before succeeding?

So for example if our $R = 10\%$ then it will increase by $1\%$ each time, upto a maximum success rate of $20\%$.

I'm having a bit of trouble modifying the solution from the linked problem to account for this constraint of a maximum success rate. Could anyone provide me a method on how to calculate the average attempts and probability of failing $N$ trials?

  • $\begingroup$ You mean somerhing like this here? There is $R=0.5$ and two successes. $\endgroup$ Commented Apr 29, 2022 at 18:22
  • $\begingroup$ @callculus42 Maybe? I'm having a bit of trouble deciphering how that works. $\endgroup$
    – Rietty
    Commented Apr 29, 2022 at 20:17

1 Answer 1


If $\ r\ $ is the fraction by which $\ R\ $ increases after each failure, then the probability $\ R_n\ $ of success in the next trial after $\ n\ $ failures is given by $$ R_n=\min\big( (1+r)^nR, 2R\big)\ , $$ —that is, $\ R_n=(1+r)^nR\ $ if $\ n\le\left\lfloor\frac{\log(2)}{\log(1+r)}\right\rfloor\ $, or $\ R_n=2R\ $ otherwise. The probability of success after after exactly $\ N\ $ failures is therefore $$ (1+r)^NR\prod_{n=0}^N\big(1-(1+r)^nR\big) $$ if $\ N\le d=\left\lfloor\frac{\log(2)}{\log(1+r)}\right\rfloor\ $, or $$ 2R\left((1-2R)^{N-d}\prod_{n=0}^d\big(1-(1+r)^nR\big)\right) $$ if $\ N>d\ $.

The expected number of failures occurring before success is terefore \begin{align} \sum_{N=1}^dN(1+r)^NR&\prod_{n=0}^N\big(1-(1+r)^nR\big)\\ &\hspace{2em}+2R\prod_{n=0}^d\big(1-(1+r)^nR\big)\sum_{N=d+1}^\infty N(1-2R)^{N-d}\\ &=\sum_{N=1}^dN(1+r)^NR\prod_{n=0}^N\big(1-(1+r)^nR\big)\\ &\hspace{1em}+2R\prod_{n=0}^d\big(1-(1+r)^nR\big)\sum_{m=1}^\infty (m+d)(1-2R)^m\\ &=\sum_{N=1}^dN(1+r)^NR\prod_{n=0}^N\big(1-(1+r)^nR\big)\\ &\hspace{3em}+(1+2Rd)\prod_{n=0}^d\big(1-(1+r)^nR\big)\ , \end{align} and the expected number of attempts to success will be one more than this.


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