# Expected value with recursive probability

First of all, I'm aware of an existing question with a similar title, but my problem is fundamentally different.

I'm trying to solve this: in the first try, there is a 5% probability of obtaining a success. In the second try, if the first one was a success, the probability resets to 5%, but if the first one was a failure, now there is 5% more probability (10%). This goes on an on for an arbitrary number of trials, what I'm trying to calculate is the expected number of trials to obtain a single success.

What I got is the probability of obtaining a success in try number $n$, which obviously depends on the probability of the previous try: $$p(n) = p(n-1)\times 0.05 + (1-p(n-1))\times(p(n-1)+0.05),$$ with $p(1)=0.05$. Is it possible to calculate what I want?

In a more general way, what I'm asking is how to calculate the expected number of trials when the probability of success in a single trial depends on the probability of the previous trial, $p(n)=f(p(n-1))$.

• Model this with a Markov model. The rest is straightforward. – msm Aug 12 '18 at 9:25
• The probability on try $n$ does not depend only on the probability on the previous try. It really depends on the actual outcome! Perhaps you should make a tree of the first few steps. – Arnaud Mortier Aug 12 '18 at 9:40
• Also note that $n$ can't be arbitrarily large. $20$ at most. – Arnaud Mortier Aug 12 '18 at 9:43
• You are right, so in my case I can just calculate the probabilities of the first 20 steps and from that get the expected value that I want. I guess I was thinking in the general case in which $E(n)$ could be arbitrarily large. Thank you. – Julian L Aug 12 '18 at 9:50

Let $p(n)$ be the probability of having a success within the first $n$ trials, and let $q(n)$ be the probability of having the FIRST success in trial $n$. Then it holds that $p(1)=q(1)=0.05$, $$q(n)=0.05n[1-p(n-1)],$$ and $$p(n)=p(n-1)+q(n).$$ Generate the sequences $p(n)$ and $q(n)$ for as long as they satisfy $0\leq p(n)\leq1$ and $0\leq q(n)\leq1$. It turns out that $p(21)=1$ and this is where you can stop. Then calculate $$\sum_{n=1}^{21}q(n)n.$$
Not a complete answer. Let the probability of success on the first try be $1/m$, where $m$ is a positive integer. If the first try fails, the probability of success on the second try is $2/m$, and so on, as explained by the OP for $m = 20$. Success requires at most $m$ tries, since the probability of success on the $m$th try is $1$. The probability of success on the $r$th try and not before is $p(r)$ where $$p(r) = {{m-1}\over{m}} \times {{m-2}\over{m}} \times \cdots \times {{m-r+1}\over{m}} \times {{r}\over{m}}.$$ The expected number of tries is $$p(1) + 2p(2) + 3p(3) + \cdots + mp(m).$$ The values for $m = 1, \dots\ 6$ work out to $$1,\quad3/2^1,\quad17/3^2,\quad142/4^3,\quad1569/5^4,\quad21576/6^5.$$ The numerators can be found in the OEIS at https://oeis.org/search?q=1%2C3%2C17%2C142%2C1569&language=english&go=Search Unfortunately the table doesn't go as far as $m = 20$ but the references may help.