Assume that $n$ people want to achieve a task T. One person can try, and is successful with probability $p$. But when a person try all the other have to do an other trial to have the right to challenge $A$ with probability of success $q$ (we can assume that $p=q$ if it's easier).
More formally I have a Markov chain on $\mathbb{N}\times\mathbb{N}$ with initial state $(n,0)$ ($n$ is the number of person that want to try and 0 is the number of success for T).
And from $(n',r)$ the probability in one step to reach $(k,r+1)$ is $$p{n'-1 \choose k} q^k(1-q)^{n'-1-k}$$ And the probability in one step to reach $(k,r)$ is $$(1-p){n'-1 \choose k} q^k(1-q)^{n'-1-k}$$
I want to know the expected number of person achieving T depending on $n$ the number of people present in the beginning. I have no idea on how to compute that.
Even a limit or even better matching at the limit upper and lower bound would do.
Thanks
I hope I'm understanding you correctly. If $X(t)$ is the number of people who want to try at stage $t$, then $X(0) = n$ and the conditional distribution of $X(t+1)$ given $X(t)$ is binomial with parameters $X(t)-1$ and $q$. If $T$ is the first $t$ where $X(t) = 0$, then conditional on $T$ the number of successes $S$ is binomial with parameters $T$ and $p$. You want $E[S] = p E[T]$.
By a first-step analysis, $E[T] = f(n)$ where $f(0) = 0$ and for $n \ge 1$, $$f(n) = 1 + \sum_{j=0}^{n-1} {n-1 \choose j} q^j (1-q)^{n-1-j} f(j)$$ It seems $f(n)$ is a polynomial of degree $n(n-1)/2$ in $q$. The first few values are
$$ \eqalign{ f(0) &= 0\cr f(1) &= 1\cr f(2) &= q+1\cr f(3) &= {q}^{3}-{q}^{2}+2\,q+1\cr f(4) &= {q}^{6}-{q}^{5}-{q}^{4}+4\,{q}^{3}-3\,{q}^{2}+3\,q+1\cr f(5) &= {q}^{10}-{q}^{9}-{q}^{8}+5\,{q}^{6}-3\,{q}^{5}-5\,{q}^{4}+10\,{q}^{3}- 6\,{q}^{2}+4\,q+1\cr}$$ I don't see a closed form.
The coefficients of the first few powers of $q$ have a tantalizing regularity: it looks like for $n \ge 1$ $$f(n) = 1 + {n-1 \choose 1} q - {n-1 \choose 2} q^2 + {n \choose 3} q^3 - {n \choose 4} q^4 + \ldots $$ But the coefficient of $q^5$ breaks the pattern: it is $(n-1)(n-2)(n-3)(n^2-4n-20)/120$
If $g(z) = \sum_{n=0}^\infty f(n) z^n$ is the generating function of $f(n)$, I get a functional equation: $$ \eqalign{g(z) &= \sum_{n=1}^\infty z^n + \sum_{j=0}^\infty \sum_{n=j+1}^\infty {n-1 \choose j} q^j (1-q)^{n-1-j} f(j) z^n\cr &= \dfrac{z}{1-z} + \sum_{j=0}^\infty \dfrac{ z^{j+1} q^j}{(1-(1-q) z)^{j+1}} f(j)\cr &= \dfrac{z}{1-z} + \dfrac{ z}{1 - (1-q) z} g\left(\dfrac{qz}{1-(1-q)z}\right)}$$ but again I don't see a way to a closed form.