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Let us try to avoid an infinite series

Let the probability of a person in line succeeding when her trial comes be $p$, failure $(1-p) = q$

Let $A$ = probability that person $S$ in line ultimately succeeds, then either she succeeds in the first round, or with everyone failing the round, we are back to start

Thus $A = pq^\left(S-1\right) + q^N\cdot A$

which yields $A = \boxed{\dfrac {pq^\left(S-1\right)}{1-q^N}}$

Let us try to avoid an infinite series

Let the probability of a person in line succeeding when her trial comes be $p$, failure $(1-p) = q$

Let $A$ = probability that person $N$ in line ultimately succeeds, then either she succeeds in the first round, or with everyone failing the round, we are back to start

Thus $A = pq^\left(N-1\right) + q^S\cdot A$

which yields $A = \boxed{\dfrac {pq^\left(N-1\right)}{1-q^S}}$

Let us try to avoid an infinite series

Let the probability of a person in line succeeding when her trial comes be $p$, failure $(1-p) = q$

Let $A$ = probability that person $S$ in line ultimately succeeds, then either she succeeds in the first round, or we are back to start

Thus $A = pq^{S-1} + q^N\cdot A$

which yields $A = \boxed{\dfrac {pq^\left(S-1\right)}{1-q^N}}$

Let us try to avoid an infinite series

Let the probability of a person in line succeeding when her trial comes be $p$, failure $(1-p) = q$

Let $A$ = probability that person $S$ in line ultimately succeeds, then either she succeeds in the first round, or with everyone failing the round, we are back to start

Thus $A = pq^\left(S-1\right) + q^N\cdot A$

which yields $A = \boxed{\dfrac {pq^\left(S-1\right)}{1-q^N}}$

Let us try to avoid an infinite series

Let the probability of the first person in line succeeding in her trial in the first round be $p$, failure $(1-p) = q$

$P(S_{th}$ person succeeding in first round) $= pq^{S-1}$

$P(\texttt{any}$ person succeeding in first round)
$= p(1+q+q^2+...q^{N-1}) = \frac{p(1-q^N)}{1-q}$

Rounds will just repeat the same pattern

Thus $P(S_{th}$ person succeeding in game) $=\dfrac{q^\left(S-1\right)*(1-q)}{1-q^N}$

Let us try to avoid an infinite series

Let the probability of a person in line succeeding when her trial comes be $p$, failure $(1-p) = q$

Let $A$ = probability that person $S$ in line ultimately succeeds, then either she succeeds in the first round, or we are back to start

Thus $A = pq^{S-1} + q^N\cdot A$

which yields $A = \boxed{\dfrac {pq^\left(S-1\right)}{1-q^N}}$