0
$\begingroup$

I have the following problem:

Two events, $A$ and $B$, occur alternately. This stops when one of the events is successful. Each event has a probability of success given by: $P(A) = \alpha; P(B) = \beta$.

Assuming A is attempted first, what is the probability that A is successful?

From this I have arrived at the following recurrence relation, showing the probability that $A$ is successful on or before trial $n$.

$P(t_{n+1}) = P(t_n) + \alpha(1-\beta)(1-P(t_n))$

With $P(t_1) = \alpha$

Then, to solve this problem I need to know the value of $P(t_{\infty})$ in terms of $\alpha$ and $\beta$.

I don't have much experience with recurrence relations and so do not know how to proceed from here. I'm not sure if this is even the right way to approach this problem. Any help would be appreciated.

$\endgroup$
1
  • $\begingroup$ It is not the case of recurrence it is a case of probability of independent events. $\endgroup$ Dec 2, 2019 at 17:39

1 Answer 1

2
$\begingroup$

Let $P_A$ be the probability that $A$ is (eventually) successful. Then consider the possible outcomes of the first two trials. Either $A$ is successful on the first trial, or $A$ fails on the first trial but $B$ succeeds, or $A,B$ both fail on their first trial in which case the game restarts. Thus $$P_A=\alpha \times 1 + (1-\alpha)\times \beta \times 0 + (1-\alpha)\times (1-\beta)\times P_A$$

Which implies that $$P_A=\frac {\alpha}{1-(1-\alpha)\times (1-\beta)}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .