Two players shooting a target alternately > Two players **A** and **B** shoot a target alternately, until somebody hits it and wins the game. Player **A** takes the first shot. Each time they take a shot, their probabilities of hitting the target are $p$ and $q$ respectively. Which is the probability $\mathbb{P}$ that **A** wins? **Is the following solution correct ?** **A** wins if : he wins with the $1$st shot **or** they both fail to hit it during the first $2$ shots, and A hits it with the $3$rd shot **or** they both fail to hit it during the first $4$ shots, and A hits it with the $5$rd shot **or** ... So, $$ \mathbb{P} = p + (1-p)(1-q)p + (1-p)^2(1-q)^{2}p + ... = p\sum_{n=0}^{\infty}[(1-p)(1-q)]^n = \cfrac{p}{p+q-pq} $$ Thank you in advance

yes, it is correct.

Alternatively, you can condition on the first outcome. If $A$ doesn't win in the first outcome, the first two trials must be failure and then it is a renewal process.

$$\mathbb{P}=p + (1-p)(1-q)\mathbb{P}$$ $$(1-(1-p)(1-q))\mathbb{P}=p$$ $$\mathbb{P}=\frac{p}{1-(1-p)(1-q)}$$