# Thread: what is the appropriate approach?

1. ## what is the appropriate approach?

In a game, two players A and B shoot a target (both to the same target). Suppose that shooters are independent and the probability of (successful) shooting to target is relatively, pA and pB. If A shoots the target (successfully) before B, how much is it probable that A is the starter of the game?

PS. would you please first give your opinion about the appropriate approach and why you choose that?

probability models(distributions)

Markov chains
non-frequentist approaches like Bayesian solutions, or other things!

2. Originally Posted by alv
In a game, two players A and B shoot a target (both to the same target). Suppose that shooters are independent and the probability of (successful) shooting to target is relatively, pA and pB. If A shoots the target (successfully) before B, how much is it probable that A is the starter of the game?

PS. would you please first give your opinion about the appropriate approach and why you choose that?

probability models(distributions)
Markov chains
non-frequentist approaches like Bayesian solutions, or other things!
You want Pr(A starts | A wins).

This suggests Bayes Theorem: Pr(A starts | A wins) = Pr(A starts and A Wins)/Pr(A wins).

Pr(A wins) = Pr(A starts and A wins) + Pr(B starts and A wins).

The calculation of two probabilities is suggested. Let $q = (1 - p_A)(1 - p_B)$. Then:

Pr(A starts and A wins) = $p_A + q p_A + q^2 p_A + ..... = p_A (1 + q + q^2 + ....) = p_A \left( \frac{1}{1-q}\right) = \frac{p_A}{1-q}$.

Pr(B starts and A wins) = $(1-p_B)p_A + q (1-p_B) p_A + q^2 (1-p_B) p_A + ..... = (1-p_B) p_A (1 + q + q^2 + ....)$ $= \frac{(1-p_B)p_A}{1-q}$.

Therefore Pr(A starts | A wins) = $\frac{1}{2 - p_B}$.

Note: If $p_B = 0$ then Pr(A starts | A wins) = 1/2, as you'd expect. If $p_B = 1$ then Pr(A starts | A wins) = 1, as you'd expect. If $p_B = \frac{1}{2}$ then Pr(A starts | A wins) = $\frac{2}{3}$, as you'd expect.