Thread: 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!

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 . Then:

Pr(A starts and A wins) = .

Pr(B starts and A wins) = .

Therefore Pr(A starts | A wins) = .


Note: If then Pr(A starts | A wins) = 1/2, as you'd expect. If then Pr(A starts | A wins) = 1, as you'd expect. If then Pr(A starts | A wins) = , as you'd expect.