# Theorem of total probability with basketball players

Suppose you have a player from team A and a player from team B. You know that one of them has a 60% chance to make a shot and other has 40% chance to make a shot, but you are not sure which is which.

Suppose you choose 1 of the 2 players to shoot, how would you go about calculating his probability of missing.

My train of thought was that I could calculate it using law of total probability, but I got stuck along the way.

let $$G$$ = event that player is the better of the two

My equation was

$$P(miss) = P(miss | G) * P(G) + (miss | \overline G) * P (\overline G)$$, but I was unsure of how to find $$P(G)$$

Alternatively, I thought maybe you could just average out the two chances and it would end up being 50/50, but that seemed off the mark.

Is there a better way of going about solving this or did I just miss something along the way?

Thanks

There are two sources of randomness here, one is your team choice and the other is if your choice is the good player or not. If the choices are made in total randomness, the effective probability will be $$1/2$$. Even if you insist on choosing Team A player, the probability of choosing the better player will be again $$1/2$$. So, assuming total randomness even on only one step here results in $$P(G)=1/2$$. When you substitute, you’ll get $$(0.6+0.4)/2=0.5$$. So your 50-50 guess is not that off.