Crossbar Challenge - Who has the greater chance of winning? You and your friend each get 2 attempts to kick a ball and hit the crossbar of the goal posts. Whoever hits it the most wins. The probability of you hitting it on your 1st attempt is 30% and on your second attempt is 50%. Your friend has a 40% chance of hitting it on both attempts.
Who has the higher probability of winning?
 
My initial thought to solve this was to write out all the possibilities of me winning and all the possibilities of my friend winning and compare the probabilities:
I WIN:
I hit shot 1, miss shot 2, he misses both
-> ((0.3) + (0))((0) + (0)) = 0

I miss shot 1, hit shot 2, he misses both
-> ((0) + (0.5))((0) + (0)) = 0

I hot both, he misses both
-> ((0.3) + (0.5))((0) + (0)) = 0

I hit both, he hits shot 1, misses shot 2
-> ((0.3) + (0.5))((0.4) + (0)) = 0.32

I hit both, he misses shot 1, hits shot 2
-> ((0.3) + (0.5))((0) + (0.4)) = 0.32

Total probability of me winning 
= 0 + 0 + 0 + 0.32 + 0.32 = 0.64 = 64%

I then did the same for all the scenarios in which my friend wins and got 64% as well. Is this correct? Do we both have the same chance of winning?
 A: Using upper case for your wins/losses, possibilities for your winning are:
$WW-ll: 0.3*0.5*0.6*0.6 = 0.054$
$WW-(wl \;or\; lw): 0.3*0.5*0.4*0.6\times2 = 0.072$
$(WL \;or\; LW)-ll: 0.3*0.5*0.6*0.6 + 0.7*0.5*0.6*0.6 = 0.18$
so P(you win) $= 0.306$
Compute similarly for your friend
A: I hit shot 1, miss shot 2, he misses both
 ((0.3) + (0))((0) + (0)) = 0

This should immediatelly send a warning bell through your head. What your result is saying is that the probability of this event is $0$, in other words, it is impossible for this to happen. But how can it be impossible? It's certainly possible for you to hit shot $1$, and it's possible for you to miss shot $2$, and it's certainly possible for him to miss both shots, right? So the end result, intuitivelly, should not be $0$.
You have no reason to be adding anything here. You are talking about $4$ distinct independent events: You hit shot 1, you miss shot 2, and he misses shots one and two.
So, calculate the probability again, and take into account:


*

*Since the events are independent, the probability of them happening all together is the product of their individual probabilities.

*The probability of you missing shot $2$ is not equal to $0$. Remember, the probability of you hitting shot $2$ plus the probability of you missing the shot must be $1$ (since one of these two things certainily happened).

