You and a friend play a shooting game. The chances of you hitting 1st and 2nd shot are 0.3 and 0.5 respectively. The chances for your friend are 0.4 and 0.4 respectively. Who has a better probability of hitting at least one shot? If you get 10 Euros for hitting the shot at least once, what is the fair value of the game?
My approach:
let P(A) be the probability for me hitting the target and P(B) the probability for my friend hitting the target.
$P(A) = 0.3 + 0.5 - (0.3*0.5) = \frac{13}{20}$
$P(B) = 0.4 + 0.4 - (0.4*0.4) = \frac{16}{25}$
since $\frac{13}{20}$ is bigger than $\frac{16}{25}$ then $P(A) > P(B)$
Now, for the fair value of the game, I suppose we should calculate the probability of both of us hitting the target, P(A,B).
$P(A,B) = \frac{13}{20} + \frac{16}{25} - (\frac{13}{20}\frac{16}{25}) = \frac{437}{500}$
So the fair value should be $10*\frac{437}{500} = 8.74\$$
I am not very convinced because, in way for a game to be fair, shouldn't the expected win equal the expected loss?
So in this case it would be $10*\frac{437}{500} -\frac{63}{500}*X = 0$ but this gives a result for X that higher than the expected win so it is not good...
as the question is worded, your income is a bernulli rv like the following
$$\text{Gain} = \begin{cases} -X, & p=0.35 \\ 10€, & p=0.65 \end{cases}$$
thus the game is fair if you pay 18.57 € to your friend when you do not hit the target. In fact, in this situation
$$\mathbb{E}[G]=0$$
You wrote
in fact the expected win is $10\times 0.65=6.5$
To be fair, the expected loss must be $L\times0.35=6.5$ and thus
$$L=18.57$$