Assume you're playing a game with your sibling (S.) and a friend (F.), in which S. will throw a coin and if the coin shows heads F. will give you 1 dollar. If the coin shows tails, you will give F. $z$ dollars. S. throws a coin which may be fair (probability of heads 0.5) or loaded (probability of heads 0.7). Because you know S. pretty well, you belief that with 0.6 probability she will use the loaded coin. Using your prior probability, determine $z$ that would make the game fair (that is neither you nor F. looses or wins money on average).
Maybe I get the question wrong, but I think $z$ should be the solution to the following linear equation
$$ 0.6\cdot(0.7\cdot 1 \$- 0.3 \cdot z)+ 0.4\cdot(0.5 \cdot 1 \$ – 0.5 \cdot z) = 0$$
which would result in approximately $z\approx 1.63 \$ $. But the solution claims that $z$ is the solution to $0.6\cdot 1\$ – 0.4 \cdot z =0$ which is $z=1.5 \$$. Is my reasoning wrong?
Best Answer
If your sibling $S$ chooses the coin prior to every toss, then the answer you provided is correct. If your sibling chooses the coin prior to starting the game, and then once the coin is chosen, you use that coin only, then the problem becomes quite different.
This answer assumes your sibling chooses the coin prior to starting the game, and then uses that same coin for as long as you play the game.
Suppose you choose $z = \dfrac{31}{19}$ as in your answer. Suppose your sibling chooses the loaded coin. Now, for the entire game, you have:
$$0.7\cdot 1 -0.3\cdot \dfrac{31}{19} = \dfrac{4}{19}$$
You will make more than you lose.
Suppose your sibling chooses the fair coin. Now, for the entire game, you have:
$$0.5\cdot 1 - 0.5\cdot \dfrac{31}{19} = -\dfrac{6}{19}$$
So, you will lose more than you gain. Thus, on average, the game is unfair no matter which coin your sibling chose.
Instead, suppose we choose $z = \dfrac{7}{3}$. If your sibling chooses the loaded coin:
$$0.7\cdot 1 - 0.3\cdot \dfrac{7}{3} = 0$$
The game is fair.
If your sibling chooses the fair coin:
$$0.5\cdot 1 - 0.5\cdot \dfrac{7}{3} = -\dfrac{2}{3}$$
On average, you lose more than you win.
However, now, $\dfrac{3}{5}$ of the time, your sibling chooses the loaded coin, and the game is fair while $\dfrac{2}{5}$ of the time, your sibling chooses the fair coin, and the game is unfair. By this metric, "on average" could imply the median. The median option here is that the game is fair more than fifty percent of the time.
In other words, you need to define "on average" in this context, as it could have different meanings.