In going over a question about expectation and payoff, I am a bit confused by one aspect of the solution.
Briefly, we have six sided fair die that is rolled three times. We are asked, considering the following rules, for what value of payoff $k$ is the game fair?
Rule 1: if no sixes are rolled, we lose 1 dollar
Rule 2: if exactly 1 six is rolled, we gain 1 dollar
Rule 3: if two sixes are rolled, we gain 2 dollars
Rule 4: if three sixes are rolled, we gain $k$ dollars
Since we are looking for fairness, it seems to make sense (though I'm not really sure why this is the case) to add up all cases times their probabilities and set it equal to zero.
Thus $pr(zero six)(-1) + pr(one six)(1) + pr(two six)(2) + pr(three six)(k) = 0$
However, in the solution to the problem, it is written this way:
$pr(zero six)(-1) + 3 * pr(one six)(1) + 3 * pr(two six)(2) + pr(three six)(k) = 0$
I am wondering what those "threes" are for in the middle of the equation, and also why setting the equation equal to zero gives you some notion of "fairness".
The probability of one six is $3 \cdot \frac 16 \cdot (\frac 56)^2=\frac {75}{216}$ because there are three choices for the position of the six. That said, $pr(onesix)$ should already reflect this. You should check the definition of $pr(onesix)$
The reason this equation gives you a fair game is that you are calculating the long run expectation. You sum up the probability of a result times the value of that result. If you had a biased coin that came heads $\frac 13$ of the time, won $1$ for a tail, and wanted to calculate the payoff $k$ for a head that made the game fair, you would say $\frac 13k+\frac 23\cdot 1=0$ and find $k=-2$,so you should pay your opponent 2 for a head. As you win twice as often as you lose, if you pay twice as much in a loss the game is fair.