I am having trouble with this question with regards to calculating expected values:
Peter and Maria players each rolling one die and earns \$1 the player with the best score (Each player rolls one six-sided die), but the maria's die is illegal. Thus, $P(1)=P(6)=\frac16$, $P(2)=P(5)=\frac14$ and $P(3)=P(4)=\frac{1}{12}$.
Expected value of each earnings ?
answer: Peter and Mary earns \$ 0.5
my idea is to calculate the odds of winning maria, ie, P (2 to maria, 1 Peter) = 1/24.
P (3 to maria, 1 Peter) and P(3 to maria, 2 Peter)= 1/36
P (4 to maria, 1 Peter) and P(4 to maria, 2 Peter) and P(5 to maria , 4 to peter)= 1/24.
but I think made a mistake.
please help me
It isn't clear to me what happens in the event of a tie. Let's say there is $0$ payout in that case (though this will not give $.5$ as the common expectation).
To analyze the game, let's just take Peter's possible tosses one at a time. Of course each of these has probability $\frac 16$. In each case we compute the (conditional) probability of a tie, a win by Peter, and a win by Maria. We'll write the results as a triple. Thus, the triple corresponding to a toss of $1$ is $(\frac 16,0,\frac 56)$ which means that the probability of a tie is $\frac 16$, the probability Peter wins is $0$, and the probability that Maria wins is $\frac 56$ (all conditional on Peter having thrown a $1$).
$1]$ $(\frac 16,0,\frac 56)$
$2]$ $(\frac 14,\frac 16,\frac {14}{24})$
$3]$ $(\frac 1{12},\frac {10}{24},\frac {12}{24})$
$4]$ $(\frac 1{12},\frac {12}{24},\frac {10}{24})$
$5]$ $(\frac 1{4},\frac {14}{24},\frac {4}{24})$
$6]$ $(\frac 1{12},\frac {20}{24},0)$
Inspection shows that the probability of Peter's victory equals that of Maria's victory, so if we ignore ties the answer makes sense. However, the probability of a tie is $\frac 16^*\frac 1{12}^*(2+3+1+1+3+2)=\frac 16$. It follows that Peter and Maria each have a $\frac 5{12}$ chance at victory so (assuming that ties pay nothing) the expected value is $.41666...$ .
Note: if you assume that, in the event of a tie the players each get $.50$ then the reported answer would be correct, but I can't see any reason why this assumption should be inevitable.
Note II: from the comments it seems we are to simply reject ties. That's fine, it just means that those events are removed from the sample space. The above calculation shows that Peter and Maria have the same chance of winning, hence with this assumption the expected winnings of each must be $.5$