Odd or Even

Problem

Players I and II simultaneously call out one of the numbers one or two. Player I’s name is Odd; he wins if the sum of the numbers if odd. Player II’s name is Even; she wins if the sum of the numbers is even. The amount paid to the winner by the loser is always the sum of the numbers in dollars. To put this game in strategic form we must specify X, Y and L. Here we may choose X = {1, 2}, Y = {1, 2}, and L as given in the following table.

It turns out that one of the players has a distinct advantage in this game. Can you tell which one it is?

Solution and thoughts

As there is no sandle point in this matrix, neither of the player can have the best choice. Instead, a probability of selecting which choice will be critical in this problem. By reassigning the matrix and use a,b,c,d to represent (x1,y1),(x1,y2),(x2,y1),(x2,y2), the following graph will represent the meaning of all possible outcome.
The x axis represent the probability of Player I picking number 1, and y axis represent the outcome score. The green line represent the case when Player II picked number 1, and orange is 2. We noticed that whenever we pick a value X which can correspond to two different Y value on the green and orange line, Player II can optimize by choicing the smaller value one. So, in order to optimize the gain of Player I, the cross point is the best solution. (Same case with Player II)

By the formula of interpolation, we know that the value of the cross point is equal to
\(\frac{(c-d)\times a + (b-a)\times c}{(b-a)+(c-d)}\)
and also equals to
\(\frac{(c-d)\times b + (b-a)\times d}{(b-a)+(c-d)}\)
As we know that the value from green line and orange line are the same Which the outcome value in this case is equal to 1/12

Same rule applies, the possiblity of Player I picking number 1 should be
\(\frac{b-a}{(b-a)+(c-d)}\)
And the value will be 5/12