Need the answer to Part (d). Thank you! Let X, Y, Z be discrete random such that

Question

Need the answer to Part (d). Thank you!

Let X, Y, Z be discrete random such that, with probability 1, X, Y, and Z are all unequal. Let a = P(X > Y), b = (Y > Z), and c = P(Z > X). Show that min {a, b, c} lessthanorequalto 2/3 [This part is related to the observation that in an election, it is possible for more than half the voters to prefer candidate A to candidate B, more than half B to C, and more than half C to A.] Show that if X, Y, and Z are independent and identically distributed then a = b = c = 1/2. Assume P(X = 0) = 1, Y and Z are independent, P(Z = 1) = P(Y = -1) = p, and P(Z = -2) = P(Y = 2) = 1 - p. Compute max {a, b, c} as a function of p, and maximize with respect to p. Construct an example where a = b = c = 2/3. Applications of statistics along these lines are used to model "mathematical games" which are used in political decision making all the time. Arguably, game theory, not nuclear weapons, won WWII.

Explanation / Answer

I came up with this example going with the given hint:

So let us consider three dice X, Y & Z such that

Now we calculate the probabilities.

P(Y=3)=1

P(X>Y) denotes the probability that the die X takes the value 6 which is 4/6 = 2/3.

Therefore, a = P(X>Y) =2/3

Further, P(Y>Z) denotes the probability that the die Z takes a value less than 3 and according to the die we have choosen the die Z has 1,1, 2,2,written on four of its sides respectively,

So, b=P(Y>Z)=4/6=2/3

Now, P(Z>X)

**This is possible only when the die Z shows 2,2,5,5 and the die X shows 1, which happens in exactly 4 out of 6 case.

*** Note that Z and X cannot be equal.

So, c=P(Z>X)=2/3

Therefore, we have a=b=c=2/3

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