Question

Whenever there is a political campaign, political leaders face a series of difficult decisions. One of the most interesting is the problem of deciding whom to support and when in a primary campaign. Suppose that a particular party leader faces a decision in a primary election. He can support Smith or Brown or he can avoid a commitment, simply declaring himself neutral in the primary. The consequences stemming from these three alternatives depend, of course, on whether it is Smith or Brown who wins the nomination. Let us suppose that our leader is mostly concerned about the number of patronage jobs that will be allocated to him and that the following table reflects the actual situation:

	If Smith Wins	If Brown Wins
If I am neutral	10 jobs	15 jobs
If I am for Smith	30 jobs	0 job
If I am for Brown	5 jobs	25 jobs

What would the party leader do if he wants to maximize expected value and the probability of Smith’s winning is equal to p?

Answer 1

Solution

Given the probability of Smith’s winning is equal to p, the probability of Brown’s winning is equal to (1 – p) since there are only two candidates and one must win.

Let E_N, E_S and E_B be the expected number of patronage jobs the party leader would get on being neutral, being for Smith and being for Brown respectively, then

E_N = 10p + 15(1 - p) = 15 – 5p …………………………………………………(1)

E_S = 30p + 0(1 - p) = 30p ………………………………………………………(2)

E_B = 5p + 25(1 - p) = 25 – 20p …………………………………………………(3)

If (1) < (3), then 15 – 5p < 25 – 20p or 15p < 10 or p < 2/3 …………………….. (4)

If (1) < (2), then 15 – 5p < 30p or 35p > 15 or p > 3/7 …………………………..(5)

If (2) < (3), then 30p < 25 – 20p or 50p < 25 or p < 1/2 …………………………..(6)

Now, p < ½ => p < 2/3 and hence (4) and (6) => (3) > both (1) and (2) => E_B is maximum.

Similarly, if ½ < p < 3/7, E_S is maximum.

And by elimination, if p > 3/7, E_N is maximum.

Thus, the optimum strategy would be: be

neutral if p > 3/7, be for Smith if ½ < p < 3/7, and be for Brown if p < ½. ANSWER