A random strategy is called for, to prevent the other two logicians from exploiting a non-random choice. For example, if logician A had a firm strategy to choose the number 1, B and C could gain an advantage by choose 1 with probability 75%, and 2 with probability 25%. The following would be the possible outcomes under that scenario:
B picks 1, C picks 1: probability 0.75×0.75 = 0.5625, A gets $1, B gets $1, C gets $1.
B picks 1, C picks 2: probability 0.75×0.25 = 0.1875, A gets $0, B gets $0, C gets $3.
B picks 2, C picks 1: probability 0.25×0.75 = 0.1875, A gets $0, B gets $3, C gets $1.
B picks 2, C picks 2: probability 0.25×0.25 = 0.0625, A gets $3, B gets $0, C gets $0.
Expected value for A = 0.5625×1 + 0.0625×3 = $0.75.
Expected value for B or C = 0.5625×1 + 0.1875×3 = $1.125.
The optimal strategy is the one that can not be defeated by any other strategy used by both the other logicians. Now, this is where I use some hand waiving logic, but I contend that if two logicians use this same strategy, it won't make any difference what the third logician does, the expected value will be equal for all three.
Let's start by consider the case where the only choices are 1 and 2. Obviously, the only way to win would be to be the singleton in a split game. There is no difference between 1 and 2. So the correct strategy would be to pick 1 or 2 randomly, with 50% each.
Next, let's introduce a third choice. If at least one logician picks 1, the game will be decided for that reason. If none pick number 1, then it reverts to the two-number game of the last step. Let p be the probability that logicians A or B pick number 1. Logician C should be indifferent to picking 1 or not. If he doesn't, then he should revert to the 50/50 strategy of the 2-number game, because it will only come into play if nobody picks 1. The expected value of logician C picking 1 is $1×p^2 + $3×(1-p)2. The expected value of C not picking 1 is $1×(1-p)^2 + $3×p^2. So, we have $1×p^2 + $3×(1-p)^2 = $1×(1-p)^2 + $3×p^2. Doing some simply algebra, we get p=0.5. So, in the 3-number case, the probability of picking 1 is 50%, 2 is 25%, and 3 is 25%.
Next, let's introduce a fourth choice. Again, if at least one logician picks 1, the game will be decided for that reason. If none pick number 1, then it reverts to the three-number game of the last step. Let p be the probability that logicians A or B pick number 1. Logician C should be indifferent to picking 1 or not. If he doesn't, then he should revert to the 50/25/25 strategy of the 3-number game, because it will only come into play if nobody picks 1. Again, the expected value of logician C picking 1 is $1×p^2 + $3×(1-p)^2. The expected value of C not picking 1 is $1×(1-p)^2 + $3×p^2. So, we have $1×p^2 + $3×(1-p)^2 = $1×(1-p)^2 + $3×p^2. The equation is the same as the 3-number game, so p will still be 0.5. So, in the 4-number case, the probability of picking 1 is 50%, 2 is 25%, 3 is 12.5%, and 4 is 12.5%.
Repeating this logic, the probability of number 1 will always be 50%, and the other numbers will shift down 1, and be divided by 2. Done infinitely, the probability of x (where x>1) will be the probability half of the probability of x-1. So, the probability of picking n is (1/2)^n.
Michael Shackleford, A.S.A.