I suppose it would be something like the probability of a change times the magnitude of the change times the number of bits of control you have over the direction.

Number of bits is a property of the choice, not a property of the chosen option. So it is the same for both candidates. Probability of change is what precisely? Some change is pretty certain. The most charitable and still reasonably literal interpretation is to take the probability distribution for each magnitude of change and integrate over, which means that you suggest to vote for the candidate whose victory causes higher expected change. It may be sensible (if you are actually trying to maximise change), but I don't see what it has to do with your influence.

If, on the other hand, by "probability of change" you mean "probability of victory of that candidate", it makes more sense. But it's still imperfect.

Compare with utilitarian answer: Suppose U(third party candidate X wins) > U(lizard Y wins) > U(lizard Z wins). Then I am deciding between voting for X or Y (there is clearly no sense in voting for Z and let's assume I must vote). We compare

1. U(X) p(X | I vote for X) + U(Y) p(Y | vote for X) + U(Z) p(Z | vote for X) to
2. U(X) p(X | vote for Y) + U(Y) p(Y | vote for Y) + U(Z) p(Z | vote for Y)

p(X | vote for X) is equal to p(X | I don't vote) + p(X has exactly as many votes as Y and more votes than Z | I don't vote) + p(X has exactly as many votes as Z and more votes than Y | I don't vote). Similarly, p(X | vote for Y) is equal to p(X | I don't vote) - p(X has one more vote than Y and more votes than Z | I don't vote) - p(X has one more vote than Z and more votes than Y | I don't vote). The p(X | don't vote) cancel out (similarly for Y and Z) and we are left to estimate the probability that my vote is the decisive one.

To find out this probability we must have a distribution over exact vote counts, not only over winning candidates. It is not clear that p(draw between X and some other candidate) is proportional to p(X wins).