By Jonathan Livengood
Elections
Now, consider three voting scenarios: two-candidate, simple-majority elections without abstentions; two-candidate, simple-majority elections with abstentions; and three-candidate, simple-plurality elections with or without abstentions. Given the vote cast by a specific voter in some election and the result of that election, I ask whether the vote is an actual cause of the result according to Woodward’s proposal.
All three election scenarios share the structure pictured in Figure 2 below. In this figure, each vote (or voter) is labeled with a “Vi,” and the outcome is labeled “Win.” This greatly simplifies our work, since for any vote, there is only one path to the outcome and that path contains a single directed edge.

Figure 2
This is not meant to be a particularly realistic description of actual voting scenarios. It assumes, unreasonably, that the votes are independent, whereas they might be causes of one another (say, due to voters influencing one another by political argument) or they might be effects of a common cause (say, due to the influence of a demagogue, like Rush Limbaugh). This more complicated scenario is pictured in Figure 3.

Figure 3
It also assumes, sometimes reasonably and sometimes not, that there are direct connections between the voters (or their votes cast) and the outcome of an election, whereas some elections process votes through voting machines that (either by accident or by design) might distort the influence of a vote on the outcome of an election from what we want that influence to be. This scenario is pictured in Figure 4.

Figure 4
On the other hand, there are interesting cases that correspond to the simple picture in Figure 2. For example, weighing collections of one-gram masses in a simple, two-pan balance corresponds to Figure 2 for two-candidate, simple-majority elections. Moreover, if the accounts of actual causation under consideration cannot get these simple cases right, it seems unreasonable to hold out hope that they will get more complicated variations right.
Before getting into the details, let me summarize the results. For two-candidate, simple-majority elections without abstentions, every vote for the winning candidate (or proposition) is an actual cause of the result, and no vote for the losing candidate (or proposition) is an actual cause of the result. For two-candidate, simple-majority elections with abstentions, every vote for the winning candidate (or proposition) is an actual cause of the result, every abstention is an actual cause of the result, and no vote for the losing candidate (or proposition) is an actual cause of the result. For three-candidate, simple-plurality elections with or without abstentions, every vote is an actual cause of the result.
Two-candidate, simple-majority
The scenario envisioned here is very simple. Everyone must cast a vote. Each vote cast is for exactly one of the two candidates.[1] If both candidates receive the same number of votes, then the election results in a tie. Otherwise, whichever candidate receives the most votes wins the election. Thus, the election may end in a victory for one or the other of the two candidates, or it may end in a tie.
Consider an election in which there are 2k voters. (I leave it to the reader to show that elections with an odd number of voters produce identical results.) Suppose that j votes are cast for candidate A, and suppose without loss of generality that 2k – j < j. Thus, candidate A is the actual winner of the election.
Is a vote for candidate A an actual cause of candidate A’s victory? To decide, choose a vote VA for candidate A. There is only one path from VA to the result of the election, and no other vote is on this path. Hence, we are free to change any of the other votes, so long as candidate A wins the election after the changes. Distribute the votes such that there are k + 1 votes for A and k – 1 votes for B. Now, change the value of VA from A to B. Since such a change results in a tie (k votes for A against k votes for B), VA is an actual cause of A’s victory. Because VA was chosen arbitrarily, the same reasoning applies to every vote for candidate A. Hence, every vote for candidate A is an actual cause of candidate A’s victory.
On the other hand, a vote for candidate B is not an actual cause of candidate A’s victory. To see this, choose a vote VB for candidate B. Again, there is only one path from VB to the result of the election, and no other vote is on this path. Hence, we are free to change any of the other votes, so long as candidate A wins the election after the changes. However, there is no redistribution of the votes such that A is the winner of the election but would not have been the winner had VB not voted for candidate B. Let r be the redistributed votes for candidate A. Since candidate A must be the winner after any redistribution, 2k – r < r. For vB to be an actual cause of candidate A’s election, there must be k, r ≥ 0 such that 2k – r – 1 ≥ r + 1. That is, a change in vote VB must result in a change in the election, either to a tie or to a victory for B. But 2k – r < r2k – r < r + 22k – r – 1 < r + 1. So, VB is not an actual cause of candidate A’s election. Because VB was chosen arbitrarily, the same reasoning applies to every vote for candidate B. Hence, no vote for candidate B is an actual cause of candidate A’s victory.
Two-candidate, simple-majority with abstentions
In this scenario, every vote cast is for exactly one of the two candidates (just like in the previous scenario described in Section 2.1). However, in this scenario, voters are not obligated to cast a vote. As before, if both candidates receive the same number of votes, then the election results in a tie. Otherwise, whichever candidate receives the most votes wins the election. So again, the election may end in a victory for one or the other of the two candidates, or it may end in a tie.
Consider an election in which there are 2k voters. (I leave it to the reader to show that elections with an odd number of voters produce identical results.) Suppose that i votes are cast for candidate A, j votes are cast for candidate B, and l voters abstain. Further, suppose without loss of generality that i > j. Hence, candidate A is the actual winner of the election.
Is a vote for candidate A an actual cause of candidate A’s victory? To decide, choose a vote VA for candidate A. There is only one path from VA to the result of the election, and no other vote is on this path. Hence, we are free to change what any of the other voters do, so long as candidate A wins the election after the changes. Distribute the votes such that there is one vote for candidate A, zero votes for candidate B, and 2k – 1 abstentions. Now, change the value of VA from A to B. Since such a change results in a victory for candidate B (zero votes for A against one vote for B), VA is an actual cause of A’s victory. Because VA was chosen arbitrarily, the same reasoning applies to every vote for candidate A. Hence, every vote for candidate A is an actual cause of candidate A’s victory.
Is a vote for candidate B an actual cause of candidate A’s victory? No. To see this, choose a vote VB for candidate B. Again, there is only one path from VB to the result of the election, and no other vote is on this path. Hence, we are free to change what any of the other voters do, so long as candidate A wins the election after the changes. However, there is no redistribution of the votes such that A is the winner of the election but would not have been the winner had VB not voted for candidate B. Let r be the redistributed votes for candidate A and a be the redistributed abstentions. Since candidate A must be the winner after any redistribution, 2k – r – a < r. For VB to be an actual cause of candidate A’s election, there must be a, k, r ≥ 0 such that 2k – r – a – 1 ≥ r. That is, a change in vote VB must result in a change in the election, either to a tie or to a victory for B. (Since the change need not be in favor of candidate A, we do not add one to the right hand side of the inequality as we did in the previous case.) But 2k – r – a < r2k – r – a < r + 12k – r – a – 1 < r. So, VB is not an actual cause of candidate A’s election. Because VB was chosen arbitrarily, the same reasoning applies to every vote for candidate B. Hence, no vote for candidate B is an actual cause of candidate A’s victory. (Notice that if we had changed the vote to favor A instead of making it an abstention, VB still would not have been counted as an actual cause of A’s victory.)
And, now, what about abstentions? Are abstentions actual causes of candidate A’s victory? Yes, they are. Choose an abstention, call it Vnone. There is only one path from Vnone to the result of the election, and no other vote is on this path. Hence, we are free to change what any of the other voters do, so long as candidate A wins the election after the changes. Distribute the votes such that there is one vote for candidate A, zero votes for candidate B, and 2k – 1 abstentions. Now, change the value of Vnone from an abstention to a vote for B. Since such a change results in a tie (one vote for A against one vote for B), Vnone is an actual cause of A’s victory. Because Vnone was chosen arbitrarily, the same reasoning applies to every abstention. Hence, every abstention is an actual cause of candidate A’s victory.
Three-candidate, Simple-plurality
In this scenario, every vote cast is for exactly one of the three candidates. (It should be noted that allowing abstentions changes nothing. Furthermore, adding additional candidates leaves the result here unchanged as well.) If all three candidates receive the same number of votes or if two candidates have the same number of votes as each other and more votes than the third candidate, then the election results in a tie. Otherwise, whichever candidate receives the most votes wins the election. (In other words, a candidate need not receive the majority of the votes, and there are no run-offs.) So again, the election may end in a victory for exactly one of the three candidates, or it may end in a tie.
Consider an election in which there are 2k voters. (I leave it to the reader to show that elections with an odd number of voters produce identical results.) Suppose that i votes are cast for candidate A, j votes are cast for candidate B, and l votes are cast for candidate C. Further, suppose without loss of generality that i > j ≥ l. Hence, candidate A is the actual winner of the election.
To see that every vote for candidate A is an actual cause of candidate A’s victory, we proceed as before. Choose a vote VA for candidate A. There is only one path from VA to the result of the election, and no other vote is on this path. Hence, we are free to change any of the other votes, so long as candidate A wins the election after the changes. Distribute the votes such that there are k + 1 votes for candidate A, k – 1 votes for candidate B, and no votes for candidate C. Now, change the value of VA from A to B. Since such a change results in a tie (k votes for A against k votes for B), VA is an actual cause of A’s victory. Because VA was chosen arbitrarily, the same reasoning applies to every vote for candidate A. Hence, every vote for candidate A is an actual cause of candidate A’s victory.
To see that every vote for candidates B is an actual cause of candidate A’s victory, choose a vote VB for candidate B. There is only one path from VB to the result of the election, and no other vote is on this path. Hence, we are free to change any of the other votes, so long as candidate A wins the election after the changes. Distribute the votes such that there are k votes for candidate A, one vote for candidate B, and k – 1 votes for candidate C. Now, change the value of VB from B to C. Since such a change results in a tie (k votes for A against k votes for C), VB is an actual cause of A’s victory. Because VB was chosen arbitrarily, the same reasoning applies to every vote for candidate B. Hence, every vote for candidate B is an actual cause of candidate A’s victory.
Similarly, every vote for candidates C is an actual cause of candidate A’s victory. Choose a vote VC for candidate C. There is only one path from VC to the result of the election, and no other vote is on this path. Hence, we are free to change any of the other votes, so long as candidate A wins the election after the changes. Distribute the votes such that there are k votes for candidate A, k – 1 votes for candidate B, and one vote for candidate C. Now, change the value of VC from C to B. Since such a change results in a tie (k votes for A against k votes for B), VC is an actual cause of A’s victory. Because VC was chosen arbitrarily, the same reasoning applies to every vote for candidate C. Hence, every vote for candidate C is an actual cause of candidate A’s victory.
Thus, on Woodward’s account, every vote cast in an election having three (or more) candidates is an actual cause of the result of the election! Notice that this result does not in any way depend on the actual number of votes cast for each candidate. Even if no one votes for candidate C, every vote for candidate B is an actual cause of candidate A’s victory (assuming we have fixed candidate A as the winning candidate as we did above).
[1] One may also think of two-candidate elections in terms of “yes” or “no” votes respecting a proposition. In the three-candidate case, the analogy only holds when there are three propositions (or, more wildly, three truth-values respecting a single proposition).