The Probability That I Alone Will Decide an Election October 03, 2006

With the midterm election fast approaching, I pondered recently why the act of voting is significant at all. To anyone reading this, have no illusions. I will be voting November 7th. But, why? In 2002, 14,921 voters showed up at the polls to decide who would represent them in the State Assembly for the 97th District. Ignoring the Libertarian Party candidate and "scattering" votes, 14,419 citizens here chose either the Democratic or Republican Party candidate. The only outcome contingent upon any one voter deciding between Roger Danielsen or Ann Nischke was 7,209 votes registered for both candidates. In that case, an individual could decide the election - the outcome would be 7,210 to 7,209. In order to locate the approximate likelihood of an event like this occurring, we need two pieces of mathematics - one relatively straightforward, the other somewhat obscure. It turns out that solving this problem involves multiplying an extremely small number by an equally extreme large number.

Logarithmic Functions

A tied vote is no different from tossing a coin 14,418 times. I ignore the last vote because I assume that that voter (the deciding vote) walks to the polls with a firm mind of his/her choice. The rest vote entirely at random. Note that this is a necessary condition for one voter to decide an election. Now, in the real world, of course, voters do not behave this way. But the result is the same as if they behaved randomly.

The probability of tossing a coin and it showing a head or tail is .5, or one-half. Why? There are only two possible events resulting from tossing a coin. It can only show either a head or a tail. Since these events are equally likely, assuming a fair coin, a coin toss over many trials will yield half of the events as heads and half as tails. All trials are independent, furthermore, of other trials. That is, the likelihood of any one event is not contingent upon the results of any other trials. The coin has no memory.

I assume now that the final vote tally (including my vote) is an odd number, in order to avoid a tie. First, suppose that only three voters show up on election day. One votes for Kramer, while the other votes for Schmuki. The probability of my vote alone electing Schmuki to the Assembly is .5. To see this, note that we now have four possible outcomes (Kramer-Kramer, Kramer-Schmuki, Schmuki-Kramer, and Schmuki-Schmuki). Thus, the probability of any one outcome is 1/4, or .25. But, two of the events (Kramer-Schmuki and Schmuki-Kramer) would permit my vote alone to determine the outcome. Therefore, the probability of the event that I decide the election is 1/4 + 1/4 = 1/2, or .5.

Now, extend this analysis, to 1,001 voters - five-hundred votes for Kramer and five-hundred votes (and mine) for Schmuki. This result is exactly the same as if a thousand voters tossed a coin to determine their choice before driving to the polls. What is the likelihood of this outcome - 1,000 coin tosses and 500 show heads while 500 show tails? Look at the number of possible outcomes. Above, we trialed only twice, so, only four outcomes were possible. With a thousand trials, the number of possible outcomes is, of course, much larger. In fact, the number of possible outcomes if one were to toss a coin a thousand times is approximately 10.72³⁰⁰!

We can now express this relationship in the following form:

y = 2^x

This simple equation expresses the base two logarithm of x on y, such that x ~ number of trials and y ~ number of possible outcomes. For our purposes, the following is the number of possible outcomes from 14,418 tosses of a fair coin:

2^14,418

The probability of any particular event, say, 7,209 Kramer votes immediately followed by 7,209 Schmuki votes, is merely its reciprocal:

2^-14,418=(1/2)^14,418

This number, of course, is vanishingly small.

Combinatorics

We cannot conclude, however, that only one combination of 14,418 voters for either candidate show up at the polls. For example, 7,209 Kramer votes could show, followed by 7,209 for Schmuki. Or, they could follow a staggered pattern: one vote for Kramer, followed by one vote for Schmuki, followed by one vote for Kramer, and so on. This pattern could occur for any block of votes provided that the number of votes within each block can be divided into 14,418 with no remainder. Lastly, Kramer and Schmuki votes could be randomized altogether, with three for Kramer here, six for Schmuki there, two for Kramer here, and so on. The only constraint is that, in the end, Kramer receives 7,209 votes. There are an enormous number of combinations that could all potentially align to a tie vote, thus allowing the last vote to decide the election. Therefore, we must now turn our attention to a somewhat obscure piece of mathematics known as combinatorics.

Combinatorics sounds harder than it really is. A combination is basically the number of ways one can draw a subset of objects from a larger set of objects when the order of that subset is irrelevant. A classic example is the number of ways a person can draw 50 marbles from a jar that contains 100 marbles (roughly 10²⁸ ways). Now, imagine that the jar is the election; each marble is a voter; and our problem is to calculate how many ways we can draw 7,209 marbles from a jar that contains 14,418 of them. The result is the total number of events in which Kramer and Schmuki tie before my vote is counted. We will then multiply it by the probability of any given event's occurence and the result will be the probability of a tied vote in which my vote decides the outcome.

As might seem obvious, 14,418 choose 7,209, in the subfield's language, is extremely large. There are approximately 2^14,410.77 ways in which 14,418 voters can arrange themselves such that 7,209 vote for each candidate. So, we now know the following:

The probability of any particular event's occurence is 2^-14,418.

The number of occurences among 14,418 voters in which exactly 7,209 of them decide to vote for each candidate is approximately 2^14,410.77.

Multiplying both together, we calculate the probability that a voter can decide the outcome of an election in which 14,418 votes split precisely down the middle.

~2^-14,418 * 2^14,410.77 = 2^-7.23 = 1/150.12

~149:1 odds

In other words, tied elections, even among thousands of voters, are not usual occurences, but neither are they freak events. Naturally the probability of having a direct, individual impact on an election outcome steadily falls as the voter pool expands.

Number of Voters	Probability of Decisive Vote
10	1/4.06
100	1/12.55
1,000	1/39.67
10,000	1/125.67
100,000	1/388.02
1,000,000	1/1,251.98
10,000,000	1/3,956.48

What we can conclude from this is that the odds of directly influencing the outcome of an election, even at the national level, are considerably greater than getting struck by lightning, dying in a terrorist attack, or winning the Powerball.