Could Your Vote Matter? Over the past couple of months, there has been a lot of hype over the presidential election. Certainly this is by no means nonsensical. After all, the country is going forward with its four year tradition of democratically deciding who the next most powerful man on Earth will be. Upon an interesting discussion with my professor and a few peers before one of my Math 311 lectures, an interesting question was posed to me: what is the probability that a single person’s vote will be the deciding factor in the election? Initially, I neglected the Electoral College and assumed that every voter in America had an equal probability of voting Republican or Democrat and thus a simple binomial distribution with parameters (p=.5, n= total number of votes minus the potentially deciding one) could be used to calculate the probability that there would be a tie before the final vote and this would indeed become the influential factor. In considering this further, I believe it is necessary to establish a certain set of axioms which will be assumed and utilized to go about finding a solution. Although these may not be completely true in the real world, they provide a reasonably plausible and objective basis on which to analyze the problem in question. The first of these was previously implied and the second directly stated. All seven are listed here: A vote can only go one of two ways. The chance of a vote going either of these two ways is ½. The potentially decisive vote (PDV) will be considered the last one counted. The potentially decisive state (PDS) will be considered the last one counted. The distribution of electoral votes without the PDS is approximately normal (Gaussian) with parameters mean = 239 - EVPDS/2 and variance = 4864 - (EVPDS2 - EVPDS)/2. (EVPDS = electoral votes possessed by the potentially decisive state) The 50 other vote holders (including DC) have a 50% chance of choosing either candidate. There can be no tie in the Electoral College. Unfortunately, the Electoral College system does make deriving a solution more complex even with these assumptions. In theory, the probability that a single vote would decide the election is equal to the chance that the vote would decide its state multiplied by the chance that this state would decide the Electoral College. This is of course since these are two independent events which both have to take place in order for the theoretical deciding vote to occur. The problem becomes even more convoluted when the possibility of an even number of in-state voters is considered. In this case, the final vote could eliminate a state tie by opting for the nominee with one more vote; or it could force a tie by selecting the candidate behind. In the event of an (in-state) tie, other means would be used as a tiebreaker and the vote would not be explicitly decisive. However, given this particular instance, there is still a 50% chance that because of the forced (in-state) tie, the candidate who would have otherwise lost will now become victorious. In this way the last vote could still be implicitly decisive. Likewise, a non-tying vote also has a 50% chance of being irrelevant because a forced tie could still have favored the otherwise preferred candidate. Therefore, the probability of one vote deciding an election given that it will decide its state and also that this state can decide the Electoral College is only 3/4 because of the chance that a tie will favor the unselected candidate or that a broken tie will be trivial. With this in mind, let’s start by calculating the chance that a vote will decide a state of an odd no number of voters. In this case the state needs to be tied before the final vote. The probability of this is equal to [no-1 Combination (no-1)/2]*(1/2)n-1, a simple binomial distribution. Contrarily, in the event of an even n¬e number of voters, the probability of a state deciding vote is equal to 2!*[ne-1 Combination ne/2]*(1/2)n-1 since either candidate could potentially have one more than half the counted votes. Although this would theoretically calculate the precise probability of a PDV, the combinatory factorial term for the quantity of voters in each state is many magnitudes of order too large to calculate for any modern computer. Therefore the normal distribution will be used to approximate the probability of a PDV by utilizing the binomial mean (np) and variance (npq). It should be noted that in this instance, the normal distribution is an extremely approximation of the binomial distribution because of the extremely large population of state voters. Unfortunately, the problem of determining the chance that a state will decide the Electoral College is many times more tedious to calculate due to the lack of mathematical structure in which the electoral votes are allocated to each state. 538 electoral votes are distributed amongst the 50 states plus Washington DC based on a disproportionate system in which every state receives an automatic two votes in the form of its Senators plus a number of Congressmen which is quasi-proportional to the population of the state. Since electoral votes do not come in fractions, two states with the same amount of votes are extremely likely to have different populations, although they will be relatively comparable and thus almost proportional. Fortunately, the 51 states and their electoral votes make for a pretty good sample of data to be approximated by the normal distribution function, (although not nearly as good as the set of in-state voters). In addition to the fact that the sample includes the entire data set minus the potentially decisive one, the np/nq>5 test is satisfied because the 50 states times the probability of a candidate winning one of them is 50*.5= 25>5. This is one of the most important conditions to be satisfied for the distribution to be justified as a decent approximation of data because it ensures the data set is sufficiently large to account for a proportional occurrence of all possible outcomes. Moving forth, the probability of any state opting for candidate A = gs(x) = ½, x € (0,1), (x can equal 0 or 1 where x=0 implies a lost state and x=1 implies a won state), and the approximating probability density function that candidate A will win Y electoral votes = f(y) and is normally distributed with mean (μ) = ∑_(s=1)^50▒/gs*EVs = 238 - EVPDS/2 and variance (σ2) = ∑_(s=1)^50▒〖1/2*〗EVs2 - (∑_(s=1)^50▒〖1/2*〗EVs)2 = 4864 - (EVPDS2 - EVPDS)/2 where EVs = the amount of votes possessed by state s. Therefore, the probability of a state being decisive is approximately equal to the positive area bounded by the bell curve with limits -t and t given by: F(z) = ∫_(-t)^t▒f(z)dz = ∫_(-t)^t▒〖1/√2π〗* e^(-z^2/2)dz where z = (Y- μ)/ σ, t € [min(z), max(z)] and Y € (μ-EVPDS+.5, μ+EVPDS-.5). Note that the plus and minus point five in the limits of change for Y is because the area desired has a base of 1, (and the continuous normal function is being used to approximate the discrete binomial case.) Putting everything together, here is an example of how this model can be used to approximate the probability that a single vote in Pennsylvania would have decided the recent election. Approximately 5,950,000 votes were cast, and Pennsylvania contains 21 electoral votes. Therefore, the probability that a candidate received X votes in PA was approximately normally distributed with mean = np = 5,950,000*.5 = 2,975,000 and variance = npq = 5,950,000*.5*.5 = 1,487,500. As a result, the theoretical probability that there was a decisive vote in PA was about .00033. Additionally, the distribution of electoral votes won by a given candidate was normally distributed with mean = 227.5 and variance = 4,654, and the chance that PA was the decisive state would have been about .24179 (by means identical to those used above with the electoral college). Therefore, the chance that a single PA voter could have decided the 2008 election is approximately equal to .00033*.24179 = .00007979. This means that if the United States were to hold together as a nation indefinitely, then on average a decisive vote would occur approximately once in every 12,500 elections, or every 50,000 years. It should be noted that the above probabilities are only theoretical and only hold true under certain axioms which are logical to assume in the general case, but may be far from actual reality in any given election. The periodic political pendulum swings back and forth every couple of decades as people get fed up with the party in office. This might very well interfere with the assumption that a randomly chosen voter had 50-50 odds of voting for a particular party. For example: the probability that a randomly chosen participant in the 1932 election voted for Herbert Hoover is far less than one half, just as the probability that another randomly chosen participant voted for FDR is substantially greater than one half. Another point to consider is the existence of third parties. In recent elections, the Democrats have lost more votes to the Green Party than the Republicans and this in and of itself could have had unforeseen consequences. Nevertheless it is quite clear that under any reasonable set of assumptions, a given individual has virtually zero chance of deciding the US presidential election. Despite this, it makes perfect sense to go out and vote, much like playing the lottery makes sense; since it turns out a very small amount of your time has the slightest potential to decide something of such importance.