Do not vote, one vote will not reverse election results. What is wrong with this reasoning? 
Do not vote, one vote will not reverse the election result. What's
  more, the probability of injury in a traffic collision on the way to the
  ballot box is much higher than your vote reversing the election
  result. What is even more, the probability that you would win grand
  prize of lottery game is higher than that you would reverse election
  result.

What is wrong with this reasoning, if anything? Is it possible to statistically prove that one vote matters?
I know that there are some arguments like "if everybody thought like that, it would change the election result". But everybody will not think like that. Even if 20% of electorate copy you, always a great number of people will go, and the margin of victory of winning candidate will be counted in hundreds of thousands. Your vote would count only in case of a tie.
Judging it with game theory gains and costs, it seems that more optimal strategy for Sunday is horse race gambling than going to the ballot box. 
Update, March 3.
I am grateful for providing me with so much material and for keeping the answers related to statistical part of the question. Not attempting to solve the stated problem but rather to share and validate my thinking path I posted an answer. I have formulated there few assumptions. 


*

*two candidates 

*unknown number of voters 

*each voter can cast a random vote on either candidate


I have showed there a solution for 6 voters (could be a case in choosing a captain on a fishing boat). I would be interested in knowing what are the odds for each additional milion of voters. 
Update, March 5.
I would like to make it clear that I am interested in more or less realistic assumptions to calculating the probability of a decisive vote. More or less because I do not want to sacrifice simplicity for precision. I have just understood that my update of March 3 formulated unrealistic assumptions. These assumptions probably formulate the highest possible probability of a decisive vote but I would be grateful if you could confirm it. 
Yet still unknown for me thing is what is meant by the number of voters in the provided formulas. Is it a maximum pool of voters or exact number of voters. Say we have 1 milion voters, so is the probability calculated for all the cases from 1 to milion voters taking part in election?   
Adding more fuel to the discussion heat
In the USA, because president is elected indirectly, your vote would be decisive if only one vote, your vote, were to reverse the electors of your state, and then, owing to the votes of your electors, there was a tie at Electoral College. Of course, breaking this double tie condition hampers the chances that a single vote may reverse election result, even more than discussed here so far. I have opened a separate thread about that here.
 A: It is easy to construct situations, where voting matters, e.g. the population consists of 3 people (including myself), one votes red, one votes blue, then clearly my vote matters. 
Of course in your quote, not such trivial quotes are meant, but real-life situations with maybe millions of voters. 
So let us extend my trivial example:
Let $X=1$ indicates, if the count of every other voter results in a tie (thus $X=0$ means no tie ). 
$Y=1$ indicates, if my vote "matters". My vote only matters all the other votes result in a tie. Otherwise it does not matter. 
Therefore $P\left(Y=1 \vert X = 1\right) = 1$ and $P\left(Y=1 \vert X = 0\right) = 0$. 
This means, there is no universal answer. If your vote "matters", completely depends on the actions of all other voters. 
Your question is already solved (with the answer: it depends how the others act), but you can ask follow-up questions: Across different elections, how often does my vote matter on average?
Or in mathematical terms: $P\left(Y=1 \right) = ?$
$P\left(Y=1 \right) = P\left(Y=1 \vert X = 1\right) P\left( X = 1\right)  + P\left(Y=1 \vert X = 0\right) P\left( X = 0\right) = P\left( X= 1\right)$. 
$P\left( X= 1\right)$ depends on the election and the situation, which I denote as $\theta$: $P\left( X= 1\right) = \int P\left( X= 1 \vert \Theta = \theta \right) f \left(\theta\right)\,d\theta$, where $f$ is the sampling distribution of the election. Realistically, for the overwhelming majority of $\theta$, $P\left( X= 1 \vert \Theta = \theta \right)$ will be very close to zero.
Now comes my critique to whuber's solution: $f$ represents the votes, you might participate in your whole lifetime. It will include elections on different candidates, different years different topics and so on. This variability is underrepresented in whuber's solution because it implicitely assumes, there are only elections with a supporters tie (meaning $f$ is a point mass on an unbelievebly improbable event) and $P\left( X= 1 \vert \theta \right)$ is simply a binomial probability of a tie from voters, that are undecided. 
$f$ should reflect the whole election variability. To say it is deterministic at the particular situation of equality between the parties is clearly an under-complex representation of reality, and even in this artificial case the probability is $\frac{1}{10000}$. If I vote 10 times in a lifetime, I need 1000 lifes, that finally my vote matters.
PS: I strongly believe, that voting matters, but not in a statistically describable way. It is a different discussions on a philosophical topic, not a statistical one. 
A: I must disappoint you: current economic theory cannot explain why people keep showing up in elections, because it appears to be irrational. See a survey of literature on this subject on pages 16-35 of  Geys, Benny (2006) - "‘Rational’ Theories of Voter Turnout: A Review". The voter turnout is a percentage of voters that showed up at the poll of a total voting eligible pool. In layman's words it appears that indeed your vote won't make a difference.
As in @whuber answer the analysis is closely related to the probability of casting a pivotal vote, i.e. making or breaking a tie. However, I think @whuber is making the question look simpler than it is, and also suggesting much higher probability of pivotal vote than US and European election data analysis suggests. A voter turnout is a paradox indeed. It must be zero according to theory, yet it's in close to 50% range in USA.
The answer cannot be derived from pure statistics point of view in my opinion. It belongs to behavioral aspects of human actions, which rational choice models explore, albeit in unsatisfactory way because people keep voting while the theory says they shouldn't.
Instrumental Voting
The instrumental voting approach that I mentioned earlier (see earlier reference) is the idea that your vote becomes tie breaking, and thus deices whether you gain benefits from electinng your favorite candidate. It is described with an equation for the expected utility R:
$$R=PB-C>0$$
Here, P is the probability your vote is tie breaking, B benefits you get from you candidate and C associated with voting. The costs C vary and are split into roughly two categories: research of candidates and things dealing with voter registration, driving to polling stations etc. People looked at these components and came to conclusion that P is so low that any positive cost C outweighs the product PB.
Probability P has been considered by many researchers, e,g, see the authorative treatment by Gelman here: Gelman, A., King, G. and Boscardin, J. W. (1998) ‘Estimating the Probability of Events That Have Never Occurred:When Is Your Vote Decisive?' 
You can find a calculation similar to the setup in @whuber's answer here in NBER paper: THE EMPIRICAL FREQUENCY OF A PIVOTAL VOTE, Casey B. Mulligan, Charles G. Hunter. Note, that this is the empirical research of voting bulletins. However, they have the independent binomial voter setup in theoretical part, see Eq.3. Their estimate is drastically different from @whuber, who came up with $\sim 1/\sqrt{n}$ while this paper derives $P=O(\frac 1 n)$, which renders very low probabilities. The treatment of probabilities is very interesting, and takes into account many non obvious considerations such whether a voter realizes what are the tie probabilities or not
A simple intuitive explanation follows, from Edlin, Aaron, Andrew Gelman, and Noah Kaplan. "Voting as a rational choice: Why and how people vote to improve the well-being of others." Rationality and society 19.3 (2007): 293-314.

Let f(d) be the predictive or forecast uncertainty distribution of the
  vote differential d (the difference in the vote proportions received
  by the two leading candidates). If n is not tiny, f(d) can be written,
  in practice, as a continuous distribution (e.g., a normal distribution
  with mean 0.04 and standard deviation 0.03). The probability of a
  decisive vote is then half the probability that a single vote can make
  or break an exact tie, or f(0)/n.
The assumption here is that an exact tie vote will be decided by a
  coin flip. 

Empirical results
Empirical results suggest that for 20,000 voters, the probability of a tie is $\frac 1 {6000}$, which is significantly lower than @whuber's model results $\frac 1 {2\sqrt{20000\pi}}=\frac 1 {500}$

Another empirical study is Gelman, Andrew, Katz, Jonathan and Bafumi, Joseph, (2004), Standard Voting Power Indexes Do Not Work: An Empirical Analysis, British Journal of Political Science, 34, issue 4, p. 657-674. Its main conclusion was first cited in @user76284's answer.
Authors show that $O(1/\sqrt{n}$ doesn't fit the reality. They analyzed a massive amount of electoral data, election held on many different levels in USA and outside. 
For instance, here's the plot from US presidential elections, 1960-2000, state vote data. They show the square root n fit vs. lowes (non-parametric) fit. It's clear that square root doesn't fit the data.

Here's another plot which also includes European election data. Again square root of n relation doesn't fit the data.

Section 2.2.2 in the paper explains the basic underlying assumption of square root result, which helps understand @whuber's approach. Section 5.1 has theoretical discussion.
A: I'm going to take a different tack than the other answers, and argue both sides of the question.
First, let's show that voting is a pointless waste of time.
The function of an election is to derive a single outcome, called "the will of the electorate", from many samples of the individual wills of individual electors.  Presumably that number of electors is large; we're not concerned here with cases of dozens or hundreds of electors.
When deciding whether you should vote, there are two possibilities. Either, as you note, there is a strong preference -- say, 51% or better -- in the electorate for one outcome. In such a scenario the probability that you will cast the "deciding" vote is minuscule, and so no matter which side of the issue you are on, you're better off staying home and not entailing all the costs of voting.
Now suppose the other possibility: the electorate is so narrowly divided that even a small number of voters choosing to vote or not vote could completely change the outcome. But in this scenario, there is no "will of the electorate" at all!  In this scenario you might as well call off the election and flip a coin, saving the expense of the election entirely.

It seems like on rational grounds there is no reason to vote. Suppose a large fraction of the electorate reasons this way -- and, why shouldn't they?  I live in the 43rd district of Washington State, one of the most "blue" districts in the United States. No matter which candidate I support in the district election, I can tell you right now what the party affiliation of the winner will be in my district, so why should I vote?
The reason to vote is to consider the strategic consequences of "a large fraction of the electorate considers it pointless and does not vote" upon small groups of ideologues.  This attitude hands power to comparatively small, well-organized blocs who may show up en masse when not expected; if the number of voters is greatly reduced by a large fraction "rationally" deciding to stay home and not vote, then the size of a bloc required to swing an election against the clear will of the majority is greatly reduced.
Voting when "not rationally necessary" decreases the probability that an effort to swing the election by a relatively small group will succeed, and thereby increases the probability that the actual will of the majority can be determined.
A: The analysis presented in whuber's answer reflects the Penrose square root law, which states that, under certain assumptions, the probability that a given vote is decisive scales like $1/\sqrt{N}$. The assumptions underlying that analysis, however, are too strong to be realistic in most real-world scenarios. In particular, it assumes that the fractions of decided voters for each outcome are virtually identical, as we'll see below.
Below is a graph showing the probability of a tie against the fraction of decided voters for one outcome, given the fraction of decided voters for the other outcome (assuming the rest vote uniformly at random) and the total number of voters:

The Mathematica code used to create the graph was
fractionYes = 0.45;
total = 1000000;
Plot[
 With[
  {
   y = Round[fractionYes*total],
   n = Round[fractionNo*total],
   u = Round[(1 - fractionYes - fractionNo)*total]
   },
  NProbability[y + yu == n + u - yu, 
   yu \[Distributed] BinomialDistribution[u, 1/2]]
  ],
 {fractionNo, 0, 1 - fractionYes},
 AxesLabel -> {"fraction decided no", "probability of tie"},
 PlotLabel -> 
  StringForm["total = ``, fraction decided yes = ``", total, 
   fractionYes],
 PlotRange -> All,
 ImageSize -> Large
 ]

As the graph shows, whuber's analysis (like the Penrose square root law) is a knife-edge phenomenon: In the limit of growing population size, it requires the fractions of decided voters for each outcome to be exactly equal. Even tiny deviations from this assumption make the probability of a tie very close to zero.
This might explain its discrepancy with the empirical results presented in Aksakal's answer. For example, Standard voting power indexes do not work: An empirical analysis (Cambridge University Press, 2004) by Gelman, Katz, and Bafumi says:

Voting power indexes such as that of Banzhaf are derived, explicitly
  or implicitly, from the assumption that all votes are equally likely
  (i.e., random voting). That assumption implies that the probability of
  a vote being decisive in a jurisdiction with $n$ voters is proportional
  to $1/\sqrt{n}$. In this article the authors show how this hypothesis
  has been empirically tested and rejected using data from various US
  and European elections. They find that the probability of a decisive
  vote is approximately proportional to $1/n$. The random voting model
  (and, more generally, the square-root rule) overestimates the
  probability of close elections in larger jurisdictions. As a result,
  classical voting power indexes make voters in large jurisdictions
  appear more powerful than they really are. The most important
  political implication of their result is that proportionally weighted
  voting systems (that is, each jurisdiction gets a number of votes
  proportional to $n$) are basically fair. This contradicts the claim in
  the voting power literature that weights should be approximately
  proportional to $\sqrt{n}$.

See also Why the square-root rule for vote allocation is a bad idea by Gelman.
A: It's wrong in part because it's based on a mathematical fallacy.  (It's even more wrong because it's such blatant voter-suppression propaganda, but that's not a suitable topic for discussion here.)
The implicit context is one in which an election looks like it's on the fence.  One reasonable model is that there will be $n$ voters (not including you) of whom approximately $m_1\lt n/2$ will definitely vote for one candidate and approximately $m_2\approx m_1$ will vote for the other, leaving $n-(m_1+m_2)$ "undecideds" who will make up their minds on the spot randomly, as if they were flipping coins.
Most people--including those with strong mathematical backgrounds--will guess that the chance of a perfect tie in this model is astronomically small.  (I have tested this assertion by actually asking  undergraduate math majors.)  The correct answer is surprising.
First, figure there's about a $1/2$ chance $n$ is odd, which means a tie is impossible.  To account for this, we'll throw in a factor of $1/2$ in the end.
Let's consider the remaining situation where $n=2k$ is even.  The chance of a tie in this model is given by the Binomial distribution as 
$$\Pr(\text{Tie}) = \binom{n - m_1 - m_2}{k - m_1} 2^{m_1+m_2-n}.$$
When $m_1\approx m_2,$ let $m = (m_1+m_2)/2$ (and round it if necessary).  The chances don't depend much on small deviations between the $m_i$ and $m,$ so writing $N=k-m,$ an excellent approximation of the Binomial coefficient is
$$\binom{n - m_1-m_2}{k - m_1} \approx \binom{2(k-m)}{k-m} = \binom{2N}{N} \approx \frac{2^{2N}}{\sqrt{N\pi}}.$$
The last approximation, due to Stirling's Formula, works well even when $N$ is small (larger than $10$ will do).
Putting these results together, and remembering to multiply by $1/2$ at the outset, gives a good estimate of the chance of tie as
$$\Pr(\text{Tie}) \approx \frac{1}{2\sqrt{N\pi}}.$$
In such a case, your vote will tip the election.  What are the chances?  In the most extreme case, imagine a direct popular vote involving, say, $10^8$ people (close to the number who vote in a US presidential election).  Typically about 90% of people's minds a clearly decided, so we might take $N$ to be on the order of $10^7.$  Now
$$\frac{1}{2\sqrt{10^7\pi}} \approx 10^{-4}.$$
That is, your participation in a close election involving one hundred million people still has about a $0.01\%$ chance of changing the outcome!
In practice, most elections involve between a few dozen and a few million voters.  Over this range, your chance of affecting the results (under the foregoing assumptions, of course) ranges from about $10\%$ (with just ten undecided voters) to $1\%$ (with a thousand undecided voters) to $0.1\%$ (with a hundred thousand undecided voters).

In summary, the chance that your vote swings a closely-contested election tends to be inversely proportional to the square root of the number of undecided voters.  Consequently, voting is important even when the electorate is large.


The history of US state and national elections supports this analysis.  Remember, for just one recent example, how the 2000 US presidential election was decided by a plurality in the state of Florida (with several million voters) that could not have exceeded a few hundred--and probably, if it had been checked more closely, would have been even narrower.
If (based on recent election outcomes) it appears there is, say, a few percent chance that an election involving a few million people will be decided by at most a few hundred votes, then the chance that the next such election is decided by just one vote (intuitively) must be at least a hundredth of one percent.  That is about one-tenth of what this inverse square root law predicts.  But that means the history of voting and this analysis are in good agreement, because this analysis applies only to close races--and most are not close.
For more (anecdotal) examples of this type, across the world, see the Wikipedia article on close election results.  It includes a table of about 200 examples.  Unfortunately, it reports the margin of victory as a proportion of the total.  As we have seen, regardless of whether all (or even most) assumptions of this analysis hold, a more meaningful measure of the closeness of an election would be the margin divided by the square root of the total.

By the way, your chance of an injury due to driving to the ballot box (if you need to drive at all) can be estimated as the rate of injuries annually (about one percent) divided by the average number of trips (or distance-weighted trips) annually, which is several hundred.  We obtain a number well below $0.01\%.$ 
Your chance of winning the lottery grand prize?  Depending on the lottery, one in a million or less.
The quotation in the question is not only scurrilous, it is outright false.
