Try to answer the question in the title for yourself. An election has n votes for candidate A, m votes for candidate B, and no third candidate. Candidate A wins, so n > m. Assuming that the votes were equally likely to be counted in any order, what’s the probability that candidate A was always ahead? (Note that “always ahead” precludes the vote being tied at any point.)
My friend showed me this puzzle a couple of days ago and told me that there’s an extremely simple way to solve it. After trying a bunch of complicated things (including some wild Pascal’s triangle variants), I eventually relented and asked for the simple solution. Not only is there a short proof of the answer, but the answer is itself incredibly simple and elegant. See if you can do better than I did!
(Read on only after you’ve attempted the problem!)
So, here’s the solution. We start by considering the opposite of A always being ahead, which is that A is either behind or that the vote is tied at some point. Since we know that A eventually wins, A being behind at some point implies that at a later point A and B are tied. So the opposite of A always being ahead is really just that there is a tie at some point.
Pr(A is always ahead | n votes for A, m votes for B)
= 1 – Pr(A is behind at some point or tied at some point | n, m)
= 1 – Pr(There’s a tie at some point | n, m)
Now, let’s consider the probability of a tie at some point. There are two ways for this to happen: either the first vote is for A or the first vote is for B. The first vote being for B entails that there must be a tie at some point, since A must eventually pull ahead to win. This allows us to say the following:
Pr(Tie at some point | n, m)
= Pr(Tie at some point & A is first vote | n, m) + Pr(Tie at some point & B is first vote | n, m)
= Pr(Tie at some point & A is first vote | n, m) + Pr(B is first vote | n, m)
= Pr(Tie at some point & A is first vote | n, m) + m/(n+m)
Now, the final task is to figure out Pr(Tie at some point & A is first vote | n, m). If you haven’t yet figured it out, I encourage you to pause for a minute and think it over.
Alright, so here’s the trick. If there’s a tie at some point, then up to and including that point there are an equal number of votes for A and B. But this means that there are the same number of possible worlds in which A votes first as there are possible worlds in which B votes first! And this means that we can say the following:
Pr(Tie at some point & A is first vote | n, m)
= Pr(Tie at some point & B is first vote | n, m)
And this is exactly the probability we’ve already solved!
Pr(Tie at some point & B is first vote | n, m)
= Pr(B is first vote | n, m)
And now we’re basically done!
Pr(Tie at some point | n, m)
= m/(n+m) + m/(n+m)
Pr(A is always ahead | n, m)
= 1 – Pr(Tie at some point | n, m)
= 1 – 2m/(n+m)
= (n – m) / (n + m)
And there we have it: The probability that A is always ahead is just the difference in votes over the total number of votes! Beautiful, right?
We can even more elegantly express this as simply the percent of people that voted for A minus the percent that voted for B.
(n – m) / (n + m)
= n/(n+m) – m/(n+m)
=%A – %B
This tells us that even in the case where candidate A gets 75% of the vote, there’s still a 50/50 chance that they fall behind at some point!
An example of this: the recent election had Biden with 81,283,485 votes and Trump with 74,223,744 votes. Imagining that there were no third candidates, this would mean that Biden had 52.27% of the popular vote and Trump had 47.73%. And if we now pretend that the votes were equally likely to be counted in any order, then this tells us that there would only be a 9.54% chance that Biden would be ahead the entire time! Taking into account the composition of mail-in ballots, which were counted later, this means that Trump having an early lead was in fact exactly what we should have expected. The chance that Biden would have fallen behind at some point was likely quite a bit higher than 90.5%!