What’s the probability that an election winner leads throughout the entire vote?

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!

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(Read on only after you’ve attempted the problem!)

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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.

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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)
= m/(n+m)

And now we’re basically done!

Pr(Tie at some point | n, m)
= m/(n+m) + m/(n+m)
= 2m/(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%!

Quantum Chess

Recently I’ve been thinking about both quantum mechanics and chess, and came up with a chess variant that blends the two. The difference between ordinary chess and quantum chess is the ability to put your pieces in superpositions. Here’s how it works!

Movement

There are five modes of movement you can choose between in quantum chess: Move, Split, Merge, Collapse, and Branchmove.

Modes 1 and 2 are for non-superposed pieces, and modes 3, 4, and 5 are for non-superposed pieces.

Mode 1: Move

This mode allows you to move just as you would in an ordinary chess game.

Mode 2: Split

In the split mode, you can choose a non-superposed piece and split it between two positions. You can’t choose a position that is occupied, even by a superposed piece, meaning that splitting moves can never be attacks.

One powerful strategy is to castle into a superposition, bringing out both rooks and forcing your opponent to gamble on which side of the board to stage an attack on.

Mode 3: Merge

In mode 3, you can merge the two branches of one of your superposed pieces, recombining them onto a square that’s accessible from both branches.

You can’t merge to a position that’s only accessible from one of the two branches, and you can’t merge onto a square that’s occupied by one of your own superposed pieces, but merge moves can be attacks.

Mode 4: Collapse

Mode 4 is the riskiest mode. In this mode, you choose one of your superposed pieces and collapse it. There are two possible outcomes: First, it might collapse to the position you clicked on. In this case, you now have a choice to either perform an ordinary move…

… or to split it into a new superposition.

But if you get unlucky, then it will collapse to the position you didn’t select. In this case, your turn is over and it goes to your opponent.

Mode 5: Branch Move

Finally, in a branch move, you relocate just one branch of the superposition, without collapsing the wave-function or affecting the other branch.

Piece Interactions

Attacking a Superposed Piece

What happens if you attack a superposed piece? The result is that the superposed piece collapses. If the piece collapses onto the square you attacked, then you capture it.

But if it collapses onto the other branch of the superposition, then it is safe, and your piece moves harmlessly into the square you just attacked.

This means that attacking a superposed piece is risky! It’s possible for your attack to backfire, resulting in the attacker being captured next turn by the same piece it attacked.

It’s also possible for a pawn to move diagonally without taking a piece, in a failed attack.

Line of Sight

Superposed pieces block the lines of sight of both your pieces and your opponent’s pieces. This allows you to defend your pieces or block open files, without fully committing to defensive positions.

Winning the Game

To win quantum chess, you must actually take the opponent’s king. Let’s see why it’s not enough to just get the opponent into a position that would ordinarily be checkmate:

It’s blue’s turn now, and things look pretty lost. But look what blue can do:

Now the red queen has to choose one of the two targets to attack, and there’s a 50% chance that she gets it wrong, in which case the blue king can freely take the red queen, turning a sure loss into a draw!

So how can red get a guaranteed win? It takes patience. Rather than trying to attack one of the two squares, the red queen can hang back and wait for a turn.

Now the blue king has two choices: leave superposition, after which they can be taken wherever they go. Or move one branch of the superposition, but any possible branch move results in a safe shot at the king with the queen. This can be repeated until the king is taken.

And that’s quantum chess! I’ve played several games with friends, and each time have noticed interesting and surprising strategies arising. Let me leave you with a quantum chess puzzle. Here’s a position I found myself in, playing red:

What do you think was my best move here?