Since its announcement, there has been intense debate over the new ‘best of 1’ system for elimination matches. The Vex community has extensively debated the pros and cons of ‘best of 1’ and ‘best of 3’. However, I believe that these debates have overlooked some basic mathematics that can help compare the two systems. I’d like to change that. In this post, I will analyze the probability that a ‘better’ team wins under a ‘best of 1’ system and under a ‘best of 3’ system.

Before we can begin this discussion, I need to establish a few things. By a probability, p, I mean some real number in the range (0,1). When I say ‘there is a probability p something takes place’ I mean that if we were to repeat that thing an arbitrarily large number of times, then the proportion of times in which the thing occurs would be p.

To clarify this, suppose that I flip a fair (unweighted) coin; Let ‘the coin lands on heads’ be the thing that we’re finding the probability of. The probability that the coin lands on heads is 0.5, since the coin will land on heads half - or 50% - of the time and tails the other half.

For this discussion, I will consider a matchup between two teams, team A and team B. I will assume that team A is the ‘better’ team, meaning that team A is more likely to win a match than team B is. I will assume that for any given match the probability that team A wins is pA, where pA is some real number between .5 and 1 (for those with a background in mathematics, pA ∈ { ℝ | .5 ≤ pA < 1}). For simplicity, I will assume that for any given match either team A wins or team B wins. This means that draws cannot occur. Therefore, the probability that B wins is

pB = 1- pA.

Finally, I will assume that the outcome of each match is independent. This means that either team’s chance of winning a given match is not affected by the outcome of previous matches. In other words, I will not attempt to model team morale.

The goal of this analysis is to compare the probability that team A wins under a ‘best of 1’ system and under a ‘best of 3’ system.

To begin, let’s consider a ‘best of 1’ system. The winner, in this case, is the first team to win a match. The probability that team A wins the first match is

pA.

The probability that team B wins is

1-pA.

Let p(team A wins)Best of 1 denote the probability that team A wins in a ‘best of 1’ system. Therefore,

p(team A wins)Best of 1 = pA.

Let us now consider a ‘Best of 3’ system. The winner of a ‘best of 3’ matchup is the first team to win two matches. There are two possible outcomes of the first match. Either team A wins, which happens with a probability of pA, or team B wins, which happens with a probability of

1-pA

Regardless of the outcome of the first match, a second match will occur. Once again, there are two possible outcomes of this match: Team A wins (probability pA) or team B wins (probability 1-pA). If team A wins both of the first two matches, which happens with a probability of

pA*pA = pA^2,

then A is the winner. Likewise, if team B wins both of the first two matches, which happens with a probability of

(1-pA)(1-pA) = (1-pA)^2,

then B is the winner. If team A wins the first match but team B wins the second match, or if team B wins the first match and team A wins the second match, then we go on to the third match. To get to either of these situations, both teams A and B must win a match. Therefore, the probability that either of these outcomes happens is

pA(1-pA).

Since there are two ways to get to a third match, the probability that a third match is played is

2*pA(1-pA).

Whoever wins the third match, if there is one, wins overall. Assuming there is a third match, there is once again a pA probability that team A wins and a (1-pA) probability that team B wins. By the third match, there must be a winner (since we don’t allow draws).

That may be a lot to take in at once. Luckily, the analysis above can be summarized in the following diagram:

Using this diagram, we can easily see that there are 3 outcomes in which team A wins and three outcomes in which team B wins. By adding the probabilities of the 3 outcomes in which team A wins, we can easily see that the probability that team A wins is:

p(team A wins) Best of 3 = pa^2 + 2*pa2(1-pa)

To understand what these results mean, let us consider the graphs of the two functions that we derived,

Likewise, it is informative to consider the absolute difference between the two curves. The absolute difference is given by,

P(team A wins)[Best of 3] - P(team A wins)[Best of 1].

Plotting this difference gives the following graph,

Likewise, we can consider the relative difference between the two curves. The relative difference between the two functions is given by,

{P(team A wins)[Best of 3] - P(team A wins)[Best of 1]} / P(team A wins)[Best of 1])

Plotting this function gives the following graph,

Based on these results, we can clearly see that a ‘best of 1’ system does, in fact, make it less likely that the ‘better’ team wins. The effect is most dramatic, in a relative sense, when team A has a 75% chance of winning (pa = .75). When pa = .75 team A is 12.5% more likely to win under a ‘best of 3’ system than under a ‘best of 1’ setting.

This analysis naturally leads us to an important conclusion: a ‘best of 3’ system gives good teams a better chance of winning; less is left up to chance. It is true that I made many assumptions to get my results. However, it is undeniable - from a purely mathematical standpoint - that a ‘best of 3’ system is superior at making the ‘better’ team win than a ‘best of 1’ system. Under the ‘best of 1’ system, top teams are less likely to win competitions. I do not know if hurting top teams and rewarding luck was one of the goals of switching from ‘best of 3’ to ‘best of 1’, but this is an indisputable consequence. The ‘best of 1’ system does a disservice to good teams, and that’s a shame.