Best of 1 hurts good teams, and there's math to prove it

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


The probability that team B wins is


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


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


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


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.


I agree, the Bo1 system sucks, but instead of discussing why it sucks, it would be more efficient to attempt to find ways to do well in spite of it, and possibly even use it to our advantage.

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I disagree. Best of 1 is not a done deal until for another month or so. It can still be changed.

It CAN be changed.

But the odds of that are incredibly low.

I hate this so much, it’s a shame we will never have the energy of it breaking to a tie breaker match. That was always the most exciting thing to happen at competition in my opinion.

This can still happen in finals

Finals is still bo1 though

Did they change it from BO1 to BO3 in finals or where is it in the rules? I thought I looked over it and did not see it.

Never mind I confused myself with worlds. Worlds was best of 3 finals

If the score is a tie then it is not. This is what @MasterCole was attempting to say.

Also, are you guys sure that Bo1 is not a done deal? @punkduckable @Leeoon

I hope you are telling the truth, but according to Paul Copioli:

The rules are not finalized until sometime in August. Until then, the rules can change. The RECF/Vex will change their rules if there is enough pressure to do so. Even if the best of 1 system is kept, I think that people should at least be aware of issues with the new system.

How long are we going to argue and present data on BO1? It’s getting kind of old.

There’s been a lot of discussion on Best of 1, but I had not seen any formal mathematical analysis of Best of 1. I think it’s important to rigorously analyze the new system. I can only hope that RECF went through a similar analysis before deciding to try best of 1

I think it might keep going on until the rule is changed or there are not enough people to continue the debate. I think the latter would be sad to see. As others have said, this is the windows 8 of vex. It is worse than the previous system, but there is potential to become even better. I think it is important to become even better rather than just go back to the old system, and it would certainly not be best to keep the current system. Hopefully next year, or even this year with a bit of luck, the even better system will be put in place.

This thread was created to present an argument against Best of 1 from a purely mathematical standpoint. Cluttering this thread with comments that merely state your opinion of the new system is pointless; there are countless other threads that have reached hundreds of posts because of this.

In the other threads, people have presented data from worlds that show that the better alliance did not always win. I really appreciate that somebody took the time to actually present a well thought, professional, argument from a mathematical perspective. It seems to me that vex made the decision to sacrifice the probability that the better alliance wins a match/tournament to get rid of 3 team alliances and to allow more teams into the elimination rounds at worlds. Personally, I see no reason to have more teams in eliminations. From my experience at worlds, the lower seeded teams do not do enough, if any scouting, because it is impossible to tell if they are going to be high enough to be a picking seed. This resulted in the lower seeds just selecting whatever the highest ranked unpicked team was left. It was pretty much pointless except to give a few teams (many of whom were undeserving of being there in my opinion) the experience. This does not seem like a good justification for lowering the odds of the better team winning.

As for 3 team alliances vs 2 team alliances, I am all for 2 team alliances at the local level because it prevents entry level teams from being carried to regionals/states/worlds. However, at worlds, there is always a few good teams in each division who get an unfortunate schedule and end up sorely under ranked. I firmly believe that they have a higher chance of making eliminations under the old system, because the higher ranked, 1,2 and 3 seeded alliances are usually experienced enough to know how to look at the whole field of teams and pick out those diamonds in the rough.

I would really like to see the Best of 1 / 2 team alliances system implemented at the local level, but bring back the traditional Best of 3 system for worlds.

I don’t think people understand how effective an EP strike would be. VEX relies on EPs to support the system.

Plain and simple forget the math and everything my opinion on the matter is that in high level competitions students work too hard on their robot for just a best of 1. I believe if a student gets beaten in a match it would be better for everyone if the team that loses gets beat hard twice rather than just once. Same with the teams that win. The team that wins the match should be able to play more matches to prove the functionality and durability of their robot through the competition. So the teams that loses will go home and be disappointed and maybe then be inspired to improve for next time rather than pushing the loss off as a small mishap. Nothing is a better teacher than losing. Fail, fail again fail better… that’s vex

Probably not the most appropriate place to say this in.

The point of this thread is to show the math behind how Bo1 hurts the good teams, and this kid put so much work into creating it just to prove it through the usage of formulas, graphs, and general equations.

There are so many other threads regarding this issue that making a post disregarding someones efforts to show a more practical approach in a post such as this one seems somewhat rude and unjust.

But it would never happen. Most EPs I know are neutral or in favor of Bo1 and wouldn’t want to risk having their event disqualified over something like that

Not to mention if we don’t host events. You guys don’t play robots. Bo1 > nothing by my math.

Plus our team wouldn’t stay funded without the registration fees we receive.