Now that I no longer play open cups I can finally share my queue order strategy for best of three Conquest and the reasoning behind it.

Buckle up cause this is gonna be a long one

So let’s talk winrates. After the ban both player will have 2 decks which can be paired into 4 possible matchups, each with their own winrate as I’ve shown on the left. On the right are the *set* winrates if the corresponding matchup is queued 1st.

So for these particular match winrates your chance of winning the whole set is 50.3% if you and your opponent both queue your 1st deck for game 1.
If you queue your 1st deck but your opponent queues their 2nd deck then your set winrate will be 51.3% instead.

Notice that despite there being no obvious symmetry in the match winrates the set winrates seem to be diagonally symmetric.

Incredibly, this symmetry appears *regardless* of what the match winrates are

Why does this symmetry seem to appear though? In order to answer that we first need to understand how winrates like the ones shown on the right are calculated.

Let’s start with a simple example where only two games are played before moving on to the best of three case.

Let’s say you’re about to play two HS games, which we’ll refer to as match A and match B respectively, and you want to know your chance of winning *at least* one. Your chances of winning A are exactly 50% and winning B is exactly 60%.

What's the right way to calculate this?

Your intuition might be to simply add the winrates, however this results in an answer greater than 100% which obviously can’t be correct.

As it turns out adding the winrates forces you to “double-count” the outcome where you win both matches. That's why that answer is too high.

This can be shown visually by comparing with the problem of finding the area of a Venn diagram.
When you add the areas of regions A and B the overlapping region gets counted twice so to get the correct answer you need to subtract the area of the overlap from the total sum.

Going back to the original example, the probability of winning both matches is simply the product of the two winrates which is 30% (remember that percentages are secretly fractions).
Subtracting this from the sum we get 50% + 60% - 30% = 80% which is indeed the right answer!

This method of finding all the ways to succeed, summing them, and finally subtracting the chances of succeeding in multiple ways will extrapolate to the best of three case, as will the Venn diagram analogy.

For this part we’ll use variables to make it more general.

Let’s start with the case where both players queue their 1st deck game 1 (matchup A). In the case where you win game 1 you can succeed by winning matches A&C or A&D. Otherwise you have to win matches B&D.

Just like in the previous example summing the winrates cause us to double-count the outcomes where you ‘win’ all 3 games.
Looking at the Venn Diagram it’s clear the overlaps happen at ACD and ABD. Subtract those regions and we get the correct formula of AC + AD + BD - ACD - ABD.

Through the same process we can figure out the winrates for when the set begins with any of the other 3 matchups.

And this is the part where we reach our key insight; the formulas when beginning with match A or match D are *exactly the same* (as are the formulas for B and C)

This is true because when beginning with match D the three ways you can succeed are winning A & D, B & D or, if you lose game 1, A & C.
These are the exact same three ways you can win when beginning with match A!

Now we've proven for a fact that the diagonal symmetry we noticed at the beginning is not a coincidence but actually a fundamental aspect of best of 3 Conquest

The implications for queueing strategy are that if you can correctly predict what deck your opponent will choose game 1 then you can *always* choose the matchup that will give you the highest possible winrate.

This makes being able to predict your opponent very advantageous.

However the exact same goes for your opponent; if they can predict what you will do then they can always pick the matchup that gives you the worst winrate.

This means that being unpredictable is just as important as predicting your opponent.

In conclusion valid strategies are:
Correctly predicting what your opponent will queue and queuing the 'counter' deck
http://random.org 

Strategies that are *not* valid are:
Always queueing the same deck 1st
Queuing using weighted randomness

Thanks for reading

“But wait, does this apply to best of 5?” you might be wondering.

My understanding is that while the diagonally symmetry isn’t present in Bo5 knowing what your opponent will queue still gives you a big advantage so you should avoid being predictable at all costs.