*This is Part 2 of my series on Red side vs. Blue side. **The first article was intended for general audiences to understand the major differences between each side’s in-game performance.** This article goes much, much deeper into the planning of professional draft and is intended for advanced audiences.*

# Intro

In every meta, there exists a set of outstanding champions. Some champs are simply better than others. Because of this, each team cares about getting ahold of those superior picks while preventing their opponent from doing the same.

Instead of arguing about who will be impactful in *this specific* year of Worlds, I wanted to analyze and solve the abstract problem of picks and bans. If each team were to iteratively consider both their initial desired choice, followed by the expected intelligent response of their opponent, how would drafts play out?

# The Setup

I’ll first remind you that this is the order of bans and picks. In Phase 1, Blue side gets bans 1, 3, and 5; Red side gets bans 2, 4, and 6. Blue gets the first pick, followed by Red getting back to back picks. We will only need to look at Phase 1 for the purposes of this article.

We’re going to do a theory crafting exercise to consider how each team should plan their picks and bans. This is, to be clear, deeply stoked in Game Theory, and it does not consider many important factors like the complementary composition of roles.

Suppose we have a meta with some number *n* of overpowered champions. Let’s call them **Skews**. They are considered practically make or break. Teams want them, and they want their opponent to not get them. More Skews on your team is better than fewer.

How do teams organize as we increase the number of Skews available?

## 1-Skew

From Blue’s perspective, they want the Skew, and they know they get the first pick. Blue won’t ban them, but Red will some time along their three bans.

**Conclusion: No Skews make it to gameplay. Red has one forced ban.**

## 2-Skews

Blue knows they’ll get first pick, but Red will get second pick. Blue tries to plan how they can ban one Skew but not the other, and Red wants to ban both Skews or neither. Let’s walk backward in the decision tree.

**Ban 6 (Red)** – If we get to Red’s final ban and both Skews are still available, Red will ban neither. However, if one is available, Red will ban that.

**Ban 5 (Blue) **– If one Skew is available, blue will not ban it. If both Skews are available, Blue will ban one.

We can see that **Ban 5** is the key. There will either be no Skews going *into* **Ban 5**, or there will be one Skew *exiting* **Ban 5**. Since **Ban 6** turns one Skew into zero, we know that no Skews will make it past this ban phase.

**Conclusion: No Skews make it to gameplay. Each team has one forced ban.**

## 3-Skews

This one gets interesting because the Red side has a chance to obtain two Skews (which is preferred). It’s also the first time when a team has to consider their own future bans. Let’s walk backward in the decision tree again.

**Ban 6 (Red)** – If there are three Skews available, Red bans none. If there are two Skews available, Red bans none. If there is one Skew available, Red bans it.

**Ban 5 (Blue)** – If there are three Skews available, Blue bans one. If there are two Skews available, Blue bans one. If there is one Skew available, Blue does not ban it.

**Ban 4 (Red)** – (*This is where it gets complex*) Red wants there to be three Skews available here. They will let them all pass, Blue will ban one, and each team gets a Skew. If there are two Skews available, Red can let them both pass, but Blue will ban one, forcing Red to ban the other during **Ban 6**. If there is one Skew available, Red can ban it here or at **Ban 6**.

We can reduce the complexity here by knowing that exiting **Ban 4 **with less than three Skews available becomes the same situation as the 2-Skew problem. Now we just need to find under what circumstances we’ll reach three skews going into **Ban 4**.

**Ban 3 (Blue)** – Blue knows that leaving three Skews will necessarily result in Red not banning any in **Ban 4**, followed by Blue banning one in **Ban 5**. Thus, if Blue bans here or there are fewer Skews going into this round, it reduces to the 2-Skew problem.

We’ve now found an equilibrium point based on preference wherein either two or none of the Skews will make it to gameplay.

**Conclusion: Zero or two Skews make it to gameplay. Red has one forced ban. If Red uses another ban on a Skew, then Blue has one forced ban.**

## 4-Skews

This is the first time both teams can end up with two Skews. I’m going to preface, though, that Blue will not be getting a second Skew. Red has too many opportunities to easily prevent that, which allows us to use the same problem reduction.

**Ban 6 (Red) **– If there are four Skews available, Red bans one. Anything less reduces to the 3-Skew problem.

**Ban 5 (Blue)** – If there are four Skews available, Blue doesn’t want to ban one, but Red is in the driver’s seat to force an outcome of three Skews. This is the ban that signals back to Blue earlier in this phase that it has to reduce the number of Skews.

**Ban 4 (Red)** – If there are four Skews available, Red is happy because it can force a two or three Skew outcome later. If there are three Skews available, it reduces down to the 3-Skew problem.

**Ban 3 (Blue) **– Blue must ban a Skew here or sooner. This reduces the situation down to the 3-Skew problem.

Thus, we either see Blue ban aggressively, resulting in no Skew making it to gameplay, or we see two Skews make it.

**Conclusion: Zero or two Skews make it to gameplay. Red has one forced ban. Blue has one forced ban and can choose to use three on Skews.**

## 5-Skews

This is when we have to plan both forwards and backwards, but we can reduce certain segments by our previous findings.

Suppose both teams ban a Skew in **Ban 1** and **Ban 2**. We thus get to the 3-Skew problem, which means zero or two Skews make it to gameplay.

If one team bans a Skew in either of those first bans, we reduce to the 3-Skew problem again.

Now, suppose neither team bans a Skew early. Let’s do our decision tree walkback again.

**Ban 6 (Red) **– Suppose five Skews are available. Red doesn’t want Blue to get three Skews to their two, so Red bans a Skew. If there are four Skews, Red also bans one to force a three Skew outcome, which is in their favor.

**Ban 5 (Blue)** – If there are five Skews available, Blue is happy because they know Red will ban one, evening it to four. If there are four Skews, Blue is in a bad spot because Red will reduce it to three. Thus, this signals to not allow four Skews to make it out of **Ban 4**.

**Ban 4 (Red)** – If five Skews make it to this point, Red can necessarily force it to a three Skew outcome. Thus, we reduce that segment to the 4-Skew problem.

**Ban 3 (Blue) **– If five Skews make it to this point, Blue knows that Red can force a three Skew outcome. Thus, Blue has to ban a skew in **Ban 1** to prevent that.

**Ban 2 (Red) **– We know that exactly four Skews will make it to this point. Whether Red bans one here does not matter as either option reduces to the 4-Skew problem in **Ban 3**. This means Red is free for a preferential target ban on the other team.

Thus, we can conclude that Red will pass on four or three Skews to Blue for the **Ban 2 - Ban 3 **link, and we will ultimately see exactly two Skews make it to gameplay.

**Conclusion: Two Skews make it to gameplay. Red has no forced bans. Blue has three forced bans.**

## 6-Skews

We have all of the tools to conclude how this will go down.

**Ban 1 (Blue) **– We know from the 5-Skew problem that Blue cannot pass five Skews to **Ban 2**. Thus, it must pass on all six.

**Ban 2 (Red)** – As we showed while building our case in the 5-Skew problem, if five Skews make it to **Ban 3**, then Red has necessarily won the draft for Skew count. Red can and will ban one Skew to force that, meaning Blue’s efforts in **Ban 1** were futile. This is a strictly dominant strategy, meaning Red will not allow six to pass.

In the end, three Skews will make it into gameplay, and **Red has won**.

**Conclusion: Three Skews make it to gameplay. Blue has two forced bans. Red has one forced ban.**

# Conclusion

What we’ve proved here is that if there are up to five broken flex champions, each team will end up with an even number of them. Six Skews is the minimum number to force an uneven outcome.

Also of significant importance is how many bans are forced to be used by each team. For smaller numbers of Skews, Red has to use more of their bans on Skews than Blue. For higher numbers of Skews, Blue has to use most of their bans on Skews. This shows that not only is each team affected by the number of Skews they receive as picks, they are also advantaged or disadvantaged by the flexibility of their available bans. This matters against teams that have specialty picks on their roster.

Although this analysis was limited in scope and excludes the context of roles, I hope it was insightful for how coaches and analysts can assess and strategize picks and bans. There’s a lot more work to be done, but this is an abstract framework from which to start.