# 8-4's, 4-3-2-2's, and Swiss (Part 1)

Earlier this week, my buddy Daniel Duterte posed the following question on Facebook:

**If you normally play 4-3-2-2 draft queues, at what point (what does your win rate have to be) does it become better value to play 8-4's in the same format?**

Being a bit of a math nerd, I decided to figure it out...

*BEWARE! There be maths here! I hope you like formulas!*

The probability of a player winning any given match is $p$ where $p$ is a value between 0 (never winning) and 1 (always winning). To determine the expected value of an 8-4 vs. a 4-3-2-2, we need to figure out the probability of each outcome and the value of that outcome. Multiply each outcome's probability by its value and sum them all up to get the expected value across the entire event.

For a single elimination draft, there are 4 possible outcomes:

- Lose in the quarterfinals
- Lose in the semifinals
- Lose in the finals
- Win the finals

The probability of each event is as follows:

**Lose in the quarterfinals**: $1-p$. The probability of losing a single match is $1-p$.**Lose in the semifinals**: $p-p^2$. To lose in the semifinals, you have to win the quarterfinals. The probability of winning the first match, $p$, is multiplied by the probability of losing the second match, $1-p$. $p(1-p)=p-p^2$.**Lose in the finals**: $p^2-p^3$. To lose in the finals, you have to win the quarterfinals and semifinals. Multiply the probability of winning the first match, $p$, with the second, $p$, and losing the third, $1-p$. $p$ times $p$ is $p^2$, $p^2(1-p)=p^2-p^3$.**Win the finals**: $p^3$. You win three matches to win the draft, and the probability of winning each match is $p$. $p$ times $p$ times $p$ is $p^3$.

Now that we have the expected probability of each outcome, we can multiple the values of those outcomes for each draft type to determine the expected value of that draft.

**8-4:**

**Lose in the quarterfinals**: 0 return.**Lose in the semifinals**: 0 return.**Lose in the finals**: $4p^2-4p^3$.**Win the finals**: $8p^3$.

$$4p^2-4p^3+8p^3=4p^3+4p^2$$

**4-3-2-2:**

**Lose in the quarterfinals**: 0 return.**Lose in the semifinals**: $2p-2p^2$.**Lose in the finals**: $3p^2-3p^3$.**Win the finals**: $4p^3$.

$$2p-2p^2+3p^2-3p^3+4p^3=p^3+p^2+2p$$

So now we have the expected returns on both events. Now we need to figure out where the two expected values are equal.

$$4p^3+4p^2=p^3+p^2+2p$$

$$3p^3+3p^2-2p=0$$

$$p(3p^2+3p-2)=0$$

That equation will equal 0 if either factor equals 0. So if $p$ is 0, the EV is equal because you can't win a match in either case. Not particularly useful. But we can figure out where $3p^2+3p-2$ equals 0. It's a quadratic equation, and we can use the quadratic formula.

For $ap^2+bp+c$:

$$p=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

$$p=\frac{-3\pm\sqrt{3^2-4(3)(-2)}}{2(3)}$$

$$p=\frac{-3\pm\sqrt{9+24}}{6}$$

$$p=\frac{-3\pm\sqrt{33}}{6}$$

We can discard the $-3-\sqrt{33}$ case as $p$ would be less than 0 in that case. So that leaves us with:

$$p=\frac{\sqrt{33}-3}{6}$$

This value is approximately .457 and we have two valid roots for which the EV of both drafts are equal, 0 and .457. For all values between those two roots, only one side of the equation will be greater than the other, and for all values above .457, only one side of the equation will be greater than the other. We wouldn't expect someone with a $p$ of .2 and someone with a $p$ of .3 to have better EV with different draft types, and the same with $p$s of .5 and .6. So we just need to test one value in each range to show which is better EV in each range.

For $p=.1$:

$$4p^3+4p^2\,?\,p^3+p^2+2p$$

$$4(.1)^3+4(.1)^2\,?\,.1^3+.1^2+2(.1)$$

$$4(.001)+4(.01)\,?\,.001+.01+2(.1)$$

$$.004+.04\,?\,.001+.01+.2$$

$$.044\,<\,.211$$

So under the .457 probability, 4-3-2-2s are better EV. How about over that probability?

For $p=.5$:

$$4p^3+4p^2\,?\,p^3+p^2+2p$$

$$4(.5)^3+4(.5)^2\,?\,.5^3+.5^2+2(.5)$$

$$4(.125)+4(.25)\,?\,.125+.25+2(.5)$$

$$.5+1\,?\,.125+.25+1$$

$$1.5\,>\,1.375$$

Over the .457 probability, 8-4's are better EV. So if you feel you're better than 45% against the field, you should play 8-4 drafts. But there's another option, Swiss drafts, that we'll explore in the next article.

As someone who doesn’t play online, here’s something that’s always bugged me. Why does anyone play 4-3-2-2s when only 11 packs are given out in that format, while 12 packs are given out in 8-4s? Why does Wizards skew it like that at all (instead of making them 5-3-2-2)?

I guess the answer has to be player sorting. If everyone who is 46% or better moves to the 8-4s, now the 8-4s will be significantly harder, while the 4-3-2-2s will be significantly easier. That gives players the ability not only to control their expected prize payout, but also the level of challenge.

Of course, in practice, once players sort, the people who were 46% against the entire field will be way worse against the higher skill 8-4 field; they’ll want to move back (especially since the 4-3-2-2 group just got a lot softer). For your another analysis, you could assume a distribution of player skills, and figure out the equilibrium choices for players of each tier (although you might have to do the computation numerically with a computer).