Yes, you can win more than 50% of your games and go down in MMR.

I see Riot repeatedly saying, "As long as you win more games than you lose then you'll keep going up in MMR." I'm going to show this is not the case, I'm going to show it right. This is based off one assumption:

• If you win as purple you're "expected" to win and gain less MMR than you would have if you were blue, but if you lose you'll lose more MMR. Likewise, a loss on blue causes you to lose less MMR while a win grants more MMR. This is because the blue team has a lower average MMR than purple. This is because MMR is basically the ELO system. Which side is higher or lower MMR is actually irrelevant, but helps for organizational purposes.

Let X be the amount of MMR you gain from winning on purple where X > 0

Let X+ε be the amount of MMR you lose from losing on purple where ε > 0

Let N be the number of games played where N > 0 and N is odd.

Let W be the number of wins on purple and L be the number of losses on purple where:

W = N / 2 + 0.5

L = N / 2 - 0.5

Thus the MMR gained is:

δMMR = WX - L(X+ε)

δMMR = (N/2+0.5)X - (N/2-0.5)(X+ε)

δMMR = X + 0.5ε - Nε/2

Thus the following statement is true:

2X/ε + ε < N implies δMMR < 0

Since X and ε are constant while N is not, the above statement will eventually be true for sufficiently large values of N. This means that you are winning more than half your games and going down in MMR.

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But wait, that's an unrealistic scenario! You don't really play all purple games so that doesn't work. Except the argument can be extended to every day scenarios. Here's that proof:

Let X be the amount of MMR you gain from winning on purple and lose from losing on blue where X > 0

Let X+ε be the amount of MMR you lose from losing on purple and gain from winning on blue where ε > 0

Let Wp be the number of wins on purple

Let Lp be the number of losses on purple

Let Wb be the number of wins on blue

Let Lb be the number of losses on blue

Thus the MMR gained is:

δMMR = WpX - Lp(x+ε) + Wb(x+ε) - Lb(X)

Now let's say Wp = Lp = Wb = Lb. This is saying you have a perfectly even distribution of wins and losses among the sides. Now let's say you play one more game on purple and win it. Thus:

Wp - 1 = Lp = Wb = Lb

At this point:

• You've won more games than you've lost
• δMMR = X

What we want to do now is keep the total number of games won and lost the same while reducing δMMR. We can do this by either taking a win out of blue side wins (Wb) and into purple side wins (Wp) or by taking a loss out blue side losses (Lb) and into purple side losses (Lp). Doing so will subtract ε from δMMR, represented by:

δMMR = X - εS where S is the number of times you've "shifted" a win or loss.

X/ε < S implies δMMR < 0

So we just need to do this enough times so that X/ε < S. But we have to be certain that we actually can do this enough times:

In the scenario where Wb = 0 or Lb = 0 we can increment all counters by 1 (i.e., play 2 games on each side, winning one and losing the other). The MMR gained by the purple side win will cancel with the blue side loss, and the MMR gained by the blue side win will cancel with the purple side loss. As a result of the 4 new games your MMR hasn't changed. Now we can shift the distribution of wins, hit 0, add 4 more games, rinse and repeat as many times as needed.

Again, this means that your MMR gains are negative, even though you've still won more games than lost.