Git Gud — Chapter 1

8 min readJun 25, 2024

What if… (I tried harder ?)

Step 0.

Defining the question.

An argument in forums that we are very familiar with goes something like this, “…you are a NOOB. In a 5:5 PVP setting, if you play just as well according to your MMR or even slightly better, on a consistent basis, you will win more and therefore CLIMB. This is because given that you do so and do not ‘int’(not be a troll), you only have four potential troll candidates whereas the enemy team has five players meaning five potential trolls. ….therefore Git Gud”. The mathematics is simple as 5 is bigger than 4. Also, I personally feel that this is true as stated in the introduction of this story. But how true is it ? Let us find out.

For what is about to be tested, I must establish a set of assumptions and conditions as below.

You are consistent.
You always play at the same level as your MMR.
More better players win.
Given your MMR, the team that has more players who are above it wins.
50:50.
If both teams have the same number of players who are above the given MMR, there is a 50% chance for any one of the two winning.
Equal impact.
There is no distinction amongst the importances of roles. A win in toplane will be just as influential as one in midlane.
Nobody as good as you.
For sake of simplicity, in a given game, no one is exactly the same MMR as you. Everyone will be at least ever so slightly worse or better than you.
You are the only one (playing on the main account).
You are the only person in the game where, with 100% confidence, you are as good as the MMR says you are. The MMRs of other players’ accounts may not truly reflect their proficiencies in the game. This is due to the fact that, although the number may vary across regions, it has become easy to obtain fresh accounts(sub-accounts). In other words, a top 0.0001% may be in your game on your team just because the account being used is around the level of your MMR.

Let us start with somewhat of an average player, Paul, currently Gold 3 (Split 1 Season 2023). His percentile is 57%(0.57), meaning that he is better than 57% of the entire solo ranked queue playerbase (Split 1, Season 2023). Forty three percent(43%) of the said population is better than him. He plays ranked every day, always in a perfectly consistent style because he is cold-blooded, keeps a cool mind and is willing to learn the game as a long term goal.

Paul queues up, enters a ranked game of 5v5, with 4 strangers on his team and another set of 5 strangers on the enemy team. With a firm belief that the matchmaking system is perfect, if not near perfect, he wonders what the chances of him winning this particular game is. Anything is possible because there are so many people, including people from Diamonds to Challengers, who are better than him playing(aka smurfing) on a sub-account or a friend’s account. He is well seasoned and expects this. Yes, on the surface, as his game loads, the maths sounds sound. But how much of a ‘shot’ has he ?

Step 1.

Calculating Possible Outcomes

Let us count the number of combinations. For the 4 players on Paul’s team, there is only one case where none of them are better than him. This can be expressed as 4C0, as in 4 choose 0. If there is only one, there are four possible cases or 4C1. To finish the rest of the iteration, there are 6, 4 and 1 or 4C2, 4C3 and 4C4 possible case(s) for when there are 2, 3 and 4 trustworthy players who are above his MMR. To sum up, 1, 4, 6, 4, 1 possible cases for when there are 0, 1, 2, 3, 4 players, respectively, who are better than Paul.

The story is similar for the opposing team. There is only one case where none of them are better than Paul, namely 5C0. For there to be only one superior player, five possibilities exist or 5C1. The rest is 10, 10, 5, and 1 for 2, 3, 4 and 5 higher MMR users respectively or 5C2, 5C3, 5C4 and 5C5.

Step 2.

Calculating Probabilities — The Intermediate Phase

Next we must calculate the probabilities for these cases for each team. I admit that this confounded me at first, but if we calculate it one step at a time, hopefully more than a few people can understand. It is not hard, but just complicated like a ball of wires stashed in a box and forgotten for an extended period of time.

For Paul’s team, if there is no ‘smurf’, it is safe to assume that it is the equivalent of flipping a coin for heads with a probability of 43% for four times and getting none. That is, in numbers, 0.57 x 0.57 x 0.57 x 0.57 or 0.430 x 0.574. In the case of there being only one ‘CHAD’ on his team, it is the same as flipping the same coin and getting only one heads or 0.43 x 0.57 x 0.57 x 0.57 or 0.431 x 0.573. For two, 0.43 x 0.43 x 0.57 x 0.57 or 0.432 x 0.572. The equation is 0.43 x 0.43 x 0.43 x 0.57 or 0.433 x 0.571 for three players better than Paul. Finally, 0.43 x 0.43 x 0.43 x 0.43 or 0.434 x 0.570 for all of them to be better than Paul. To summarise, 0.430 x 0.574, 0.431 x 0.573, 0.432 x 0.572, 0.433 x 0.571 and 0.434 x 0.570 for 0, 1, 2, 3 and 4 players to be above Paul, respectively. Notice how the exponents in the smaller font (superscripts) always add up to 4, the number of players on Paul’s team besides Paul.

For Paul’s enemy team, it is again very similar with only the number of players getting allotted changes from four to five. In this case, we can imagine flipping the same coin 5 times. For there to be no better-than-Paul users on this team, 0.57 x 0.57 x 0.57 x 0.57 x 0.57 or 0.430 x 0.575 is what we can expect. For a sole player to turn out to be a ‘heads’, it would be 0.43 x 0.57 x 0.57 x 0.57 x 0.57 or 0.431 x 0.574. The rest of the iteration is 0.432 x 0.573, 0.433 x 0.572, 0.434 x 0.571 and 0.435 x 0.570 for 2, 3, 4, 5 better players on the opposing end, respectively. Here, just as well as above, the exponents always add up to 5, the number of players on the enemy team.

Step 3.

Combining Possible Outcomes and Probabilities by Multiplication

Now all we have to do is use the results we have above and insert them in the right place in the equation to multiply the probability for each case. In short, the number of cases for a certain number of ‘better player(s) times the probability of it happening.

For example, because there is only one possible case where there is no one better than Paul, the final probability is 1 times 0.10556001 or 4C0 x 0.574 x 0.430. The next where there is only one such player is 4 times 0.07963299 or 4C1 x 0.573 x 0.431. Following this logic, the table below demonstrates how the rest of the final probabilities will be.

In order to ensure that all cases have been accounted for, the final probabilities for each team have been summed up. As the two SUMs of ‘Final Probs’ both add up to ‘1’ in their own tables, it may be safe to assume that every possible outcome has been considered.

Step 4.

Combining the results

Now that we have the two tables, we need to multiply each case on Paul’s team (Ally Team) to each case on the other team (Enemy Team). For example, if there is no one better than Paul on the two competing teams, it would equate to 0.10556001 times 0.0601692057. If there is no one better than Paul amongst the allies, but there is only one such user amongst the opponents, 0.10556001 times 0.2269540215 is what we would need to calculate. The table below shows what this procedure will yield.

Step 5.

Making Sense of It All

We now need to interpret the resulting probabilities. As it has been stated in Step 0, the team that has more players better than Paul, wins and if both teams have the same number of such players, both teams have a 50% chance of winning that particular match. Therefore, the probability of the two teams having an even number is to be divided in half and added to the chances of Ally Team winning and the chances of Enemy Team winning. Below is the resulting table with matching colours as the table above, for the purpose of making the interpretation easier which aggregates the probabilities.