League of Legends

There are 42 467 328 possible different pickems (331 776 group combinations). If we included play-in, the number would be 9784472371200

I made a similar post last Worlds (https://www.reddit.com/r/leagueoflegends/comments/j3yjne/there_are_42_467_328_possible_different_pickems/) but I would like to expand with more maths and more info. This has a Q/A format.

How do you get the first number?

For every group at Group Stage, there are 4! possible permutations (4 * 3 * 2 = 24) and there are four different groups so the number of different pickems at group stage is 244 or 331776.

After the group stage, there are 7 different matches with only 2 possible outcomes (Team A wins or Team B wins) so there are 27 or 128 different knockout combinations

Multiplying 331776 * 128 gives 42467328 different pickems


How do you get the second number?

Play-in groups have 5! possible permutations and there are 2 groups.The number of possible combinations is (5!)2 or 1202 or 14400.

After that, 4 Bo5 are played so there are 24 or 16 knockout combinations.

With that consideration there are 14400*16= 230400 possible theorical play-in pickems.

Combining 230400 play ins and 42467328 main stages we get 230400*42467328 = 9784472371200 options.

According to Wolfram Alpha, that number is 32 times the number of stars in our galaxy. Sure, the stars would need to align if we want Red Canids to win the finals against Beyond Gaming with C9 finishing 5th at play in while all chinese teams end 4th at group stage. But sadly G2 cant win Worlds 2021 in our observable universe.


Why is it 244 when calculating groups and not 24*4?

I will copy the comment I did last year regarding that statement:

You got the 96 by multiplying the 24 cases for each group by 4 groups, right? But to find the total you have to rise the number of cases to the number of groups, not multiply.

Simpler example. Imagine you have three numbers 1 2 3 and three letters A B C

You can sort the numbers with 6 combinations (123 / 132 / 213 / 231 / 312 / 321) and the letters with 6 combinations too (ABC / ACB / BAC / BCA / CAB / CBA)

Now think that you want a password that starts with one number group and continues with one letters group.

You have 6 combinations that start with 123 (123ABC 123ACB … ) ; 6 combinations with 132 (132ABC 132ACB … ) ; 6 combinations with 213 (213ABC 213ACB … ) ; … for a total of 6 * 6 or 62 or 36 different passwords, not 6 * 2

Shouldnt the number of combinations be lower since the knockout results depend of the group stage?

They don't, since when you complete the knockouts you are voting for team A or team B. The permutations let you say "im going to bet team A every game" or "Team B in quarters, Team A in semifinals and Team B in finals" or "I bet team A team B team A team B team B team B team A", etc… and that is independent of who are the specific teams (DFM-FPX, FNC-C9 and SKT T1 2015-My tier IV Clash team have a binary outcome, even if the likeliness of that matchup appearing in quarters is different)

After you have your groups filled, your results, and the knockouts are announced, you have 128 ways of filling the 7 games. If you couldnt change them between stages you would have a 0.78125% chance of being perfect. Thankfully, Riot let us change our semifinals so we can vote for korean or chinese teams after Fnatic goes 0-3 in quarters against them.

A little off-topic, but is it harder to be perfect in group stage or getting 0 points?

Out of the 24 group permutations, only 1 of them gives perfect points but 4 give 0 points. You don't get points as long as you put the top 2 teams out of the top 2. You get perfect points if you do 1-2-3-4 and the 4 combinations that give 0 points are 3-4-1-2, 3-4-2-1, 4-3-1-2, 4-3-2-1. Combining that with the fact that this can happen in the 4 groups, you are 256 times more likely to earn 0 points than being perfect. At least on paper. On practice few people put korean teams 4th and Rogue 1st.

Theres no way Galatasaray wins Worlds against UOL while every korean team fails to get top 8

This calculations take only into account the mathematical number of possibilities, not the likeliness of the outcome.

This only takes into account group order and BO5 results, not every single game

Yes, because thats what you can choose in pickems, this is not the different amount of "Worlds parallel universes". That number is harder to calculate, if there werent tiebreakers you could simply do 2group games * 20BO5 played (BO5 have 20 possible outcomes if I'm not wrong)

Please give me a comparison to put the numbers into perspective:

-> Main stage: We could get all the combinations with less than half the number of active league players (total number estimated at 115 million) or close to Uganda population

-> Including play-in number: In order to get all the combinations, every single HUMAN on Earth should do 1224 pickems without repetition.


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