Calculating total possible outcomes in the League of Legends 2018 World Championship

Question

I am wondering how to calculate the total number of possible combinations in the current League of Legends world championship. I am interested after seeing how few people managed to predict the outcomes.

For those who don't know here is how it's played:

16 teams enter the group stage. They are divided into four pools (A, B, C and D) of four teams each. Every team plays 2 matches against the other teams in its pool and the top 2 teams go to the knockout stage.

After the pool matches are completed:

• the winner of pool A plays the 2nd placed team of pool B.
• the winner of pool B plays the 2nd placed team of pool A.
• the winner of pool C plays the 2nd placed team of pool D.
• the winner of pool D plays the 2nd placed team of pool C.

The winners of these 4 matches then can play semi-finals, and the winner of the semi-finals play in the finals.

Answer 1

If you have to pick the finish order in each group, there are $4!=24$ orders the group could finish in assuming all ties are somehow broken in the final standings. Then there are seven games in the knockout stage, with $2$ results possible in each game. The total number of results then is $24^4\cdot 2^7=42\ 467\ 328$. If you make a random prediction the chance you are right is $1$ in this. Of course, if you know some teams are better than others, you can improve your odds, but even so the odds are against you.

Answer 2

For the knockout stage, I'll assume you don't care about 3rd and 4th places. So, you have 2^4 options for the quarter finals, 2^2 for the semifinals, and 2 for the finals. All in all, there are 2^7=128 possible outcomes.

Why is it like this? For each game, one of two teams can win, and in the same stage, the games are independent (sort of. we can assume RNG vs G2 is independent from KT vs IG even though there are factors that may influence the outcomes. SPOILERS: After IG won, G2 might have felt pressured to win as well, being the underdog (even though they still managed to win, yay!)). So there are 4 games in the quarter finals, hence the 2^4 so on and so forth.

If you only care about the place each team got, it's (4!)^4, as there are 4 groups and inside each group, team A might get 1st,2nd,3rd or 4th, so 4 possibilites, but after this team is locked in, team B only has 3, etc. (4! = 4*3*2*1). Since you have 4 groups, (4!)^4.

(2^7 * 4*4!) = 12288 <- incorrect

So if I solved the problem correctly (which is very likely that I didn't), if you chose your pick'ems randomly, there would be aprox. 0.008% chance to win. The EUphoria is real :)

EDIT: This is wrong since I added the 4!, instead of multiplying. The result is 2^7 * (4!)^4 = 42467328 and there is actually aprox. 0.00000235475% chance to win