Intresting combination & probability problem

I am a DOTA 2 player and I came across this intriguing problem which I do not know how to solve. In this game 1 match is played between two teams, each consisting of 5 players. Each player has to pick a hero/character from a pool of total 107 heroes, i.e 10 heroes picked. Moreover both team's captain bans 5 heroes for other team to pick. i.e 10 heroes banned.

Now, in a tournament of 16 teams, played on a knockout format with each contest between two teams consisting of 3 matches (best of 3 winner) and grand final of 5 matches. What is the probability that atleast X (lets say 5) heroes will not be picked or banned in whole tournament.

Note by Musabbir Hussain
5 years, 6 months ago

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I don't know of any simple formula for this one, but here's what I came up with:

In the whole tournament, there will be (8+4+2)3+15=47 (8+4+2) \cdot 3 + 1 \cdot 5 = 47 matches. That translates to 4720=940 47 \cdot 20 = 940 hero-selections (either for picking or banning, that isn't our concern).

If you don't want y y players to be selected in the whole tournament, then, effectively in every match you have Ny=107y \displaystyle N_y = 107 - y heroes. So, in each match, you can have (Ny20)\displaystyle {N_y \choose 20} selections. And since there are 940 940 matches, totally there are (Ny20)940 \displaystyle {N_y \choose 20}^{940} possible selections.

Thus, the probability of y y heroes being completely neglected in the whole tournament is P(y)=(Ny20)940(10720)940\displaystyle P(y) = \dfrac{ {N_y \choose 20}^{940} } { {107 \choose 20}^{940} } .

So, the probability that at least X X are totally neglected in the whole tournament is:

10yX1P(y) \displaystyle 1 - \sum_{0 \le y \le {X-1} } P(y)

Parth Thakkar - 5 years, 5 months ago

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And since there are 11401140 matches,

Isn't there only 14 matches. Other than that, good solution

Siddhartha Srivastava - 5 years, 5 months ago

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Thank you Parth, But i think ( 8 + 4 + 2 ) x 3 + 1 x 5 = 47...So, there will be 47 matches in total 15 contests..that would mean 47 x 20 = 940 hero-selections. Right?

A good solution though.

Musabbir Hussain - 5 years, 5 months ago

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Aah that was so dumb! :D Thanks for pointing it out. Dumb Dumb Dumb! Sometimes its fun to do dumb things though ;) :D Edited.

Parth Thakkar - 5 years, 5 months ago

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@Parth Thakkar No Problem..I also do such dumb things alot :D..! another thing I would like to understand is that why did u put ( 940 20 ) in denominator..Can you please explain the equation in words. I am sorry if that is something too obvious. Sadly, I am much interested but not so good in Probability, Combination & Permutation :-S

Musabbir Hussain - 5 years, 5 months ago

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@Musabbir Hussain That's another mistake I made! Well, it should be (10720) {107 \choose 20} and not (94020) {940 \choose 20} . That's the total number of possible selections, without any player being neglected.

Parth Thakkar - 5 years, 5 months ago

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One question. Can both teams ban the same hero?

Siddhartha Srivastava - 5 years, 5 months ago

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No. A hero can be picked/banned only once.

Musabbir Hussain - 5 years, 5 months ago

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