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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
3 years, 8 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 $$(8+4+2) \cdot 3 + 1 \cdot 5 = 47$$ matches. That translates to $$47 \cdot 20 = 940$$ hero-selections (either for picking or banning, that isn't our concern).

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

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

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

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

- 3 years, 8 months ago

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.

- 3 years, 8 months ago

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

- 3 years, 8 months ago

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

- 3 years, 8 months ago

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

- 3 years, 8 months ago

And since there are $$1140$$ matches,

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

- 3 years, 8 months ago

One question. Can both teams ban the same hero?

- 3 years, 8 months ago

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

- 3 years, 8 months ago