What is the probability of meeting a copy of your hero?

Question

Mathematics

Help, correct the solution of the problem, if I'm wrong.
There are 27 different heroes in the game.
The game has a team of 6 players.
Each player can only choose one hero. All heroes in a team must be unique.
Those. The first player chose hero #1, which means that the second player was left with the choice of heroes from 2 to 27.
Each hero has standard skins. Only 10 skins for each hero.
All players have all standard skins.
Each hero has special skins. A total of 12 skins for each hero.
Each player purchased only one special skin. (one - unknown for which hero) I.e. in the enemy team, of 6 players, will there be both my hero and my unique appearance on it? PS
What is the probability that I will buy and select a special appearance and meet in the game a copy of my hero in the same appearance in the enemy team?
The decisions of the opponent player are equiprobable, random - i.e. he absolutely randomly chooses a hero and appearance.
[U]My decision. Not exact, approximate.[/U]
The probability of choosing my hero.
First player 1/27, second player 1/26, third player 1/25... Probability
of player buying my special skin 1/(27*12)=1/324
special appearance 1/(10+1)=1/11 (after all, he can choose a normal appearance)
Total for the first player - the probability that I will see my hero + my unique appearance
Approximately
1/(1/27*1/324* 1/11)=1/96228=0.00001.
Because 6 players (we will not count 1/26..1/25), then 0.00001*6=0.00006.
Total probability 0.006%
Perhaps (and most likely) I'm wrong. Please correct.

Answer 1

To meet a copy of "you", you need three things: for one of the enemies to choose your hero, for him to have your paid skin, and for him to choose to wear your skin. We consider:
The probability of meeting your hero in the opposing team = 6/27 (6 heroes are necessarily different out of 27).
The probability that a player with this hero bought a skin that matches yours = 1/12*27 (it is known that he bought exactly one, and there are 12 skins for each of the 27 heroes).
The probability that he "put it on" in the conditions of the problem is not given, but if you read the last sentence as "the player chooses the skin with equal probability from those available to him for this hero", then it will be 1/11 (conditional).
Since it is assumed that all choices are independent, therefore P(A and B)==P(A)*P(B), therefore, we can multiply these probabilities, we get 6/12*27*27*11 = 1/16038 .
That is, you decided that you were correct in essence, and the main mistake is that you considered the probability of each next player choosing your hero as independent of previous choices, but this is not so - if the first player rolled his 1/27 and took your hero , the rest of the players will choose it with a probability of 0, and if not, then the probability of choosing the second player will be 26/27*1/26==1/27, where the first part is the probability that the first player did NOT choose your hero, and the second - that the second player chose him from the remaining 26. The formula here is P(A and B) = P(A)*P(B|A) where P(B|A) is the probability of event B given the occurrence of event A.

Answer 2

Not a solution, but an algorithm
1) The probability of meeting an opponent's team where there is the same hero (6 \ 27)
multiplied by
2) The probability that the same character has a standard appearance 1 / (27 * 10) \ u003d 1/270
multiplied by
3) Probability purchases of the same special appearance by the player 1/(12)=1/12
and this is the answer.

Answer 3

The problem does not have an exact mathematical solution, since it will include too many uncertain factors - making decisions about choosing a character, buying a paid skin, choosing a skin - they cannot be considered equally probable.
For example, a player who has bought a paid skin will most likely choose it, otherwise why buy.