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My Example:

Players 1 through n are playing a game and in each round there is one winner and after the winner is determined, one of the non-winners gets randomly taken out of the total players pool (i.e. round 1 = n total players, round 2 = n-1, round 3 = n-2, etc).

I am trying to show that if we look at a large sample of these games, a strong player (potentially the winner) should typically have a higher amount of competition round wins at the end of each game. However, I do not understand how to go about adjusting for the increased probability that a player might win each round when a non-winner is eliminated [i.e. round 1 = $\frac{1}{n}$ % chance of winning that round, round 2 = $\frac{1}{n-1}$, round 3 = $\frac{1}{n-2}$, etc]. This is assuming of course that each player has an equal chance of winning each round, which is obviously what i'm trying to get around.

Does anyone know how I might account for this increased probability and possibly how to also take into consideration as the game goes on that a stronger player who has won in the first round (or other future rounds) is given a higher probability than the rest to win the next rounds?

Basically, how do I calculate the strength of a player and add it into a predictive model on how they will do based on past games with other players data.

Thank you for any help in advance.

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  • $\begingroup$ If players have strength, then the probability of winning is not 1 over the number of players. Right? $\endgroup$
    – YJT
    Commented Aug 5, 2020 at 17:58
  • $\begingroup$ Yes, sorry for the confusion. That is what I am trying to figure out how to calculate over a large sample of games. I want to try and figure out what each player's % chance of winning is per round if they won a previous round. $\endgroup$ Commented Aug 5, 2020 at 18:09

1 Answer 1

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There are multiple ways to solve this, but you can give each player the following score :

$\mathbf{S}(x) = \underset{\{i\ :\ \text{x won round i}\}}{\sum \frac{1}{p_i}}=\underset{\{i\ :\ \text{x won round i}\}}{\sum (n- i)}$

With $p_i = \frac{1}{n -i}$ being the probability of winning the round $i$. So the earlier rounds count for more than the later ones.

If you want the effect of the round won to be smaller, you can take

$\mathbf{S}(x) = \underset{\{i\ :\ \text{x won round i}\}}{\sum 1 + \frac{1}{k}\frac{1}{p_i}}= \text{number of wins} + \underset{\{i\ :\ \text{x won round i}\}}{\sum \frac{1}{k}(n- i)}$

So you are still counting all the wins of the player, and then adding extra points for winning in the earlier rounds.

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