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I'm tracking stats for my board game group. I'm using a modified ELO system where I consider each game of multiple players be a group of 1v1 matches between all pairs of players as seen here: http://elo-norsak.rhcloud.com/3.php

Once I have an ELO for each player, For any two player I can calculate the percent chance A will beat B via 1 / (1 + Math.pow(10, EloDifference / 400))

Assuming everybody plays to win, no ganging up, etc, How do I calculate the expected finishing order of the game?

This question calculates who is expected to win. Can I assume the player 2nd most likely to win is most likely to come 2nd, etc?

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FigBug
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  • "Can I assume the player 2nd most likely to win is most likely to come 2nd, etc?" -- consider a player who cares nothing at all for coming second, regarding it as exactly the same as finishing last. They may be inclined to try more risky strategies that increase both the probability of a win (if the risk comes off) and coming last (if it doesn't), when compared to a player who would prefer to come second than last. Indeed, depending on the sort of game it is, coming last may be the most likely outcome for any player who is the second-most-likely to come first. It's very hard to generalize. – Glen_b Dec 21 '15 at 22:57

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There are many kind of board games. Some of them have final ranking (e.g. Ticket to Ride), some don't (one player wins, the others lose, e.g. Villa Paletti), some have just a loser (e.g. Jenga), some have more than a winner, and more.

Providing that you're playing a game with ordered output and you're using a ranking system that rewards a 2nd place better than other non-winning position (making more risky strategies suitable, but most risky), the answer is yes, the player 2nd most likely to win is most likely to come 2nd.

In the scheme you're using, it comes from multi-match structure, but considering each game of multiple players be a group of 1v1 matches between all pairs of players is not the best option, indeed. Rankade, our post-Elo ranking system, manage this issue in a specific way.

Tomaso Neri
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  • Is your Ree algorithm available for others to use? – FigBug Feb 23 '16 at 18:37
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    Rankade's algorithm (*ree* algorithm) is extremely complex (here's a [comparison](https://rankade.com/ree#ranking-system-comparison) between most known ranking systems) and not in the public domain, for the time being, but our webapp and mobile apps are free-to-use, if you want to give it a try. – Tomaso Neri Feb 23 '16 at 23:42