Finding a pro disc golfer's chance of winning a tournament, knowing their chance of beating each other player I'm working on a strength of field metric for disc golf tournaments, and I'd first like to come up with a way to determine a player's likelihood of winning an event, given their player rating and the ratings of other players in the event.
For anyone not familiar (probably most reading this), competitive disc golf is played very similar to golf. The competitors play a set number of rounds on the tournament course, and the player with the lowest stroke total at the end of the event wins. The governing body calculates player ratings based on performance at these competitive events, which allows any player with enough rated rounds to be compared to another player, even if they haven't competed directly.
I compiled data from every tournament in the 2022 disc golf pro tour, including the entrants, their finishes, and their ratings at the start of the event. I separated every player at every tournament into buckets of 5 ratings points each, and for all players at all tournaments calculated the probability of an average player of rating x beating an average player of rating y (beating defined simply as placing higher/scoring lower in the tournament). For instance, a player who enters a tournament with a rating between 1040-1045 has a 33% chance of placing higher than a 1045-1050 rated player - but a 74% chance of placing higher than a player rated 1025-1030.
My thought was that I could use this matrix to find a player's likelihood of winning a tournament. Once I know their likelihood of placing higher than each other player in the event individually, I can multiply all of those probabilities to find the chance of that player finishing higher than every other player at the event. These should all be independent events I assume, as each player is playing against the course - their individual scores have no effect on one another's scores.
However, after running this calculation, the resulting event-win probabilities I get are way too low. They do make sense directionally - higher rated players in my sample had higher win probabilities - but they are still far too low. For example, one of these events had 131 entrants - so if I were to assume every player entered had the same skill, they should all have about a 0.8% chance of winning. In my calculation, with skill difference taken into account, the highest rated player at the event had a .0004% likelihood of winning.
I thought floating point error from my probability matrix might be to blame, so I tried rounding any value below .005 in the matrix to 0. This did not turn out to be the source of the issue however.
Am I missing something in my approach, or my assumptions? I'm a bit of an amateur at all of this, so I wouldn't be surprised if the math just doesn't work out the way I think it does.
Thanks!
 A: If I understand you correctly, the problem is that "Player A beats player B" and "Player A beats player C" are definitely not independent events, even though the players are playing against the course. There are underlying variables -- namely each player's actual score -- which are correlated with who beats whom.
As an extreme example, imagine that one of your players is Ursula the Unreliable. In half of the tournaments Ursula plays in, she gets a hole in 1 on every hole. In the other half, she quintuple-bogeys every hole.
Ursula's probability of beating any other player head-to-head is going to be very close to $\frac{1}{2}$. But her probability of winning an $n$-player tournament is also very close to $\frac{1}{2}$: if you know that she beats a single player, it's almost certainly one of her good tournaments and so she's probably beating everyone else as well. But your method would say that she has a win probability of $\frac{1}{2^{n-1}}$.
Basically, when you multiply all the win probabilities, you're computing the probability that a particular player would win every single game in a head-to-head round robin tournament. This is a much less likely event than that player having the lowest stroke total in a standard golf-style tournament! In particular, in a large head-to-head tournament, it's very likely that everyone will lose at least one game, which is why your probabilities don't add up to $1$.
I think the right way to do what you're doing is not to try to compute head-to-head win probabilities based on rating, but to try to come up with some probability distribution of scores based on rating.
A: I would approach the problem in a different way.
Assume that the players are indexed $n_1, n_2, \cdots, n_r$.
Since the ratings of each of the $r$ players are known, given any two players $~n_i, n_j~: ~i \neq j$, the following probabilites should be known:

*

*$P(i,j)$ which denotes the probability that $n_i$ should finish higher than $n_j.$


*$Q(i,j)$ which denotes the probability that $n_i$ should finish lower than $n_j.$


*$R(i,j)$ which denotes the probability that $n_i$ should finish equal to $n_j$.
Here, it is assumed that for any $~i,j ~: i \neq j,~$ 
that $P(i,j) + Q(i,j) + R(i,j) = 1.$
For simplicity it is also assumed that there is a clear winner.
Then, the probability that (for example) $n_1$ is the clear winner, given that there is a clear winner, is equal to
$$\frac{p\left(n_1 ~\text{is the clear winner}\right)}{p\left(\text{there is a clear winner}\right)}. \tag1 $$
In (1) above, the numerator may be computed as
$$\prod_{j=2}^n P(1,j)$$
and the denominator may be computed as
$$\sum_{i=1}^n \left[\prod_{j \neq i} P(i,j)\right].$$
