Question about simulating finishing order

This isn't really for anything important, but every now and then my brain gets locked on something and won't let go.... you know how it is, I'm sure.


I started thinking about how video game rating systems work, and set out to devise my own simple version of one, just so I could test it and simulate some games.


So say it's a racing game, with 4 players. Video game rating systems are meant to attempt to measure the "true skill" of a player, so for my simple version I'm going to assume that it does accurately enough.


I want to simulate the finishing order of 4 players with the following ratings of "true skill." The rating numbers themselves are arbitrary for now, but I've tried to disperse them in a way that there should be an obvious trend in how they finish, leaving the last two closer together to attempt to see more "flip-flopping" of positions than the first two.


Player A - 1500
Player B - 1250
Player C - 1000
Player D - 900

To clarify, these numbers just represent relative skill. ie - Player A is ~18% better player than Player B (so they should be scalable). I want to be able to calculate the probability of the different outcomes so I can reward players more for doing better than they "should." For example, with the ratings above, Player C would get a bigger rating boost for winning the race than Player A would, because it is expected that A would win. Conversely, Player A would get a huge rating penalty if he came in last, relative to the penalty Player D would get for the same result.

In case my question wasn't clear. I'm trying to calculate the odds of each possible outcome. Knowing there are 24 different orders they can finish in, a % chance that a particular outcome happens. I imagine this must use some sort of weighted permutation calc, but I don't really know where to begin on it.

pearldrumbum said:

In case my question wasn't clear. I'm trying to calculate the odds of each possible outcome. Knowing there are 24 different orders they can finish in, a % chance that a particular outcome happens. I imagine this must use some sort of weighted permutation calc, but I don't really know where to begin on it.

Click to expand...

Can you make an Excel spreadsheet? Let there be 24 rows for the permutations, and let there be 4 columns representing 1sr, 2nd, 3rd, 4th positions. I'll call them columns A-D and rows 1-24, but they can be anywhere convenient in the sheet. Cells A1:A6 represent A finishing first, and have probability A/(A+B+C+D) = 0.3226 for the example given.
Then cells A7:A12 represent B finishing first, probability B/(A+B+C+D),
.......cells A:13:18 = C/(A+B+C+D)
.......cells A:19:24 = D/(A+B+C+D)

Now for second column:
Cells B1:B2 are the conditional probability that B is second given A was first: B/(B+C+D)
.......B3:B4 are the conditional probability that C is second given A was first: C/(B+C+D)
.......B5:B6 are the conditional probability that D is second given A was first: D/(B+C+D)
.......B7:B8 are the conditional probability that A is second given B was first: A/(A+C+D)
.......B9:B10 are the conditional probability that C is second given B was first: C/(A+C+D)
.......B11:B12 are the conditional probability that D is second given B was first: D/(A+C+D)
.......B13:B14 are the conditional probability that A is second given C was first: A/(A+B+D)
. . .
Third column has conditional probabilities for finishing 3rd.
Cell C1 = P(C | AB) = C/(C+D)
......C2 = P(D | AB) = D/(C+D)
......C3 = P(B | AC) = B/(B+D)
. . .

For the fourth column, the conditional probabilities would all be 1 because there is only one choice left. Instead, use the final column to compute the product of the first three columns, which will be the probability of that particular permutation.
D1 = A1*B1*C1
copy formula to D224

For a check, the sum of col D should be 1.

Hey, that's a pretty a good idea. I'm actually doing this whole simulation in Excel, so it makes sense to do it this way. I'll have some time later tonight to play with it so I'll let you know how it works out. Seems pretty straight-forward, if not a little tedious, haha.