Say you have two sports teams. To keep it simple, assume they have each played the same number of games against exactly the same teams.
If you know only the winning percentages of each team, what is the probability that the team with the better record will beat the other team?
Disregard considerations of handicapping, like how well they scored against which team, whether one team won more games against teams with better records, etc. Treat this like a marbles-in-a-bag probability problem.
Let’s try this: A team’s probability of winning is their winning percentage divided by the sum of the winning percentages of the two teams. Does that work?
There’s no magic correct probability. A probability is only defined relative to a probability distribution. You haven’t specified any particular probability distribution.
To illustrate the point:
Suppose the game is “Who’s first in the alphabet?”: Bob plays against Adam, Alex, Carl, and Joseph twenty times each, and, as it so happens, ends up winning 25% of these games. Brian plays against Adam, Alex, Carl, and Joseph twenty times each, and, as it so happens, ends up winning 50% of these games. What’s the probability that Brian will beat Bob? Well, 100% of course…
Suppose the game is “Flip a fair coin to determine the winner”: Bob plays against Adam, Alex, Carl, and Joseph twenty times each, and, as it so happens, ends up winning 25% of these game. Brian plays against Adam, Alex, Carl, and Joseph twenty times each, and, as it so happens, ends up winning 50% of these games. What’s the probability that Brian will beat Bob? Well, 50% of course…
There’s no magic formula that you can feed 25% and 50% into to determine the probability that Brian will beat Bob without further information about the probability distribution of interest.
And if you really want to make an assumption about a completely unknown distribution, you could use the ratio of each team’s wins vs. other teams (as you did). But as Indistinguishable said, it’ll be meaningless. Analogous to expecting your children to be born with roughly one testicle and one ovary, since that’s the average.
First, it has been about 30 years since my last statistics class so I don’t remember much about distributions. I do remember a few things about the normal distribution. But I thought distributions had to do with a statistic on a sample of a population (or the whole population), so I’m not completely sure what you’re telling me. Help me out here.
Why would my assumption be meaningless? It implies that a team with the same record as their opponent would have a .500 chance of winning. The record is a surrogate for their skill level. That passes the horse-sense test but I don’t know if the math works. And if a team that has won only .250 of their games plays a team that has won .500 of their games, it seems to me that their chance of winning is only half that of their opponent, or .333, because they are only winning half as many games against the same opponents.
Why is it meaningless and what information about a distribution would you need to make it meaningful? And how would you use it?
One thing that occurs to me is the boundary condition. Regardless of your record and your opponent’s records, your chances of winning never become 1 and they never to go 0, although using my method, you could generate those results, pitting a 1.000 team against a .000 team.
I don’t understand your extension to ovaries and testicles at all, at least not for this application. However, I did hear about a statistician who drowned in a river with an average depth of 6". 
The difference between this illustration and my situation is that my events are not random, they are games of skill, and the relative skill of each player is known in the form of their record. Suppose instead of flipping a coin they are arm wrestling. Would you still stay that the chances that Brian will beat Bob are 50%?
I not saying that I think you’re wrong, I’m trying to draw out the salient points so I can understand why I’m wrong.
This is all highly relevant information. Why on earth would you disregard it?
It depends (among other things) on the sport, which is unspecified in the OP. In some sports form is a much better predictor of performance than in others.
Can’t. Here’s why:
Nobody has ever quantified “luck,” “will,” and “being clutch.” Luck, for example, would be the Immaculate Reception. “Will” would be Michael Jordan. “Clutch” would be defined as athletes who raise their performance under specific stressful conditions (e.g. runners in scoring position, last minute of a football game, etc.) The reason this is so hard is that it doesn’t happen consistently nor predictably.
We also know that physical characteristics of speed, strength, endurance, etc. aren’t absolute. A typical NFL wr, for example, is not running at their top speed for the entire route. Basketball players are famous for being notoriously poor long-distance runners. Things we can measure in a lab aren’t transferable to on-field performance.
Team sports have too many variables to predict. A simple case would be injuries. If a particular player gets injured, the sample size is too small to predict it’s effect on team performance. Sometimes, the replacement player is Joe Dimaggio, sometimes it’s somebody less skilled. Weather, off-field issues, coaching changes, number of practices, etc. etc. etc. all could affect the outcome. It’s much easier to predict a one-on-one matchup vs a team game.
Sample Size: the closest thing you have to predictability is a 7 game series. Better teams usually win more games, over time, against the same team. However, for single games, any result is possible, especially at the professional level.
Because I don’t have it.
Your statement is true but let’s make a substitution. “For a single flip of two coins, any result is possible.” So it’s possible that both could be heads, but we know there’s only a 1/4 chance.
I’m not looking for a “prediction,” I’m looking for probability.
Not remotely helpful. Here’s why:
The OP is asking a probability question, not a promise for a guaranteed result. If a team is playing .600 baseball, and their opponents is playing .400, one (dumb) way to look at the outcome is to average the percentages together and say (absurdly) that they each have a .500 chance of winning a given game. Slightly smarter is to conclude that each team compiles its record vs. a basically .500 league so the .600 team will improve against a .400 opponent, and the .400 oppenent will do worse vs. a .600 team–the real question the OP is asking in my view, is “OK but how MUCH?” Will the .600 team win 70% of its games vs. a .400 team? 80%? 65%?
Bill James have devised a formula for making the exact probabilty (somewhere I don’t have handy at the moment) but some statgeek (not inconceivably me) can point you in the right direction. I’m just here to say that it’s a fine question (given a sufficent sample size, of course), not only capable of being answered but useful as well.
So there are four types of team, the ones that
I suspect the outcome is always that this is a very uncertain way to predict the result. Otherwise every TV-commentator would quote the probabilities in every match.
Still too many variables. A team may have a great record so far, then loses its two best pitchers before they two teams play. How does that affect anything?
Any statistics are based on previous games. But a sports team “starts over” statistically for each game; a batter hitting .250 may or may not go 1 for 4. A batter may hit three home runs, but the pitcher – who is usually an ace – has a terrible outing and gives up four.
It’s just not something you can come up with an actual probability. At best, you can run a simulation and then say, “these are the results of a simulation” (all simulations are flawed).
Heiseinberg had it easy finding the location and speed of an atom compared to this (and Heiseinberg found that impossible).
If you are willing to impose some sort of transitivity like
if “P(A beats B) > 0.5” and “P(B beats C) > 0.5” then “P(A beats C) > 0.5”
(which needn’t be true) and if you are willing to assume that the collection of teams they have both played is a sufficiently large ensemble to establish the two teams’ relative ordering (unlikely, since leagues aren’t very large), then even still the most you can say is that the probability is somewhere between 0.5 and 1, depending on the nature of the game. Some games (like chess) have low variance – the better player usually wins. Some (like rock-paper-scissors) have high variance – the better player may or may not win.
You might feel that this would be captured in how different the winning percentages are, but that just isn’t enough information. To see this, construct two games:
(1) Give every team a biased coin. The coin will come up heads with probability p, where p is different for each team, uniformly drawn from (0,1). Each game consists of flipping the two competitors’ coins. If one teams gets H and the other T, the ‘H’ team wins. If the result is HH or TT, reflip. There is some team (Team P60) that wins this game 60% of the time against the large field of opponents. There is also some team that wins 40% of the time. If they play each other, Team P60 wins 77% of the time (…if I did my quick math right.)
(2) Give every team a piece of paper. The paper has a number p written on it, where p is different for each team, uniformly drawn from (0,1). Each game consists of comparing the numbers of the two competitors’ papers. The higher number wins. There is some team (Team P60) that wins this game 60% of the time against the large field of opponents. There is also some team that wins 40% of the time. If they play each other, Team P60 wins 100% of the time.
You are looking for something that says “Since Team P60 won 60% of its games and Team P40 won 40% of its games, then Team P60 has x chance of beating Team P40.” The above examples show that the answer depends very much on the game. There is no general answer.
No, TV commentators don’t like commenting on math very much. Joe Morgan’s head would actually explode if he ever had to think.
Look at my example. A .600 team playing a .400 team obviously will have a probability greater than .600 of winning a given game, but how much greater? Obviously not greater than 100% and almost as obviously not even close. Figuring out whether it’s more like 65% or 80% is doable, though, and James has done it. I’ll try to track down his writing–it’s more of a technical question than it is a baseball one (and James has extrapolated his findings to other sports, where they worked as well as in baseball.)
Please note. My question is not, “How can I accurately determine the probability of a given team winning against another given team on any given day?” The answer to that indeed involves many variables, as has been pointed out.
My question is, “Given certain assumptions and limited data to make a simplified model, what is the probability of a given event?” I accept that the model will be flawed. But after all, all models are flawed–it’s just a matter of how much flaw you can tolerate.
We make assumptions based on limited data all the time. Weather is complex with a huge number of variables but that doesn’t stop meteorologists from developing models and giving the probability of rain in a very localized area.
Do we at least know who is the home team and who is the away team? Home field advantage is real, and pertains to all sports, some to a greater degree than others. We might assume a .600 home team owns a massive advantage over a .400 visitor, but the same match-up could be a close call if played in the other stadium.
Normally this would matter, and we would know. However, in my case it’s house little league and although there is a nominal home-team designation for each game, fields are chosen arbitrarily so there is no home-field advantage.
I don’t know if there has been any study that shows whether a team batting in the bottom of the inning has any advantage, since that isn’t normally independent of the home field. (My own pet theory is that there is a psychological advantage in batting last in that people who are attempting to reach a clear goal are generally more successful than people trying to “do their best.”)