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# WinnerVsLoser

last edited by 13 years, 7 months ago

# What's the Probability?

This seemed like it would have been a pretty simple question, but it puzzled a number of pretty smart people. Here's the question:

What is the probability of a win for a team that wins 70% of its games when it plays a team that wins 30% of its games?

Assume that the teams have played enough games against everyone else in the league that the winning percentages are an accurate reflection of their abilities. Assume also that we have no other information about the teams. Assume no "trick" parts to the question or answer.

If you are new to this question you may want to try puzzling it out yourself before you read the responses below. These were offered with varying caveats about the author's confidence and level of guessing. If/when you do check the responses, I'm thinking the one from I.J. is correct.

## Responses

D.B. says:

The two teams play a game. Assuming no possibility of a tie, there are only two possible outcomes. Either A wins or B wins. Assume A has 70% winning probability, B has a 30% winning probability.

Draw a square. Split it 70/30 both horizontally and vertically. You now have four rectangles of differing sizes. The biggest one corresponds to the chances of A winning (A wins 70% of the time, B loses 30% of the time). The smallest one is B winning (30% chance to win, A loses 30% of the time). Take the area of the big one, add .5 the sum of the areas of the remainders, that's A's winning percentage for this game. Obviously, 1-(A's winning percentage) is B's winning percentage, which should work out to be the same as the area of the smallest rectangle plus .5 the sum of the remainders.

I.J. says:

You can assume the average for the entire league is .500, since for every game one team wins and one team loses. The probability of a team winning can be modeled as a normal distribution. The entire area under a normal curve = 1, so the integral area from -infinity to a given value x from the mean is the probability of that team beating a team with a .500 average (this probability is a z-score).

So, for the team with a .700 average, z = .70, and you can look up in a z table the value of x in standard deviations above the mean, which turns out to be .53. Similarly, the .300 team's x value is .53 (since .7 and .3 are both equidistant from the mean).

We don't know the standard deviation for the entire league, but we do know the number of standard deviations apart the two teams are (.53 + .53 = 1.06). So, we can set the mean of the normal curve at .30, and say that the .70 team falls 1.06 standard deviations above the mean. Then you can look up the z-score for x = 1.06 sd, which turns out to be z = .8554. This should be the probability of the .700 team winning.

R.P. says:

OK. To assist in my pedestrian calculations, I've changed the assumptions somewhat. Let's assume a league of 4 teams, where each team plays 120 games. That means that there have been 240 wins and losses

registered (120 games each, with each team registering one of two outcomes per game).

One team has won 75% games. One team has won 25%. For the sake of argument, assume the other 2 teams have won 50% of their games. You have 4 teams, ranked thusly:

1. 75% of 120 or 90-30

2. 50% of 120 or 60-60

3. 50% of 120 or 60-60

4. 25% of 120 or 30-90

Assume also that each team has played each other team an equal amount of times. So the 75% team has played 40 games against each .500 team, and 40 games against the .250 team. The same goes for all teams: they've played 40 games against each opponent. The sum of wins vs. opponent 1 + sum of wins vs. opponent 2 + sum of wins vs. opponent 3 = P winning.

Assume the two .500 teams have played each other 40 times, and split. So of team 2 and 3's 60 wins, 20 of those came from each other, and 20 of the losses came from each other. That leaves another 40 wins from

the .750 team and the .250 teams combined. Assume the .500 teams won more games against the .250 team than against the .750 team. Even so, the .750 team couldn't have won 75% of the games against the .500 team.

If they had, then the .750 team would have an average higher than .750 overall, because it's more likely to beat the .250 team than the .500 team, and if it beat the .500 team 75% of the time, it's going to have

an greater winning record against the .250 team, and there are no other teams against which it competed to lower the average back to .750.

Let's further assume both .500 teams performed exactly the same against the .750 team. Of the .750 team's wins, X = number of wins the .500 teams allowed; Y = number of wins the .250 team allowed. Let's assume

that Y has to be greater than X, because Y represents wins against the suckier team. Let's also assume that team 1 does not beat team 4 100% of the time.

x+x+y = 90, or 2x + y = 90, or y=90-2x

y = 2(45-x)

Also,

Y > X

X <= 40, because that's all the number of games each .500 team has left to lose, while still remaining at .500. Remember, we've assumed that the .500 teams went 20-20 against each other, meaning they went 40-40 against the rest of the league.

Y < 40 , because we're assuming that the .250 team won at least one game against the .750 team.

Assuming the .750 plays the .500 teams at a .750 pace, the .750 team has gotten 60 of its 90 wins from the 2 .500 teams (75% of 40 games = 30 wins against each team). This means that it also won 30 games from the .250 team, to get to 90 wins. This would break our assumption that Y > X, and therefore fails. This mathematically matches our assumption above: that the .750 team beat the .500 team less than 75% of the time. We still don't know how much less, though.

So x < 30.

Now, let's assume that the .500 teams win 50% of its games against the .750 team. Its win-loss record would look like this:

Record against team

1: 20-20

3: 20-20

4: 20-20

This can't happen, because even if team 4 lost all 40 games to the .750 team, it still would have amassed 40 wins from the two .500 teams (assuming that 2 and 3 play each of their opponents equally) and would have a record in exceess of at least .333, instead of .250. So X > 20

Let's try 25 to solve for X:

The .750 teams record would have been reached thusly:

Record against team

2: 25-15

3: 25-15

4: 40-0

Under this scenario, the .500 teams would have lost 20 to each other, 25 to the .750 team, and 15 to the .250 team. But in this scenario, Y = 40, which breaks our assumption that the probability of team 4 beating team at least once exceeds 0.

So x > 25 and less than 30 . Because x can only be an integer (you can't win .5 of a game in baseball), x must be 26, 27, 28, or 29. Let's try those numbers out.

If x is 26:

y=2*(45-x)

y=2*(45-26)

y=2*19

y=38

following the same approach, if x is 27:

y=36

and,

if x is 28:

y=34

and,

if x is 29

y=32

let's see the outcome with these numbers:

If y = 38, then team 1's winning percentage is 95%

if y = 36, then it's 90%,

if y = 34, then it's 85%,

and if y = 32 then it's 80%.

Each of these outcomes of wins leads to the same winning percentage, so if you normalize and assume each outcome is equally probable, then. . . . the answer is 88%. If you include the likelihood of team 4 having zero wins, then the number increases to 90%.

I'm no math dude; I just used algebra to figure this out. I bet there's some cool calculus that would do a better job.