# Probability Q: Bayes Theorem?

Gold Member

## Homework Statement

Team 0 and Team 1 have played 1000 games and Team 0 has won 900 of them.[/B]
When the two teams play next, knowing only this information, which team is more likely to win?

## Homework Equations

P(X,Y) = P(YlX) x P(X) = P(XIY) x P(Y) (Not Sure)

## The Attempt at a Solution

Hi,
I am trying to solve it using Bayes Theorem. Kindly guide me is it correct or not?
Probability of Team 1 winning = 100/1000 = 0.1

Probability of Team 2 winning = 900/1000 = 0.9

Probability of Team 1 losing = 900/1000 = 0.1

Probability of Team 2 losing = 100/1000 = 0.1

Let X represents the Probability of Team winning the match

Let Y represents the Probability of Team losing the match

P(X=1|Y=0) = P(X=0|Y=1)

Zulfi.

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Stephen Tashi

## Homework Statement

Team 0 and Team 1 have played 1000 games and Team 0 has won 900 of them.
When the two teams play next, knowing only this information, which team is more likely to win?[/B]

Were you instructed to use Bayes Theorem to compute the probabilities of each team winning the next game? If so, exactly what did those instructions say? Were you told to assume a prior probability distribution?

Gold Member
Hi,
Not instructed but the topic was Bayes theorem so i thought it can be solved with it. Thats why i evaluated a prior probability & then used in Bayes theorem. Kindly tell me if its correct or not. Should i be using Bayes theorem or it can be solved using some other method? If it can be solved using Bayes theorem am i right or wrong?

Zulfi.

Stephen Tashi
If it can be solved using Bayes theorem am i right or wrong?

As far as I can see, you didn't answer the question you stated, which was "Which team is more likely to win?".

It isn't clear from your statement of the question whether the two teams are necessarily playing against each other.

You wrote down the equation $P(X=1|Y=0) = P(X=0|Y=1)$ which presumably refers to the probability of team X winning or losing conditioned on whether team Y wins or loses. If the two teams are playing against each other, both those probabilities are equal to 1.

Gold Member
Hi,
<It isn't clear from your statement of the question whether the two teams are necessarily playing against each other.>
Lets us suppose two teams are playing against each other.
<As far as I can see, you didn't answer the question you stated, which was "Which team is more likely to win?".>
Shoud i write:
P(X=1|Y=1) = P(X=0|Y=0)?
By which I mean: Conditional Probability of Team1 winning the match given the probability of Team1 losing the match =
Conditional Probability of Team0 winning the match given the probability of Team 0 losing the match?

Zulfi.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Team 0 and Team 1 have played 1000 games and Team 0 has won 900 of them.
When the two teams play next, knowing only this information, which team is more likely to win?[/B]

## Homework Equations

P(X,Y) = P(YlX) x P(X) = P(XIY) x P(Y) (Not Sure)

## The Attempt at a Solution

Hi,
I am trying to solve it using Bayes Theorem. Kindly guide me is it correct or not?
Probability of Team 1 winning = 100/1000 = 0.1

Probability of Team 2 winning = 900/1000 = 0.9

Probability of Team 1 losing = 900/1000 = 0.1

Probability of Team 2 losing = 100/1000 = 0.1

Let X represents the Probability of Team winning the match

Let Y represents the Probability of Team losing the match

P(X=1|Y=0) = P(X=0|Y=1)

Zulfi.
Hi,
Not instructed but the topic was Bayes theorem so i thought it can be solved with it. Thats why i evaluated a prior probability & then used in Bayes theorem. Kindly tell me if its correct or not. Should i be using Bayes theorem or it can be solved using some other method? If it can be solved using Bayes theorem am i right or wrong?

Zulfi.
What you have done so far has little, if anything, to do with Bayes' theorem. Unless you tell us what type of prior "win" probability distribution you are using, nothing you have done relates to Bayes.

For example, one type of prior will probability distribution you could use would be a uniform distribution between 0 and 1; that is, before observing any actual games you could say that you have no reason at all to favor one team over the other, so the probability $p$ that team 0 will will a game is unknown to you. You have no reason to prefer the "guess" $p = 0.1$ over the "guess" $p = 0.7642$ over the guess $p = 0.01$ or over the guess $p = 0.9999$, or over any other number between 0 and 1.

After observing 1000 games in which team 0 has won 900 times and team 1 has won 100 times, your best single guess now is that $p = 0.9$. However, other values of $p$ are still quite likely to occur, and (if you are a "Bayesian") the question is really asking for the posterior probability that $p > 1/2.$

Stephen Tashi
Lets us suppose two teams are playing against each other.
Ok, let's assume that.

One way to interpret the question is:

Which of the following probabilites is greater?:
1) p_0 = the probability that team 0 wins the match given it has won 900 of 1000 other matches.
2) p_1 = the probability that team 1 wins the match given it has won 100 of 1000 other matches.

If the problem is intended as a very elementary question, the answer would be that p_0 is greater than p_1 because we would use the results of the other games as an approximation for p_0 and p_1.

If the problem expects you to demonstrate a detailed Bayesian approach, it is a sophisticated problem and there are several different ways to do it. We need to know if you are taking an elementary course in probability or an advanced course.

Gold Member
Hi,

Thanks.
God bless you.

Zulfi.

Ray Vickson
Homework Helper
Dearly Missed
Hi,

Thanks.
God bless you.

Zulfi.
In an advanced course an interpretation like that in #6 is also quite likely. Both types of models are used in practice by Bayesians.

Stephen Tashi
Hi,