Probability that a team that loses in a best of series is the better team
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Say we have two teams $A$, $B$ that will play in a best of $n$ series, which to win you must win $w$ games, which is the ceiling of $fracn2$. Say that when the two teams play each other, $A$ has a probability $p$ of winning a single game, and $B$ has a probability $q$ of winning a single game.
Say that in our (for example) best of $5$ match, $B$ wins $3-2$ over $A$, what is the probability that $A$ was actually the better team, or that $p>q$?
Rephrased: what is
$$
P(p>qmid Btext won 3-2)
$$
My attempt: Since $p+q=1$, we know $q = 1-p$ so in order for $p>q implies p>frac12$ via substitution. The probability will then be
$$
P(p>qmid Btext won 3-2) = int_.5^1 P(p=xmid Btext won 3-2)dx
$$
which we can expand with Bayes Theorem to get
$$
int_.5^1 P(p=xmid Btext won 3-2)dx = int_.5^1 dfracP(Btext won 3-2mid p=x)P(Btext won 3-2)P(p=x)dx
$$
Which is where I get stuck. Any help would be appreciated!
probability integration statistics recreational-mathematics bayesian
asked Aug 26 at 5:46
wjmccann
598117
 |Â
show 1 more comment
up vote
1
down vote
favorite
Say we have two teams $A$, $B$ that will play in a best of $n$ series, which to win you must win $w$ games, which is the ceiling of $fracn2$. Say that when the two teams play each other, $A$ has a probability $p$ of winning a single game, and $B$ has a probability $q$ of winning a single game.
Say that in our (for example) best of $5$ match, $B$ wins $3-2$ over $A$, what is the probability that $A$ was actually the better team, or that $p>q$?
Rephrased: what is
$$
P(p>qmid Btext won 3-2)
$$
My attempt: Since $p+q=1$, we know $q = 1-p$ so in order for $p>q implies p>frac12$ via substitution. The probability will then be
$$
P(p>qmid Btext won 3-2) = int_.5^1 P(p=xmid Btext won 3-2)dx
$$
which we can expand with Bayes Theorem to get
$$
int_.5^1 P(p=xmid Btext won 3-2)dx = int_.5^1 dfracP(Btext won 3-2mid p=x)P(Btext won 3-2)P(p=x)dx
$$
Which is where I get stuck. Any help would be appreciated!
probability integration statistics recreational-mathematics bayesian
asked Aug 26 at 5:46
wjmccann
598117
haven't checked the logic, but if you have that right it will be much easier if you replace the integral with a summation, since these are discrete events.
– eSurfsnake
Aug 26 at 5:53
I think in this case it would be an integral because we are adding all possible probabilities but I don't know, I did get stuck
– wjmccann
Aug 26 at 5:55
This may not help you much... but this problem reminds me of one I saw in the book Sheldon Ross in chapter 6 section 5, example 5d (at least in my edition). Punch line is that the probability p that team A wins follows a Beta distribution.
– HJ_beginner
Aug 26 at 5:56
To follow up with this possibly wrong approach... let $X sim uniform(0,1)$ and $N sim Binomial(n+m,p)$ Then $f_N(x|n) = cx^2(1-x)^3$ then you integrate from $0$ to $1$ to figure out what $c$ is, then with the conditional pdf you can get the cum cdf and see what the probability that $X|N > 0.5$
– HJ_beginner
Aug 26 at 6:02
1
There's not enough information to answer the question; you need a prior for $p$.
– joriki
Aug 27 at 5:18
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Say we have two teams $A$, $B$ that will play in a best of $n$ series, which to win you must win $w$ games, which is the ceiling of $fracn2$. Say that when the two teams play each other, $A$ has a probability $p$ of winning a single game, and $B$ has a probability $q$ of winning a single game.
Say that in our (for example) best of $5$ match, $B$ wins $3-2$ over $A$, what is the probability that $A$ was actually the better team, or that $p>q$?
Rephrased: what is
$$
P(p>qmid Btext won 3-2)
$$
My attempt: Since $p+q=1$, we know $q = 1-p$ so in order for $p>q implies p>frac12$ via substitution. The probability will then be
$$
P(p>qmid Btext won 3-2) = int_.5^1 P(p=xmid Btext won 3-2)dx
$$
which we can expand with Bayes Theorem to get
$$
int_.5^1 P(p=xmid Btext won 3-2)dx = int_.5^1 dfracP(Btext won 3-2mid p=x)P(Btext won 3-2)P(p=x)dx
$$
Which is where I get stuck. Any help would be appreciated!
probability integration statistics recreational-mathematics bayesian
asked Aug 26 at 5:46
wjmccann
598117
Say we have two teams $A$, $B$ that will play in a best of $n$ series, which to win you must win $w$ games, which is the ceiling of $fracn2$. Say that when the two teams play each other, $A$ has a probability $p$ of winning a single game, and $B$ has a probability $q$ of winning a single game.
Say that in our (for example) best of $5$ match, $B$ wins $3-2$ over $A$, what is the probability that $A$ was actually the better team, or that $p>q$?
Rephrased: what is
$$
P(p>qmid Btext won 3-2)
$$
My attempt: Since $p+q=1$, we know $q = 1-p$ so in order for $p>q implies p>frac12$ via substitution. The probability will then be
$$
P(p>qmid Btext won 3-2) = int_.5^1 P(p=xmid Btext won 3-2)dx
$$
which we can expand with Bayes Theorem to get
$$
int_.5^1 P(p=xmid Btext won 3-2)dx = int_.5^1 dfracP(Btext won 3-2mid p=x)P(Btext won 3-2)P(p=x)dx
$$
Which is where I get stuck. Any help would be appreciated!
probability integration statistics recreational-mathematics bayesian
asked Aug 26 at 5:46
wjmccann
598117
asked Aug 26 at 5:46
wjmccann
598117
asked Aug 26 at 5:46
wjmccann
598117
asked Aug 26 at 5:46
wjmccann
598117
598117
haven't checked the logic, but if you have that right it will be much easier if you replace the integral with a summation, since these are discrete events.
– eSurfsnake
Aug 26 at 5:53
I think in this case it would be an integral because we are adding all possible probabilities but I don't know, I did get stuck
– wjmccann
Aug 26 at 5:55
This may not help you much... but this problem reminds me of one I saw in the book Sheldon Ross in chapter 6 section 5, example 5d (at least in my edition). Punch line is that the probability p that team A wins follows a Beta distribution.
– HJ_beginner
Aug 26 at 5:56
To follow up with this possibly wrong approach... let $X sim uniform(0,1)$ and $N sim Binomial(n+m,p)$ Then $f_N(x|n) = cx^2(1-x)^3$ then you integrate from $0$ to $1$ to figure out what $c$ is, then with the conditional pdf you can get the cum cdf and see what the probability that $X|N > 0.5$
– HJ_beginner
Aug 26 at 6:02
1
There's not enough information to answer the question; you need a prior for $p$.
– joriki
Aug 27 at 5:18
 |Â
show 1 more comment
haven't checked the logic, but if you have that right it will be much easier if you replace the integral with a summation, since these are discrete events.
– eSurfsnake
Aug 26 at 5:53
I think in this case it would be an integral because we are adding all possible probabilities but I don't know, I did get stuck
– wjmccann
Aug 26 at 5:55
This may not help you much... but this problem reminds me of one I saw in the book Sheldon Ross in chapter 6 section 5, example 5d (at least in my edition). Punch line is that the probability p that team A wins follows a Beta distribution.
– HJ_beginner
Aug 26 at 5:56
To follow up with this possibly wrong approach... let $X sim uniform(0,1)$ and $N sim Binomial(n+m,p)$ Then $f_N(x|n) = cx^2(1-x)^3$ then you integrate from $0$ to $1$ to figure out what $c$ is, then with the conditional pdf you can get the cum cdf and see what the probability that $X|N > 0.5$
– HJ_beginner
Aug 26 at 6:02
1
There's not enough information to answer the question; you need a prior for $p$.
– joriki
Aug 27 at 5:18
haven't checked the logic, but if you have that right it will be much easier if you replace the integral with a summation, since these are discrete events.
– eSurfsnake
Aug 26 at 5:53
haven't checked the logic, but if you have that right it will be much easier if you replace the integral with a summation, since these are discrete events.
– eSurfsnake
Aug 26 at 5:53
I think in this case it would be an integral because we are adding all possible probabilities but I don't know, I did get stuck
– wjmccann
Aug 26 at 5:55
I think in this case it would be an integral because we are adding all possible probabilities but I don't know, I did get stuck
– wjmccann
Aug 26 at 5:55
This may not help you much... but this problem reminds me of one I saw in the book Sheldon Ross in chapter 6 section 5, example 5d (at least in my edition). Punch line is that the probability p that team A wins follows a Beta distribution.
– HJ_beginner
Aug 26 at 5:56
This may not help you much... but this problem reminds me of one I saw in the book Sheldon Ross in chapter 6 section 5, example 5d (at least in my edition). Punch line is that the probability p that team A wins follows a Beta distribution.
– HJ_beginner
Aug 26 at 5:56
To follow up with this possibly wrong approach... let $X sim uniform(0,1)$ and $N sim Binomial(n+m,p)$ Then $f_N(x|n) = cx^2(1-x)^3$ then you integrate from $0$ to $1$ to figure out what $c$ is, then with the conditional pdf you can get the cum cdf and see what the probability that $X|N > 0.5$
– HJ_beginner
Aug 26 at 6:02
To follow up with this possibly wrong approach... let $X sim uniform(0,1)$ and $N sim Binomial(n+m,p)$ Then $f_N(x|n) = cx^2(1-x)^3$ then you integrate from $0$ to $1$ to figure out what $c$ is, then with the conditional pdf you can get the cum cdf and see what the probability that $X|N > 0.5$
– HJ_beginner
Aug 26 at 6:02
1
1
There's not enough information to answer the question; you need a prior for $p$.
– joriki
Aug 27 at 5:18
There's not enough information to answer the question; you need a prior for $p$.
– joriki
Aug 27 at 5:18
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
0
down vote
I think this is possible in a Bayesian context. Suppose $A$ wins $m$ out of $n$ matches.
We place an "uninformative" uniform $textrmBeta(1,1)$ priors on $p$, and get the posterior distribution:
beginequation
p| D sim textrmBeta(1+m,1+n-m)
endequation
Where $D$ denotes the data. Then we want $P[p>q] = P[p > 1-p] = P[p > .5]$.
This becomes evaluating the Beta cdf, but to check it note that $1-p$, when $p sim textrmBeta(a,b)$, has distribution:
beginequation
q sim textrmBeta(b,a)
endequation
This gives us that $q sim textrmBeta(1+n-m,1+m)$.
In your example this gives us that $P[p>.5] = .34375$ and $P[q>.5] = .65625$.
answered Sep 3 at 0:35
Ryan Warnick
1,25367
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I think this is possible in a Bayesian context. Suppose $A$ wins $m$ out of $n$ matches.
We place an "uninformative" uniform $textrmBeta(1,1)$ priors on $p$, and get the posterior distribution:
beginequation
p| D sim textrmBeta(1+m,1+n-m)
endequation
Where $D$ denotes the data. Then we want $P[p>q] = P[p > 1-p] = P[p > .5]$.
This becomes evaluating the Beta cdf, but to check it note that $1-p$, when $p sim textrmBeta(a,b)$, has distribution:
beginequation
q sim textrmBeta(b,a)
endequation
This gives us that $q sim textrmBeta(1+n-m,1+m)$.
In your example this gives us that $P[p>.5] = .34375$ and $P[q>.5] = .65625$.
answered Sep 3 at 0:35
Ryan Warnick
1,25367
add a comment |Â
up vote
0
down vote
I think this is possible in a Bayesian context. Suppose $A$ wins $m$ out of $n$ matches.
We place an "uninformative" uniform $textrmBeta(1,1)$ priors on $p$, and get the posterior distribution:
beginequation
p| D sim textrmBeta(1+m,1+n-m)
endequation
Where $D$ denotes the data. Then we want $P[p>q] = P[p > 1-p] = P[p > .5]$.
This becomes evaluating the Beta cdf, but to check it note that $1-p$, when $p sim textrmBeta(a,b)$, has distribution:
beginequation
q sim textrmBeta(b,a)
endequation
This gives us that $q sim textrmBeta(1+n-m,1+m)$.
In your example this gives us that $P[p>.5] = .34375$ and $P[q>.5] = .65625$.
answered Sep 3 at 0:35
Ryan Warnick
1,25367
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think this is possible in a Bayesian context. Suppose $A$ wins $m$ out of $n$ matches.
We place an "uninformative" uniform $textrmBeta(1,1)$ priors on $p$, and get the posterior distribution:
beginequation
p| D sim textrmBeta(1+m,1+n-m)
endequation
Where $D$ denotes the data. Then we want $P[p>q] = P[p > 1-p] = P[p > .5]$.
This becomes evaluating the Beta cdf, but to check it note that $1-p$, when $p sim textrmBeta(a,b)$, has distribution:
beginequation
q sim textrmBeta(b,a)
endequation
This gives us that $q sim textrmBeta(1+n-m,1+m)$.
In your example this gives us that $P[p>.5] = .34375$ and $P[q>.5] = .65625$.
answered Sep 3 at 0:35
Ryan Warnick
1,25367
I think this is possible in a Bayesian context. Suppose $A$ wins $m$ out of $n$ matches.
We place an "uninformative" uniform $textrmBeta(1,1)$ priors on $p$, and get the posterior distribution:
beginequation
p| D sim textrmBeta(1+m,1+n-m)
endequation
Where $D$ denotes the data. Then we want $P[p>q] = P[p > 1-p] = P[p > .5]$.
This becomes evaluating the Beta cdf, but to check it note that $1-p$, when $p sim textrmBeta(a,b)$, has distribution:
beginequation
q sim textrmBeta(b,a)
endequation
This gives us that $q sim textrmBeta(1+n-m,1+m)$.
In your example this gives us that $P[p>.5] = .34375$ and $P[q>.5] = .65625$.
answered Sep 3 at 0:35
Ryan Warnick
1,25367
edited Sep 3 at 1:37
answered Sep 3 at 0:35
Ryan Warnick
1,25367
answered Sep 3 at 0:35
Ryan Warnick
1,25367
answered Sep 3 at 0:35
Ryan Warnick
1,25367
1,25367
add a comment |Â
add a comment |Â
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haven't checked the logic, but if you have that right it will be much easier if you replace the integral with a summation, since these are discrete events.
– eSurfsnake
Aug 26 at 5:53
I think in this case it would be an integral because we are adding all possible probabilities but I don't know, I did get stuck
– wjmccann
Aug 26 at 5:55
This may not help you much... but this problem reminds me of one I saw in the book Sheldon Ross in chapter 6 section 5, example 5d (at least in my edition). Punch line is that the probability p that team A wins follows a Beta distribution.
– HJ_beginner
Aug 26 at 5:56
To follow up with this possibly wrong approach... let $X sim uniform(0,1)$ and $N sim Binomial(n+m,p)$ Then $f_N(x|n) = cx^2(1-x)^3$ then you integrate from $0$ to $1$ to figure out what $c$ is, then with the conditional pdf you can get the cum cdf and see what the probability that $X|N > 0.5$
– HJ_beginner
Aug 26 at 6:02
1
There's not enough information to answer the question; you need a prior for $p$.
– joriki
Aug 27 at 5:18