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Probability to win a series of three games, with a given advantage for the home team
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In a game with two equal teams, the home team wins any game with probability $p>frac12$. In a best of three playoff series, a team with a home advantage has play at home, followed by a game away, followed by a home play if necessary. The series is over as soon as one team wins two games.
What is $P[T]$, the probability that the team with home advantage wins the series? Prove that $P[T]>p$ for all $p>frac12$.
In the first sentence, I think there's a typo. I think it meant to say $p=frac12$ (I don't know how you can get a number as your answer if its not equal to $frac12$)
I drew a tree diagram as shown below, and calculated the probability that Team A wins.
$.5^2 + .5^3 + .5^3 = .5$
probability
edited Jul 20 at 16:32
Did
242k23208443
asked Jul 20 at 16:05
David
603
add a comment |Â
up vote
2
down vote
favorite
In a game with two equal teams, the home team wins any game with probability $p>frac12$. In a best of three playoff series, a team with a home advantage has play at home, followed by a game away, followed by a home play if necessary. The series is over as soon as one team wins two games.
What is $P[T]$, the probability that the team with home advantage wins the series? Prove that $P[T]>p$ for all $p>frac12$.
In the first sentence, I think there's a typo. I think it meant to say $p=frac12$ (I don't know how you can get a number as your answer if its not equal to $frac12$)
I drew a tree diagram as shown below, and calculated the probability that Team A wins.
$.5^2 + .5^3 + .5^3 = .5$
probability
edited Jul 20 at 16:32
Did
242k23208443
asked Jul 20 at 16:05
David
603
There's no typo. You are using $P=frac12$ but $P>frac12$. This is a binomial distribution where some of the outcomes never happen (If A wins all it's games, the series is shorter).
– David Diaz
Jul 20 at 16:08
1
I think you are to calculate an answer in terms of $P$, not a numerical answer. Your tree is a good way to start.
– Ethan Bolker
Jul 20 at 16:11
if p>1/2, wouldn't P(T) be less than that?
– David
Jul 20 at 16:14
It's often easier to imagine all possible sequences of three games (even though the winner may well be decided before the third game is played). The ultimate winner in all cases is the one that wins at least two of the three. Doing it this way lets you just use the properties of the binomial distribution.
– lulu
Jul 20 at 16:18
@David You're actually right. The sense of the last inequality in the question should be inverted (see my answer for working).
– Parcly Taxel
Jul 20 at 16:55
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In a game with two equal teams, the home team wins any game with probability $p>frac12$. In a best of three playoff series, a team with a home advantage has play at home, followed by a game away, followed by a home play if necessary. The series is over as soon as one team wins two games.
What is $P[T]$, the probability that the team with home advantage wins the series? Prove that $P[T]>p$ for all $p>frac12$.
In the first sentence, I think there's a typo. I think it meant to say $p=frac12$ (I don't know how you can get a number as your answer if its not equal to $frac12$)
I drew a tree diagram as shown below, and calculated the probability that Team A wins.
$.5^2 + .5^3 + .5^3 = .5$
probability
edited Jul 20 at 16:32
Did
242k23208443
asked Jul 20 at 16:05
David
603
In a game with two equal teams, the home team wins any game with probability $p>frac12$. In a best of three playoff series, a team with a home advantage has play at home, followed by a game away, followed by a home play if necessary. The series is over as soon as one team wins two games.
What is $P[T]$, the probability that the team with home advantage wins the series? Prove that $P[T]>p$ for all $p>frac12$.
In the first sentence, I think there's a typo. I think it meant to say $p=frac12$ (I don't know how you can get a number as your answer if its not equal to $frac12$)
I drew a tree diagram as shown below, and calculated the probability that Team A wins.
$.5^2 + .5^3 + .5^3 = .5$
probability
edited Jul 20 at 16:32
Did
242k23208443
asked Jul 20 at 16:05
David
603
edited Jul 20 at 16:32
Did
242k23208443
edited Jul 20 at 16:32
Did
242k23208443
edited Jul 20 at 16:32
Did
242k23208443
242k23208443
asked Jul 20 at 16:05
David
603
asked Jul 20 at 16:05
David
603
asked Jul 20 at 16:05
David
603
603
There's no typo. You are using $P=frac12$ but $P>frac12$. This is a binomial distribution where some of the outcomes never happen (If A wins all it's games, the series is shorter).
– David Diaz
Jul 20 at 16:08
1
I think you are to calculate an answer in terms of $P$, not a numerical answer. Your tree is a good way to start.
– Ethan Bolker
Jul 20 at 16:11
if p>1/2, wouldn't P(T) be less than that?
– David
Jul 20 at 16:14
It's often easier to imagine all possible sequences of three games (even though the winner may well be decided before the third game is played). The ultimate winner in all cases is the one that wins at least two of the three. Doing it this way lets you just use the properties of the binomial distribution.
– lulu
Jul 20 at 16:18
@David You're actually right. The sense of the last inequality in the question should be inverted (see my answer for working).
– Parcly Taxel
Jul 20 at 16:55
add a comment |Â
There's no typo. You are using $P=frac12$ but $P>frac12$. This is a binomial distribution where some of the outcomes never happen (If A wins all it's games, the series is shorter).
– David Diaz
Jul 20 at 16:08
1
I think you are to calculate an answer in terms of $P$, not a numerical answer. Your tree is a good way to start.
– Ethan Bolker
Jul 20 at 16:11
if p>1/2, wouldn't P(T) be less than that?
– David
Jul 20 at 16:14
It's often easier to imagine all possible sequences of three games (even though the winner may well be decided before the third game is played). The ultimate winner in all cases is the one that wins at least two of the three. Doing it this way lets you just use the properties of the binomial distribution.
– lulu
Jul 20 at 16:18
@David You're actually right. The sense of the last inequality in the question should be inverted (see my answer for working).
– Parcly Taxel
Jul 20 at 16:55
There's no typo. You are using $P=frac12$ but $P>frac12$. This is a binomial distribution where some of the outcomes never happen (If A wins all it's games, the series is shorter).
– David Diaz
Jul 20 at 16:08
There's no typo. You are using $P=frac12$ but $P>frac12$. This is a binomial distribution where some of the outcomes never happen (If A wins all it's games, the series is shorter).
– David Diaz
Jul 20 at 16:08
1
1
I think you are to calculate an answer in terms of $P$, not a numerical answer. Your tree is a good way to start.
– Ethan Bolker
Jul 20 at 16:11
I think you are to calculate an answer in terms of $P$, not a numerical answer. Your tree is a good way to start.
– Ethan Bolker
Jul 20 at 16:11
if p>1/2, wouldn't P(T) be less than that?
– David
Jul 20 at 16:14
if p>1/2, wouldn't P(T) be less than that?
– David
Jul 20 at 16:14
It's often easier to imagine all possible sequences of three games (even though the winner may well be decided before the third game is played). The ultimate winner in all cases is the one that wins at least two of the three. Doing it this way lets you just use the properties of the binomial distribution.
– lulu
Jul 20 at 16:18
It's often easier to imagine all possible sequences of three games (even though the winner may well be decided before the third game is played). The ultimate winner in all cases is the one that wins at least two of the three. Doing it this way lets you just use the properties of the binomial distribution.
– lulu
Jul 20 at 16:18
@David You're actually right. The sense of the last inequality in the question should be inverted (see my answer for working).
– Parcly Taxel
Jul 20 at 16:55
@David You're actually right. The sense of the last inequality in the question should be inverted (see my answer for working).
– Parcly Taxel
Jul 20 at 16:55
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
This is not a typo. $p$ is a variable and you are supposed to include it in your answer.
There are three sequences of results leading to the home advantage team winning, as you found out: WW ($p(1-p)$), WLW ($p^3$) and LWW ($(1-p)^2p$). Thus
$$P[T]=p(1-p)+p(1-p)^2+p^3=(p-p^2)(2-p)+p^3=2p-3p^2+2p^3=p(2-3p+2p^2)$$
and if $p=frac12+d$ with $d>0$
$$P[T]=pleft(2-3left(frac12+dright)+2left(frac12+dright)^2right)$$
$$=pleft(2d^2-d+1right)$$
$$=pleft(1+d(2d-1)right)<p$$
where the last inequality is because $2d-1<0$ for $0<d<frac12$. Indeed, the inequality is reversed compared to what is printed: conversely, $P[T]>p$ when $p<frac12$.
answered Jul 20 at 16:27
Parcly Taxel
33.6k136588
So d is between 0 and 1/2, and P(T) is less than, not greater than p, because if you multiply any p by what you get in this equation - p(2d^2−d+1) - you get something smaller than p?
– David
Jul 20 at 17:49
@David yes. And that is the real typo in the question.
– Parcly Taxel
Jul 20 at 17:50
add a comment |Â
up vote
1
down vote
I'll add an answer to point out that you can approach the analysis somewhat differently by extending the series to three games even when the same team wins the first two. This is permissible because the third game doesn't change the outcome in those cases; that is, after all, why they then don't usually play the third game. We'll have them play it just to simplify things a bit.
Then the team with home advantage wins its two home games with probability $p^3$ (the home team wins all three games) plus $2p(1-p)^2$ (it splits its home games and wins away). It wins all three games with probability $p^2(1-p)$, for a total of
$$
P(texthome team wins series) = 2p-3p^2+2p^3
$$
To see that this is actually less than $p$, we write
beginalign
2p-3p^2+2p^3-p & = p(2p^3-3p+1) \
& = p(2p-1)(p-1)
endalign
This shows that for the interval $0 leq p leq 1$, the left-hand side is positive when $0 < p < 1/2$, and negative when $1/2 < p < 1$. The latter is our situation, and since $(2p-3p^2+2p^3)-p > 0$, it must be the case that $P(texthome team wins the series) = 2p-3p^2+2p^3 > p$.
answered Jul 20 at 16:59
Brian Tung
25.2k32453
Oops, totally misread the rules. Thanks, I'll delete and edit! :-o
– Brian Tung
Jul 20 at 17:06
Thanks @littleO for pointing out my misinterpretation (I read too quickly, is what really happened). Unfortunately, my original approach won't work, and what remains is not as simple as I'd hoped, but I'll leave the answer here just because it's a little different from Parcly Taxel's, and maybe the difference will be helpful.
– Brian Tung
Jul 20 at 17:19
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
This is not a typo. $p$ is a variable and you are supposed to include it in your answer.
There are three sequences of results leading to the home advantage team winning, as you found out: WW ($p(1-p)$), WLW ($p^3$) and LWW ($(1-p)^2p$). Thus
$$P[T]=p(1-p)+p(1-p)^2+p^3=(p-p^2)(2-p)+p^3=2p-3p^2+2p^3=p(2-3p+2p^2)$$
and if $p=frac12+d$ with $d>0$
$$P[T]=pleft(2-3left(frac12+dright)+2left(frac12+dright)^2right)$$
$$=pleft(2d^2-d+1right)$$
$$=pleft(1+d(2d-1)right)<p$$
where the last inequality is because $2d-1<0$ for $0<d<frac12$. Indeed, the inequality is reversed compared to what is printed: conversely, $P[T]>p$ when $p<frac12$.
answered Jul 20 at 16:27
Parcly Taxel
33.6k136588
So d is between 0 and 1/2, and P(T) is less than, not greater than p, because if you multiply any p by what you get in this equation - p(2d^2−d+1) - you get something smaller than p?
– David
Jul 20 at 17:49
@David yes. And that is the real typo in the question.
– Parcly Taxel
Jul 20 at 17:50
add a comment |Â
up vote
3
down vote
This is not a typo. $p$ is a variable and you are supposed to include it in your answer.
There are three sequences of results leading to the home advantage team winning, as you found out: WW ($p(1-p)$), WLW ($p^3$) and LWW ($(1-p)^2p$). Thus
$$P[T]=p(1-p)+p(1-p)^2+p^3=(p-p^2)(2-p)+p^3=2p-3p^2+2p^3=p(2-3p+2p^2)$$
and if $p=frac12+d$ with $d>0$
$$P[T]=pleft(2-3left(frac12+dright)+2left(frac12+dright)^2right)$$
$$=pleft(2d^2-d+1right)$$
$$=pleft(1+d(2d-1)right)<p$$
where the last inequality is because $2d-1<0$ for $0<d<frac12$. Indeed, the inequality is reversed compared to what is printed: conversely, $P[T]>p$ when $p<frac12$.
answered Jul 20 at 16:27
Parcly Taxel
33.6k136588
So d is between 0 and 1/2, and P(T) is less than, not greater than p, because if you multiply any p by what you get in this equation - p(2d^2−d+1) - you get something smaller than p?
– David
Jul 20 at 17:49
@David yes. And that is the real typo in the question.
– Parcly Taxel
Jul 20 at 17:50
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This is not a typo. $p$ is a variable and you are supposed to include it in your answer.
There are three sequences of results leading to the home advantage team winning, as you found out: WW ($p(1-p)$), WLW ($p^3$) and LWW ($(1-p)^2p$). Thus
$$P[T]=p(1-p)+p(1-p)^2+p^3=(p-p^2)(2-p)+p^3=2p-3p^2+2p^3=p(2-3p+2p^2)$$
and if $p=frac12+d$ with $d>0$
$$P[T]=pleft(2-3left(frac12+dright)+2left(frac12+dright)^2right)$$
$$=pleft(2d^2-d+1right)$$
$$=pleft(1+d(2d-1)right)<p$$
where the last inequality is because $2d-1<0$ for $0<d<frac12$. Indeed, the inequality is reversed compared to what is printed: conversely, $P[T]>p$ when $p<frac12$.
answered Jul 20 at 16:27
Parcly Taxel
33.6k136588
This is not a typo. $p$ is a variable and you are supposed to include it in your answer.
There are three sequences of results leading to the home advantage team winning, as you found out: WW ($p(1-p)$), WLW ($p^3$) and LWW ($(1-p)^2p$). Thus
$$P[T]=p(1-p)+p(1-p)^2+p^3=(p-p^2)(2-p)+p^3=2p-3p^2+2p^3=p(2-3p+2p^2)$$
and if $p=frac12+d$ with $d>0$
$$P[T]=pleft(2-3left(frac12+dright)+2left(frac12+dright)^2right)$$
$$=pleft(2d^2-d+1right)$$
$$=pleft(1+d(2d-1)right)<p$$
where the last inequality is because $2d-1<0$ for $0<d<frac12$. Indeed, the inequality is reversed compared to what is printed: conversely, $P[T]>p$ when $p<frac12$.
answered Jul 20 at 16:27
Parcly Taxel
33.6k136588
edited Jul 20 at 16:53
answered Jul 20 at 16:27
Parcly Taxel
33.6k136588
answered Jul 20 at 16:27
Parcly Taxel
33.6k136588
answered Jul 20 at 16:27
Parcly Taxel
33.6k136588
33.6k136588
So d is between 0 and 1/2, and P(T) is less than, not greater than p, because if you multiply any p by what you get in this equation - p(2d^2−d+1) - you get something smaller than p?
– David
Jul 20 at 17:49
@David yes. And that is the real typo in the question.
– Parcly Taxel
Jul 20 at 17:50
add a comment |Â
So d is between 0 and 1/2, and P(T) is less than, not greater than p, because if you multiply any p by what you get in this equation - p(2d^2−d+1) - you get something smaller than p?
– David
Jul 20 at 17:49
@David yes. And that is the real typo in the question.
– Parcly Taxel
Jul 20 at 17:50
So d is between 0 and 1/2, and P(T) is less than, not greater than p, because if you multiply any p by what you get in this equation - p(2d^2−d+1) - you get something smaller than p?
– David
Jul 20 at 17:49
So d is between 0 and 1/2, and P(T) is less than, not greater than p, because if you multiply any p by what you get in this equation - p(2d^2−d+1) - you get something smaller than p?
– David
Jul 20 at 17:49
@David yes. And that is the real typo in the question.
– Parcly Taxel
Jul 20 at 17:50
@David yes. And that is the real typo in the question.
– Parcly Taxel
Jul 20 at 17:50
add a comment |Â
up vote
1
down vote
I'll add an answer to point out that you can approach the analysis somewhat differently by extending the series to three games even when the same team wins the first two. This is permissible because the third game doesn't change the outcome in those cases; that is, after all, why they then don't usually play the third game. We'll have them play it just to simplify things a bit.
Then the team with home advantage wins its two home games with probability $p^3$ (the home team wins all three games) plus $2p(1-p)^2$ (it splits its home games and wins away). It wins all three games with probability $p^2(1-p)$, for a total of
$$
P(texthome team wins series) = 2p-3p^2+2p^3
$$
To see that this is actually less than $p$, we write
beginalign
2p-3p^2+2p^3-p & = p(2p^3-3p+1) \
& = p(2p-1)(p-1)
endalign
This shows that for the interval $0 leq p leq 1$, the left-hand side is positive when $0 < p < 1/2$, and negative when $1/2 < p < 1$. The latter is our situation, and since $(2p-3p^2+2p^3)-p > 0$, it must be the case that $P(texthome team wins the series) = 2p-3p^2+2p^3 > p$.
answered Jul 20 at 16:59
Brian Tung
25.2k32453
Oops, totally misread the rules. Thanks, I'll delete and edit! :-o
– Brian Tung
Jul 20 at 17:06
Thanks @littleO for pointing out my misinterpretation (I read too quickly, is what really happened). Unfortunately, my original approach won't work, and what remains is not as simple as I'd hoped, but I'll leave the answer here just because it's a little different from Parcly Taxel's, and maybe the difference will be helpful.
– Brian Tung
Jul 20 at 17:19
add a comment |Â
up vote
1
down vote
I'll add an answer to point out that you can approach the analysis somewhat differently by extending the series to three games even when the same team wins the first two. This is permissible because the third game doesn't change the outcome in those cases; that is, after all, why they then don't usually play the third game. We'll have them play it just to simplify things a bit.
Then the team with home advantage wins its two home games with probability $p^3$ (the home team wins all three games) plus $2p(1-p)^2$ (it splits its home games and wins away). It wins all three games with probability $p^2(1-p)$, for a total of
$$
P(texthome team wins series) = 2p-3p^2+2p^3
$$
To see that this is actually less than $p$, we write
beginalign
2p-3p^2+2p^3-p & = p(2p^3-3p+1) \
& = p(2p-1)(p-1)
endalign
This shows that for the interval $0 leq p leq 1$, the left-hand side is positive when $0 < p < 1/2$, and negative when $1/2 < p < 1$. The latter is our situation, and since $(2p-3p^2+2p^3)-p > 0$, it must be the case that $P(texthome team wins the series) = 2p-3p^2+2p^3 > p$.
answered Jul 20 at 16:59
Brian Tung
25.2k32453
Oops, totally misread the rules. Thanks, I'll delete and edit! :-o
– Brian Tung
Jul 20 at 17:06
Thanks @littleO for pointing out my misinterpretation (I read too quickly, is what really happened). Unfortunately, my original approach won't work, and what remains is not as simple as I'd hoped, but I'll leave the answer here just because it's a little different from Parcly Taxel's, and maybe the difference will be helpful.
– Brian Tung
Jul 20 at 17:19
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I'll add an answer to point out that you can approach the analysis somewhat differently by extending the series to three games even when the same team wins the first two. This is permissible because the third game doesn't change the outcome in those cases; that is, after all, why they then don't usually play the third game. We'll have them play it just to simplify things a bit.
Then the team with home advantage wins its two home games with probability $p^3$ (the home team wins all three games) plus $2p(1-p)^2$ (it splits its home games and wins away). It wins all three games with probability $p^2(1-p)$, for a total of
$$
P(texthome team wins series) = 2p-3p^2+2p^3
$$
To see that this is actually less than $p$, we write
beginalign
2p-3p^2+2p^3-p & = p(2p^3-3p+1) \
& = p(2p-1)(p-1)
endalign
This shows that for the interval $0 leq p leq 1$, the left-hand side is positive when $0 < p < 1/2$, and negative when $1/2 < p < 1$. The latter is our situation, and since $(2p-3p^2+2p^3)-p > 0$, it must be the case that $P(texthome team wins the series) = 2p-3p^2+2p^3 > p$.
answered Jul 20 at 16:59
Brian Tung
25.2k32453
I'll add an answer to point out that you can approach the analysis somewhat differently by extending the series to three games even when the same team wins the first two. This is permissible because the third game doesn't change the outcome in those cases; that is, after all, why they then don't usually play the third game. We'll have them play it just to simplify things a bit.
Then the team with home advantage wins its two home games with probability $p^3$ (the home team wins all three games) plus $2p(1-p)^2$ (it splits its home games and wins away). It wins all three games with probability $p^2(1-p)$, for a total of
$$
P(texthome team wins series) = 2p-3p^2+2p^3
$$
To see that this is actually less than $p$, we write
beginalign
2p-3p^2+2p^3-p & = p(2p^3-3p+1) \
& = p(2p-1)(p-1)
endalign
This shows that for the interval $0 leq p leq 1$, the left-hand side is positive when $0 < p < 1/2$, and negative when $1/2 < p < 1$. The latter is our situation, and since $(2p-3p^2+2p^3)-p > 0$, it must be the case that $P(texthome team wins the series) = 2p-3p^2+2p^3 > p$.
answered Jul 20 at 16:59
Brian Tung
25.2k32453
edited Jul 20 at 17:18
answered Jul 20 at 16:59
Brian Tung
25.2k32453
answered Jul 20 at 16:59
Brian Tung
25.2k32453
answered Jul 20 at 16:59
Brian Tung
25.2k32453
25.2k32453
Oops, totally misread the rules. Thanks, I'll delete and edit! :-o
– Brian Tung
Jul 20 at 17:06
Thanks @littleO for pointing out my misinterpretation (I read too quickly, is what really happened). Unfortunately, my original approach won't work, and what remains is not as simple as I'd hoped, but I'll leave the answer here just because it's a little different from Parcly Taxel's, and maybe the difference will be helpful.
– Brian Tung
Jul 20 at 17:19
add a comment |Â
Oops, totally misread the rules. Thanks, I'll delete and edit! :-o
– Brian Tung
Jul 20 at 17:06
Thanks @littleO for pointing out my misinterpretation (I read too quickly, is what really happened). Unfortunately, my original approach won't work, and what remains is not as simple as I'd hoped, but I'll leave the answer here just because it's a little different from Parcly Taxel's, and maybe the difference will be helpful.
– Brian Tung
Jul 20 at 17:19
Oops, totally misread the rules. Thanks, I'll delete and edit! :-o
– Brian Tung
Jul 20 at 17:06
Oops, totally misread the rules. Thanks, I'll delete and edit! :-o
– Brian Tung
Jul 20 at 17:06
Thanks @littleO for pointing out my misinterpretation (I read too quickly, is what really happened). Unfortunately, my original approach won't work, and what remains is not as simple as I'd hoped, but I'll leave the answer here just because it's a little different from Parcly Taxel's, and maybe the difference will be helpful.
– Brian Tung
Jul 20 at 17:19
Thanks @littleO for pointing out my misinterpretation (I read too quickly, is what really happened). Unfortunately, my original approach won't work, and what remains is not as simple as I'd hoped, but I'll leave the answer here just because it's a little different from Parcly Taxel's, and maybe the difference will be helpful.
– Brian Tung
Jul 20 at 17:19
add a comment |Â
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There's no typo. You are using $P=frac12$ but $P>frac12$. This is a binomial distribution where some of the outcomes never happen (If A wins all it's games, the series is shorter).
– David Diaz
Jul 20 at 16:08
1
I think you are to calculate an answer in terms of $P$, not a numerical answer. Your tree is a good way to start.
– Ethan Bolker
Jul 20 at 16:11
if p>1/2, wouldn't P(T) be less than that?
– David
Jul 20 at 16:14
It's often easier to imagine all possible sequences of three games (even though the winner may well be decided before the third game is played). The ultimate winner in all cases is the one that wins at least two of the three. Doing it this way lets you just use the properties of the binomial distribution.
– lulu
Jul 20 at 16:18
@David You're actually right. The sense of the last inequality in the question should be inverted (see my answer for working).
– Parcly Taxel
Jul 20 at 16:55