Series and probability Games
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Suppose we have Series 1 and Series 2, and four teams $A,B,C,D$. $A$ and $B$ play in Series 1 and $C$ and $D$ play in Series 2. In each series, the first team to win 3 games wins the series. The home team has a $3/4$ chance of winning in each game. The winner of each series plays each other in a similar format. What is the probability of $A$ defeating $D$ in the final series $3-2$?
The home team is based on:
Game # Home Team
Game 1 A/C
Game 2 A/C
Game 3 B/D
Game 4(if necessary) B/D
Game 5(if necessary) A/C
Can the probability be decomposed to something like: $$P(text{A defeats D 3-2}) = P(text{A defeats D 3-2}|text{A wins Series 1 and D wins Series 2})$$
Added In the $A/C$ notation, Game 1 of Series 1 is played in A's hometown and Game 1 of Series 2 is played in C's hometown, etc.
Games 1,2, and 5 will be played in the Series 1 winner's hometown, Games 3 and 4 will be played in Series 2 winner hometown.
probability
asked Jan 16 at 20:14
DamienDamien
2,07332032
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|
show 3 more comments
$begingroup$
Suppose we have Series 1 and Series 2, and four teams $A,B,C,D$. $A$ and $B$ play in Series 1 and $C$ and $D$ play in Series 2. In each series, the first team to win 3 games wins the series. The home team has a $3/4$ chance of winning in each game. The winner of each series plays each other in a similar format. What is the probability of $A$ defeating $D$ in the final series $3-2$?
The home team is based on:
Game # Home Team
Game 1 A/C
Game 2 A/C
Game 3 B/D
Game 4(if necessary) B/D
Game 5(if necessary) A/C
Can the probability be decomposed to something like: $$P(text{A defeats D 3-2}) = P(text{A defeats D 3-2}|text{A wins Series 1 and D wins Series 2})$$
Added In the $A/C$ notation, Game 1 of Series 1 is played in A's hometown and Game 1 of Series 2 is played in C's hometown, etc.
Games 1,2, and 5 will be played in the Series 1 winner's hometown, Games 3 and 4 will be played in Series 2 winner hometown.
probability
asked Jan 16 at 20:14
DamienDamien
2,07332032
$endgroup$
$begingroup$
Your notation for the Home Team isn't clear. In, say, the $A/C$ series...is $A$ always the home team? Does it alternate? In the final series between the two winners...how does home team advantage sort out?
$endgroup$
– lulu
Jan 16 at 20:18
$begingroup$
To your closing question: No. What you want isn't that conditional probability since the answer you want depends (heavily) on the probability that $A$ and $D$ win their respective series. You have to multiply your conditional by the probability that both $A,D$ win.
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– lulu
Jan 16 at 20:21
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@lulu: Games 1,2, and 5 will be played in the Series 1 winner's hometown, Games 3 and 4 will be played in Series 2 winner hometown.In the $A/C$ notation, Game 1 of Series 1 is played in A's hometown and Game 1 of Series 2 is played in C's hometown, etc..
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– Damien
Jan 16 at 20:30
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Regardless of the confusion: because of the home team advantage, it's probably easiest to simply write out all the cases. It's tedious, but not at all difficult.
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– lulu
Jan 16 at 20:33
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@lulu: See my addition.
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– Damien
Jan 16 at 20:34
|
show 3 more comments
$begingroup$
Suppose we have Series 1 and Series 2, and four teams $A,B,C,D$. $A$ and $B$ play in Series 1 and $C$ and $D$ play in Series 2. In each series, the first team to win 3 games wins the series. The home team has a $3/4$ chance of winning in each game. The winner of each series plays each other in a similar format. What is the probability of $A$ defeating $D$ in the final series $3-2$?
The home team is based on:
Game # Home Team
Game 1 A/C
Game 2 A/C
Game 3 B/D
Game 4(if necessary) B/D
Game 5(if necessary) A/C
Can the probability be decomposed to something like: $$P(text{A defeats D 3-2}) = P(text{A defeats D 3-2}|text{A wins Series 1 and D wins Series 2})$$
Added In the $A/C$ notation, Game 1 of Series 1 is played in A's hometown and Game 1 of Series 2 is played in C's hometown, etc.
Games 1,2, and 5 will be played in the Series 1 winner's hometown, Games 3 and 4 will be played in Series 2 winner hometown.
probability
asked Jan 16 at 20:14
DamienDamien
2,07332032
$endgroup$
Suppose we have Series 1 and Series 2, and four teams $A,B,C,D$. $A$ and $B$ play in Series 1 and $C$ and $D$ play in Series 2. In each series, the first team to win 3 games wins the series. The home team has a $3/4$ chance of winning in each game. The winner of each series plays each other in a similar format. What is the probability of $A$ defeating $D$ in the final series $3-2$?
The home team is based on:
Game # Home Team
Game 1 A/C
Game 2 A/C
Game 3 B/D
Game 4(if necessary) B/D
Game 5(if necessary) A/C
Can the probability be decomposed to something like: $$P(text{A defeats D 3-2}) = P(text{A defeats D 3-2}|text{A wins Series 1 and D wins Series 2})$$
Added In the $A/C$ notation, Game 1 of Series 1 is played in A's hometown and Game 1 of Series 2 is played in C's hometown, etc.
Games 1,2, and 5 will be played in the Series 1 winner's hometown, Games 3 and 4 will be played in Series 2 winner hometown.
probability
probability
asked Jan 16 at 20:14
DamienDamien
2,07332032
asked Jan 16 at 20:14
DamienDamien
2,07332032
edited Jan 16 at 20:33
Damien
asked Jan 16 at 20:14
DamienDamien
2,07332032
asked Jan 16 at 20:14
DamienDamien
2,07332032
asked Jan 16 at 20:14
DamienDamien
2,07332032
2,07332032
$begingroup$
Your notation for the Home Team isn't clear. In, say, the $A/C$ series...is $A$ always the home team? Does it alternate? In the final series between the two winners...how does home team advantage sort out?
$endgroup$
– lulu
Jan 16 at 20:18
$begingroup$
To your closing question: No. What you want isn't that conditional probability since the answer you want depends (heavily) on the probability that $A$ and $D$ win their respective series. You have to multiply your conditional by the probability that both $A,D$ win.
$endgroup$
– lulu
Jan 16 at 20:21
$begingroup$
@lulu: Games 1,2, and 5 will be played in the Series 1 winner's hometown, Games 3 and 4 will be played in Series 2 winner hometown.In the $A/C$ notation, Game 1 of Series 1 is played in A's hometown and Game 1 of Series 2 is played in C's hometown, etc..
$endgroup$
– Damien
Jan 16 at 20:30
$begingroup$
Regardless of the confusion: because of the home team advantage, it's probably easiest to simply write out all the cases. It's tedious, but not at all difficult.
$endgroup$
– lulu
Jan 16 at 20:33
$begingroup$
@lulu: See my addition.
$endgroup$
– Damien
Jan 16 at 20:34
|
show 3 more comments
$begingroup$
Your notation for the Home Team isn't clear. In, say, the $A/C$ series...is $A$ always the home team? Does it alternate? In the final series between the two winners...how does home team advantage sort out?
$endgroup$
– lulu
Jan 16 at 20:18
$begingroup$
To your closing question: No. What you want isn't that conditional probability since the answer you want depends (heavily) on the probability that $A$ and $D$ win their respective series. You have to multiply your conditional by the probability that both $A,D$ win.
$endgroup$
– lulu
Jan 16 at 20:21
$begingroup$
@lulu: Games 1,2, and 5 will be played in the Series 1 winner's hometown, Games 3 and 4 will be played in Series 2 winner hometown.In the $A/C$ notation, Game 1 of Series 1 is played in A's hometown and Game 1 of Series 2 is played in C's hometown, etc..
$endgroup$
– Damien
Jan 16 at 20:30
$begingroup$
Regardless of the confusion: because of the home team advantage, it's probably easiest to simply write out all the cases. It's tedious, but not at all difficult.
$endgroup$
– lulu
Jan 16 at 20:33
$begingroup$
@lulu: See my addition.
$endgroup$
– Damien
Jan 16 at 20:34
$begingroup$
Your notation for the Home Team isn't clear. In, say, the $A/C$ series...is $A$ always the home team? Does it alternate? In the final series between the two winners...how does home team advantage sort out?
$endgroup$
– lulu
Jan 16 at 20:18
$begingroup$
Your notation for the Home Team isn't clear. In, say, the $A/C$ series...is $A$ always the home team? Does it alternate? In the final series between the two winners...how does home team advantage sort out?
$endgroup$
– lulu
Jan 16 at 20:18
$begingroup$
To your closing question: No. What you want isn't that conditional probability since the answer you want depends (heavily) on the probability that $A$ and $D$ win their respective series. You have to multiply your conditional by the probability that both $A,D$ win.
$endgroup$
– lulu
Jan 16 at 20:21
$begingroup$
To your closing question: No. What you want isn't that conditional probability since the answer you want depends (heavily) on the probability that $A$ and $D$ win their respective series. You have to multiply your conditional by the probability that both $A,D$ win.
$endgroup$
– lulu
Jan 16 at 20:21
$begingroup$
@lulu: Games 1,2, and 5 will be played in the Series 1 winner's hometown, Games 3 and 4 will be played in Series 2 winner hometown.In the $A/C$ notation, Game 1 of Series 1 is played in A's hometown and Game 1 of Series 2 is played in C's hometown, etc..
$endgroup$
– Damien
Jan 16 at 20:30
$begingroup$
@lulu: Games 1,2, and 5 will be played in the Series 1 winner's hometown, Games 3 and 4 will be played in Series 2 winner hometown.In the $A/C$ notation, Game 1 of Series 1 is played in A's hometown and Game 1 of Series 2 is played in C's hometown, etc..
$endgroup$
– Damien
Jan 16 at 20:30
$begingroup$
Regardless of the confusion: because of the home team advantage, it's probably easiest to simply write out all the cases. It's tedious, but not at all difficult.
$endgroup$
– lulu
Jan 16 at 20:33
$begingroup$
Regardless of the confusion: because of the home team advantage, it's probably easiest to simply write out all the cases. It's tedious, but not at all difficult.
$endgroup$
– lulu
Jan 16 at 20:33
$begingroup$
@lulu: See my addition.
$endgroup$
– Damien
Jan 16 at 20:34
$begingroup$
@lulu: See my addition.
$endgroup$
– Damien
Jan 16 at 20:34
|
show 3 more comments
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$begingroup$
Your notation for the Home Team isn't clear. In, say, the $A/C$ series...is $A$ always the home team? Does it alternate? In the final series between the two winners...how does home team advantage sort out?
$endgroup$
– lulu
Jan 16 at 20:18
$begingroup$
To your closing question: No. What you want isn't that conditional probability since the answer you want depends (heavily) on the probability that $A$ and $D$ win their respective series. You have to multiply your conditional by the probability that both $A,D$ win.
$endgroup$
– lulu
Jan 16 at 20:21
$begingroup$
@lulu: Games 1,2, and 5 will be played in the Series 1 winner's hometown, Games 3 and 4 will be played in Series 2 winner hometown.In the $A/C$ notation, Game 1 of Series 1 is played in A's hometown and Game 1 of Series 2 is played in C's hometown, etc..
$endgroup$
– Damien
Jan 16 at 20:30
$begingroup$
Regardless of the confusion: because of the home team advantage, it's probably easiest to simply write out all the cases. It's tedious, but not at all difficult.
$endgroup$
– lulu
Jan 16 at 20:33
$begingroup$
@lulu: See my addition.
$endgroup$
– Damien
Jan 16 at 20:34