If a sports team is up $2-1$ in a best of $7$ series, what's the probability that they will win the series?...
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This question already has an answer here:
If a sports team is down by 1-2 in a best of 7 series, what is their chance of winning (if both teams have a 1/2 chance of winning every game)?
2 answers
Assuming that each team has an equal $frac{1}{2}$ probability of winning each game, and the probability of winning each game is independent.
I solved this using a tree diagram, and got the answer of $frac{11}{16}$, but I want to figure out a way of solving this without relying on a tree.
I tried diving this up to cases where the series ends in 2 games, 3 games and 4 games, but am having difficulty.
Any help would be greatly appreciated!
combinatorics statistics permutations
edited Jan 16 at 19:05
Key Flex
8,58571233
asked Jan 16 at 18:19
PanasoPanaso
143
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marked as duplicate by David G. Stork, lulu, Key Flex, N. F. Taussig combinatorics
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Jan 16 at 19:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
If a sports team is down by 1-2 in a best of 7 series, what is their chance of winning (if both teams have a 1/2 chance of winning every game)?
2 answers
Assuming that each team has an equal $frac{1}{2}$ probability of winning each game, and the probability of winning each game is independent.
I solved this using a tree diagram, and got the answer of $frac{11}{16}$, but I want to figure out a way of solving this without relying on a tree.
I tried diving this up to cases where the series ends in 2 games, 3 games and 4 games, but am having difficulty.
Any help would be greatly appreciated!
combinatorics statistics permutations
edited Jan 16 at 19:05
Key Flex
8,58571233
asked Jan 16 at 18:19
PanasoPanaso
143
$endgroup$
marked as duplicate by David G. Stork, lulu, Key Flex, N. F. Taussig combinatorics
Users with the combinatorics badge can single-handedly close combinatorics questions as duplicates and reopen them as needed.
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Jan 16 at 19:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
An alternative to dividing this up into cases where the series ends in $2$ games, $3$ games and $4$ games, is to play all $4$ remaining games (some unnecessarily): this team need to win $2$, $3$ or $4$ of them to win the series
$endgroup$
– Henry
Jan 16 at 18:23
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I checked that thread out, but the solution given uses a tree or assuming all of the remaining 4 games get played out. I'm trying to use cases where not all 4 games necessarily get played out.
$endgroup$
– Panaso
Jan 16 at 18:33
add a comment |
$begingroup$
This question already has an answer here:
If a sports team is down by 1-2 in a best of 7 series, what is their chance of winning (if both teams have a 1/2 chance of winning every game)?
2 answers
Assuming that each team has an equal $frac{1}{2}$ probability of winning each game, and the probability of winning each game is independent.
I solved this using a tree diagram, and got the answer of $frac{11}{16}$, but I want to figure out a way of solving this without relying on a tree.
I tried diving this up to cases where the series ends in 2 games, 3 games and 4 games, but am having difficulty.
Any help would be greatly appreciated!
combinatorics statistics permutations
edited Jan 16 at 19:05
Key Flex
8,58571233
asked Jan 16 at 18:19
PanasoPanaso
143
$endgroup$
This question already has an answer here:
If a sports team is down by 1-2 in a best of 7 series, what is their chance of winning (if both teams have a 1/2 chance of winning every game)?
2 answers
Assuming that each team has an equal $frac{1}{2}$ probability of winning each game, and the probability of winning each game is independent.
I solved this using a tree diagram, and got the answer of $frac{11}{16}$, but I want to figure out a way of solving this without relying on a tree.
I tried diving this up to cases where the series ends in 2 games, 3 games and 4 games, but am having difficulty.
Any help would be greatly appreciated!
This question already has an answer here:
If a sports team is down by 1-2 in a best of 7 series, what is their chance of winning (if both teams have a 1/2 chance of winning every game)?
2 answers
combinatorics statistics permutations
combinatorics statistics permutations
edited Jan 16 at 19:05
Key Flex
8,58571233
asked Jan 16 at 18:19
PanasoPanaso
143
edited Jan 16 at 19:05
Key Flex
8,58571233
asked Jan 16 at 18:19
PanasoPanaso
143
edited Jan 16 at 19:05
Key Flex
8,58571233
edited Jan 16 at 19:05
Key Flex
8,58571233
edited Jan 16 at 19:05
Key Flex
8,58571233
8,58571233
asked Jan 16 at 18:19
PanasoPanaso
143
asked Jan 16 at 18:19
PanasoPanaso
143
asked Jan 16 at 18:19
PanasoPanaso
143
143
marked as duplicate by David G. Stork, lulu, Key Flex, N. F. Taussig combinatorics
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Jan 16 at 19:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by David G. Stork, lulu, Key Flex, N. F. Taussig combinatorics
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Jan 16 at 19:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
An alternative to dividing this up into cases where the series ends in $2$ games, $3$ games and $4$ games, is to play all $4$ remaining games (some unnecessarily): this team need to win $2$, $3$ or $4$ of them to win the series
$endgroup$
– Henry
Jan 16 at 18:23
$begingroup$
I checked that thread out, but the solution given uses a tree or assuming all of the remaining 4 games get played out. I'm trying to use cases where not all 4 games necessarily get played out.
$endgroup$
– Panaso
Jan 16 at 18:33
add a comment |
$begingroup$
An alternative to dividing this up into cases where the series ends in $2$ games, $3$ games and $4$ games, is to play all $4$ remaining games (some unnecessarily): this team need to win $2$, $3$ or $4$ of them to win the series
$endgroup$
– Henry
Jan 16 at 18:23
$begingroup$
I checked that thread out, but the solution given uses a tree or assuming all of the remaining 4 games get played out. I'm trying to use cases where not all 4 games necessarily get played out.
$endgroup$
– Panaso
Jan 16 at 18:33
$begingroup$
An alternative to dividing this up into cases where the series ends in $2$ games, $3$ games and $4$ games, is to play all $4$ remaining games (some unnecessarily): this team need to win $2$, $3$ or $4$ of them to win the series
$endgroup$
– Henry
Jan 16 at 18:23
$begingroup$
An alternative to dividing this up into cases where the series ends in $2$ games, $3$ games and $4$ games, is to play all $4$ remaining games (some unnecessarily): this team need to win $2$, $3$ or $4$ of them to win the series
$endgroup$
– Henry
Jan 16 at 18:23
$begingroup$
I checked that thread out, but the solution given uses a tree or assuming all of the remaining 4 games get played out. I'm trying to use cases where not all 4 games necessarily get played out.
$endgroup$
– Panaso
Jan 16 at 18:33
$begingroup$
I checked that thread out, but the solution given uses a tree or assuming all of the remaining 4 games get played out. I'm trying to use cases where not all 4 games necessarily get played out.
$endgroup$
– Panaso
Jan 16 at 18:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that the winning team necessarily wins the last game of the series. So there are three cases to consider:
- The series ends in $5$ games. This occurs if the current leading team wins games $4$ and $5$, which occurs with probability $frac{1}{4}$
- The series ends in $6$ games. Since they must win game $6$, they also must win exactly one of games $4,5$. The outcomes look like WLW, LWW. There are $2$, and each has probability $frac{1}{8}$ so this outcome occurs with probability $frac{1}{4}$
- The series ends in $7$ games. Since they must win game $7$, they also must win exactly one of games $4,5,6$. The outcomes look like WLLW, LWLW, LLWW. There are $3$ outcomes and each has probability $frac{1}{16}$ so this situation occurs with probability $frac{3}{16}$
Therefore the answer is
$$frac{1}{4}+frac{1}{4}+frac{3}{16}=frac{11}{16}$$
answered Jan 16 at 18:29
pwerthpwerth
3,340417
$endgroup$
$begingroup$
Is there a way to calculate this with using combinations?
$endgroup$
– Panaso
Jan 16 at 19:29
$begingroup$
What do you mean by "combinations"? I simply wrote out all possible outcomes that satisfy the condition in your question.
$endgroup$
– pwerth
Jan 16 at 19:31
$begingroup$
I understand, but I'm required to put in combinations, such as 2C2.
$endgroup$
– Panaso
Jan 16 at 19:49
$begingroup$
I see. That wasn't included in your question so I took a different route. For future reference, if you need a question solved in a specific manner you should include that in your question.
$endgroup$
– pwerth
Jan 16 at 19:52
$begingroup$
Sorry about that, I'm not sure how to get the denominators using combinations. Could you give me a hint?
$endgroup$
– Panaso
Jan 16 at 20:04
|
show 5 more comments
$begingroup$
You can take a Markov chain approach. The possible states that describe the series are of the form $(m,n)$, where $m$ is the number of games won so far by team A and $n$ is the number of games won so far by team B. In your example, the series is in state $(2,1)$.
Let $P(m,n)$ denote the probability that team A will win the series, given it is in state $(m,n)$. Let's consider your specific example of $(2,1)$. Two outcomes are possible starting from this state (each with probability ${1 over 2}$. Team A can win game 4 and the series could go to state $(3,1)$ or Team B could win game 4 and the series could go the state $(2,2)$. You can express the probability relation as:
$$P(2,1) = {1 over 2}P(3,1) + {1 over 2}P(2,2)$$
By the same logic:
$$P(3,1) = {1 over 2}(1) + {1 over 2}P(3,2)$$
$$P(3,2) = {1 over 2}(1) + {1 over 2}P(3,3)$$
$$P(3,3) = {1 over 2}(1) + {1 over 2}(0)$$
$$P(2,2) = {1 over 2}P(3,2) + {1 over 2}P(2,3)$$
$$P(2,3) = {1 over 2}P(3,3) + {1 over 2}(0)$$
You have the necessary equations above to solve for $P(2,1)$ for the probability of winning from any state, for that matter.
Also, you can use matrix notation to express state transitions instead of the tedious equations I wrote above.
answered Jan 16 at 18:35
Aditya DuaAditya Dua
1,15418
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that the winning team necessarily wins the last game of the series. So there are three cases to consider:
- The series ends in $5$ games. This occurs if the current leading team wins games $4$ and $5$, which occurs with probability $frac{1}{4}$
- The series ends in $6$ games. Since they must win game $6$, they also must win exactly one of games $4,5$. The outcomes look like WLW, LWW. There are $2$, and each has probability $frac{1}{8}$ so this outcome occurs with probability $frac{1}{4}$
- The series ends in $7$ games. Since they must win game $7$, they also must win exactly one of games $4,5,6$. The outcomes look like WLLW, LWLW, LLWW. There are $3$ outcomes and each has probability $frac{1}{16}$ so this situation occurs with probability $frac{3}{16}$
Therefore the answer is
$$frac{1}{4}+frac{1}{4}+frac{3}{16}=frac{11}{16}$$
answered Jan 16 at 18:29
pwerthpwerth
3,340417
$endgroup$
$begingroup$
Is there a way to calculate this with using combinations?
$endgroup$
– Panaso
Jan 16 at 19:29
$begingroup$
What do you mean by "combinations"? I simply wrote out all possible outcomes that satisfy the condition in your question.
$endgroup$
– pwerth
Jan 16 at 19:31
$begingroup$
I understand, but I'm required to put in combinations, such as 2C2.
$endgroup$
– Panaso
Jan 16 at 19:49
$begingroup$
I see. That wasn't included in your question so I took a different route. For future reference, if you need a question solved in a specific manner you should include that in your question.
$endgroup$
– pwerth
Jan 16 at 19:52
$begingroup$
Sorry about that, I'm not sure how to get the denominators using combinations. Could you give me a hint?
$endgroup$
– Panaso
Jan 16 at 20:04
|
show 5 more comments
$begingroup$
Note that the winning team necessarily wins the last game of the series. So there are three cases to consider:
- The series ends in $5$ games. This occurs if the current leading team wins games $4$ and $5$, which occurs with probability $frac{1}{4}$
- The series ends in $6$ games. Since they must win game $6$, they also must win exactly one of games $4,5$. The outcomes look like WLW, LWW. There are $2$, and each has probability $frac{1}{8}$ so this outcome occurs with probability $frac{1}{4}$
- The series ends in $7$ games. Since they must win game $7$, they also must win exactly one of games $4,5,6$. The outcomes look like WLLW, LWLW, LLWW. There are $3$ outcomes and each has probability $frac{1}{16}$ so this situation occurs with probability $frac{3}{16}$
Therefore the answer is
$$frac{1}{4}+frac{1}{4}+frac{3}{16}=frac{11}{16}$$
answered Jan 16 at 18:29
pwerthpwerth
3,340417
$endgroup$
$begingroup$
Is there a way to calculate this with using combinations?
$endgroup$
– Panaso
Jan 16 at 19:29
$begingroup$
What do you mean by "combinations"? I simply wrote out all possible outcomes that satisfy the condition in your question.
$endgroup$
– pwerth
Jan 16 at 19:31
$begingroup$
I understand, but I'm required to put in combinations, such as 2C2.
$endgroup$
– Panaso
Jan 16 at 19:49
$begingroup$
I see. That wasn't included in your question so I took a different route. For future reference, if you need a question solved in a specific manner you should include that in your question.
$endgroup$
– pwerth
Jan 16 at 19:52
$begingroup$
Sorry about that, I'm not sure how to get the denominators using combinations. Could you give me a hint?
$endgroup$
– Panaso
Jan 16 at 20:04
|
show 5 more comments
$begingroup$
Note that the winning team necessarily wins the last game of the series. So there are three cases to consider:
- The series ends in $5$ games. This occurs if the current leading team wins games $4$ and $5$, which occurs with probability $frac{1}{4}$
- The series ends in $6$ games. Since they must win game $6$, they also must win exactly one of games $4,5$. The outcomes look like WLW, LWW. There are $2$, and each has probability $frac{1}{8}$ so this outcome occurs with probability $frac{1}{4}$
- The series ends in $7$ games. Since they must win game $7$, they also must win exactly one of games $4,5,6$. The outcomes look like WLLW, LWLW, LLWW. There are $3$ outcomes and each has probability $frac{1}{16}$ so this situation occurs with probability $frac{3}{16}$
Therefore the answer is
$$frac{1}{4}+frac{1}{4}+frac{3}{16}=frac{11}{16}$$
answered Jan 16 at 18:29
pwerthpwerth
3,340417
$endgroup$
Note that the winning team necessarily wins the last game of the series. So there are three cases to consider:
- The series ends in $5$ games. This occurs if the current leading team wins games $4$ and $5$, which occurs with probability $frac{1}{4}$
- The series ends in $6$ games. Since they must win game $6$, they also must win exactly one of games $4,5$. The outcomes look like WLW, LWW. There are $2$, and each has probability $frac{1}{8}$ so this outcome occurs with probability $frac{1}{4}$
- The series ends in $7$ games. Since they must win game $7$, they also must win exactly one of games $4,5,6$. The outcomes look like WLLW, LWLW, LLWW. There are $3$ outcomes and each has probability $frac{1}{16}$ so this situation occurs with probability $frac{3}{16}$
Therefore the answer is
$$frac{1}{4}+frac{1}{4}+frac{3}{16}=frac{11}{16}$$
answered Jan 16 at 18:29
pwerthpwerth
3,340417
answered Jan 16 at 18:29
pwerthpwerth
3,340417
answered Jan 16 at 18:29
pwerthpwerth
3,340417
answered Jan 16 at 18:29
pwerthpwerth
3,340417
3,340417
$begingroup$
Is there a way to calculate this with using combinations?
$endgroup$
– Panaso
Jan 16 at 19:29
$begingroup$
What do you mean by "combinations"? I simply wrote out all possible outcomes that satisfy the condition in your question.
$endgroup$
– pwerth
Jan 16 at 19:31
$begingroup$
I understand, but I'm required to put in combinations, such as 2C2.
$endgroup$
– Panaso
Jan 16 at 19:49
$begingroup$
I see. That wasn't included in your question so I took a different route. For future reference, if you need a question solved in a specific manner you should include that in your question.
$endgroup$
– pwerth
Jan 16 at 19:52
$begingroup$
Sorry about that, I'm not sure how to get the denominators using combinations. Could you give me a hint?
$endgroup$
– Panaso
Jan 16 at 20:04
|
show 5 more comments
$begingroup$
Is there a way to calculate this with using combinations?
$endgroup$
– Panaso
Jan 16 at 19:29
$begingroup$
What do you mean by "combinations"? I simply wrote out all possible outcomes that satisfy the condition in your question.
$endgroup$
– pwerth
Jan 16 at 19:31
$begingroup$
I understand, but I'm required to put in combinations, such as 2C2.
$endgroup$
– Panaso
Jan 16 at 19:49
$begingroup$
I see. That wasn't included in your question so I took a different route. For future reference, if you need a question solved in a specific manner you should include that in your question.
$endgroup$
– pwerth
Jan 16 at 19:52
$begingroup$
Sorry about that, I'm not sure how to get the denominators using combinations. Could you give me a hint?
$endgroup$
– Panaso
Jan 16 at 20:04
$begingroup$
Is there a way to calculate this with using combinations?
$endgroup$
– Panaso
Jan 16 at 19:29
$begingroup$
Is there a way to calculate this with using combinations?
$endgroup$
– Panaso
Jan 16 at 19:29
$begingroup$
What do you mean by "combinations"? I simply wrote out all possible outcomes that satisfy the condition in your question.
$endgroup$
– pwerth
Jan 16 at 19:31
$begingroup$
What do you mean by "combinations"? I simply wrote out all possible outcomes that satisfy the condition in your question.
$endgroup$
– pwerth
Jan 16 at 19:31
$begingroup$
I understand, but I'm required to put in combinations, such as 2C2.
$endgroup$
– Panaso
Jan 16 at 19:49
$begingroup$
I understand, but I'm required to put in combinations, such as 2C2.
$endgroup$
– Panaso
Jan 16 at 19:49
$begingroup$
I see. That wasn't included in your question so I took a different route. For future reference, if you need a question solved in a specific manner you should include that in your question.
$endgroup$
– pwerth
Jan 16 at 19:52
$begingroup$
I see. That wasn't included in your question so I took a different route. For future reference, if you need a question solved in a specific manner you should include that in your question.
$endgroup$
– pwerth
Jan 16 at 19:52
$begingroup$
Sorry about that, I'm not sure how to get the denominators using combinations. Could you give me a hint?
$endgroup$
– Panaso
Jan 16 at 20:04
$begingroup$
Sorry about that, I'm not sure how to get the denominators using combinations. Could you give me a hint?
$endgroup$
– Panaso
Jan 16 at 20:04
|
show 5 more comments
$begingroup$
You can take a Markov chain approach. The possible states that describe the series are of the form $(m,n)$, where $m$ is the number of games won so far by team A and $n$ is the number of games won so far by team B. In your example, the series is in state $(2,1)$.
Let $P(m,n)$ denote the probability that team A will win the series, given it is in state $(m,n)$. Let's consider your specific example of $(2,1)$. Two outcomes are possible starting from this state (each with probability ${1 over 2}$. Team A can win game 4 and the series could go to state $(3,1)$ or Team B could win game 4 and the series could go the state $(2,2)$. You can express the probability relation as:
$$P(2,1) = {1 over 2}P(3,1) + {1 over 2}P(2,2)$$
By the same logic:
$$P(3,1) = {1 over 2}(1) + {1 over 2}P(3,2)$$
$$P(3,2) = {1 over 2}(1) + {1 over 2}P(3,3)$$
$$P(3,3) = {1 over 2}(1) + {1 over 2}(0)$$
$$P(2,2) = {1 over 2}P(3,2) + {1 over 2}P(2,3)$$
$$P(2,3) = {1 over 2}P(3,3) + {1 over 2}(0)$$
You have the necessary equations above to solve for $P(2,1)$ for the probability of winning from any state, for that matter.
Also, you can use matrix notation to express state transitions instead of the tedious equations I wrote above.
answered Jan 16 at 18:35
Aditya DuaAditya Dua
1,15418
$endgroup$
add a comment |
$begingroup$
You can take a Markov chain approach. The possible states that describe the series are of the form $(m,n)$, where $m$ is the number of games won so far by team A and $n$ is the number of games won so far by team B. In your example, the series is in state $(2,1)$.
Let $P(m,n)$ denote the probability that team A will win the series, given it is in state $(m,n)$. Let's consider your specific example of $(2,1)$. Two outcomes are possible starting from this state (each with probability ${1 over 2}$. Team A can win game 4 and the series could go to state $(3,1)$ or Team B could win game 4 and the series could go the state $(2,2)$. You can express the probability relation as:
$$P(2,1) = {1 over 2}P(3,1) + {1 over 2}P(2,2)$$
By the same logic:
$$P(3,1) = {1 over 2}(1) + {1 over 2}P(3,2)$$
$$P(3,2) = {1 over 2}(1) + {1 over 2}P(3,3)$$
$$P(3,3) = {1 over 2}(1) + {1 over 2}(0)$$
$$P(2,2) = {1 over 2}P(3,2) + {1 over 2}P(2,3)$$
$$P(2,3) = {1 over 2}P(3,3) + {1 over 2}(0)$$
You have the necessary equations above to solve for $P(2,1)$ for the probability of winning from any state, for that matter.
Also, you can use matrix notation to express state transitions instead of the tedious equations I wrote above.
answered Jan 16 at 18:35
Aditya DuaAditya Dua
1,15418
$endgroup$
add a comment |
$begingroup$
You can take a Markov chain approach. The possible states that describe the series are of the form $(m,n)$, where $m$ is the number of games won so far by team A and $n$ is the number of games won so far by team B. In your example, the series is in state $(2,1)$.
Let $P(m,n)$ denote the probability that team A will win the series, given it is in state $(m,n)$. Let's consider your specific example of $(2,1)$. Two outcomes are possible starting from this state (each with probability ${1 over 2}$. Team A can win game 4 and the series could go to state $(3,1)$ or Team B could win game 4 and the series could go the state $(2,2)$. You can express the probability relation as:
$$P(2,1) = {1 over 2}P(3,1) + {1 over 2}P(2,2)$$
By the same logic:
$$P(3,1) = {1 over 2}(1) + {1 over 2}P(3,2)$$
$$P(3,2) = {1 over 2}(1) + {1 over 2}P(3,3)$$
$$P(3,3) = {1 over 2}(1) + {1 over 2}(0)$$
$$P(2,2) = {1 over 2}P(3,2) + {1 over 2}P(2,3)$$
$$P(2,3) = {1 over 2}P(3,3) + {1 over 2}(0)$$
You have the necessary equations above to solve for $P(2,1)$ for the probability of winning from any state, for that matter.
Also, you can use matrix notation to express state transitions instead of the tedious equations I wrote above.
answered Jan 16 at 18:35
Aditya DuaAditya Dua
1,15418
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You can take a Markov chain approach. The possible states that describe the series are of the form $(m,n)$, where $m$ is the number of games won so far by team A and $n$ is the number of games won so far by team B. In your example, the series is in state $(2,1)$.
Let $P(m,n)$ denote the probability that team A will win the series, given it is in state $(m,n)$. Let's consider your specific example of $(2,1)$. Two outcomes are possible starting from this state (each with probability ${1 over 2}$. Team A can win game 4 and the series could go to state $(3,1)$ or Team B could win game 4 and the series could go the state $(2,2)$. You can express the probability relation as:
$$P(2,1) = {1 over 2}P(3,1) + {1 over 2}P(2,2)$$
By the same logic:
$$P(3,1) = {1 over 2}(1) + {1 over 2}P(3,2)$$
$$P(3,2) = {1 over 2}(1) + {1 over 2}P(3,3)$$
$$P(3,3) = {1 over 2}(1) + {1 over 2}(0)$$
$$P(2,2) = {1 over 2}P(3,2) + {1 over 2}P(2,3)$$
$$P(2,3) = {1 over 2}P(3,3) + {1 over 2}(0)$$
You have the necessary equations above to solve for $P(2,1)$ for the probability of winning from any state, for that matter.
Also, you can use matrix notation to express state transitions instead of the tedious equations I wrote above.
answered Jan 16 at 18:35
Aditya DuaAditya Dua
1,15418
answered Jan 16 at 18:35
Aditya DuaAditya Dua
1,15418
answered Jan 16 at 18:35
Aditya DuaAditya Dua
1,15418
answered Jan 16 at 18:35
Aditya DuaAditya Dua
1,15418
1,15418
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An alternative to dividing this up into cases where the series ends in $2$ games, $3$ games and $4$ games, is to play all $4$ remaining games (some unnecessarily): this team need to win $2$, $3$ or $4$ of them to win the series
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– Henry
Jan 16 at 18:23
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I checked that thread out, but the solution given uses a tree or assuming all of the remaining 4 games get played out. I'm trying to use cases where not all 4 games necessarily get played out.
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– Panaso
Jan 16 at 18:33