A and B play until one scores 2 points in a row.
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A and B play until one scores 2 points in a row, who will win. Probability of A or B scoring a point is $a$ or $b$, respectively. What is the probability that A wins?
I know the correct answer is $fraca^21-2ab$, based on this.
I was wondering what is wrong with the following approach. We form the following tree. The green/red rectangles are where A/B has won.
Tree showing possible sequences leading to the winning of A.
We can see the following pattern for the top and bottom rows (green boxes where A has won):
Top: $a^2, a^3b, a^4b^2, ldots$
Bottom: $a^2b, a^3b^2, a^4b^3, ldots$
When we add the probabilities, both top and bottom rows are geometric series with a factor of $ab$. So,
Sum of the top row: $fraca^21-ab$
Sum of the bottom row: $fraca^2b1-ab$
Total probability: $fraca^2(1+b)1-ab$
Questions: This not equal to $fraca^21-2ab$. What am I missing?
probability sequences-and-series
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A and B play until one scores 2 points in a row, who will win. Probability of A or B scoring a point is $a$ or $b$, respectively. What is the probability that A wins?
I know the correct answer is $fraca^21-2ab$, based on this.
I was wondering what is wrong with the following approach. We form the following tree. The green/red rectangles are where A/B has won.
Tree showing possible sequences leading to the winning of A.
We can see the following pattern for the top and bottom rows (green boxes where A has won):
Top: $a^2, a^3b, a^4b^2, ldots$
Bottom: $a^2b, a^3b^2, a^4b^3, ldots$
When we add the probabilities, both top and bottom rows are geometric series with a factor of $ab$. So,
Sum of the top row: $fraca^21-ab$
Sum of the bottom row: $fraca^2b1-ab$
Total probability: $fraca^2(1+b)1-ab$
Questions: This not equal to $fraca^21-2ab$. What am I missing?
probability sequences-and-series
Hi. I'm assuming that either $A$ or $B$ gets the point, so $a+b=1?$
â Jason Kim
Aug 30 at 4:36
I think the question you linked (ie "based on this") is not the exact same the way you worded it. In the linked question, it's about accumulating 2 more points than the other. For example, if the outcome of the first 3 trials is ABB the score is 1 to 2 and the game is not over but the way you wrote it in this question B just won.
â HJ_beginner
Aug 30 at 5:30
Also just curious OP but how did you make that tree? Was it some program or just like MS Paint?
â HJ_beginner
Aug 30 at 5:33
1
@HJ_beginner: Thanks a lot. Now I guess I understand. I used the following to make the tree: smartdraw.com
â user1022959
Aug 30 at 14:48
@user1022959 thanks for letting me know!
â HJ_beginner
Aug 30 at 17:26
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A and B play until one scores 2 points in a row, who will win. Probability of A or B scoring a point is $a$ or $b$, respectively. What is the probability that A wins?
I know the correct answer is $fraca^21-2ab$, based on this.
I was wondering what is wrong with the following approach. We form the following tree. The green/red rectangles are where A/B has won.
Tree showing possible sequences leading to the winning of A.
We can see the following pattern for the top and bottom rows (green boxes where A has won):
Top: $a^2, a^3b, a^4b^2, ldots$
Bottom: $a^2b, a^3b^2, a^4b^3, ldots$
When we add the probabilities, both top and bottom rows are geometric series with a factor of $ab$. So,
Sum of the top row: $fraca^21-ab$
Sum of the bottom row: $fraca^2b1-ab$
Total probability: $fraca^2(1+b)1-ab$
Questions: This not equal to $fraca^21-2ab$. What am I missing?
probability sequences-and-series
A and B play until one scores 2 points in a row, who will win. Probability of A or B scoring a point is $a$ or $b$, respectively. What is the probability that A wins?
I know the correct answer is $fraca^21-2ab$, based on this.
I was wondering what is wrong with the following approach. We form the following tree. The green/red rectangles are where A/B has won.
Tree showing possible sequences leading to the winning of A.
We can see the following pattern for the top and bottom rows (green boxes where A has won):
Top: $a^2, a^3b, a^4b^2, ldots$
Bottom: $a^2b, a^3b^2, a^4b^3, ldots$
When we add the probabilities, both top and bottom rows are geometric series with a factor of $ab$. So,
Sum of the top row: $fraca^21-ab$
Sum of the bottom row: $fraca^2b1-ab$
Total probability: $fraca^2(1+b)1-ab$
Questions: This not equal to $fraca^21-2ab$. What am I missing?
probability sequences-and-series
probability sequences-and-series
asked Aug 30 at 4:21
user1022959
31
31
Hi. I'm assuming that either $A$ or $B$ gets the point, so $a+b=1?$
â Jason Kim
Aug 30 at 4:36
I think the question you linked (ie "based on this") is not the exact same the way you worded it. In the linked question, it's about accumulating 2 more points than the other. For example, if the outcome of the first 3 trials is ABB the score is 1 to 2 and the game is not over but the way you wrote it in this question B just won.
â HJ_beginner
Aug 30 at 5:30
Also just curious OP but how did you make that tree? Was it some program or just like MS Paint?
â HJ_beginner
Aug 30 at 5:33
1
@HJ_beginner: Thanks a lot. Now I guess I understand. I used the following to make the tree: smartdraw.com
â user1022959
Aug 30 at 14:48
@user1022959 thanks for letting me know!
â HJ_beginner
Aug 30 at 17:26
add a comment |Â
Hi. I'm assuming that either $A$ or $B$ gets the point, so $a+b=1?$
â Jason Kim
Aug 30 at 4:36
I think the question you linked (ie "based on this") is not the exact same the way you worded it. In the linked question, it's about accumulating 2 more points than the other. For example, if the outcome of the first 3 trials is ABB the score is 1 to 2 and the game is not over but the way you wrote it in this question B just won.
â HJ_beginner
Aug 30 at 5:30
Also just curious OP but how did you make that tree? Was it some program or just like MS Paint?
â HJ_beginner
Aug 30 at 5:33
1
@HJ_beginner: Thanks a lot. Now I guess I understand. I used the following to make the tree: smartdraw.com
â user1022959
Aug 30 at 14:48
@user1022959 thanks for letting me know!
â HJ_beginner
Aug 30 at 17:26
Hi. I'm assuming that either $A$ or $B$ gets the point, so $a+b=1?$
â Jason Kim
Aug 30 at 4:36
Hi. I'm assuming that either $A$ or $B$ gets the point, so $a+b=1?$
â Jason Kim
Aug 30 at 4:36
I think the question you linked (ie "based on this") is not the exact same the way you worded it. In the linked question, it's about accumulating 2 more points than the other. For example, if the outcome of the first 3 trials is ABB the score is 1 to 2 and the game is not over but the way you wrote it in this question B just won.
â HJ_beginner
Aug 30 at 5:30
I think the question you linked (ie "based on this") is not the exact same the way you worded it. In the linked question, it's about accumulating 2 more points than the other. For example, if the outcome of the first 3 trials is ABB the score is 1 to 2 and the game is not over but the way you wrote it in this question B just won.
â HJ_beginner
Aug 30 at 5:30
Also just curious OP but how did you make that tree? Was it some program or just like MS Paint?
â HJ_beginner
Aug 30 at 5:33
Also just curious OP but how did you make that tree? Was it some program or just like MS Paint?
â HJ_beginner
Aug 30 at 5:33
1
1
@HJ_beginner: Thanks a lot. Now I guess I understand. I used the following to make the tree: smartdraw.com
â user1022959
Aug 30 at 14:48
@HJ_beginner: Thanks a lot. Now I guess I understand. I used the following to make the tree: smartdraw.com
â user1022959
Aug 30 at 14:48
@user1022959 thanks for letting me know!
â HJ_beginner
Aug 30 at 17:26
@user1022959 thanks for letting me know!
â HJ_beginner
Aug 30 at 17:26
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
0
down vote
accepted
Your game is not the same as the game in the linked question.
I did not look at your infinite tree. Anyway: The result is correct.
Apart from the initial state $O$ my graph (it's not a tree) has just two nonterminal states: last game won by $A$, and last game won by $B$. Denote by $p|_A$, $p|_B$, and $p|_O$ the probabilities that $A$ wins when we are in state $A$, $B$, and $O$ respectively. Then we have the equations
$$p|_A=acdot 1+ bcdot p|_B>,qquad p|_B= a cdot p|_A+ bcdot 0 ,$$
resulting in
$$p|_A=aover 1-ab,qquad p|_B=a^2over 1-ab .$$
It follows that
$$p|_O=acdot p|_A+bcdot p|_B=a^2(1+b)over 1-ab .$$
add a comment |Â
up vote
0
down vote
Let $P(A)=textProbability of A winning, P(B)=textProbability of B winning.$
We are actually supposed to find $dfractextSuccess ProbabilitytextSuccess Probability + Failure Probability,$ (we are assuming that this game will end) so we need to find the failure probability, as well as the success probability.
This is true because it will either have $A$ win or $B$ win at some point, so the solution has a chance of $fractextP(A)textP(Acup B)=fractextP(A)textP(A)+textP(B)$ since both are mutually exclusive ($A$ and $B$ cannot both win).
add a comment |Â
up vote
0
down vote
The question you linked is different from the question you have asked.
In the linked question, they play until one player has two more points than the other. In your question, they play until one player has won two points in a row.
This is not the same game--for example, consider the sequence $abb$.
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Your game is not the same as the game in the linked question.
I did not look at your infinite tree. Anyway: The result is correct.
Apart from the initial state $O$ my graph (it's not a tree) has just two nonterminal states: last game won by $A$, and last game won by $B$. Denote by $p|_A$, $p|_B$, and $p|_O$ the probabilities that $A$ wins when we are in state $A$, $B$, and $O$ respectively. Then we have the equations
$$p|_A=acdot 1+ bcdot p|_B>,qquad p|_B= a cdot p|_A+ bcdot 0 ,$$
resulting in
$$p|_A=aover 1-ab,qquad p|_B=a^2over 1-ab .$$
It follows that
$$p|_O=acdot p|_A+bcdot p|_B=a^2(1+b)over 1-ab .$$
add a comment |Â
up vote
0
down vote
accepted
Your game is not the same as the game in the linked question.
I did not look at your infinite tree. Anyway: The result is correct.
Apart from the initial state $O$ my graph (it's not a tree) has just two nonterminal states: last game won by $A$, and last game won by $B$. Denote by $p|_A$, $p|_B$, and $p|_O$ the probabilities that $A$ wins when we are in state $A$, $B$, and $O$ respectively. Then we have the equations
$$p|_A=acdot 1+ bcdot p|_B>,qquad p|_B= a cdot p|_A+ bcdot 0 ,$$
resulting in
$$p|_A=aover 1-ab,qquad p|_B=a^2over 1-ab .$$
It follows that
$$p|_O=acdot p|_A+bcdot p|_B=a^2(1+b)over 1-ab .$$
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Your game is not the same as the game in the linked question.
I did not look at your infinite tree. Anyway: The result is correct.
Apart from the initial state $O$ my graph (it's not a tree) has just two nonterminal states: last game won by $A$, and last game won by $B$. Denote by $p|_A$, $p|_B$, and $p|_O$ the probabilities that $A$ wins when we are in state $A$, $B$, and $O$ respectively. Then we have the equations
$$p|_A=acdot 1+ bcdot p|_B>,qquad p|_B= a cdot p|_A+ bcdot 0 ,$$
resulting in
$$p|_A=aover 1-ab,qquad p|_B=a^2over 1-ab .$$
It follows that
$$p|_O=acdot p|_A+bcdot p|_B=a^2(1+b)over 1-ab .$$
Your game is not the same as the game in the linked question.
I did not look at your infinite tree. Anyway: The result is correct.
Apart from the initial state $O$ my graph (it's not a tree) has just two nonterminal states: last game won by $A$, and last game won by $B$. Denote by $p|_A$, $p|_B$, and $p|_O$ the probabilities that $A$ wins when we are in state $A$, $B$, and $O$ respectively. Then we have the equations
$$p|_A=acdot 1+ bcdot p|_B>,qquad p|_B= a cdot p|_A+ bcdot 0 ,$$
resulting in
$$p|_A=aover 1-ab,qquad p|_B=a^2over 1-ab .$$
It follows that
$$p|_O=acdot p|_A+bcdot p|_B=a^2(1+b)over 1-ab .$$
edited Aug 31 at 20:25
answered Aug 30 at 7:50
Christian Blatter
165k7109311
165k7109311
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $P(A)=textProbability of A winning, P(B)=textProbability of B winning.$
We are actually supposed to find $dfractextSuccess ProbabilitytextSuccess Probability + Failure Probability,$ (we are assuming that this game will end) so we need to find the failure probability, as well as the success probability.
This is true because it will either have $A$ win or $B$ win at some point, so the solution has a chance of $fractextP(A)textP(Acup B)=fractextP(A)textP(A)+textP(B)$ since both are mutually exclusive ($A$ and $B$ cannot both win).
add a comment |Â
up vote
0
down vote
Let $P(A)=textProbability of A winning, P(B)=textProbability of B winning.$
We are actually supposed to find $dfractextSuccess ProbabilitytextSuccess Probability + Failure Probability,$ (we are assuming that this game will end) so we need to find the failure probability, as well as the success probability.
This is true because it will either have $A$ win or $B$ win at some point, so the solution has a chance of $fractextP(A)textP(Acup B)=fractextP(A)textP(A)+textP(B)$ since both are mutually exclusive ($A$ and $B$ cannot both win).
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $P(A)=textProbability of A winning, P(B)=textProbability of B winning.$
We are actually supposed to find $dfractextSuccess ProbabilitytextSuccess Probability + Failure Probability,$ (we are assuming that this game will end) so we need to find the failure probability, as well as the success probability.
This is true because it will either have $A$ win or $B$ win at some point, so the solution has a chance of $fractextP(A)textP(Acup B)=fractextP(A)textP(A)+textP(B)$ since both are mutually exclusive ($A$ and $B$ cannot both win).
Let $P(A)=textProbability of A winning, P(B)=textProbability of B winning.$
We are actually supposed to find $dfractextSuccess ProbabilitytextSuccess Probability + Failure Probability,$ (we are assuming that this game will end) so we need to find the failure probability, as well as the success probability.
This is true because it will either have $A$ win or $B$ win at some point, so the solution has a chance of $fractextP(A)textP(Acup B)=fractextP(A)textP(A)+textP(B)$ since both are mutually exclusive ($A$ and $B$ cannot both win).
answered Aug 30 at 5:05
Jason Kim
53016
53016
add a comment |Â
add a comment |Â
up vote
0
down vote
The question you linked is different from the question you have asked.
In the linked question, they play until one player has two more points than the other. In your question, they play until one player has won two points in a row.
This is not the same game--for example, consider the sequence $abb$.
add a comment |Â
up vote
0
down vote
The question you linked is different from the question you have asked.
In the linked question, they play until one player has two more points than the other. In your question, they play until one player has won two points in a row.
This is not the same game--for example, consider the sequence $abb$.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The question you linked is different from the question you have asked.
In the linked question, they play until one player has two more points than the other. In your question, they play until one player has won two points in a row.
This is not the same game--for example, consider the sequence $abb$.
The question you linked is different from the question you have asked.
In the linked question, they play until one player has two more points than the other. In your question, they play until one player has won two points in a row.
This is not the same game--for example, consider the sequence $abb$.
answered Aug 30 at 5:28
mathmandan
1,4251712
1,4251712
add a comment |Â
add a comment |Â
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Hi. I'm assuming that either $A$ or $B$ gets the point, so $a+b=1?$
â Jason Kim
Aug 30 at 4:36
I think the question you linked (ie "based on this") is not the exact same the way you worded it. In the linked question, it's about accumulating 2 more points than the other. For example, if the outcome of the first 3 trials is ABB the score is 1 to 2 and the game is not over but the way you wrote it in this question B just won.
â HJ_beginner
Aug 30 at 5:30
Also just curious OP but how did you make that tree? Was it some program or just like MS Paint?
â HJ_beginner
Aug 30 at 5:33
1
@HJ_beginner: Thanks a lot. Now I guess I understand. I used the following to make the tree: smartdraw.com
â user1022959
Aug 30 at 14:48
@user1022959 thanks for letting me know!
â HJ_beginner
Aug 30 at 17:26