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?










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  • Hi. I'm assuming that either $A$ or $B$ gets the point, so $a+b=1?$
    – 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.
    – 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?
    – 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
    – user1022959
    Aug 30 at 14:48










  • @user1022959 thanks for letting me know!
    – HJ_beginner
    Aug 30 at 17:26














up vote
0
down vote

favorite
1












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?










share|cite|improve this question





















  • Hi. I'm assuming that either $A$ or $B$ gets the point, so $a+b=1?$
    – 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.
    – 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?
    – 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
    – user1022959
    Aug 30 at 14:48










  • @user1022959 thanks for letting me know!
    – HJ_beginner
    Aug 30 at 17:26












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





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?










share|cite|improve this question













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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asked Aug 30 at 4:21









user1022959

31




31











  • Hi. I'm assuming that either $A$ or $B$ gets the point, so $a+b=1?$
    – 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.
    – 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?
    – 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
    – user1022959
    Aug 30 at 14:48










  • @user1022959 thanks for letting me know!
    – HJ_beginner
    Aug 30 at 17:26
















  • Hi. I'm assuming that either $A$ or $B$ gets the point, so $a+b=1?$
    – 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.
    – 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?
    – 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
    – user1022959
    Aug 30 at 14:48










  • @user1022959 thanks for letting me know!
    – HJ_beginner
    Aug 30 at 17:26















Hi. I'm assuming that either $A$ or $B$ gets the point, so $a+b=1?$
– Jason Kim
Aug 30 at 4:36





Hi. I'm assuming that either $A$ or $B$ gets the point, so $a+b=1?$
– 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.
– 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.
– 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?
– 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?
– 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
– 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
– user1022959
Aug 30 at 14:48












@user1022959 thanks for letting me know!
– HJ_beginner
Aug 30 at 17:26




@user1022959 thanks for letting me know!
– HJ_beginner
Aug 30 at 17:26










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
$$p|_O=acdot p|_A+bcdot p|_B=a^2(1+b)over 1-ab .$$






share|cite|improve this answer





























    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).






    share|cite|improve this answer



























      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 $abb$.






      share|cite|improve this answer




















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        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
        $$p|_O=acdot p|_A+bcdot p|_B=a^2(1+b)over 1-ab .$$






        share|cite|improve this answer


























          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
          $$p|_O=acdot p|_A+bcdot p|_B=a^2(1+b)over 1-ab .$$






          share|cite|improve this answer
























            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
            $$p|_O=acdot p|_A+bcdot p|_B=a^2(1+b)over 1-ab .$$






            share|cite|improve this answer














            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
            $$p|_O=acdot p|_A+bcdot p|_B=a^2(1+b)over 1-ab .$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 31 at 20:25

























            answered Aug 30 at 7:50









            Christian Blatter

            165k7109311




            165k7109311




















                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).






                share|cite|improve this answer
























                  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).






                  share|cite|improve this answer






















                    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).






                    share|cite|improve this answer












                    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).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Aug 30 at 5:05









                    Jason Kim

                    53016




                    53016




















                        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 $abb$.






                        share|cite|improve this answer
























                          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 $abb$.






                          share|cite|improve this answer






















                            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 $abb$.






                            share|cite|improve this answer












                            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 $abb$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 30 at 5:28









                            mathmandan

                            1,4251712




                            1,4251712



























                                 

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