Difficult Coin Problem












4












$begingroup$



Two players, A and B, alternately and independently flip a coin and
the first player to obtain a head wins. Player A flips first. What is
the probability that A wins?



Official answer: 2/3, but I cannot arrive at it.




Thought process: Find the probability that a head comes on the nth trial. That's easy to do, it's just a Geometric random variable. Then find the probability that nth turn is player's A turn. Finally, multiply both probabilities.



When I came up with each probability, both of them depended on the amount of trials n, so my answer was a non-constant function of n.



However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.






probability







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$endgroup$












  • $begingroup$
    What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
    $endgroup$
    – Pakk
    9 hours ago










  • $begingroup$
    I have understood the flaw in my method by understanding the correct answer, but thank you anyways
    $endgroup$
    – Victor S.
    8 hours ago
















4












$begingroup$



Two players, A and B, alternately and independently flip a coin and
the first player to obtain a head wins. Player A flips first. What is
the probability that A wins?



Official answer: 2/3, but I cannot arrive at it.




Thought process: Find the probability that a head comes on the nth trial. That's easy to do, it's just a Geometric random variable. Then find the probability that nth turn is player's A turn. Finally, multiply both probabilities.



When I came up with each probability, both of them depended on the amount of trials n, so my answer was a non-constant function of n.



However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.






probability







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
    $endgroup$
    – Pakk
    9 hours ago










  • $begingroup$
    I have understood the flaw in my method by understanding the correct answer, but thank you anyways
    $endgroup$
    – Victor S.
    8 hours ago














4












4








4





$begingroup$



Two players, A and B, alternately and independently flip a coin and
the first player to obtain a head wins. Player A flips first. What is
the probability that A wins?



Official answer: 2/3, but I cannot arrive at it.




Thought process: Find the probability that a head comes on the nth trial. That's easy to do, it's just a Geometric random variable. Then find the probability that nth turn is player's A turn. Finally, multiply both probabilities.



When I came up with each probability, both of them depended on the amount of trials n, so my answer was a non-constant function of n.



However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.






probability







share|cite|improve this question









$endgroup$





Two players, A and B, alternately and independently flip a coin and
the first player to obtain a head wins. Player A flips first. What is
the probability that A wins?



Official answer: 2/3, but I cannot arrive at it.




Thought process: Find the probability that a head comes on the nth trial. That's easy to do, it's just a Geometric random variable. Then find the probability that nth turn is player's A turn. Finally, multiply both probabilities.



When I came up with each probability, both of them depended on the amount of trials n, so my answer was a non-constant function of n.



However, what I find quite fantastic is that the answer is a constant, so the amount of tries until a head comes doesn't seem to matter.






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 14 hours ago









Victor S.Victor S.

897




897












  • $begingroup$
    What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
    $endgroup$
    – Pakk
    9 hours ago










  • $begingroup$
    I have understood the flaw in my method by understanding the correct answer, but thank you anyways
    $endgroup$
    – Victor S.
    8 hours ago


















  • $begingroup$
    What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
    $endgroup$
    – Pakk
    9 hours ago










  • $begingroup$
    I have understood the flaw in my method by understanding the correct answer, but thank you anyways
    $endgroup$
    – Victor S.
    8 hours ago
















$begingroup$
What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
$endgroup$
– Pakk
9 hours ago




$begingroup$
What is your question? Do you want to know the answer to the problem, or do you want to know the mistake in your method? If it is the second one, you should make that clear, because all answers so far ignore that.
$endgroup$
– Pakk
9 hours ago












$begingroup$
I have understood the flaw in my method by understanding the correct answer, but thank you anyways
$endgroup$
– Victor S.
8 hours ago




$begingroup$
I have understood the flaw in my method by understanding the correct answer, but thank you anyways
$endgroup$
– Victor S.
8 hours ago










5 Answers
5






active

oldest

votes


















11












$begingroup$

So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
$$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$



where $p$ is probability that head comes in one toss and $q=1-p$.






share|cite|improve this answer









$endgroup$





















    22












    $begingroup$

    Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$






    share|cite|improve this answer









    $endgroup$





















      6












      $begingroup$

      Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
      $$p = .5+ .25p$$
      which yields $p=2/3$.






      share|cite|improve this answer









      $endgroup$





















        6












        $begingroup$

        Here's another approach.



        Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.



        In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).



        But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.



        This gives A a $frac23$ chance of winning to $frac13$ for B.






        share|cite|improve this answer









        $endgroup$





















          0












          $begingroup$

          P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)




          A loses first round, B loses second round, A wins the third round (0.5)^3




          And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2




          P(A winning) = 0.5/(1-0.25)







          share|cite|improve this answer








          New contributor




          Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            11












            $begingroup$

            So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
            $$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$



            where $p$ is probability that head comes in one toss and $q=1-p$.






            share|cite|improve this answer









            $endgroup$


















              11












              $begingroup$

              So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
              $$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$



              where $p$ is probability that head comes in one toss and $q=1-p$.






              share|cite|improve this answer









              $endgroup$
















                11












                11








                11





                $begingroup$

                So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
                $$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$



                where $p$ is probability that head comes in one toss and $q=1-p$.






                share|cite|improve this answer









                $endgroup$



                So $A$ wins if head comes first time in odd number of tosses. Say $P_k$ is probability that head comes first time in $k$-th toss then $P_k = q^{k-1}p$ and
                $$ P = P_1+P_3+P_5+... =$$$$=p+q^2p+q^4p+q^6p+....= {pover 1-q^2} = {1over 1+q} = {2over 3}$$



                where $p$ is probability that head comes in one toss and $q=1-p$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 14 hours ago









                greedoidgreedoid

                41.8k1152105




                41.8k1152105























                    22












                    $begingroup$

                    Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$






                    share|cite|improve this answer









                    $endgroup$


















                      22












                      $begingroup$

                      Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$






                      share|cite|improve this answer









                      $endgroup$
















                        22












                        22








                        22





                        $begingroup$

                        Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$






                        share|cite|improve this answer









                        $endgroup$



                        Let $p$ be the probability that the first player wins. A wins half the time when he tosses a head. The other half the time, B becomes the first tosser, and A wins if the first player doesn't win.$$p=frac12+frac12(1-p)implies p=frac23$$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered 14 hours ago









                        saulspatzsaulspatz

                        15k21330




                        15k21330























                            6












                            $begingroup$

                            Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
                            $$p = .5+ .25p$$
                            which yields $p=2/3$.






                            share|cite|improve this answer









                            $endgroup$


















                              6












                              $begingroup$

                              Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
                              $$p = .5+ .25p$$
                              which yields $p=2/3$.






                              share|cite|improve this answer









                              $endgroup$
















                                6












                                6








                                6





                                $begingroup$

                                Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
                                $$p = .5+ .25p$$
                                which yields $p=2/3$.






                                share|cite|improve this answer









                                $endgroup$



                                Let $p$ denote the probability that $A$ wins. There is a $.5$ probability $A$ wins on the first flip. If he flips tails, then in order to win $B$ must then flip tails. This situation occurs with probability $.25$ ($1/2 * 1/2$ for both tails) and in this case we are back to where we started. So
                                $$p = .5+ .25p$$
                                which yields $p=2/3$.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 14 hours ago









                                pwerthpwerth

                                3,233417




                                3,233417























                                    6












                                    $begingroup$

                                    Here's another approach.



                                    Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.



                                    In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).



                                    But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.



                                    This gives A a $frac23$ chance of winning to $frac13$ for B.






                                    share|cite|improve this answer









                                    $endgroup$


















                                      6












                                      $begingroup$

                                      Here's another approach.



                                      Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.



                                      In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).



                                      But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.



                                      This gives A a $frac23$ chance of winning to $frac13$ for B.






                                      share|cite|improve this answer









                                      $endgroup$
















                                        6












                                        6








                                        6





                                        $begingroup$

                                        Here's another approach.



                                        Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.



                                        In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).



                                        But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.



                                        This gives A a $frac23$ chance of winning to $frac13$ for B.






                                        share|cite|improve this answer









                                        $endgroup$



                                        Here's another approach.



                                        Consider a round of this game as A flipping and then B flipping. Of course the game may end in the middle of a round if A flips heads.



                                        In the first round A has twice as great a chance to win as B. (A wins half the time by flipping heads); and B wins with tails-heads which occurs one quarter of the time).



                                        But this two-to-one ratio of A winning to B winning holds in every round. So overall, A is twice as likely to win as B.



                                        This gives A a $frac23$ chance of winning to $frac13$ for B.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered 13 hours ago









                                        paw88789paw88789

                                        29.1k12349




                                        29.1k12349























                                            0












                                            $begingroup$

                                            P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)




                                            A loses first round, B loses second round, A wins the third round (0.5)^3




                                            And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2




                                            P(A winning) = 0.5/(1-0.25)







                                            share|cite|improve this answer








                                            New contributor




                                            Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                            Check out our Code of Conduct.






                                            $endgroup$


















                                              0












                                              $begingroup$

                                              P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)




                                              A loses first round, B loses second round, A wins the third round (0.5)^3




                                              And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2




                                              P(A winning) = 0.5/(1-0.25)







                                              share|cite|improve this answer








                                              New contributor




                                              Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.






                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)




                                                A loses first round, B loses second round, A wins the third round (0.5)^3




                                                And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2




                                                P(A winning) = 0.5/(1-0.25)







                                                share|cite|improve this answer








                                                New contributor




                                                Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.






                                                $endgroup$



                                                P(A wins) can be thought of a sum of infinite number of cases, starting with >A wins the first round(0.5)




                                                A loses first round, B loses second round, A wins the third round (0.5)^3




                                                And so on. This forms an infinite gp with first term 0.5 and common product 0.5^2




                                                P(A winning) = 0.5/(1-0.25)








                                                share|cite|improve this answer








                                                New contributor




                                                Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.









                                                share|cite|improve this answer



                                                share|cite|improve this answer






                                                New contributor




                                                Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.









                                                answered 14 hours ago









                                                Shreyas Shreyas

                                                812




                                                812




                                                New contributor




                                                Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.





                                                New contributor





                                                Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.






                                                Shreyas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.






























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                                                    Халкинская богословская школа

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                                                    Халкинская богословская школа тур. Rum Ortodoks Ruhban Okulu греч. Ιερά Θεολογική Σχολή της Χάλκης Главное здание Халкинской школы Международное название Theological School of Halki Год основания 1844 Конфессия православие Церковь Константинопольская Ректор Елпидифор (Ламбриниадис) Расположение о. Хейбелиада, Турция Юридический адрес Ümit tepesi No 5, 34973 Heybeliada, Istanbul, Turkey Сайт www.greece.org  [[commons:Category:Halki seminary|Халкинская богословская школа тур. Rum Ortodoks Ruhban Okulu греч. Ιερά Θεολογική Σχολή της Χάλκης на Викискладе]] Ха́лкинская богосло́вская шко́ла (или Духовная семинария на Халки , тур. Rum Ortodoks Ruhban Okulu , греч. Ιερά Θεολογική Σχολή της Χάλκης ) — недействующее среднее специальное учебное заведение Константинопольской православной церкви, расположенное на территории Свято-Троицкого мужского монастыря на вершине холма Надежды (тур. Ümit Tepesi ) острова Хейбелиада, в Турции....
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                                                    Высокополье (Харьковская область)

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                                                    У этого термина существуют и другие значения, см. Высокополье. Село Высокополье укр. Високопілля Герб 49°55′07″ с. ш. 35°25′13″ в. д. H G Я O L Страна Украина   Украина Область Харьковская Район Валковский Сельский совет Высокопольский История и география Основан 1670 Прежние названия Хмелевое Площадь 55,56 км² Часовой пояс UTC+2, летом UTC+3 Население Население 1957 человек ( 2001 ) Плотность 35,220 чел./км² Цифровые идентификаторы Телефонный код +380 5753 Почтовый индекс 63020 Автомобильный код AX, КХ / 21 КОАТУУ 6321281501 Высокополье Высокополье Высокополье  Аудио, фото и видео на Викискладе Высокопо́лье (укр. Високопілля ) — село, Высокопольский сельский совет, Валковский район, Харьковская область, Украина. Код КОАТУУ — 6321281501. Население по переписи 2001 г. составляет 1957 (866/1091 м/ж) человек. Является административным центр...
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