a coin tossing problem












0












$begingroup$


A and B are playing coin tossing games. If there is a 2 continuous heads, then A win. If there is a 3 continuous tails, then B win. They keep tossing the coin until one of them win. what's the probability of A wins the game?






probability







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    A and B are playing coin tossing games. If there is a 2 continuous heads, then A win. If there is a 3 continuous tails, then B win. They keep tossing the coin until one of them win. what's the probability of A wins the game?






    probability







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      A and B are playing coin tossing games. If there is a 2 continuous heads, then A win. If there is a 3 continuous tails, then B win. They keep tossing the coin until one of them win. what's the probability of A wins the game?






      probability







      share|cite|improve this question









      $endgroup$




      A and B are playing coin tossing games. If there is a 2 continuous heads, then A win. If there is a 3 continuous tails, then B win. They keep tossing the coin until one of them win. what's the probability of A wins the game?






      probability




      probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 7 '18 at 5:35









      Mango SickMango Sick

      91




      91






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          Hint:



          For A to win, it needs to get $2H$ before B gets $3T$.



          So, possible cass may be: $$(H,H),(T,H,H), (T,T,H,H)$$
          $$P(HH)=frac{1}{2^2}$$
          $$P(THH)=frac{1}{2^3}$$
          $$P(TTHH)=frac{1}{2^4}$$



          There can be more possibilities such as $$(H,T,T,H,H),(T,T,H,T,H,H) text{ etc. }$$



          Try to generalize for $n$ tosses, and then solve for $nto infty$






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            The probability vector $[mathbb{P}(text{A wins}), mathbb{P}(text{B wins})]$ is given by the average of the first and third row of (matrices typed like when inputted to for example to WolframAlpha)




            { inverse (1/2*[[2,-1,-1,0,0], [0,2,-1,0,0],[-1,0,2,-1,0], [-1,0,0,2,-1], [-1,0,0,0,2]] ) } [[0,0], [1/2,0], [0,0], [0,0], [0,1/2]]




            and is $[frac{15}{22}, frac{7}{22}]$.



            In this solution we are calculating the absorption probabilities of a Markov chain with states $H, HH, T, TT, TTT, A, B$ (with obvious meanings and transition probabilities).



            Here' how the chain looks like (all transitions have probability $frac{1}{2}$ except the absorbing winning states $A$ and $B$ of course have self-transition with probability $1$) I included the starting state $S$ here but not in the calculations where this is accounted for by averaging whether you start from $H$ or $T$ (whether the first toss is heads or tails):



            enter image description here






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
              $endgroup$
              – ploosu2
              Dec 7 '18 at 12:12











            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029517%2fa-coin-tossing-problem%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            Hint:



            For A to win, it needs to get $2H$ before B gets $3T$.



            So, possible cass may be: $$(H,H),(T,H,H), (T,T,H,H)$$
            $$P(HH)=frac{1}{2^2}$$
            $$P(THH)=frac{1}{2^3}$$
            $$P(TTHH)=frac{1}{2^4}$$



            There can be more possibilities such as $$(H,T,T,H,H),(T,T,H,T,H,H) text{ etc. }$$



            Try to generalize for $n$ tosses, and then solve for $nto infty$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Hint:



              For A to win, it needs to get $2H$ before B gets $3T$.



              So, possible cass may be: $$(H,H),(T,H,H), (T,T,H,H)$$
              $$P(HH)=frac{1}{2^2}$$
              $$P(THH)=frac{1}{2^3}$$
              $$P(TTHH)=frac{1}{2^4}$$



              There can be more possibilities such as $$(H,T,T,H,H),(T,T,H,T,H,H) text{ etc. }$$



              Try to generalize for $n$ tosses, and then solve for $nto infty$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Hint:



                For A to win, it needs to get $2H$ before B gets $3T$.



                So, possible cass may be: $$(H,H),(T,H,H), (T,T,H,H)$$
                $$P(HH)=frac{1}{2^2}$$
                $$P(THH)=frac{1}{2^3}$$
                $$P(TTHH)=frac{1}{2^4}$$



                There can be more possibilities such as $$(H,T,T,H,H),(T,T,H,T,H,H) text{ etc. }$$



                Try to generalize for $n$ tosses, and then solve for $nto infty$






                share|cite|improve this answer











                $endgroup$



                Hint:



                For A to win, it needs to get $2H$ before B gets $3T$.



                So, possible cass may be: $$(H,H),(T,H,H), (T,T,H,H)$$
                $$P(HH)=frac{1}{2^2}$$
                $$P(THH)=frac{1}{2^3}$$
                $$P(TTHH)=frac{1}{2^4}$$



                There can be more possibilities such as $$(H,T,T,H,H),(T,T,H,T,H,H) text{ etc. }$$



                Try to generalize for $n$ tosses, and then solve for $nto infty$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 7 '18 at 5:59

























                answered Dec 7 '18 at 5:53









                ideaidea

                2,15841125




                2,15841125























                    0












                    $begingroup$

                    The probability vector $[mathbb{P}(text{A wins}), mathbb{P}(text{B wins})]$ is given by the average of the first and third row of (matrices typed like when inputted to for example to WolframAlpha)




                    { inverse (1/2*[[2,-1,-1,0,0], [0,2,-1,0,0],[-1,0,2,-1,0], [-1,0,0,2,-1], [-1,0,0,0,2]] ) } [[0,0], [1/2,0], [0,0], [0,0], [0,1/2]]




                    and is $[frac{15}{22}, frac{7}{22}]$.



                    In this solution we are calculating the absorption probabilities of a Markov chain with states $H, HH, T, TT, TTT, A, B$ (with obvious meanings and transition probabilities).



                    Here' how the chain looks like (all transitions have probability $frac{1}{2}$ except the absorbing winning states $A$ and $B$ of course have self-transition with probability $1$) I included the starting state $S$ here but not in the calculations where this is accounted for by averaging whether you start from $H$ or $T$ (whether the first toss is heads or tails):



                    enter image description here






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
                      $endgroup$
                      – ploosu2
                      Dec 7 '18 at 12:12
















                    0












                    $begingroup$

                    The probability vector $[mathbb{P}(text{A wins}), mathbb{P}(text{B wins})]$ is given by the average of the first and third row of (matrices typed like when inputted to for example to WolframAlpha)




                    { inverse (1/2*[[2,-1,-1,0,0], [0,2,-1,0,0],[-1,0,2,-1,0], [-1,0,0,2,-1], [-1,0,0,0,2]] ) } [[0,0], [1/2,0], [0,0], [0,0], [0,1/2]]




                    and is $[frac{15}{22}, frac{7}{22}]$.



                    In this solution we are calculating the absorption probabilities of a Markov chain with states $H, HH, T, TT, TTT, A, B$ (with obvious meanings and transition probabilities).



                    Here' how the chain looks like (all transitions have probability $frac{1}{2}$ except the absorbing winning states $A$ and $B$ of course have self-transition with probability $1$) I included the starting state $S$ here but not in the calculations where this is accounted for by averaging whether you start from $H$ or $T$ (whether the first toss is heads or tails):



                    enter image description here






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
                      $endgroup$
                      – ploosu2
                      Dec 7 '18 at 12:12














                    0












                    0








                    0





                    $begingroup$

                    The probability vector $[mathbb{P}(text{A wins}), mathbb{P}(text{B wins})]$ is given by the average of the first and third row of (matrices typed like when inputted to for example to WolframAlpha)




                    { inverse (1/2*[[2,-1,-1,0,0], [0,2,-1,0,0],[-1,0,2,-1,0], [-1,0,0,2,-1], [-1,0,0,0,2]] ) } [[0,0], [1/2,0], [0,0], [0,0], [0,1/2]]




                    and is $[frac{15}{22}, frac{7}{22}]$.



                    In this solution we are calculating the absorption probabilities of a Markov chain with states $H, HH, T, TT, TTT, A, B$ (with obvious meanings and transition probabilities).



                    Here' how the chain looks like (all transitions have probability $frac{1}{2}$ except the absorbing winning states $A$ and $B$ of course have self-transition with probability $1$) I included the starting state $S$ here but not in the calculations where this is accounted for by averaging whether you start from $H$ or $T$ (whether the first toss is heads or tails):



                    enter image description here






                    share|cite|improve this answer











                    $endgroup$



                    The probability vector $[mathbb{P}(text{A wins}), mathbb{P}(text{B wins})]$ is given by the average of the first and third row of (matrices typed like when inputted to for example to WolframAlpha)




                    { inverse (1/2*[[2,-1,-1,0,0], [0,2,-1,0,0],[-1,0,2,-1,0], [-1,0,0,2,-1], [-1,0,0,0,2]] ) } [[0,0], [1/2,0], [0,0], [0,0], [0,1/2]]




                    and is $[frac{15}{22}, frac{7}{22}]$.



                    In this solution we are calculating the absorption probabilities of a Markov chain with states $H, HH, T, TT, TTT, A, B$ (with obvious meanings and transition probabilities).



                    Here' how the chain looks like (all transitions have probability $frac{1}{2}$ except the absorbing winning states $A$ and $B$ of course have self-transition with probability $1$) I included the starting state $S$ here but not in the calculations where this is accounted for by averaging whether you start from $H$ or $T$ (whether the first toss is heads or tails):



                    enter image description here







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 8 '18 at 10:32

























                    answered Dec 7 '18 at 12:09









                    ploosu2ploosu2

                    4,6451024




                    4,6451024












                    • $begingroup$
                      For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
                      $endgroup$
                      – ploosu2
                      Dec 7 '18 at 12:12


















                    • $begingroup$
                      For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
                      $endgroup$
                      – ploosu2
                      Dec 7 '18 at 12:12
















                    $begingroup$
                    For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
                    $endgroup$
                    – ploosu2
                    Dec 7 '18 at 12:12




                    $begingroup$
                    For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
                    $endgroup$
                    – ploosu2
                    Dec 7 '18 at 12:12


















                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029517%2fa-coin-tossing-problem%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Ellipse (mathématiques)

                    Image
                    Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                    Read more

                    Quarter-circle Tiles

                    Image
                    up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                    Read more

                    Mont Emei

                    Image
                    Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                    Read more