Count conditional probability of winning a game

Multi tool use
Multi tool use












1












$begingroup$



In a certain game of tennis, Alice has a 60% probability to win any
given point against Bob. The player who gets to 4 points first wins
the game, and points cannot end in a tie. What is Alice's probability
to win the game?




When solving in terms of a random walk it gives a probability around 83%, but the actual solution for this particular problem is around 71%.



What's the difference between the methods used and how correctly solve such kind of problems?






probability probability-theory random-walk







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    In a certain game of tennis, Alice has a 60% probability to win any
    given point against Bob. The player who gets to 4 points first wins
    the game, and points cannot end in a tie. What is Alice's probability
    to win the game?




    When solving in terms of a random walk it gives a probability around 83%, but the actual solution for this particular problem is around 71%.



    What's the difference between the methods used and how correctly solve such kind of problems?






    probability probability-theory random-walk







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      0



      $begingroup$



      In a certain game of tennis, Alice has a 60% probability to win any
      given point against Bob. The player who gets to 4 points first wins
      the game, and points cannot end in a tie. What is Alice's probability
      to win the game?




      When solving in terms of a random walk it gives a probability around 83%, but the actual solution for this particular problem is around 71%.



      What's the difference between the methods used and how correctly solve such kind of problems?






      probability probability-theory random-walk







      share|cite|improve this question











      $endgroup$





      In a certain game of tennis, Alice has a 60% probability to win any
      given point against Bob. The player who gets to 4 points first wins
      the game, and points cannot end in a tie. What is Alice's probability
      to win the game?




      When solving in terms of a random walk it gives a probability around 83%, but the actual solution for this particular problem is around 71%.



      What's the difference between the methods used and how correctly solve such kind of problems?






      probability probability-theory random-walk




      probability probability-theory random-walk






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 2 at 23:41









      Henno Brandsma

      113k348123




      113k348123










      asked Jan 14 at 8:24









      Most WantedMost Wanted

      1306




      1306






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          There are 4 possible outcomes:




          1. Alice wins $4-0$. There is only one way this happens, i.e. Alice wins 4 points. The probability of that is $0.6^4$.

          2. Alice wins $4-1$. There are $4$ ways this can happen, Bob can win the first, second, third or fourth point. So the probability of this event is $4cdot 0.6^4cdot0.4$

          3. Alice wins $4-2$. Bob needs to win $2$ of $6$ points for this, and he cannot win the last point, so he can win $2$ of the first $5$ points, the probability here is ${5choose 2}cdot 0.6^4cdot0.4^2$

          4. Alice wins $4-3$, the probability here is ${6choose 3}cdot 0.6^4cdot 0.4^3$


          Summing those up, you get $approx0.7$. I cannot comment on the random walk method because you didn't give enough details for me to tell you where you went wrong. But the linked answer considers a completely different game in which a player needs an advantage of 2 points to win. Your game has no such condition and will always end in at most 7 points. The linked answer's game can be arbitrarily long (and if you watch tennis, they sometimes get very long)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Stewart in the book "game, set and math" analyses the game of tennis that way in one chapter. Sets then also need to be won by 1 two game difference etc. It turns out that even a very small "per point" advantage gets magnified quite a lot over a whole match.
            $endgroup$
            – Henno Brandsma
            Jan 25 at 22:27



















          0












          $begingroup$

          Just as a generalization for the answer above, formula for getting N points in a game first



          $sum_{i=0}^{N-1}{N-1+ichoose i}cdot p^Ncdot (1-p)^{i}$



          So in our case we have N=4, p=0.6, q=1-p=0.4



          $sum_{i=0}^{3}{2+ichoose i}cdot 0.6^4cdot 0.4^{i}$



          which gives us 0.71 as a result.






          share|cite|improve this answer









          $endgroup$













            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%2f3072979%2fcount-conditional-probability-of-winning-a-game%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









            2












            $begingroup$

            There are 4 possible outcomes:




            1. Alice wins $4-0$. There is only one way this happens, i.e. Alice wins 4 points. The probability of that is $0.6^4$.

            2. Alice wins $4-1$. There are $4$ ways this can happen, Bob can win the first, second, third or fourth point. So the probability of this event is $4cdot 0.6^4cdot0.4$

            3. Alice wins $4-2$. Bob needs to win $2$ of $6$ points for this, and he cannot win the last point, so he can win $2$ of the first $5$ points, the probability here is ${5choose 2}cdot 0.6^4cdot0.4^2$

            4. Alice wins $4-3$, the probability here is ${6choose 3}cdot 0.6^4cdot 0.4^3$


            Summing those up, you get $approx0.7$. I cannot comment on the random walk method because you didn't give enough details for me to tell you where you went wrong. But the linked answer considers a completely different game in which a player needs an advantage of 2 points to win. Your game has no such condition and will always end in at most 7 points. The linked answer's game can be arbitrarily long (and if you watch tennis, they sometimes get very long)






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Stewart in the book "game, set and math" analyses the game of tennis that way in one chapter. Sets then also need to be won by 1 two game difference etc. It turns out that even a very small "per point" advantage gets magnified quite a lot over a whole match.
              $endgroup$
              – Henno Brandsma
              Jan 25 at 22:27
















            2












            $begingroup$

            There are 4 possible outcomes:




            1. Alice wins $4-0$. There is only one way this happens, i.e. Alice wins 4 points. The probability of that is $0.6^4$.

            2. Alice wins $4-1$. There are $4$ ways this can happen, Bob can win the first, second, third or fourth point. So the probability of this event is $4cdot 0.6^4cdot0.4$

            3. Alice wins $4-2$. Bob needs to win $2$ of $6$ points for this, and he cannot win the last point, so he can win $2$ of the first $5$ points, the probability here is ${5choose 2}cdot 0.6^4cdot0.4^2$

            4. Alice wins $4-3$, the probability here is ${6choose 3}cdot 0.6^4cdot 0.4^3$


            Summing those up, you get $approx0.7$. I cannot comment on the random walk method because you didn't give enough details for me to tell you where you went wrong. But the linked answer considers a completely different game in which a player needs an advantage of 2 points to win. Your game has no such condition and will always end in at most 7 points. The linked answer's game can be arbitrarily long (and if you watch tennis, they sometimes get very long)






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Stewart in the book "game, set and math" analyses the game of tennis that way in one chapter. Sets then also need to be won by 1 two game difference etc. It turns out that even a very small "per point" advantage gets magnified quite a lot over a whole match.
              $endgroup$
              – Henno Brandsma
              Jan 25 at 22:27














            2












            2








            2





            $begingroup$

            There are 4 possible outcomes:




            1. Alice wins $4-0$. There is only one way this happens, i.e. Alice wins 4 points. The probability of that is $0.6^4$.

            2. Alice wins $4-1$. There are $4$ ways this can happen, Bob can win the first, second, third or fourth point. So the probability of this event is $4cdot 0.6^4cdot0.4$

            3. Alice wins $4-2$. Bob needs to win $2$ of $6$ points for this, and he cannot win the last point, so he can win $2$ of the first $5$ points, the probability here is ${5choose 2}cdot 0.6^4cdot0.4^2$

            4. Alice wins $4-3$, the probability here is ${6choose 3}cdot 0.6^4cdot 0.4^3$


            Summing those up, you get $approx0.7$. I cannot comment on the random walk method because you didn't give enough details for me to tell you where you went wrong. But the linked answer considers a completely different game in which a player needs an advantage of 2 points to win. Your game has no such condition and will always end in at most 7 points. The linked answer's game can be arbitrarily long (and if you watch tennis, they sometimes get very long)






            share|cite|improve this answer











            $endgroup$



            There are 4 possible outcomes:




            1. Alice wins $4-0$. There is only one way this happens, i.e. Alice wins 4 points. The probability of that is $0.6^4$.

            2. Alice wins $4-1$. There are $4$ ways this can happen, Bob can win the first, second, third or fourth point. So the probability of this event is $4cdot 0.6^4cdot0.4$

            3. Alice wins $4-2$. Bob needs to win $2$ of $6$ points for this, and he cannot win the last point, so he can win $2$ of the first $5$ points, the probability here is ${5choose 2}cdot 0.6^4cdot0.4^2$

            4. Alice wins $4-3$, the probability here is ${6choose 3}cdot 0.6^4cdot 0.4^3$


            Summing those up, you get $approx0.7$. I cannot comment on the random walk method because you didn't give enough details for me to tell you where you went wrong. But the linked answer considers a completely different game in which a player needs an advantage of 2 points to win. Your game has no such condition and will always end in at most 7 points. The linked answer's game can be arbitrarily long (and if you watch tennis, they sometimes get very long)







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 14 at 8:38

























            answered Jan 14 at 8:32









            5xum5xum

            91.5k394161




            91.5k394161












            • $begingroup$
              Stewart in the book "game, set and math" analyses the game of tennis that way in one chapter. Sets then also need to be won by 1 two game difference etc. It turns out that even a very small "per point" advantage gets magnified quite a lot over a whole match.
              $endgroup$
              – Henno Brandsma
              Jan 25 at 22:27


















            • $begingroup$
              Stewart in the book "game, set and math" analyses the game of tennis that way in one chapter. Sets then also need to be won by 1 two game difference etc. It turns out that even a very small "per point" advantage gets magnified quite a lot over a whole match.
              $endgroup$
              – Henno Brandsma
              Jan 25 at 22:27
















            $begingroup$
            Stewart in the book "game, set and math" analyses the game of tennis that way in one chapter. Sets then also need to be won by 1 two game difference etc. It turns out that even a very small "per point" advantage gets magnified quite a lot over a whole match.
            $endgroup$
            – Henno Brandsma
            Jan 25 at 22:27




            $begingroup$
            Stewart in the book "game, set and math" analyses the game of tennis that way in one chapter. Sets then also need to be won by 1 two game difference etc. It turns out that even a very small "per point" advantage gets magnified quite a lot over a whole match.
            $endgroup$
            – Henno Brandsma
            Jan 25 at 22:27











            0












            $begingroup$

            Just as a generalization for the answer above, formula for getting N points in a game first



            $sum_{i=0}^{N-1}{N-1+ichoose i}cdot p^Ncdot (1-p)^{i}$



            So in our case we have N=4, p=0.6, q=1-p=0.4



            $sum_{i=0}^{3}{2+ichoose i}cdot 0.6^4cdot 0.4^{i}$



            which gives us 0.71 as a result.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Just as a generalization for the answer above, formula for getting N points in a game first



              $sum_{i=0}^{N-1}{N-1+ichoose i}cdot p^Ncdot (1-p)^{i}$



              So in our case we have N=4, p=0.6, q=1-p=0.4



              $sum_{i=0}^{3}{2+ichoose i}cdot 0.6^4cdot 0.4^{i}$



              which gives us 0.71 as a result.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Just as a generalization for the answer above, formula for getting N points in a game first



                $sum_{i=0}^{N-1}{N-1+ichoose i}cdot p^Ncdot (1-p)^{i}$



                So in our case we have N=4, p=0.6, q=1-p=0.4



                $sum_{i=0}^{3}{2+ichoose i}cdot 0.6^4cdot 0.4^{i}$



                which gives us 0.71 as a result.






                share|cite|improve this answer









                $endgroup$



                Just as a generalization for the answer above, formula for getting N points in a game first



                $sum_{i=0}^{N-1}{N-1+ichoose i}cdot p^Ncdot (1-p)^{i}$



                So in our case we have N=4, p=0.6, q=1-p=0.4



                $sum_{i=0}^{3}{2+ichoose i}cdot 0.6^4cdot 0.4^{i}$



                which gives us 0.71 as a result.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 14 at 9:10









                Most WantedMost Wanted

                1306




                1306






























                    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%2f3072979%2fcount-conditional-probability-of-winning-a-game%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







                    I4t eGf,D3VN7XbYr4 yih9ZHJ9MUH4oGIL8UO0L3aALl
                    -
                    5 J,DgQp23Tb Y19,0fFn,GkWVb

                    Popular posts from this blog

                    Human spaceflight

                    Image
                    Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
                    Read more

                    Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                    Image
                    0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
                    Read more

                    Antiochus X Eusebes

                    Image
                    Antiochus X Eusebes From Wikipedia, the free encyclopedia Jump to navigation Jump to search King of Syria Antiochus X Eusebes Antiochus X's portrait on the obverse of a tetradrachm King of Syria (Seleucid Empire) Reign 95–92 or 88 BC Predecessor Seleucus VI, Demetrius III Successor Demetrius III, Philip I Born c.  113 BC Died 92 or 88 BC Spouse Cleopatra Selene Issue Detail Antiochus XIII Dynasty Seleucid Father Antiochus IX Mother Cleopatra IV ? Antiochus X Eusebes Philopator (Greek: Ἀντίοχος Εὐσεβής Φιλοπάτωρ ; c.  113 –92 or 88 BC) was a Hellenistic Seleucid monarch who reigned as the King of Syria between 95 and 92 or 89/88 BC (224 Seleucid year (SE)). [note 1] He was the son of Antiochus IX and perhaps his Egyptian wife Cleopatra IV. He lived at a period that witnessed a general disintegration of Seleucid Syria characterized by civil wars, foreign inter...
                    Read more