Count conditional probability of winning a game












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











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    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









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          2 Answers
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          $begingroup$

          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





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            2 Answers
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            2 Answers
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            $begingroup$

            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









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