Probability of a result of a game.












0












$begingroup$


Problem statement:



Two players play the following game of several rounds. If some player wins a round, the winner is given $1$ point, and loser is given $0$ points, and there are no draws. The game ends when some player gets $a$ points.



It is known that probability that first player wins a round is $p$. Find probability that difference between winner's and loser's score is equal to $b$.



My approach:



First, notice that if the difference between winner's and loser's score is equal to $b$, then the final scores must be $(a, a - b)$ or $(a - b, a)$.



Suppose that the final scores are $(a, a - b)$. Consider an integer lattice with points $(0, 0)$ and $(a, a - b)$. Number of paths from $(0, 0)$ to $(a, a - b)$ such that we move only up or right is $binom{2a - b}{a}$. Now suppose that probability of "up"-movement is $p$, then the probability of "right"-movement is $1 - p$. Hence, because any such path consists of $a - b$ "up"-movements and $a$ "right"-movements, the probability for such final scores is $$p_{(a, a - b)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} .$$ Similarly, the probability for final scores $(a - b, a)$ is $$p_{(a - b, a)} = binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$



Hence, the final probability:
$$ p' = p_{(a, a - b)} + p_{(a - b, a)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} + binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$



Is my approach correct? I've tried to simulate the game in Python and got different result ($approx 0.06$ by the simulation vs. $approx 0.105$ by the formula for $a = 11, b = 3, p = 0.3$).






probability combinatorics game-theory







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Problem statement:



    Two players play the following game of several rounds. If some player wins a round, the winner is given $1$ point, and loser is given $0$ points, and there are no draws. The game ends when some player gets $a$ points.



    It is known that probability that first player wins a round is $p$. Find probability that difference between winner's and loser's score is equal to $b$.



    My approach:



    First, notice that if the difference between winner's and loser's score is equal to $b$, then the final scores must be $(a, a - b)$ or $(a - b, a)$.



    Suppose that the final scores are $(a, a - b)$. Consider an integer lattice with points $(0, 0)$ and $(a, a - b)$. Number of paths from $(0, 0)$ to $(a, a - b)$ such that we move only up or right is $binom{2a - b}{a}$. Now suppose that probability of "up"-movement is $p$, then the probability of "right"-movement is $1 - p$. Hence, because any such path consists of $a - b$ "up"-movements and $a$ "right"-movements, the probability for such final scores is $$p_{(a, a - b)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} .$$ Similarly, the probability for final scores $(a - b, a)$ is $$p_{(a - b, a)} = binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$



    Hence, the final probability:
    $$ p' = p_{(a, a - b)} + p_{(a - b, a)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} + binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$



    Is my approach correct? I've tried to simulate the game in Python and got different result ($approx 0.06$ by the simulation vs. $approx 0.105$ by the formula for $a = 11, b = 3, p = 0.3$).






    probability combinatorics game-theory







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Problem statement:



      Two players play the following game of several rounds. If some player wins a round, the winner is given $1$ point, and loser is given $0$ points, and there are no draws. The game ends when some player gets $a$ points.



      It is known that probability that first player wins a round is $p$. Find probability that difference between winner's and loser's score is equal to $b$.



      My approach:



      First, notice that if the difference between winner's and loser's score is equal to $b$, then the final scores must be $(a, a - b)$ or $(a - b, a)$.



      Suppose that the final scores are $(a, a - b)$. Consider an integer lattice with points $(0, 0)$ and $(a, a - b)$. Number of paths from $(0, 0)$ to $(a, a - b)$ such that we move only up or right is $binom{2a - b}{a}$. Now suppose that probability of "up"-movement is $p$, then the probability of "right"-movement is $1 - p$. Hence, because any such path consists of $a - b$ "up"-movements and $a$ "right"-movements, the probability for such final scores is $$p_{(a, a - b)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} .$$ Similarly, the probability for final scores $(a - b, a)$ is $$p_{(a - b, a)} = binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$



      Hence, the final probability:
      $$ p' = p_{(a, a - b)} + p_{(a - b, a)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} + binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$



      Is my approach correct? I've tried to simulate the game in Python and got different result ($approx 0.06$ by the simulation vs. $approx 0.105$ by the formula for $a = 11, b = 3, p = 0.3$).






      probability combinatorics game-theory







      share|cite|improve this question











      $endgroup$




      Problem statement:



      Two players play the following game of several rounds. If some player wins a round, the winner is given $1$ point, and loser is given $0$ points, and there are no draws. The game ends when some player gets $a$ points.



      It is known that probability that first player wins a round is $p$. Find probability that difference between winner's and loser's score is equal to $b$.



      My approach:



      First, notice that if the difference between winner's and loser's score is equal to $b$, then the final scores must be $(a, a - b)$ or $(a - b, a)$.



      Suppose that the final scores are $(a, a - b)$. Consider an integer lattice with points $(0, 0)$ and $(a, a - b)$. Number of paths from $(0, 0)$ to $(a, a - b)$ such that we move only up or right is $binom{2a - b}{a}$. Now suppose that probability of "up"-movement is $p$, then the probability of "right"-movement is $1 - p$. Hence, because any such path consists of $a - b$ "up"-movements and $a$ "right"-movements, the probability for such final scores is $$p_{(a, a - b)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} .$$ Similarly, the probability for final scores $(a - b, a)$ is $$p_{(a - b, a)} = binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$



      Hence, the final probability:
      $$ p' = p_{(a, a - b)} + p_{(a - b, a)} = binom{2a - b}{a} (1-p)^{a} p^{a - b} + binom{2a - b}{a - b} (1-p)^{a - b} p^{a} .$$



      Is my approach correct? I've tried to simulate the game in Python and got different result ($approx 0.06$ by the simulation vs. $approx 0.105$ by the formula for $a = 11, b = 3, p = 0.3$).






      probability combinatorics game-theory




      probability combinatorics game-theory






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      edited Dec 6 '18 at 22:28







      Elman

















      asked Dec 6 '18 at 22:14









      ElmanElman

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          $begingroup$

          It seems that your calculation includes the possibility, for example, that one player wins $a$ times and then the other player wins $a-b$ times, since this would be among the counted paths from $(0,0)$ to $(a,a-b)$, right?



          Edit: To fix the calculation, I think you should consider all paths from $(0,0)$ to $(a-1, a-b)$, and then multiply by $p$ (and similarly $(a-b, a-1)$, and $1-p$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Good catch. Do you see how to fix the calculation?
            $endgroup$
            – saulspatz
            Dec 6 '18 at 22:57











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          $begingroup$

          It seems that your calculation includes the possibility, for example, that one player wins $a$ times and then the other player wins $a-b$ times, since this would be among the counted paths from $(0,0)$ to $(a,a-b)$, right?



          Edit: To fix the calculation, I think you should consider all paths from $(0,0)$ to $(a-1, a-b)$, and then multiply by $p$ (and similarly $(a-b, a-1)$, and $1-p$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Good catch. Do you see how to fix the calculation?
            $endgroup$
            – saulspatz
            Dec 6 '18 at 22:57
















          1












          $begingroup$

          It seems that your calculation includes the possibility, for example, that one player wins $a$ times and then the other player wins $a-b$ times, since this would be among the counted paths from $(0,0)$ to $(a,a-b)$, right?



          Edit: To fix the calculation, I think you should consider all paths from $(0,0)$ to $(a-1, a-b)$, and then multiply by $p$ (and similarly $(a-b, a-1)$, and $1-p$).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Good catch. Do you see how to fix the calculation?
            $endgroup$
            – saulspatz
            Dec 6 '18 at 22:57














          1












          1








          1





          $begingroup$

          It seems that your calculation includes the possibility, for example, that one player wins $a$ times and then the other player wins $a-b$ times, since this would be among the counted paths from $(0,0)$ to $(a,a-b)$, right?



          Edit: To fix the calculation, I think you should consider all paths from $(0,0)$ to $(a-1, a-b)$, and then multiply by $p$ (and similarly $(a-b, a-1)$, and $1-p$).






          share|cite|improve this answer











          $endgroup$



          It seems that your calculation includes the possibility, for example, that one player wins $a$ times and then the other player wins $a-b$ times, since this would be among the counted paths from $(0,0)$ to $(a,a-b)$, right?



          Edit: To fix the calculation, I think you should consider all paths from $(0,0)$ to $(a-1, a-b)$, and then multiply by $p$ (and similarly $(a-b, a-1)$, and $1-p$).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 7 '18 at 23:26

























          answered Dec 6 '18 at 22:43









          RobRRobR

          113




          113












          • $begingroup$
            Good catch. Do you see how to fix the calculation?
            $endgroup$
            – saulspatz
            Dec 6 '18 at 22:57


















          • $begingroup$
            Good catch. Do you see how to fix the calculation?
            $endgroup$
            – saulspatz
            Dec 6 '18 at 22:57
















          $begingroup$
          Good catch. Do you see how to fix the calculation?
          $endgroup$
          – saulspatz
          Dec 6 '18 at 22:57




          $begingroup$
          Good catch. Do you see how to fix the calculation?
          $endgroup$
          – saulspatz
          Dec 6 '18 at 22:57


















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