Expected profit of my simple board game












3
















How to play:



Use 1 host and at least 1 player



Each player has to toss fair six-sided dice to go to goal.



If the player is at the 35th cell and tosses 2 or more, he can go to goal aa same as he tosses 1.



If the player reaches the goal in 9 tossing or less, the host has to pay to that player 1$ per 1 tossing less than 10.



For example, if the player reaches the goal in 7 tossing, host have to pay 3$ to that player.



If the player reaches the goal in 11 tossing or more, that player has to pay to host 1$ per 1 tossing more than 10.



For example, if the player reaches the goal in 12 tossing, that player has to pay 2$ to host.



If the player reaches the goal in 10 tossing, no one has to pay.



Each game will end only if the player reaches the goal.



Player can't pay 1$ and start new game if he can't reaches the goal in 11th tossing.



What is expected profit of host per player for each game?



As much as I know for this game, The expected value in rolling a six-sided die is 3.5.
The expected value of distance in 10 tossing is 35-cell but the goal is at 36-cell distance so expected profit of host is positive. If the goal is at 35th-cell, expected profit of host is 0. But I have no idea to calculate.






combinatorics contest-math puzzle combinatorial-game-theory







share|cite|improve this question





























    3
















    How to play:



    Use 1 host and at least 1 player



    Each player has to toss fair six-sided dice to go to goal.



    If the player is at the 35th cell and tosses 2 or more, he can go to goal aa same as he tosses 1.



    If the player reaches the goal in 9 tossing or less, the host has to pay to that player 1$ per 1 tossing less than 10.



    For example, if the player reaches the goal in 7 tossing, host have to pay 3$ to that player.



    If the player reaches the goal in 11 tossing or more, that player has to pay to host 1$ per 1 tossing more than 10.



    For example, if the player reaches the goal in 12 tossing, that player has to pay 2$ to host.



    If the player reaches the goal in 10 tossing, no one has to pay.



    Each game will end only if the player reaches the goal.



    Player can't pay 1$ and start new game if he can't reaches the goal in 11th tossing.



    What is expected profit of host per player for each game?



    As much as I know for this game, The expected value in rolling a six-sided die is 3.5.
    The expected value of distance in 10 tossing is 35-cell but the goal is at 36-cell distance so expected profit of host is positive. If the goal is at 35th-cell, expected profit of host is 0. But I have no idea to calculate.






    combinatorics contest-math puzzle combinatorial-game-theory







    share|cite|improve this question



























      3












      3








      3









      How to play:



      Use 1 host and at least 1 player



      Each player has to toss fair six-sided dice to go to goal.



      If the player is at the 35th cell and tosses 2 or more, he can go to goal aa same as he tosses 1.



      If the player reaches the goal in 9 tossing or less, the host has to pay to that player 1$ per 1 tossing less than 10.



      For example, if the player reaches the goal in 7 tossing, host have to pay 3$ to that player.



      If the player reaches the goal in 11 tossing or more, that player has to pay to host 1$ per 1 tossing more than 10.



      For example, if the player reaches the goal in 12 tossing, that player has to pay 2$ to host.



      If the player reaches the goal in 10 tossing, no one has to pay.



      Each game will end only if the player reaches the goal.



      Player can't pay 1$ and start new game if he can't reaches the goal in 11th tossing.



      What is expected profit of host per player for each game?



      As much as I know for this game, The expected value in rolling a six-sided die is 3.5.
      The expected value of distance in 10 tossing is 35-cell but the goal is at 36-cell distance so expected profit of host is positive. If the goal is at 35th-cell, expected profit of host is 0. But I have no idea to calculate.






      combinatorics contest-math puzzle combinatorial-game-theory







      share|cite|improve this question

















      How to play:



      Use 1 host and at least 1 player



      Each player has to toss fair six-sided dice to go to goal.



      If the player is at the 35th cell and tosses 2 or more, he can go to goal aa same as he tosses 1.



      If the player reaches the goal in 9 tossing or less, the host has to pay to that player 1$ per 1 tossing less than 10.



      For example, if the player reaches the goal in 7 tossing, host have to pay 3$ to that player.



      If the player reaches the goal in 11 tossing or more, that player has to pay to host 1$ per 1 tossing more than 10.



      For example, if the player reaches the goal in 12 tossing, that player has to pay 2$ to host.



      If the player reaches the goal in 10 tossing, no one has to pay.



      Each game will end only if the player reaches the goal.



      Player can't pay 1$ and start new game if he can't reaches the goal in 11th tossing.



      What is expected profit of host per player for each game?



      As much as I know for this game, The expected value in rolling a six-sided die is 3.5.
      The expected value of distance in 10 tossing is 35-cell but the goal is at 36-cell distance so expected profit of host is positive. If the goal is at 35th-cell, expected profit of host is 0. But I have no idea to calculate.






      combinatorics contest-math puzzle combinatorial-game-theory




      combinatorics contest-math puzzle combinatorial-game-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 at 10:27









      Vee Hua Zhi

      774224




      774224










      asked Nov 22 at 6:17









      uesdto signin

      184




      184






















          1 Answer
          1






          active

          oldest

          votes


















          2














          Here is the exact computation, taking all possible games into account. Idea: Consider the polynomial
          $$p_j(x):=(x+x^2+x^3+x^4+x^5+x^6)^j .$$ The coefficient $[x^k]p_j(x)$ gives the number of $j$-tosses histories that bring the player exactly to square $k$. Since I'm not interested in squares $kgeq36$ I truncate $p_j(x)$ after the $x^{35}$ term. In this way I obtain the "truncated series" ${tt s[j]}$. The sum $sum_{k=0}^{35} [x^k]p_j(x)$ counts the number of games that are not over after $j$ tosses. Dividing this sum by $6^j$ gives the probability $p(j)$ that the game is not yet over after $j$ tosses, and $q(j):=p(j-1)-p(j)$ is the probability that the game ends with the $j^{rm th}$ toss. The expected gain for the host then is $sum_{j=1}^{36} (j-10)q(j)$.



          enter image description here



          If the goal is at square $35$ instead of $36$ the corresponding value is $0.476195$, and for $34$ it is $0.190481$ in favor of the host. In any case I suggest you write your own program and tune the parameters as desired.






          share|cite|improve this answer























          • I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
            – uesdto signin
            Nov 22 at 9:41










          • $0.761905$ is the expected profit of host per game, as demanded in the question.
            – Christian Blatter
            Nov 22 at 10:08










          • Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
            – uesdto signin
            Nov 22 at 10:47










          • Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
            – uesdto signin
            Nov 22 at 11:48












          • I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
            – Christian Blatter
            Nov 22 at 12:22











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          2














          Here is the exact computation, taking all possible games into account. Idea: Consider the polynomial
          $$p_j(x):=(x+x^2+x^3+x^4+x^5+x^6)^j .$$ The coefficient $[x^k]p_j(x)$ gives the number of $j$-tosses histories that bring the player exactly to square $k$. Since I'm not interested in squares $kgeq36$ I truncate $p_j(x)$ after the $x^{35}$ term. In this way I obtain the "truncated series" ${tt s[j]}$. The sum $sum_{k=0}^{35} [x^k]p_j(x)$ counts the number of games that are not over after $j$ tosses. Dividing this sum by $6^j$ gives the probability $p(j)$ that the game is not yet over after $j$ tosses, and $q(j):=p(j-1)-p(j)$ is the probability that the game ends with the $j^{rm th}$ toss. The expected gain for the host then is $sum_{j=1}^{36} (j-10)q(j)$.



          enter image description here



          If the goal is at square $35$ instead of $36$ the corresponding value is $0.476195$, and for $34$ it is $0.190481$ in favor of the host. In any case I suggest you write your own program and tune the parameters as desired.






          share|cite|improve this answer























          • I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
            – uesdto signin
            Nov 22 at 9:41










          • $0.761905$ is the expected profit of host per game, as demanded in the question.
            – Christian Blatter
            Nov 22 at 10:08










          • Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
            – uesdto signin
            Nov 22 at 10:47










          • Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
            – uesdto signin
            Nov 22 at 11:48












          • I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
            – Christian Blatter
            Nov 22 at 12:22
















          2














          Here is the exact computation, taking all possible games into account. Idea: Consider the polynomial
          $$p_j(x):=(x+x^2+x^3+x^4+x^5+x^6)^j .$$ The coefficient $[x^k]p_j(x)$ gives the number of $j$-tosses histories that bring the player exactly to square $k$. Since I'm not interested in squares $kgeq36$ I truncate $p_j(x)$ after the $x^{35}$ term. In this way I obtain the "truncated series" ${tt s[j]}$. The sum $sum_{k=0}^{35} [x^k]p_j(x)$ counts the number of games that are not over after $j$ tosses. Dividing this sum by $6^j$ gives the probability $p(j)$ that the game is not yet over after $j$ tosses, and $q(j):=p(j-1)-p(j)$ is the probability that the game ends with the $j^{rm th}$ toss. The expected gain for the host then is $sum_{j=1}^{36} (j-10)q(j)$.



          enter image description here



          If the goal is at square $35$ instead of $36$ the corresponding value is $0.476195$, and for $34$ it is $0.190481$ in favor of the host. In any case I suggest you write your own program and tune the parameters as desired.






          share|cite|improve this answer























          • I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
            – uesdto signin
            Nov 22 at 9:41










          • $0.761905$ is the expected profit of host per game, as demanded in the question.
            – Christian Blatter
            Nov 22 at 10:08










          • Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
            – uesdto signin
            Nov 22 at 10:47










          • Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
            – uesdto signin
            Nov 22 at 11:48












          • I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
            – Christian Blatter
            Nov 22 at 12:22














          2












          2








          2






          Here is the exact computation, taking all possible games into account. Idea: Consider the polynomial
          $$p_j(x):=(x+x^2+x^3+x^4+x^5+x^6)^j .$$ The coefficient $[x^k]p_j(x)$ gives the number of $j$-tosses histories that bring the player exactly to square $k$. Since I'm not interested in squares $kgeq36$ I truncate $p_j(x)$ after the $x^{35}$ term. In this way I obtain the "truncated series" ${tt s[j]}$. The sum $sum_{k=0}^{35} [x^k]p_j(x)$ counts the number of games that are not over after $j$ tosses. Dividing this sum by $6^j$ gives the probability $p(j)$ that the game is not yet over after $j$ tosses, and $q(j):=p(j-1)-p(j)$ is the probability that the game ends with the $j^{rm th}$ toss. The expected gain for the host then is $sum_{j=1}^{36} (j-10)q(j)$.



          enter image description here



          If the goal is at square $35$ instead of $36$ the corresponding value is $0.476195$, and for $34$ it is $0.190481$ in favor of the host. In any case I suggest you write your own program and tune the parameters as desired.






          share|cite|improve this answer














          Here is the exact computation, taking all possible games into account. Idea: Consider the polynomial
          $$p_j(x):=(x+x^2+x^3+x^4+x^5+x^6)^j .$$ The coefficient $[x^k]p_j(x)$ gives the number of $j$-tosses histories that bring the player exactly to square $k$. Since I'm not interested in squares $kgeq36$ I truncate $p_j(x)$ after the $x^{35}$ term. In this way I obtain the "truncated series" ${tt s[j]}$. The sum $sum_{k=0}^{35} [x^k]p_j(x)$ counts the number of games that are not over after $j$ tosses. Dividing this sum by $6^j$ gives the probability $p(j)$ that the game is not yet over after $j$ tosses, and $q(j):=p(j-1)-p(j)$ is the probability that the game ends with the $j^{rm th}$ toss. The expected gain for the host then is $sum_{j=1}^{36} (j-10)q(j)$.



          enter image description here



          If the goal is at square $35$ instead of $36$ the corresponding value is $0.476195$, and for $34$ it is $0.190481$ in favor of the host. In any case I suggest you write your own program and tune the parameters as desired.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 22 at 11:13

























          answered Nov 22 at 8:57









          Christian Blatter

          172k7112325




          172k7112325












          • I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
            – uesdto signin
            Nov 22 at 9:41










          • $0.761905$ is the expected profit of host per game, as demanded in the question.
            – Christian Blatter
            Nov 22 at 10:08










          • Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
            – uesdto signin
            Nov 22 at 10:47










          • Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
            – uesdto signin
            Nov 22 at 11:48












          • I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
            – Christian Blatter
            Nov 22 at 12:22


















          • I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
            – uesdto signin
            Nov 22 at 9:41










          • $0.761905$ is the expected profit of host per game, as demanded in the question.
            – Christian Blatter
            Nov 22 at 10:08










          • Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
            – uesdto signin
            Nov 22 at 10:47










          • Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
            – uesdto signin
            Nov 22 at 11:48












          • I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
            – Christian Blatter
            Nov 22 at 12:22
















          I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
          – uesdto signin
          Nov 22 at 9:41




          I don't understand the program much. But is the expected profit per player for each game really 0.761905? I think it should be around 0-0.2. If it is really 0.761905, if the goal is at 35th-cell or even 34th-cell, the expected profit will still be positive.
          – uesdto signin
          Nov 22 at 9:41












          $0.761905$ is the expected profit of host per game, as demanded in the question.
          – Christian Blatter
          Nov 22 at 10:08




          $0.761905$ is the expected profit of host per game, as demanded in the question.
          – Christian Blatter
          Nov 22 at 10:08












          Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
          – uesdto signin
          Nov 22 at 10:47




          Could you calculate the expected profit for case the goal is at 35th-cell and 34th-cell, please ?
          – uesdto signin
          Nov 22 at 10:47












          Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
          – uesdto signin
          Nov 22 at 11:48






          Thank you very much. It is weird that the expected value in rolling a six-sided die is 3.5. But if the goal is at square 34, the expected profit for host is still positive. I think it should be negative. If I write program as same as you write in mathematica, will I get answer as same as you get (Is there anything else)?
          – uesdto signin
          Nov 22 at 11:48














          I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
          – Christian Blatter
          Nov 22 at 12:22




          I hope so. I simulated $1,000,000$ games with goal $=36$ and obtained $0.762758$.
          – Christian Blatter
          Nov 22 at 12:22


















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