A Short Dice Puzzle












17












$begingroup$


There are $2$ fair dice:




  • An $11$-sided die, valued from $-5$ to $5$,

  • A $41$-sided die, valued from $-20$ to $20$.


You pick one, and I'll take the other one. We roll the dice, and whoever rolls the larger number wins (if tied, we roll again.)



Which die will you choose to maximize your winning probability?










share|improve this question











$endgroup$

















    17












    $begingroup$


    There are $2$ fair dice:




    • An $11$-sided die, valued from $-5$ to $5$,

    • A $41$-sided die, valued from $-20$ to $20$.


    You pick one, and I'll take the other one. We roll the dice, and whoever rolls the larger number wins (if tied, we roll again.)



    Which die will you choose to maximize your winning probability?










    share|improve this question











    $endgroup$















      17












      17








      17


      1



      $begingroup$


      There are $2$ fair dice:




      • An $11$-sided die, valued from $-5$ to $5$,

      • A $41$-sided die, valued from $-20$ to $20$.


      You pick one, and I'll take the other one. We roll the dice, and whoever rolls the larger number wins (if tied, we roll again.)



      Which die will you choose to maximize your winning probability?










      share|improve this question











      $endgroup$




      There are $2$ fair dice:




      • An $11$-sided die, valued from $-5$ to $5$,

      • A $41$-sided die, valued from $-20$ to $20$.


      You pick one, and I'll take the other one. We roll the dice, and whoever rolls the larger number wins (if tied, we roll again.)



      Which die will you choose to maximize your winning probability?







      probability dice






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Dec 17 '18 at 18:16









      JonMark Perry

      18k63786




      18k63786










      asked Dec 17 '18 at 16:37









      athinathin

      7,34822469




      7,34822469






















          6 Answers
          6






          active

          oldest

          votes


















          21












          $begingroup$

          The answer is




          The chances are equal




          Proof




          Suppose we consider all the possible outcomes. Then I propose, there is a one-to-one mapping from winning outcomes for you to winning outcomes for me.

          To see this, consider a scenario where you roll $i$ and I roll $j$ and you win ($i > j$). Then, in the case, that you roll $-i$ and I roll $-j$, I win (since ($-i < -j$). The mapping $(i,j) rightarrow (-i, -j)$ is clearly 1-1 for the given problem, it inverts the winner, and the set of draws maps onto itself.

          If we were to enumerate overall all possible outcomes, I would win the same number of times as you so the probability of winning must be the same.

          NB: This additionally show that the chances would be equal if we replaced $5$ and $20$ by any integer values.




          Alternative proof




          The probability of winning with the $41$-sided dice in a single go can be summed as the probability of winning given that the $11$-sided die rolls a $5,4,3,ldots$ multiplied by the probability of rolling a value in an $11$-sided die that is $$ P(41text{ wins}) = frac{1}{11}displaystyle sum_{j=15}^{25} frac{j}{41} = frac{220}{451}$$ Similarly, the probability of the $41$-sided die losing is given by calculating a similar sum $$ P(41 text{ loses}) = frac{1}{11}displaystyle sum_{j=25}^{15} frac{j}{41} = frac{220}{451}$$ where the sum is descending over the integers in this case. The probability of a single outcome being a draw is just $frac{11}{451}$.







          share|improve this answer











          $endgroup$





















            5












            $begingroup$

            It:




            Doesn't matter.




            Why?




            The D11 wins with probability $frac{220}{451}$, where $220=16+17+dots+24+25$. It's a tie with probability $frac{11}{451}=frac1{41}$, and the D41 also wins with a probability of $frac{220}{451}$.







            share|improve this answer









            $endgroup$





















              3












              $begingroup$

              I'd take the




              11-sided die




              because




              it has less sides, therefore it has less tendency to roll and thus it is easier to "aim" for a number with a controlled roll, at least after some practice. Even more so if the numbers are grouped with a cluster of positive numbers somewhere. Just like it is easier to throw a 4-sided die on a predetermined side than doing the same with a D20 which will inevitably roll away and land on a random number.







              share|improve this answer









              $endgroup$













              • $begingroup$
                An interesting take. The question talks about "fair dice" though. I'd think this implies no such tricks as well.
                $endgroup$
                – Gassa
                Dec 18 '18 at 10:13








              • 1




                $begingroup$
                @Gassa AFAIK "fair dice" means only that each of the faces have equal chance, i.e. it is not skewed, leaded, drilled, sanded or otherwise modified. A "fair roll" on the other hand would indeed imply no such tricks. As a corollary, my approach is of course not fair, but as the proper analytical answer has already been given and accepted, I ventured to offer a more unusual solution.
                $endgroup$
                – zovits
                Dec 18 '18 at 10:24



















              2












              $begingroup$

              Mapping out the 2-dimensional probability space you can see that the area in which each die wins is symmetrical, therefore it does not matter which die you choose, your odds of winning are the same.



               A -20 ... -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6 ... 20
              B
              -5 B B . A A A A A A A A A A A A
              -4 B B B . A A A A A A A A A A A
              -3 B B B B . A A A A A A A A A A
              -2 B B B B B . A A A A A A A A A
              -1 B B B B B B . A A A A A A A A
              0 B B B B B B B . A A A A A A A
              1 B B B B B B B B . A A A A A A
              2 B B B B B B B B B . A A A A A
              3 B B B B B B B B B B . A A A A
              4 B B B B B B B B B B B . A A A
              5 B B B B B B B B B B B B . A A





              share|improve this answer









              $endgroup$





















                1












                $begingroup$

                The average value of each die rolled an infinite number of times will be zero for either die. So it makes no difference as neither has an advantage over the other for gaining a larger number.






                share|improve this answer









                $endgroup$









                • 9




                  $begingroup$
                  Considering the long term average won't work in general. For example, consider the case where the first die has 40 on one side and -1 on every other side and the second die has -10 on one side and 1 on every other side. The average value is 0 for both dice but one of them is more likely to win.
                  $endgroup$
                  – hexomino
                  Dec 17 '18 at 17:12






                • 1




                  $begingroup$
                  That's not an even distribution of frequency though, OPs example is.
                  $endgroup$
                  – George Appleton
                  Dec 17 '18 at 17:17






                • 9




                  $begingroup$
                  Sure, but do you see that the first sentence of your answer is true for my choice of dice as well as the OP's?
                  $endgroup$
                  – hexomino
                  Dec 17 '18 at 17:21





















                -2












                $begingroup$

                I would take the 11 sided dice



                because
                P(getting a higher number in 20 sided dice ) = 1/11*25/41 + 1/11*24/41 + .. 1/11*15/41 = 220/451=10/41



                P(getting a higher number in 11 sided dice ) = P(getting -20 in dice2)*P(getting a higher number in dice2) + P(getting -19 in dice2)*P(getting a higher number in dice2) + ..
                =1/41*1 + 1/41*1 + .. etc = (1/41 )*15 + etc .. > 10/41



                Clearly 11 sided dice wins..






                share|improve this answer











                $endgroup$













                • $begingroup$
                  Okay but you have the exact opposite on the negative side. The numbers range from -20 to 20 not 0 to 40
                  $endgroup$
                  – George Appleton
                  Dec 21 '18 at 16:12













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                6 Answers
                6






                active

                oldest

                votes








                6 Answers
                6






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                21












                $begingroup$

                The answer is




                The chances are equal




                Proof




                Suppose we consider all the possible outcomes. Then I propose, there is a one-to-one mapping from winning outcomes for you to winning outcomes for me.

                To see this, consider a scenario where you roll $i$ and I roll $j$ and you win ($i > j$). Then, in the case, that you roll $-i$ and I roll $-j$, I win (since ($-i < -j$). The mapping $(i,j) rightarrow (-i, -j)$ is clearly 1-1 for the given problem, it inverts the winner, and the set of draws maps onto itself.

                If we were to enumerate overall all possible outcomes, I would win the same number of times as you so the probability of winning must be the same.

                NB: This additionally show that the chances would be equal if we replaced $5$ and $20$ by any integer values.




                Alternative proof




                The probability of winning with the $41$-sided dice in a single go can be summed as the probability of winning given that the $11$-sided die rolls a $5,4,3,ldots$ multiplied by the probability of rolling a value in an $11$-sided die that is $$ P(41text{ wins}) = frac{1}{11}displaystyle sum_{j=15}^{25} frac{j}{41} = frac{220}{451}$$ Similarly, the probability of the $41$-sided die losing is given by calculating a similar sum $$ P(41 text{ loses}) = frac{1}{11}displaystyle sum_{j=25}^{15} frac{j}{41} = frac{220}{451}$$ where the sum is descending over the integers in this case. The probability of a single outcome being a draw is just $frac{11}{451}$.







                share|improve this answer











                $endgroup$


















                  21












                  $begingroup$

                  The answer is




                  The chances are equal




                  Proof




                  Suppose we consider all the possible outcomes. Then I propose, there is a one-to-one mapping from winning outcomes for you to winning outcomes for me.

                  To see this, consider a scenario where you roll $i$ and I roll $j$ and you win ($i > j$). Then, in the case, that you roll $-i$ and I roll $-j$, I win (since ($-i < -j$). The mapping $(i,j) rightarrow (-i, -j)$ is clearly 1-1 for the given problem, it inverts the winner, and the set of draws maps onto itself.

                  If we were to enumerate overall all possible outcomes, I would win the same number of times as you so the probability of winning must be the same.

                  NB: This additionally show that the chances would be equal if we replaced $5$ and $20$ by any integer values.




                  Alternative proof




                  The probability of winning with the $41$-sided dice in a single go can be summed as the probability of winning given that the $11$-sided die rolls a $5,4,3,ldots$ multiplied by the probability of rolling a value in an $11$-sided die that is $$ P(41text{ wins}) = frac{1}{11}displaystyle sum_{j=15}^{25} frac{j}{41} = frac{220}{451}$$ Similarly, the probability of the $41$-sided die losing is given by calculating a similar sum $$ P(41 text{ loses}) = frac{1}{11}displaystyle sum_{j=25}^{15} frac{j}{41} = frac{220}{451}$$ where the sum is descending over the integers in this case. The probability of a single outcome being a draw is just $frac{11}{451}$.







                  share|improve this answer











                  $endgroup$
















                    21












                    21








                    21





                    $begingroup$

                    The answer is




                    The chances are equal




                    Proof




                    Suppose we consider all the possible outcomes. Then I propose, there is a one-to-one mapping from winning outcomes for you to winning outcomes for me.

                    To see this, consider a scenario where you roll $i$ and I roll $j$ and you win ($i > j$). Then, in the case, that you roll $-i$ and I roll $-j$, I win (since ($-i < -j$). The mapping $(i,j) rightarrow (-i, -j)$ is clearly 1-1 for the given problem, it inverts the winner, and the set of draws maps onto itself.

                    If we were to enumerate overall all possible outcomes, I would win the same number of times as you so the probability of winning must be the same.

                    NB: This additionally show that the chances would be equal if we replaced $5$ and $20$ by any integer values.




                    Alternative proof




                    The probability of winning with the $41$-sided dice in a single go can be summed as the probability of winning given that the $11$-sided die rolls a $5,4,3,ldots$ multiplied by the probability of rolling a value in an $11$-sided die that is $$ P(41text{ wins}) = frac{1}{11}displaystyle sum_{j=15}^{25} frac{j}{41} = frac{220}{451}$$ Similarly, the probability of the $41$-sided die losing is given by calculating a similar sum $$ P(41 text{ loses}) = frac{1}{11}displaystyle sum_{j=25}^{15} frac{j}{41} = frac{220}{451}$$ where the sum is descending over the integers in this case. The probability of a single outcome being a draw is just $frac{11}{451}$.







                    share|improve this answer











                    $endgroup$



                    The answer is




                    The chances are equal




                    Proof




                    Suppose we consider all the possible outcomes. Then I propose, there is a one-to-one mapping from winning outcomes for you to winning outcomes for me.

                    To see this, consider a scenario where you roll $i$ and I roll $j$ and you win ($i > j$). Then, in the case, that you roll $-i$ and I roll $-j$, I win (since ($-i < -j$). The mapping $(i,j) rightarrow (-i, -j)$ is clearly 1-1 for the given problem, it inverts the winner, and the set of draws maps onto itself.

                    If we were to enumerate overall all possible outcomes, I would win the same number of times as you so the probability of winning must be the same.

                    NB: This additionally show that the chances would be equal if we replaced $5$ and $20$ by any integer values.




                    Alternative proof




                    The probability of winning with the $41$-sided dice in a single go can be summed as the probability of winning given that the $11$-sided die rolls a $5,4,3,ldots$ multiplied by the probability of rolling a value in an $11$-sided die that is $$ P(41text{ wins}) = frac{1}{11}displaystyle sum_{j=15}^{25} frac{j}{41} = frac{220}{451}$$ Similarly, the probability of the $41$-sided die losing is given by calculating a similar sum $$ P(41 text{ loses}) = frac{1}{11}displaystyle sum_{j=25}^{15} frac{j}{41} = frac{220}{451}$$ where the sum is descending over the integers in this case. The probability of a single outcome being a draw is just $frac{11}{451}$.








                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Dec 17 '18 at 17:47

























                    answered Dec 17 '18 at 16:47









                    hexominohexomino

                    37.7k2111179




                    37.7k2111179























                        5












                        $begingroup$

                        It:




                        Doesn't matter.




                        Why?




                        The D11 wins with probability $frac{220}{451}$, where $220=16+17+dots+24+25$. It's a tie with probability $frac{11}{451}=frac1{41}$, and the D41 also wins with a probability of $frac{220}{451}$.







                        share|improve this answer









                        $endgroup$


















                          5












                          $begingroup$

                          It:




                          Doesn't matter.




                          Why?




                          The D11 wins with probability $frac{220}{451}$, where $220=16+17+dots+24+25$. It's a tie with probability $frac{11}{451}=frac1{41}$, and the D41 also wins with a probability of $frac{220}{451}$.







                          share|improve this answer









                          $endgroup$
















                            5












                            5








                            5





                            $begingroup$

                            It:




                            Doesn't matter.




                            Why?




                            The D11 wins with probability $frac{220}{451}$, where $220=16+17+dots+24+25$. It's a tie with probability $frac{11}{451}=frac1{41}$, and the D41 also wins with a probability of $frac{220}{451}$.







                            share|improve this answer









                            $endgroup$



                            It:




                            Doesn't matter.




                            Why?




                            The D11 wins with probability $frac{220}{451}$, where $220=16+17+dots+24+25$. It's a tie with probability $frac{11}{451}=frac1{41}$, and the D41 also wins with a probability of $frac{220}{451}$.








                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered Dec 17 '18 at 16:52









                            JonMark PerryJonMark Perry

                            18k63786




                            18k63786























                                3












                                $begingroup$

                                I'd take the




                                11-sided die




                                because




                                it has less sides, therefore it has less tendency to roll and thus it is easier to "aim" for a number with a controlled roll, at least after some practice. Even more so if the numbers are grouped with a cluster of positive numbers somewhere. Just like it is easier to throw a 4-sided die on a predetermined side than doing the same with a D20 which will inevitably roll away and land on a random number.







                                share|improve this answer









                                $endgroup$













                                • $begingroup$
                                  An interesting take. The question talks about "fair dice" though. I'd think this implies no such tricks as well.
                                  $endgroup$
                                  – Gassa
                                  Dec 18 '18 at 10:13








                                • 1




                                  $begingroup$
                                  @Gassa AFAIK "fair dice" means only that each of the faces have equal chance, i.e. it is not skewed, leaded, drilled, sanded or otherwise modified. A "fair roll" on the other hand would indeed imply no such tricks. As a corollary, my approach is of course not fair, but as the proper analytical answer has already been given and accepted, I ventured to offer a more unusual solution.
                                  $endgroup$
                                  – zovits
                                  Dec 18 '18 at 10:24
















                                3












                                $begingroup$

                                I'd take the




                                11-sided die




                                because




                                it has less sides, therefore it has less tendency to roll and thus it is easier to "aim" for a number with a controlled roll, at least after some practice. Even more so if the numbers are grouped with a cluster of positive numbers somewhere. Just like it is easier to throw a 4-sided die on a predetermined side than doing the same with a D20 which will inevitably roll away and land on a random number.







                                share|improve this answer









                                $endgroup$













                                • $begingroup$
                                  An interesting take. The question talks about "fair dice" though. I'd think this implies no such tricks as well.
                                  $endgroup$
                                  – Gassa
                                  Dec 18 '18 at 10:13








                                • 1




                                  $begingroup$
                                  @Gassa AFAIK "fair dice" means only that each of the faces have equal chance, i.e. it is not skewed, leaded, drilled, sanded or otherwise modified. A "fair roll" on the other hand would indeed imply no such tricks. As a corollary, my approach is of course not fair, but as the proper analytical answer has already been given and accepted, I ventured to offer a more unusual solution.
                                  $endgroup$
                                  – zovits
                                  Dec 18 '18 at 10:24














                                3












                                3








                                3





                                $begingroup$

                                I'd take the




                                11-sided die




                                because




                                it has less sides, therefore it has less tendency to roll and thus it is easier to "aim" for a number with a controlled roll, at least after some practice. Even more so if the numbers are grouped with a cluster of positive numbers somewhere. Just like it is easier to throw a 4-sided die on a predetermined side than doing the same with a D20 which will inevitably roll away and land on a random number.







                                share|improve this answer









                                $endgroup$



                                I'd take the




                                11-sided die




                                because




                                it has less sides, therefore it has less tendency to roll and thus it is easier to "aim" for a number with a controlled roll, at least after some practice. Even more so if the numbers are grouped with a cluster of positive numbers somewhere. Just like it is easier to throw a 4-sided die on a predetermined side than doing the same with a D20 which will inevitably roll away and land on a random number.








                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered Dec 18 '18 at 7:43









                                zovitszovits

                                1304




                                1304












                                • $begingroup$
                                  An interesting take. The question talks about "fair dice" though. I'd think this implies no such tricks as well.
                                  $endgroup$
                                  – Gassa
                                  Dec 18 '18 at 10:13








                                • 1




                                  $begingroup$
                                  @Gassa AFAIK "fair dice" means only that each of the faces have equal chance, i.e. it is not skewed, leaded, drilled, sanded or otherwise modified. A "fair roll" on the other hand would indeed imply no such tricks. As a corollary, my approach is of course not fair, but as the proper analytical answer has already been given and accepted, I ventured to offer a more unusual solution.
                                  $endgroup$
                                  – zovits
                                  Dec 18 '18 at 10:24


















                                • $begingroup$
                                  An interesting take. The question talks about "fair dice" though. I'd think this implies no such tricks as well.
                                  $endgroup$
                                  – Gassa
                                  Dec 18 '18 at 10:13








                                • 1




                                  $begingroup$
                                  @Gassa AFAIK "fair dice" means only that each of the faces have equal chance, i.e. it is not skewed, leaded, drilled, sanded or otherwise modified. A "fair roll" on the other hand would indeed imply no such tricks. As a corollary, my approach is of course not fair, but as the proper analytical answer has already been given and accepted, I ventured to offer a more unusual solution.
                                  $endgroup$
                                  – zovits
                                  Dec 18 '18 at 10:24
















                                $begingroup$
                                An interesting take. The question talks about "fair dice" though. I'd think this implies no such tricks as well.
                                $endgroup$
                                – Gassa
                                Dec 18 '18 at 10:13






                                $begingroup$
                                An interesting take. The question talks about "fair dice" though. I'd think this implies no such tricks as well.
                                $endgroup$
                                – Gassa
                                Dec 18 '18 at 10:13






                                1




                                1




                                $begingroup$
                                @Gassa AFAIK "fair dice" means only that each of the faces have equal chance, i.e. it is not skewed, leaded, drilled, sanded or otherwise modified. A "fair roll" on the other hand would indeed imply no such tricks. As a corollary, my approach is of course not fair, but as the proper analytical answer has already been given and accepted, I ventured to offer a more unusual solution.
                                $endgroup$
                                – zovits
                                Dec 18 '18 at 10:24




                                $begingroup$
                                @Gassa AFAIK "fair dice" means only that each of the faces have equal chance, i.e. it is not skewed, leaded, drilled, sanded or otherwise modified. A "fair roll" on the other hand would indeed imply no such tricks. As a corollary, my approach is of course not fair, but as the proper analytical answer has already been given and accepted, I ventured to offer a more unusual solution.
                                $endgroup$
                                – zovits
                                Dec 18 '18 at 10:24











                                2












                                $begingroup$

                                Mapping out the 2-dimensional probability space you can see that the area in which each die wins is symmetrical, therefore it does not matter which die you choose, your odds of winning are the same.



                                 A -20 ... -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6 ... 20
                                B
                                -5 B B . A A A A A A A A A A A A
                                -4 B B B . A A A A A A A A A A A
                                -3 B B B B . A A A A A A A A A A
                                -2 B B B B B . A A A A A A A A A
                                -1 B B B B B B . A A A A A A A A
                                0 B B B B B B B . A A A A A A A
                                1 B B B B B B B B . A A A A A A
                                2 B B B B B B B B B . A A A A A
                                3 B B B B B B B B B B . A A A A
                                4 B B B B B B B B B B B . A A A
                                5 B B B B B B B B B B B B . A A





                                share|improve this answer









                                $endgroup$


















                                  2












                                  $begingroup$

                                  Mapping out the 2-dimensional probability space you can see that the area in which each die wins is symmetrical, therefore it does not matter which die you choose, your odds of winning are the same.



                                   A -20 ... -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6 ... 20
                                  B
                                  -5 B B . A A A A A A A A A A A A
                                  -4 B B B . A A A A A A A A A A A
                                  -3 B B B B . A A A A A A A A A A
                                  -2 B B B B B . A A A A A A A A A
                                  -1 B B B B B B . A A A A A A A A
                                  0 B B B B B B B . A A A A A A A
                                  1 B B B B B B B B . A A A A A A
                                  2 B B B B B B B B B . A A A A A
                                  3 B B B B B B B B B B . A A A A
                                  4 B B B B B B B B B B B . A A A
                                  5 B B B B B B B B B B B B . A A





                                  share|improve this answer









                                  $endgroup$
















                                    2












                                    2








                                    2





                                    $begingroup$

                                    Mapping out the 2-dimensional probability space you can see that the area in which each die wins is symmetrical, therefore it does not matter which die you choose, your odds of winning are the same.



                                     A -20 ... -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6 ... 20
                                    B
                                    -5 B B . A A A A A A A A A A A A
                                    -4 B B B . A A A A A A A A A A A
                                    -3 B B B B . A A A A A A A A A A
                                    -2 B B B B B . A A A A A A A A A
                                    -1 B B B B B B . A A A A A A A A
                                    0 B B B B B B B . A A A A A A A
                                    1 B B B B B B B B . A A A A A A
                                    2 B B B B B B B B B . A A A A A
                                    3 B B B B B B B B B B . A A A A
                                    4 B B B B B B B B B B B . A A A
                                    5 B B B B B B B B B B B B . A A





                                    share|improve this answer









                                    $endgroup$



                                    Mapping out the 2-dimensional probability space you can see that the area in which each die wins is symmetrical, therefore it does not matter which die you choose, your odds of winning are the same.



                                     A -20 ... -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6 ... 20
                                    B
                                    -5 B B . A A A A A A A A A A A A
                                    -4 B B B . A A A A A A A A A A A
                                    -3 B B B B . A A A A A A A A A A
                                    -2 B B B B B . A A A A A A A A A
                                    -1 B B B B B B . A A A A A A A A
                                    0 B B B B B B B . A A A A A A A
                                    1 B B B B B B B B . A A A A A A
                                    2 B B B B B B B B B . A A A A A
                                    3 B B B B B B B B B B . A A A A
                                    4 B B B B B B B B B B B . A A A
                                    5 B B B B B B B B B B B B . A A






                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered Dec 19 '18 at 15:13









                                    Paul SmithPaul Smith

                                    1213




                                    1213























                                        1












                                        $begingroup$

                                        The average value of each die rolled an infinite number of times will be zero for either die. So it makes no difference as neither has an advantage over the other for gaining a larger number.






                                        share|improve this answer









                                        $endgroup$









                                        • 9




                                          $begingroup$
                                          Considering the long term average won't work in general. For example, consider the case where the first die has 40 on one side and -1 on every other side and the second die has -10 on one side and 1 on every other side. The average value is 0 for both dice but one of them is more likely to win.
                                          $endgroup$
                                          – hexomino
                                          Dec 17 '18 at 17:12






                                        • 1




                                          $begingroup$
                                          That's not an even distribution of frequency though, OPs example is.
                                          $endgroup$
                                          – George Appleton
                                          Dec 17 '18 at 17:17






                                        • 9




                                          $begingroup$
                                          Sure, but do you see that the first sentence of your answer is true for my choice of dice as well as the OP's?
                                          $endgroup$
                                          – hexomino
                                          Dec 17 '18 at 17:21


















                                        1












                                        $begingroup$

                                        The average value of each die rolled an infinite number of times will be zero for either die. So it makes no difference as neither has an advantage over the other for gaining a larger number.






                                        share|improve this answer









                                        $endgroup$









                                        • 9




                                          $begingroup$
                                          Considering the long term average won't work in general. For example, consider the case where the first die has 40 on one side and -1 on every other side and the second die has -10 on one side and 1 on every other side. The average value is 0 for both dice but one of them is more likely to win.
                                          $endgroup$
                                          – hexomino
                                          Dec 17 '18 at 17:12






                                        • 1




                                          $begingroup$
                                          That's not an even distribution of frequency though, OPs example is.
                                          $endgroup$
                                          – George Appleton
                                          Dec 17 '18 at 17:17






                                        • 9




                                          $begingroup$
                                          Sure, but do you see that the first sentence of your answer is true for my choice of dice as well as the OP's?
                                          $endgroup$
                                          – hexomino
                                          Dec 17 '18 at 17:21
















                                        1












                                        1








                                        1





                                        $begingroup$

                                        The average value of each die rolled an infinite number of times will be zero for either die. So it makes no difference as neither has an advantage over the other for gaining a larger number.






                                        share|improve this answer









                                        $endgroup$



                                        The average value of each die rolled an infinite number of times will be zero for either die. So it makes no difference as neither has an advantage over the other for gaining a larger number.







                                        share|improve this answer












                                        share|improve this answer



                                        share|improve this answer










                                        answered Dec 17 '18 at 16:47









                                        George AppletonGeorge Appleton

                                        1192




                                        1192








                                        • 9




                                          $begingroup$
                                          Considering the long term average won't work in general. For example, consider the case where the first die has 40 on one side and -1 on every other side and the second die has -10 on one side and 1 on every other side. The average value is 0 for both dice but one of them is more likely to win.
                                          $endgroup$
                                          – hexomino
                                          Dec 17 '18 at 17:12






                                        • 1




                                          $begingroup$
                                          That's not an even distribution of frequency though, OPs example is.
                                          $endgroup$
                                          – George Appleton
                                          Dec 17 '18 at 17:17






                                        • 9




                                          $begingroup$
                                          Sure, but do you see that the first sentence of your answer is true for my choice of dice as well as the OP's?
                                          $endgroup$
                                          – hexomino
                                          Dec 17 '18 at 17:21
















                                        • 9




                                          $begingroup$
                                          Considering the long term average won't work in general. For example, consider the case where the first die has 40 on one side and -1 on every other side and the second die has -10 on one side and 1 on every other side. The average value is 0 for both dice but one of them is more likely to win.
                                          $endgroup$
                                          – hexomino
                                          Dec 17 '18 at 17:12






                                        • 1




                                          $begingroup$
                                          That's not an even distribution of frequency though, OPs example is.
                                          $endgroup$
                                          – George Appleton
                                          Dec 17 '18 at 17:17






                                        • 9




                                          $begingroup$
                                          Sure, but do you see that the first sentence of your answer is true for my choice of dice as well as the OP's?
                                          $endgroup$
                                          – hexomino
                                          Dec 17 '18 at 17:21










                                        9




                                        9




                                        $begingroup$
                                        Considering the long term average won't work in general. For example, consider the case where the first die has 40 on one side and -1 on every other side and the second die has -10 on one side and 1 on every other side. The average value is 0 for both dice but one of them is more likely to win.
                                        $endgroup$
                                        – hexomino
                                        Dec 17 '18 at 17:12




                                        $begingroup$
                                        Considering the long term average won't work in general. For example, consider the case where the first die has 40 on one side and -1 on every other side and the second die has -10 on one side and 1 on every other side. The average value is 0 for both dice but one of them is more likely to win.
                                        $endgroup$
                                        – hexomino
                                        Dec 17 '18 at 17:12




                                        1




                                        1




                                        $begingroup$
                                        That's not an even distribution of frequency though, OPs example is.
                                        $endgroup$
                                        – George Appleton
                                        Dec 17 '18 at 17:17




                                        $begingroup$
                                        That's not an even distribution of frequency though, OPs example is.
                                        $endgroup$
                                        – George Appleton
                                        Dec 17 '18 at 17:17




                                        9




                                        9




                                        $begingroup$
                                        Sure, but do you see that the first sentence of your answer is true for my choice of dice as well as the OP's?
                                        $endgroup$
                                        – hexomino
                                        Dec 17 '18 at 17:21






                                        $begingroup$
                                        Sure, but do you see that the first sentence of your answer is true for my choice of dice as well as the OP's?
                                        $endgroup$
                                        – hexomino
                                        Dec 17 '18 at 17:21













                                        -2












                                        $begingroup$

                                        I would take the 11 sided dice



                                        because
                                        P(getting a higher number in 20 sided dice ) = 1/11*25/41 + 1/11*24/41 + .. 1/11*15/41 = 220/451=10/41



                                        P(getting a higher number in 11 sided dice ) = P(getting -20 in dice2)*P(getting a higher number in dice2) + P(getting -19 in dice2)*P(getting a higher number in dice2) + ..
                                        =1/41*1 + 1/41*1 + .. etc = (1/41 )*15 + etc .. > 10/41



                                        Clearly 11 sided dice wins..






                                        share|improve this answer











                                        $endgroup$













                                        • $begingroup$
                                          Okay but you have the exact opposite on the negative side. The numbers range from -20 to 20 not 0 to 40
                                          $endgroup$
                                          – George Appleton
                                          Dec 21 '18 at 16:12


















                                        -2












                                        $begingroup$

                                        I would take the 11 sided dice



                                        because
                                        P(getting a higher number in 20 sided dice ) = 1/11*25/41 + 1/11*24/41 + .. 1/11*15/41 = 220/451=10/41



                                        P(getting a higher number in 11 sided dice ) = P(getting -20 in dice2)*P(getting a higher number in dice2) + P(getting -19 in dice2)*P(getting a higher number in dice2) + ..
                                        =1/41*1 + 1/41*1 + .. etc = (1/41 )*15 + etc .. > 10/41



                                        Clearly 11 sided dice wins..






                                        share|improve this answer











                                        $endgroup$













                                        • $begingroup$
                                          Okay but you have the exact opposite on the negative side. The numbers range from -20 to 20 not 0 to 40
                                          $endgroup$
                                          – George Appleton
                                          Dec 21 '18 at 16:12
















                                        -2












                                        -2








                                        -2





                                        $begingroup$

                                        I would take the 11 sided dice



                                        because
                                        P(getting a higher number in 20 sided dice ) = 1/11*25/41 + 1/11*24/41 + .. 1/11*15/41 = 220/451=10/41



                                        P(getting a higher number in 11 sided dice ) = P(getting -20 in dice2)*P(getting a higher number in dice2) + P(getting -19 in dice2)*P(getting a higher number in dice2) + ..
                                        =1/41*1 + 1/41*1 + .. etc = (1/41 )*15 + etc .. > 10/41



                                        Clearly 11 sided dice wins..






                                        share|improve this answer











                                        $endgroup$



                                        I would take the 11 sided dice



                                        because
                                        P(getting a higher number in 20 sided dice ) = 1/11*25/41 + 1/11*24/41 + .. 1/11*15/41 = 220/451=10/41



                                        P(getting a higher number in 11 sided dice ) = P(getting -20 in dice2)*P(getting a higher number in dice2) + P(getting -19 in dice2)*P(getting a higher number in dice2) + ..
                                        =1/41*1 + 1/41*1 + .. etc = (1/41 )*15 + etc .. > 10/41



                                        Clearly 11 sided dice wins..







                                        share|improve this answer














                                        share|improve this answer



                                        share|improve this answer








                                        edited Dec 19 '18 at 8:31

























                                        answered Dec 18 '18 at 10:24









                                        Liji JoseLiji Jose

                                        12




                                        12












                                        • $begingroup$
                                          Okay but you have the exact opposite on the negative side. The numbers range from -20 to 20 not 0 to 40
                                          $endgroup$
                                          – George Appleton
                                          Dec 21 '18 at 16:12




















                                        • $begingroup$
                                          Okay but you have the exact opposite on the negative side. The numbers range from -20 to 20 not 0 to 40
                                          $endgroup$
                                          – George Appleton
                                          Dec 21 '18 at 16:12


















                                        $begingroup$
                                        Okay but you have the exact opposite on the negative side. The numbers range from -20 to 20 not 0 to 40
                                        $endgroup$
                                        – George Appleton
                                        Dec 21 '18 at 16:12






                                        $begingroup$
                                        Okay but you have the exact opposite on the negative side. The numbers range from -20 to 20 not 0 to 40
                                        $endgroup$
                                        – George Appleton
                                        Dec 21 '18 at 16:12




















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