Choosing an integer that is most likely to be the sum of die rolls












1












$begingroup$


Choose any positive integer $n$. Then roll an unbiased six-sided die as many times as needed until the sum of the results is either exactly equal to $n$ (you win) or greater than $n$ (you lose).



If you were to play this game; what number $n$ would you select in order to maximise the chance of winning and what is the probability of it?



(For example, choose $n=7$, then you would roll a die and if you get $6$ on the 1st throw and $1$ on the 2nd throw you would stop since you reached $n$).










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Choose any positive integer $n$. Then roll an unbiased six-sided die as many times as needed until the sum of the results is either exactly equal to $n$ (you win) or greater than $n$ (you lose).



    If you were to play this game; what number $n$ would you select in order to maximise the chance of winning and what is the probability of it?



    (For example, choose $n=7$, then you would roll a die and if you get $6$ on the 1st throw and $1$ on the 2nd throw you would stop since you reached $n$).










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Choose any positive integer $n$. Then roll an unbiased six-sided die as many times as needed until the sum of the results is either exactly equal to $n$ (you win) or greater than $n$ (you lose).



      If you were to play this game; what number $n$ would you select in order to maximise the chance of winning and what is the probability of it?



      (For example, choose $n=7$, then you would roll a die and if you get $6$ on the 1st throw and $1$ on the 2nd throw you would stop since you reached $n$).










      share|cite|improve this question









      $endgroup$




      Choose any positive integer $n$. Then roll an unbiased six-sided die as many times as needed until the sum of the results is either exactly equal to $n$ (you win) or greater than $n$ (you lose).



      If you were to play this game; what number $n$ would you select in order to maximise the chance of winning and what is the probability of it?



      (For example, choose $n=7$, then you would roll a die and if you get $6$ on the 1st throw and $1$ on the 2nd throw you would stop since you reached $n$).







      probability dice






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 11 at 18:30









      MilTomMilTom

      1288




      1288






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          Denote by $p_{k}$ the probability that a sum of $k$ is observed. Clearly
          $$p_{1}=frac{1}{6}$$
          For $p_{2}$, there is a $frac{1}{6}$ chance of rolling a $2$ and a $frac{1}{6}$ chance of rolling a $1$, followed by another one. This can be written as
          $$p_{2}=frac{1}{6}+frac{p_{1}}{6}=frac{1}{6}(1+p_{1})$$



          For $p_{3}$, there is a $frac{1}{6}$ chance of rolling a $3$ and winning. The only other ways to win are to roll a $1$ or $2$, each with probability $frac{1}{6}$. If we roll a $1$, then our odds of winning are $p_{2}$ and if we roll a $2$, our odds of winning are $p_{1}$. Therefore
          $$p_{3}=frac{1}{6}+frac{p_{1}}{6}+frac{p_{2}}{6}=frac{1}{6}(1+p_{1}+p_{2})$$
          Similarly we have
          $$p_{4}=frac{1}{6}(1+p_{1}+p_{2}+p_{3})$$
          $$p_{5}=frac{1}{6}(1+p_{1}+p_{2}+p_{3}+p_{4})$$
          $$p_{6}=frac{1}{6}(1+p_{1}+p_{2}+p_{3}+p_{4}+p_{5})$$
          Now, for $k>6$ we cannot win in one roll so we have
          $$p_{k}=frac{1}{6}(p_{k-1}+p_{k-2}+p_{k-3}+p_{k-4}+p_{k-5}+p_{k-6})$$
          Observe that for any $k>6$,
          $$p_{k}=frac{1}{6}sum_{m=1}^{6}p_{k-m}<frac{1}{6}cdot 6max_{m=1,2,3,4,5,6}p_{k-m}=max_{m=1,2,3,4,5,6}p_{k-m}$$
          This proves that the winning choice is not bigger than $6$, and in fact must be the max of $p_{1},p_{2},...p_{6}$. Clearly this is $p_{6}$, so we should choose $n=6$ to maximize the odds of winning.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            "Observe that for any $k$, $$sum_{m=1}^{6}p_{k-m}<1".$$ The "observation" is wrong for all $k>6$.
            $endgroup$
            – user
            Jan 11 at 23:13












          • $begingroup$
            Actually $p_k gt frac14$ for $k gt 6$, contrary to your statement $p_k lt frac16$
            $endgroup$
            – Henry
            Jan 11 at 23:15










          • $begingroup$
            You are both correct. I need to slightly reword my definition of $p_{n}$.
            $endgroup$
            – pwerth
            Jan 12 at 0:53



















          4












          $begingroup$

          In general you can reach $n$ directly from any of the six preceding numbers each with probability $frac16$ so if $p_n$ is the probability that a sum of $n$ is hit then $$p_n=frac{1}{6}left(p_{n-1}+p_{n-2}+p_{n-3}+p_{n-4}+p_{n-5}+p_{n-6}right)$$ starting with $p_0=1$ and $p_{-1}=p_{-2}=p_{-3}=p_{-4}=p_{-5}=0$



          This is increasing from $n=1$ to $n=6$ since in this range $p_n=p_{n-1}+frac16p_{n-1}=frac76p_{n-1}=frac{7^{n-1}}{6^n}$. In particular $p_6=frac{7^5}{6^n}approx 0.3602323$



          $p_6$ must then (by induction) be the highest $p_n$ for positive $n$, since $p_n$ for $n gt 6$ is the average of the preceding six values, which will be strictly less than $p_6$. The average roll is $3.5$ so about $frac{1}{3.5} =frac27$ of the values are hit, meaning $p_n to frac27 approx 0.2857143$ as $n$ increases. Actual values for smaller $n$ are:



          p_0                 1 / 1               1
          p_1 1 / 6 0.1666667
          p_2 7 / 36 0.1944444
          p_3 49 / 216 0.2268519
          p_4 343 / 1296 0.2646605
          p_5 2401 / 7776 0.3087706
          p_6 16807 / 46656 0.3602323
          p_7 70993 / 279936 0.2536044
          p_8 450295 / 1679616 0.2680940
          p_9 2825473 / 10077696 0.2803689
          p_10 17492167 / 60466176 0.2892885
          p_11 106442161 / 362797056 0.2933931
          p_12 633074071 / 2176782336 0.2908302
          p_13 3647371105 / 13060694016 0.2792632
          p_14 22219348327 / 78364164096 0.2835397
          p_15 134526474769 / 470184984576 0.2861139
          p_16 809860055095 / 2821109907456 0.2870714
          p_17 4852905842113 / 16926659444736 0.2867019





          share|cite|improve this answer









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070185%2fchoosing-an-integer-that-is-most-likely-to-be-the-sum-of-die-rolls%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Denote by $p_{k}$ the probability that a sum of $k$ is observed. Clearly
            $$p_{1}=frac{1}{6}$$
            For $p_{2}$, there is a $frac{1}{6}$ chance of rolling a $2$ and a $frac{1}{6}$ chance of rolling a $1$, followed by another one. This can be written as
            $$p_{2}=frac{1}{6}+frac{p_{1}}{6}=frac{1}{6}(1+p_{1})$$



            For $p_{3}$, there is a $frac{1}{6}$ chance of rolling a $3$ and winning. The only other ways to win are to roll a $1$ or $2$, each with probability $frac{1}{6}$. If we roll a $1$, then our odds of winning are $p_{2}$ and if we roll a $2$, our odds of winning are $p_{1}$. Therefore
            $$p_{3}=frac{1}{6}+frac{p_{1}}{6}+frac{p_{2}}{6}=frac{1}{6}(1+p_{1}+p_{2})$$
            Similarly we have
            $$p_{4}=frac{1}{6}(1+p_{1}+p_{2}+p_{3})$$
            $$p_{5}=frac{1}{6}(1+p_{1}+p_{2}+p_{3}+p_{4})$$
            $$p_{6}=frac{1}{6}(1+p_{1}+p_{2}+p_{3}+p_{4}+p_{5})$$
            Now, for $k>6$ we cannot win in one roll so we have
            $$p_{k}=frac{1}{6}(p_{k-1}+p_{k-2}+p_{k-3}+p_{k-4}+p_{k-5}+p_{k-6})$$
            Observe that for any $k>6$,
            $$p_{k}=frac{1}{6}sum_{m=1}^{6}p_{k-m}<frac{1}{6}cdot 6max_{m=1,2,3,4,5,6}p_{k-m}=max_{m=1,2,3,4,5,6}p_{k-m}$$
            This proves that the winning choice is not bigger than $6$, and in fact must be the max of $p_{1},p_{2},...p_{6}$. Clearly this is $p_{6}$, so we should choose $n=6$ to maximize the odds of winning.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              "Observe that for any $k$, $$sum_{m=1}^{6}p_{k-m}<1".$$ The "observation" is wrong for all $k>6$.
              $endgroup$
              – user
              Jan 11 at 23:13












            • $begingroup$
              Actually $p_k gt frac14$ for $k gt 6$, contrary to your statement $p_k lt frac16$
              $endgroup$
              – Henry
              Jan 11 at 23:15










            • $begingroup$
              You are both correct. I need to slightly reword my definition of $p_{n}$.
              $endgroup$
              – pwerth
              Jan 12 at 0:53
















            3












            $begingroup$

            Denote by $p_{k}$ the probability that a sum of $k$ is observed. Clearly
            $$p_{1}=frac{1}{6}$$
            For $p_{2}$, there is a $frac{1}{6}$ chance of rolling a $2$ and a $frac{1}{6}$ chance of rolling a $1$, followed by another one. This can be written as
            $$p_{2}=frac{1}{6}+frac{p_{1}}{6}=frac{1}{6}(1+p_{1})$$



            For $p_{3}$, there is a $frac{1}{6}$ chance of rolling a $3$ and winning. The only other ways to win are to roll a $1$ or $2$, each with probability $frac{1}{6}$. If we roll a $1$, then our odds of winning are $p_{2}$ and if we roll a $2$, our odds of winning are $p_{1}$. Therefore
            $$p_{3}=frac{1}{6}+frac{p_{1}}{6}+frac{p_{2}}{6}=frac{1}{6}(1+p_{1}+p_{2})$$
            Similarly we have
            $$p_{4}=frac{1}{6}(1+p_{1}+p_{2}+p_{3})$$
            $$p_{5}=frac{1}{6}(1+p_{1}+p_{2}+p_{3}+p_{4})$$
            $$p_{6}=frac{1}{6}(1+p_{1}+p_{2}+p_{3}+p_{4}+p_{5})$$
            Now, for $k>6$ we cannot win in one roll so we have
            $$p_{k}=frac{1}{6}(p_{k-1}+p_{k-2}+p_{k-3}+p_{k-4}+p_{k-5}+p_{k-6})$$
            Observe that for any $k>6$,
            $$p_{k}=frac{1}{6}sum_{m=1}^{6}p_{k-m}<frac{1}{6}cdot 6max_{m=1,2,3,4,5,6}p_{k-m}=max_{m=1,2,3,4,5,6}p_{k-m}$$
            This proves that the winning choice is not bigger than $6$, and in fact must be the max of $p_{1},p_{2},...p_{6}$. Clearly this is $p_{6}$, so we should choose $n=6$ to maximize the odds of winning.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              "Observe that for any $k$, $$sum_{m=1}^{6}p_{k-m}<1".$$ The "observation" is wrong for all $k>6$.
              $endgroup$
              – user
              Jan 11 at 23:13












            • $begingroup$
              Actually $p_k gt frac14$ for $k gt 6$, contrary to your statement $p_k lt frac16$
              $endgroup$
              – Henry
              Jan 11 at 23:15










            • $begingroup$
              You are both correct. I need to slightly reword my definition of $p_{n}$.
              $endgroup$
              – pwerth
              Jan 12 at 0:53














            3












            3








            3





            $begingroup$

            Denote by $p_{k}$ the probability that a sum of $k$ is observed. Clearly
            $$p_{1}=frac{1}{6}$$
            For $p_{2}$, there is a $frac{1}{6}$ chance of rolling a $2$ and a $frac{1}{6}$ chance of rolling a $1$, followed by another one. This can be written as
            $$p_{2}=frac{1}{6}+frac{p_{1}}{6}=frac{1}{6}(1+p_{1})$$



            For $p_{3}$, there is a $frac{1}{6}$ chance of rolling a $3$ and winning. The only other ways to win are to roll a $1$ or $2$, each with probability $frac{1}{6}$. If we roll a $1$, then our odds of winning are $p_{2}$ and if we roll a $2$, our odds of winning are $p_{1}$. Therefore
            $$p_{3}=frac{1}{6}+frac{p_{1}}{6}+frac{p_{2}}{6}=frac{1}{6}(1+p_{1}+p_{2})$$
            Similarly we have
            $$p_{4}=frac{1}{6}(1+p_{1}+p_{2}+p_{3})$$
            $$p_{5}=frac{1}{6}(1+p_{1}+p_{2}+p_{3}+p_{4})$$
            $$p_{6}=frac{1}{6}(1+p_{1}+p_{2}+p_{3}+p_{4}+p_{5})$$
            Now, for $k>6$ we cannot win in one roll so we have
            $$p_{k}=frac{1}{6}(p_{k-1}+p_{k-2}+p_{k-3}+p_{k-4}+p_{k-5}+p_{k-6})$$
            Observe that for any $k>6$,
            $$p_{k}=frac{1}{6}sum_{m=1}^{6}p_{k-m}<frac{1}{6}cdot 6max_{m=1,2,3,4,5,6}p_{k-m}=max_{m=1,2,3,4,5,6}p_{k-m}$$
            This proves that the winning choice is not bigger than $6$, and in fact must be the max of $p_{1},p_{2},...p_{6}$. Clearly this is $p_{6}$, so we should choose $n=6$ to maximize the odds of winning.






            share|cite|improve this answer











            $endgroup$



            Denote by $p_{k}$ the probability that a sum of $k$ is observed. Clearly
            $$p_{1}=frac{1}{6}$$
            For $p_{2}$, there is a $frac{1}{6}$ chance of rolling a $2$ and a $frac{1}{6}$ chance of rolling a $1$, followed by another one. This can be written as
            $$p_{2}=frac{1}{6}+frac{p_{1}}{6}=frac{1}{6}(1+p_{1})$$



            For $p_{3}$, there is a $frac{1}{6}$ chance of rolling a $3$ and winning. The only other ways to win are to roll a $1$ or $2$, each with probability $frac{1}{6}$. If we roll a $1$, then our odds of winning are $p_{2}$ and if we roll a $2$, our odds of winning are $p_{1}$. Therefore
            $$p_{3}=frac{1}{6}+frac{p_{1}}{6}+frac{p_{2}}{6}=frac{1}{6}(1+p_{1}+p_{2})$$
            Similarly we have
            $$p_{4}=frac{1}{6}(1+p_{1}+p_{2}+p_{3})$$
            $$p_{5}=frac{1}{6}(1+p_{1}+p_{2}+p_{3}+p_{4})$$
            $$p_{6}=frac{1}{6}(1+p_{1}+p_{2}+p_{3}+p_{4}+p_{5})$$
            Now, for $k>6$ we cannot win in one roll so we have
            $$p_{k}=frac{1}{6}(p_{k-1}+p_{k-2}+p_{k-3}+p_{k-4}+p_{k-5}+p_{k-6})$$
            Observe that for any $k>6$,
            $$p_{k}=frac{1}{6}sum_{m=1}^{6}p_{k-m}<frac{1}{6}cdot 6max_{m=1,2,3,4,5,6}p_{k-m}=max_{m=1,2,3,4,5,6}p_{k-m}$$
            This proves that the winning choice is not bigger than $6$, and in fact must be the max of $p_{1},p_{2},...p_{6}$. Clearly this is $p_{6}$, so we should choose $n=6$ to maximize the odds of winning.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 12 at 0:57

























            answered Jan 11 at 18:52









            pwerthpwerth

            3,265417




            3,265417












            • $begingroup$
              "Observe that for any $k$, $$sum_{m=1}^{6}p_{k-m}<1".$$ The "observation" is wrong for all $k>6$.
              $endgroup$
              – user
              Jan 11 at 23:13












            • $begingroup$
              Actually $p_k gt frac14$ for $k gt 6$, contrary to your statement $p_k lt frac16$
              $endgroup$
              – Henry
              Jan 11 at 23:15










            • $begingroup$
              You are both correct. I need to slightly reword my definition of $p_{n}$.
              $endgroup$
              – pwerth
              Jan 12 at 0:53


















            • $begingroup$
              "Observe that for any $k$, $$sum_{m=1}^{6}p_{k-m}<1".$$ The "observation" is wrong for all $k>6$.
              $endgroup$
              – user
              Jan 11 at 23:13












            • $begingroup$
              Actually $p_k gt frac14$ for $k gt 6$, contrary to your statement $p_k lt frac16$
              $endgroup$
              – Henry
              Jan 11 at 23:15










            • $begingroup$
              You are both correct. I need to slightly reword my definition of $p_{n}$.
              $endgroup$
              – pwerth
              Jan 12 at 0:53
















            $begingroup$
            "Observe that for any $k$, $$sum_{m=1}^{6}p_{k-m}<1".$$ The "observation" is wrong for all $k>6$.
            $endgroup$
            – user
            Jan 11 at 23:13






            $begingroup$
            "Observe that for any $k$, $$sum_{m=1}^{6}p_{k-m}<1".$$ The "observation" is wrong for all $k>6$.
            $endgroup$
            – user
            Jan 11 at 23:13














            $begingroup$
            Actually $p_k gt frac14$ for $k gt 6$, contrary to your statement $p_k lt frac16$
            $endgroup$
            – Henry
            Jan 11 at 23:15




            $begingroup$
            Actually $p_k gt frac14$ for $k gt 6$, contrary to your statement $p_k lt frac16$
            $endgroup$
            – Henry
            Jan 11 at 23:15












            $begingroup$
            You are both correct. I need to slightly reword my definition of $p_{n}$.
            $endgroup$
            – pwerth
            Jan 12 at 0:53




            $begingroup$
            You are both correct. I need to slightly reword my definition of $p_{n}$.
            $endgroup$
            – pwerth
            Jan 12 at 0:53











            4












            $begingroup$

            In general you can reach $n$ directly from any of the six preceding numbers each with probability $frac16$ so if $p_n$ is the probability that a sum of $n$ is hit then $$p_n=frac{1}{6}left(p_{n-1}+p_{n-2}+p_{n-3}+p_{n-4}+p_{n-5}+p_{n-6}right)$$ starting with $p_0=1$ and $p_{-1}=p_{-2}=p_{-3}=p_{-4}=p_{-5}=0$



            This is increasing from $n=1$ to $n=6$ since in this range $p_n=p_{n-1}+frac16p_{n-1}=frac76p_{n-1}=frac{7^{n-1}}{6^n}$. In particular $p_6=frac{7^5}{6^n}approx 0.3602323$



            $p_6$ must then (by induction) be the highest $p_n$ for positive $n$, since $p_n$ for $n gt 6$ is the average of the preceding six values, which will be strictly less than $p_6$. The average roll is $3.5$ so about $frac{1}{3.5} =frac27$ of the values are hit, meaning $p_n to frac27 approx 0.2857143$ as $n$ increases. Actual values for smaller $n$ are:



            p_0                 1 / 1               1
            p_1 1 / 6 0.1666667
            p_2 7 / 36 0.1944444
            p_3 49 / 216 0.2268519
            p_4 343 / 1296 0.2646605
            p_5 2401 / 7776 0.3087706
            p_6 16807 / 46656 0.3602323
            p_7 70993 / 279936 0.2536044
            p_8 450295 / 1679616 0.2680940
            p_9 2825473 / 10077696 0.2803689
            p_10 17492167 / 60466176 0.2892885
            p_11 106442161 / 362797056 0.2933931
            p_12 633074071 / 2176782336 0.2908302
            p_13 3647371105 / 13060694016 0.2792632
            p_14 22219348327 / 78364164096 0.2835397
            p_15 134526474769 / 470184984576 0.2861139
            p_16 809860055095 / 2821109907456 0.2870714
            p_17 4852905842113 / 16926659444736 0.2867019





            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              In general you can reach $n$ directly from any of the six preceding numbers each with probability $frac16$ so if $p_n$ is the probability that a sum of $n$ is hit then $$p_n=frac{1}{6}left(p_{n-1}+p_{n-2}+p_{n-3}+p_{n-4}+p_{n-5}+p_{n-6}right)$$ starting with $p_0=1$ and $p_{-1}=p_{-2}=p_{-3}=p_{-4}=p_{-5}=0$



              This is increasing from $n=1$ to $n=6$ since in this range $p_n=p_{n-1}+frac16p_{n-1}=frac76p_{n-1}=frac{7^{n-1}}{6^n}$. In particular $p_6=frac{7^5}{6^n}approx 0.3602323$



              $p_6$ must then (by induction) be the highest $p_n$ for positive $n$, since $p_n$ for $n gt 6$ is the average of the preceding six values, which will be strictly less than $p_6$. The average roll is $3.5$ so about $frac{1}{3.5} =frac27$ of the values are hit, meaning $p_n to frac27 approx 0.2857143$ as $n$ increases. Actual values for smaller $n$ are:



              p_0                 1 / 1               1
              p_1 1 / 6 0.1666667
              p_2 7 / 36 0.1944444
              p_3 49 / 216 0.2268519
              p_4 343 / 1296 0.2646605
              p_5 2401 / 7776 0.3087706
              p_6 16807 / 46656 0.3602323
              p_7 70993 / 279936 0.2536044
              p_8 450295 / 1679616 0.2680940
              p_9 2825473 / 10077696 0.2803689
              p_10 17492167 / 60466176 0.2892885
              p_11 106442161 / 362797056 0.2933931
              p_12 633074071 / 2176782336 0.2908302
              p_13 3647371105 / 13060694016 0.2792632
              p_14 22219348327 / 78364164096 0.2835397
              p_15 134526474769 / 470184984576 0.2861139
              p_16 809860055095 / 2821109907456 0.2870714
              p_17 4852905842113 / 16926659444736 0.2867019





              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                In general you can reach $n$ directly from any of the six preceding numbers each with probability $frac16$ so if $p_n$ is the probability that a sum of $n$ is hit then $$p_n=frac{1}{6}left(p_{n-1}+p_{n-2}+p_{n-3}+p_{n-4}+p_{n-5}+p_{n-6}right)$$ starting with $p_0=1$ and $p_{-1}=p_{-2}=p_{-3}=p_{-4}=p_{-5}=0$



                This is increasing from $n=1$ to $n=6$ since in this range $p_n=p_{n-1}+frac16p_{n-1}=frac76p_{n-1}=frac{7^{n-1}}{6^n}$. In particular $p_6=frac{7^5}{6^n}approx 0.3602323$



                $p_6$ must then (by induction) be the highest $p_n$ for positive $n$, since $p_n$ for $n gt 6$ is the average of the preceding six values, which will be strictly less than $p_6$. The average roll is $3.5$ so about $frac{1}{3.5} =frac27$ of the values are hit, meaning $p_n to frac27 approx 0.2857143$ as $n$ increases. Actual values for smaller $n$ are:



                p_0                 1 / 1               1
                p_1 1 / 6 0.1666667
                p_2 7 / 36 0.1944444
                p_3 49 / 216 0.2268519
                p_4 343 / 1296 0.2646605
                p_5 2401 / 7776 0.3087706
                p_6 16807 / 46656 0.3602323
                p_7 70993 / 279936 0.2536044
                p_8 450295 / 1679616 0.2680940
                p_9 2825473 / 10077696 0.2803689
                p_10 17492167 / 60466176 0.2892885
                p_11 106442161 / 362797056 0.2933931
                p_12 633074071 / 2176782336 0.2908302
                p_13 3647371105 / 13060694016 0.2792632
                p_14 22219348327 / 78364164096 0.2835397
                p_15 134526474769 / 470184984576 0.2861139
                p_16 809860055095 / 2821109907456 0.2870714
                p_17 4852905842113 / 16926659444736 0.2867019





                share|cite|improve this answer









                $endgroup$



                In general you can reach $n$ directly from any of the six preceding numbers each with probability $frac16$ so if $p_n$ is the probability that a sum of $n$ is hit then $$p_n=frac{1}{6}left(p_{n-1}+p_{n-2}+p_{n-3}+p_{n-4}+p_{n-5}+p_{n-6}right)$$ starting with $p_0=1$ and $p_{-1}=p_{-2}=p_{-3}=p_{-4}=p_{-5}=0$



                This is increasing from $n=1$ to $n=6$ since in this range $p_n=p_{n-1}+frac16p_{n-1}=frac76p_{n-1}=frac{7^{n-1}}{6^n}$. In particular $p_6=frac{7^5}{6^n}approx 0.3602323$



                $p_6$ must then (by induction) be the highest $p_n$ for positive $n$, since $p_n$ for $n gt 6$ is the average of the preceding six values, which will be strictly less than $p_6$. The average roll is $3.5$ so about $frac{1}{3.5} =frac27$ of the values are hit, meaning $p_n to frac27 approx 0.2857143$ as $n$ increases. Actual values for smaller $n$ are:



                p_0                 1 / 1               1
                p_1 1 / 6 0.1666667
                p_2 7 / 36 0.1944444
                p_3 49 / 216 0.2268519
                p_4 343 / 1296 0.2646605
                p_5 2401 / 7776 0.3087706
                p_6 16807 / 46656 0.3602323
                p_7 70993 / 279936 0.2536044
                p_8 450295 / 1679616 0.2680940
                p_9 2825473 / 10077696 0.2803689
                p_10 17492167 / 60466176 0.2892885
                p_11 106442161 / 362797056 0.2933931
                p_12 633074071 / 2176782336 0.2908302
                p_13 3647371105 / 13060694016 0.2792632
                p_14 22219348327 / 78364164096 0.2835397
                p_15 134526474769 / 470184984576 0.2861139
                p_16 809860055095 / 2821109907456 0.2870714
                p_17 4852905842113 / 16926659444736 0.2867019






                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 11 at 23:13









                HenryHenry

                101k481168




                101k481168






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070185%2fchoosing-an-integer-that-is-most-likely-to-be-the-sum-of-die-rolls%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Human spaceflight

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

                    張江高科駅