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

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 $$n$$).

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 $$k>6$$, $$p_{k}=\frac{1}{6}\sum_{m=1}^{6}p_{k-m}<\frac{1}{6}\cdot 6\max_{m=1,2,3,4,5,6}p_{k-m}=\max_{m=1,2,3,4,5,6}p_{k-m}$$ 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.

• "Observe that for any $k$, $$\sum_{m=1}^{6}p_{k-m}<1".$$ The "observation" is wrong for all $k>6$. – user Jan 11 at 23:13
• Actually $p_k \gt \frac14$ for $k \gt 6$, contrary to your statement $p_k \lt \frac16$ – Henry Jan 11 at 23:15
• You are both correct. I need to slightly reword my definition of $p_{n}$. – pwerth Jan 12 at 0:53

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