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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 $n$).

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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.

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  • $\begingroup$ "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
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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
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