Dtyjlui

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

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