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 $n$).
On
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
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.