U can throw the dice atmost n times

Probability of getting $1$ in $1^{st} throw = \frac{1}{6}$ .

Probability of getting $1$ in $2^{nd} throw =$ Probability of not getting $1$ in $1^{st} throw$ X Probability of getting $1$ in $2^{nd} throw = \frac{5}{6}.\frac{1}{6}$ .

Similarly Probability of getting $1$ in $3^{rd} throw = \frac{5}{6}.\frac{5}{6}.\frac{1}{6}$ .

Similarly Probability of getting $1$ in $n^{th} throw = {(\frac{5}{6})}^{n-1}.\frac{1}{6}$ .

$\therefore$ Final Probability of getting $1 = \frac{1}{6}.(1+\frac{5}{6}+(\frac{5}{6})^{2}+ .........+(\frac{5}{6})^{n-1}) = 1 - (\frac{5}{6})^{n}$ .

Given, Final Probability $\gt \frac{1}{2} \implies 1 - (\frac{5}{6})^{n} \gt 0.5 \implies (\frac{5}{6})^{n} \lt 0.5$ .

We find $n = 4$ as the least for this condition.

Hence answer $= 4$