I have found the similar question for Uniform Distribution i.e, U(0,1) but not for Normal distribution,
So basically the problem is, using a standard normal distribution i.e., ∼(,$^2$) random number generator, keep generating and summing until the sum is greater than .
How to find the expected number of moves you will take. I am not even sure the value is defined.
I was modelling the equation like here - Link, and now the equation also has Normal distribution term in the right hand side while picking a number, this converts it to a form where it appears, convolution theorem might be used, tho I am not able to proceed ahead or gain any useful breakthrough.
Welcome to any other methods as well
Assume $\mu$ and $r$ are positive. Let $Y_i \sim N(\mu, \sigma^2)$ be i.i.d., and let $X_r$ be the minimum number of these variables needed to sum to at least $r$. Then you have $$ E[X_r]=1+\sum_{k=1}^{\infty}P[X_r > k] = 1+\sum_{k=1}^{\infty}P[Y_1+\ldots+Y_k < r]. $$ Since $Y_1+\ldots+Y_k \sim N(k\mu, k\sigma^2)$, this reduces to $$ E[X_r]=1+\sum_{k=1}^{\infty}\Phi\left(\frac{r-k\mu}{\sigma\sqrt{k}}\right), $$ where $\Phi$ is the c.d.f. of the standard normal distribution. This sum converges; the terms (once $k\mu \gg r$) decay exponentially with $k$.