Finding the mean number of times a die is thrown if it is thrown until a number at least as high as the result of the first throw is obtained A die is thrown and the number is noted. Then the die is thrown again repeatedly until a number at least as high as the number obtained on the first throw is thrown. Find the mean number of times the die is thrown, including the first throw. ![mean_number_of_throws_calculation]( The answer is $3.45$, but I am getting $2.53$.

If your first roll was a $4$ then for each roll thereafter there is a $\frac{3}{6}$ chance to roll a number at least a large as a $4$ again. Looking at this specific case, the expected number of rolls until doing so would be $\frac{6}{3}=2$.

In general, if you have chance $p$ for success, it will take on average $\frac{1}{p}$ many independent attempts to get your first success.

Noting that having rolled a four as your first roll only accounts for $\frac{1}{6}$ of the time and calculating the rest of the related probabilities and finally accounting for the initial roll we get the final answer.

> $1+\frac{1}{6}(\frac{6}{1}+\frac{6}{2}+\frac{6}{3}+\frac{6}{4}+\frac{6}{5}+\frac{6}{6})=3.45$ The $1$ comes from the initial roll, the $\frac{1}{6}$ comes from the chance to be in each respective case, and each $\frac{6}{k}$ comes from the expected number of rolls until rolling a $7-k$ or greater.