On average, how many times does one need to roll three fair dice to get a sum of 10? Three fair dice are rolled simultaneously. On average, how many times does one need to roll three fair dice to get a sum of $10$?(All rolls are independent of each other, and a roll of the three dice counts as one time.)
So I found the probability that the three dice sum to $10$ to be $.125$. I don't know what to do to find the number of rolls. I feel like it has something to do with expected value but I'm not sure how to set it up.
thanks for the help!
 A: well .125 is equivalent to 1/8 which means 1 time out of the 8 trials you perform will get the expected value of 10. So how many trails do you think it would take to guarentee the expected value of 10...
A: The generating function for the 3 die is
$$ \frac{x^3}{216}+\frac{x^4}{72}+\frac{x^5}{36}+\frac{5 x^6}{108}+\frac{5 x^7}{72}+\frac{7 x^8}{72}+\frac{25 x^9}{216}+\frac{x^{10}}{8}+\frac{x^{11}}{8}+\frac{25 x^{12}}{216}+\frac{7 x^{13}}{72}+\frac{5 x^{14}}{72}+\frac{5 x^{15}}{108}+\frac{x^{16}}{36}+\frac{x^{17}}{72}+\frac{x^{18}}{216}$$
this looks tough but can be done by pencil and paper. From it we see that the probability is 1 / 8 for one roll of 3 die. From the formula for expectation we get:
$$\frac{1}{8}\cdot 1+\frac{1}{8}\cdot\frac{7}{8}\cdot 2+\frac{1}{8} \left(\frac{7}{8}\right)^2 \cdot 3+\frac{1}{8} \left(\frac{7}{8}\right)^3 \cdot 4 \ +...+ = \frac{1}{8} \sum _{k=1}^{\infty } k \left(\frac{7}{8}\right)^{k-1}=8$$
A: If you have a sequence of independent and identically distributed coin flips which each give success with probability $p$, the waiting time until the first success follows a geometric distribution with parameter $p$. In this case, the coin flips have success probability equal to that of getting a sum of 10.
