Sum of 3 dice rolls

Question

If we roll a fair dice 3 times, what is the probability that the sum of values of the three rolls is greater than 9?

I approached it like this: Let X be the result of a single dice roll. Let Sn = X1 + X2 + X3. We know that E(X) = 7/2 and that V(X) = 35/12 By the rules of expectation, we know that E(Sn) = 21/2 and that V(Sn) = 35/4.

Shouldn't the answer just be P(Z > (9-(21/2))/sqrt(35/4))? For some reason I keep getting a different answer than what the book has.

Answer 1

$a=$ total for the roll of the first two dice
$b=$ number of possible occurrences of $a$
$c=$ required value of the roll of the third die to exceed $9$
$d=$ number of possible occurrences of $c$
$e=b×d$

$\begin{matrix} a&b&c&d&e\\ 2&1&>7&0&0\\ 3&2&>6&0&0\\ 4&3&>5&1&3\\ 5&4&>4&2&8\\ 6&5&>3&3&15\\ 7&6&>2&4&24\\ 8&5&>1&5&25\\ 9&4&>0&6&24\\ 10&3&>-1&6&18\\ 11&2&>-2&6&12\\ 12&1&>-3&6&6\\ \end{matrix}$

$\sum{e}=135$ number of possible occurrences $>9$ for all $216$ 3-die rolls.
${{135}\over{216}}=.625$
Aleæ jactæ sunt!

Answer 2

The problem is, the distribution of $S_n$ is not normal, nor closely approximated by a normal random variable, since $3$ rolls is a small number. Note that if the first two dice give a total of at most $3$ (which can happen in $3$ ways), then no matter what the third die shows, the total will be at most $9;$ this gives $3\cdot 6=18$ ways to get a total of at most $9$ if the first two dice total at most $3$. If the first two dice give a total of $4$ (which can happen in $3$ ways), then we must get at most $5$ on the third die; this gives $3\cdot5=15$ ways to get a total of at most $9$ if the first two dice give a total of $4.$ Similarly, there are (respectively) $4\cdot4=16,$ $5\cdot3=15,$ $6\cdot2=12,$ and $5\cdot1=5$ ways to get a total of at most $9$ if the first two dice give a total of (respectively) $5,6,7,$ and $8.$ There are no other totals on the first two dice that will allow us to get a total of at most $9$ on all three dice. Hence, adding them up, there are $81$ ways to get a total of at most $9$ on the three dice. Since there are $216$ ways to roll, it follows that the probability of a total of more than $9$ is $\frac58=0.625.$ (Why?)

I leave it to you to fill in the details, but let me know if you have any questions.