Suppose we roll $4$ $6$-sided dice, what is the probability that their sum is equal to $20$ ?
Let $X_1,X_2,X_3,X_4$ denote the rolls' outcomes. I usually deal with this kind of question the same way : I use the law of total probability until reaching a sum of two-dice where I know by heart the possibilities. For example, if I am asked to find the probability that the sum of three $6-$sided die is $11$, then I do $$P(X_1+X_2+X_3=11)=\sum_{i=1}^6 P(X_2+X_3=11-i)P(X_1=i) = \frac 16 \sum_{i=1}^6 P(X_2+X_3=11-i) = \frac 16 P(5 \leq X_2+X_3 \leq 10)$$
$$= \frac 16 \frac{27}{36}=\frac{27}{6^3}$$
Here I do the same so : $$P(X_1+X_2+X_3+X_4=20)=\sum_{i=1}^6 P(X_2+X_3+X_4=20-i)P(X_1=i) $$ $$= \frac 16 \sum_{i=1}^6 P(X_2+X_3+X_4=20-i)$$ $$= \frac 16 \sum_{i=1}^6\sum_{j=1}^6 P(X_3+X_4=20-i-j)P(X_2=j) =\frac {1}{36} P(8 \leq X_3+X_4 \leq 18)=\frac{15}{6^4}$$
which is not the right answer.
Can anyone explain what is the problem here ?
Best Answer
You have got a comment pointing out your mistake, and another suggesting the use of inclusion-exclusion.
I am giving a simple stars and bars solution without inclusion-exclusion.
The mean total expected $= 3.5*4 = 14,$ and the distribution will be symmetrical around the mean, so number of outcomes will be the same for totals of $(14+6)= 20 $ and $(14-6)= 8$
And by stars and bars on the latter total, we directly get
# of otcomes = $\binom{7-1}{4-1} = 35$
and $Pr = \Large\frac{35}{6^4}$