Find $\mathbb P (X_1 + \cdots + X_n = 6n-3)$

Question

A fair die is tossed n times (for large n). Assume tosses are independent. What is the probability that the sum of the face showing is $6n-3$?

Is there a way to do this without random variables explicitly? This is in a basic probability theory reviewer, and random variables was not yet discussed. It's Larsen and Marx before Chapter 3 (where random variables starts)

One thing I tried:

$$P(X_1 = 3) = 1/6$$

$$P(\sum_{i=1}^{2} X_i = 9) = (1/6)^2 4$$

$$P(\sum_{i=1}^{3} X_i = 15) = (1/6)^3 (10)$$

$$P(\sum_{i=1}^{4} X_i = 21) = (1/6)^4 (20)$$

I'm not seeing any pattern for the multiplicand of $(1/6)^n$

$$P(\sum_{i=1}^{n} X_i = 6n-3) = (1/6)^n (?)$$

Another thing I tried:

$$P(\sum_{i=1}^{n} X_i = 6n-3)$$

$$= P(X_i = 6 \ \text{except for 3 which are 5's}) \tag{1}$$

$$+ P(X_i = 6 \ \text{except for 2 which are a 4 and a 5}) \tag{2}$$

$$+ P(X_i = 6 \ \text{except for 1 which is 3}) \tag{3}$$

where I think

$$(1) = (1/6)^{n-3}(1/6)^{3}\binom{n}{3}$$

$$(2) = (1/6)^{n-3}(1/6)^{2}\binom{n}{2}(1/6)^{1}\binom{n-1}{1}$$

$$(3) = (1/6)^{n-2}(1/6)^{1}\binom{n}{1}$$

Is any approach going somewhere? If not, please suggest

Answer 1

If $Y_i:=6-X_i$ then you are looking for $P(Y_1+\cdots+Y_n=3)$.

Observe that (preassuming the die is $6$-sided) the $Y_i$ take values in $\{0,1,2,3,4,5\}$

The answer is:

$$\left[\binom{n}3+n(n-1)+n\right]\left(\frac16\right)^n$$

$\binom{n}3$ corresponds with $n$-tuples having on $3$ spots a $1$ and on the other spot(s) a $0$.

$n(n-1)$ corresponds with $n$-tuples having on $1$ spot a $1$, on $1$ spot a $2$ and on the other spot(s) a $0$.

$n$ corresponds with $n$-tuples having on $1$ spot a $3$, and on the other spot(s) a $0$.

Using the convention that $\binom{n}3=0$ if $3\notin\{0,\dots,n\}$ the answer works for every positive integer $n$.

Answer 2

Your way is correct, but your calculations have some flaws. For (2), you choose $1$ place for $4$ and $1$ place for $5$. So, the binomial coefficients should be ${n\choose 1},{n-1\choose 1}$. So, $(2)={n\choose 1}(1/6)^1{n-1\choose 1}(1/6)^1(1/6)^{n-2}$

For three you need to have $(1/6)^{n-1}$, instead of $(1/6)^{n-2}$.

An alternative way would be defining $a_n=P(X=6n),b_n=P(X=6n-1),c_n=(X=6n-2)$, which are easier to calculate, and $d_n=P(X=6n-3)$. Then, using the recurrence relation $$d_{n+1}=1/6(d_n+c_n+b_n+a_n)$$