# A die is thrown $n$ times. What is the probability that $6$ appears even number of times?

A die is thrown $$n$$ times. What is the probability that $$6$$ appears even number of times (for the purpose of task $$0$$ is even number)?

The solution from my textbook is: We have two hypotheses $$H_1$$ that the array starts with 6 and $$H_2$$ array does not start with 6; $$p_n$$ is the probability. $$p_n=P(H_1)(1-p_{n-1})+P(H_2)p_{n-1}=\frac{1}{6}+\frac{2}{3}p_{n-1}$$. $$p_1=\frac{5}{6}$$ After solving this recursive relation we get $$p_n=\frac{1}{2}(1+(\frac{2}{3})^n)$$.

I know how to solve recursion I just don't understand how they got it (I know probabilities of hypothesis). Also, is there any other way to approach this task?

The recurrence is derived as follows. Suppose that the sequence begins with a $$6$$. Then we must roll a $$6$$ an odd number of times in the remaining $$n-1$$ rolls (which has probability $$1-p_{n-1}$$) to get $$6$$ an even number of times in $$n$$ rolls.

Similarly if the sequence does not begin with a $$6$$ then we need to roll a $$6$$ an even number of times in the remaining $$n-1$$ rolls (which has probability $$p_{n-1}$$) in order to get get $$6$$ an even number of times in $$n$$ rolls.

Together, you get the recurrence

$$p_n=P(H_1)(1-p_{n-1})+P(H_2)p_{n-1}$$

Here is a another way to proceed. Let $$N$$ be the number of times that $$6$$ appears in the $$n$$ dice rolls. Then $$P(N=k)=\binom{n}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{n-k}\quad (0\leq k\leq n)$$ by way of the binomial distribution. In particular the probability generating function of $$N$$ is given by $$G_{N}(t)=\sum_{k=0}^nP(N=k)t^k=\left(\frac{1}{6}t+\frac{5}{6}\right)^n.$$ by the binomial theorem. Now we note that $$\sum_{k=0,\, k\;\text{even}}^nP(N=k)=\frac{G_{N}(1)+G_{N}(-1)}{2}=\frac{1}{2}\left(1+(2/3)^n\right)$$

Hint: Make use of Binomial Expansion.

Assuming n to be even $$(a + b)^n = {n \choose 0}a^n + {n \choose 1}a^{n-1}b + {n \choose 2}a^{n-2}b^{2} + {n \choose 3}a^{n-3}b^{3} ... + {n \choose n}b^{n}$$

and,

$$(a - b)^n = {n \choose 0}a^n - {n \choose 1}a^{n-1}b + {n \choose 2}a^{n-2}b^{2} - {n \choose 3}a^{n-3}b^{3} ... + {n \choose n}b^{n}$$

$$(a + b)^n + (a - b)^n = 2({n \choose 0}a^n + {n \choose 2}a^{n-2}b^{2}...)$$,
thus the quantity on the right of the above equation represents the probability of getting an even number of events $$a$$. Bring the $$2$$ from RHS to LHS,
$$\frac{1}{2}((a + b)^n + (a - b)^n) = ({n \choose 0}a^n + {n \choose 2}a^{n-2}b^{2}...)$$
We know from a fair, unbiased die, we have P(Die Value = 6) = 1/6 and P(Die Value != 6) = 5/6. Then, substitute $$a$$ for 1/6 and $$b$$ for 5/6 and solve.