# Bag of $10$ balls; $8$ identical red balls,$2$ blue balls. With replacement.

Suppose you have a bag of $10$ balls; $8$ of which are red (and identical), and $2$ of which are blue (and identical). Now suppose that you take one ball out of the bag, mark down its colour, and then put it back in the bag. Then repeat this 100 times. What is the probability that you don't have consecutive blue balls?

Now, I know that there are $2^{100}$ possible outcomes, you either pick a red or a blue ball. And you also know that there is an $80\%$ chance of picking a red ball at any given instance. So my approach would be to find all the possible outcomes and then subtract the number of outcomes where you pick consecutive blue balls. That's the part where I am stuck. Any tips? Thanks.

Denote by $p_n$ the probability that after $n$ drawings you have not yet seen two consecutive blue balls. Then $p_0=p_1=1$ and $$p_n={4\over5}p_{n-1}+{1\over5}\cdot{4\over5} p_{n-2}\qquad(n\geq2)\ .\tag{1}$$ In order to prove $(1)$ we argue as follows: Assume $n\geq2$. With probability ${4\over5}$ the first drawn ball is red, in which case we have to see another $n-1$ drawings without two consecutive blue balls. With probability ${1\over5}$ the first drawn ball is blue, in which case we need the second ball to be red, and no two consecutive blue balls in the remaining $n-2$ drawings.

Now $(1)$ can be easily solved with the "Master Theorem" for constant coefficient linear difference equations. One obtains $$p_n= A u^n + B v^n\qquad(n\geq0)\ ,\tag{2}$$ whereby $$u={2\over5}(1-\sqrt{2}),\quad v={2\over5}(1+\sqrt{2}),\quad A={4-3\sqrt{2}\over8},\quad B={4+3\sqrt{2}\over8}\ .$$ This leads to $p_{100}=0.0313721$. The $Au^n$ term in $(2)$ can be neglected when $n\gg1$.

Hint:

Suppose no blue ball is drawn, there is one such choice. $0.8^{100}$

Suppose there is only $1$ blue ball. place $99$ red balls in a straight line, you have $100$ choices of location to insert the blue ball. $\binom{100}{1}(0.2)(0.8^{99}$)

Suppose there are $k$ blue balls, place $100-k$ red balls in a straight line, you have $100-k+1$ locations to choose to place the blue balls where each slot only allows one blue balls. $\binom{100-k+1}{k}(0.2)^k(0.8)^{100-k}$

Of course we need $100-k+1 \geq k$, that is $k \leq 50$.