Consider a hypothetical coin (with two sides: heads and tails) that has a $3/4$ probability of landing on the side it was before the flip (meaning, if I flip it starting heads-up, then it will have an only $1/4$ probability of landing tails-up). If it begins on heads, what is the probability that it is on tails after 10 flips? What about 100 flips? Assume that each flip starts on the same side as it landed on the previous flip.
Note: this is not a homework problem, just something I thought up myself.
Best Answer
Let $p_n$ be the probability the coin is on tails after $n$ flips. Note that $p_0=0$.
The coin can be on tails after $n+1$ flips in two different ways: (i) it was on tails after $n$ flips, and the next result was a tail or (ii) it was on heads after $n$ flips, and the next result was a tail.
The probability of (i) is $(3/4)p_n$ and the probability of (ii) is $(1/4)(1-p_n)$. Thus $$p_{n+1}=\frac{1}{4}+\frac{1}{2}p_n.\tag{1}$$ Solve this recurrence relation. The general solution of the homogeneous recurrence $p_{n+1}=\frac{1}{2}p_n$ is $A\cdot \frac{1}{2^n}$. A particular solution of the recurrence (1) is $\frac{1}{2}$. So the general solution of the recurrence (1) is $$p_{n}=A\cdot \frac{1}{2^n}+\frac{1}{2}.$$ Set $p_0=0$ to find $A$. We find that $p_n=\frac{1}{2}-\frac{1}{2^{n+1}}$.