Biased coin with a $3/4$ chance to land on the side it was before the flip 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.
 A: 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 $p_n=\frac{1}{2}-\frac{1}{2^{n+1}}$.
A: If you mean that $\Pr(H_i|H_{i-1}) = \Pr(T_i|T_{i-1}) = 3/4$ for all $i$, then this is a two-state homogeneous Markov chain.  The transition matrix is
$$
P = \begin{pmatrix}
3/4 & 1/4 \\
1/4 & 3/4
\end{pmatrix}.
$$
This should get you started....
I'd comment but I don't have the rep, strangely you require 50.
Also, with regards to the bias in a coin flip, it's a paper by Diaconis:
http://statweb.stanford.edu/~susan/papers/headswithJ.pdf
A: Let $p_n$ be the probability that the coin is heads up after $n$ tosses.  Then $p_0=1$ since it starts on heads, and
$$p_n=\frac34p_{n-1}+\frac14(1-p_{n-1})$$
which simplifies to
$$\left(p_n-\frac12\right)=\frac12\left(p_{n-1}-\frac12\right)\ .$$
Iterating,
$$p_n-\frac12=\Bigl(\frac12\Bigr)^{n+1}$$
so
$$p_n=\frac12+\Bigl(\frac12\Bigr)^{n+1}\ .$$
A: Another way of looking at it is to imagine that you toss two fair coins each time, using the second coin to choose between the first coin and the previous (originally heads). If the second coin always chooses the previous result then you end up with heads, while if at any point the second coin chooses the first coin then as this is fair the end result will have equal probability of being heads or tails. The chances of the latter being $1-\frac1{2^n}$, the chance of the final result being tails is therefore half of that.
A: $p_n(\text{heads}) = \frac 3 4p_{n-1}(\text{heads}) + \frac 1 4 (1 - p_{n-1}(\text{heads}))$
$p_n(\text{heads}) = \frac 1 4 + \frac 1 2p_{n-1}(\text{heads})$
$p_1(\text{heads}) = \frac 3 4$

Everyone else beat me too it. Was still writing the recurrence relation. :\

A: Any solution with the words "heads" and "tails" is doing extra work.
We want the chance of an odd number of reversals, each with probability $p=1/4$, to happen in $n$ independent trials.   The coin briefly remembers its current state but not whether that state was reached by reversing on the previous trial.  This makes the reversals independent of each other.
From the binomial theorem that probability is  $$\frac{((1-p)+p)^n - ((1-p) - p)^n)}{2} = \frac{1}{2} - \frac{(1 - 2p)^n}{2}$$ which is consistent with the other solutions and shows that the 
$(1/2)^{n}$ exponential decay of the "memory" is really a power of $(1-2p)$.
