TITLE: Probability Puzzle : Robot and coins
QUESTION [7 upvotes]: Someone walks into your room and dumps a huge bag of quarters all over the floor. They spread them out so no quarters are on top of any other quarters. a robot then comes into the room and is programmed such that if it sees a head, it flips it to tails. If it sees a tail, it throws it in the air. the robot moves around randomly forever. Will there be a convergence in distribution of heads vs. tails?
I am trying this puzzle for the past two days . But got lot of confusions !! As i don't know what is convergence in probability,i cannot proceed further. Please don't provide a link to wikipedia for convergence in probability.Can someone give me a simple definition of convergence in probability and explain the solution to the puzzle ?

REPLY [2 votes]: The solution to the above problem can be simply understood in terms of Markov chain theory. Let $1,0$ denote Heads, Tails respectively.
$X_n \in \{0,1\}$ be a random variable describing the no. of heads (1) and tails (0) at $n^{th}$ instant. 
Then the transition probability matrix for this Markov chain is
$$ 
\mathcal{P}= \begin{bmatrix}
1/2 & 1/2 \\
1 & 0
\end{bmatrix}
$$
Thus $[P(X_{n+1}=0) \ \ \ \ P(X_{n+1}=1)]= [P(X_{n}=0) \ \ \ \ P(X_{n}=1)] \mathcal{P}$
Limiting distribution is same as the stationary distribution is equivalent to solving for the left eigenvector of $\mathcal{P}$, i.e
$$
\bar{x}=\bar{x} \mathcal{P}
$$
which yields $[2/3 \ \ \ 1/3]$ as the required answer.