Probability Puzzle : Robot and coins 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 ? 
 A: Let $h$ be the proportion of heads among the coins and $t=1-h$ the proportion of tails.  In a stationary state, the chance of a transition heads$ \to $ tails must equal the chance of a transition tails$ \to $heads  So we have $h=\frac t2$ and the equilibrium state is  $\frac 13$ heads and $\frac 23$ tails.
A: It will reach an equilibrium of 2/3 tails and 1/3 heads.  At that point, the expected outcome of the next iteration is:


*

*1/3 of the coins, which were previously heads, will all be flipped to tails.

*1/3 of the coins, which were previously tails, will be tossed and land heads-up.

*1/3 of the coins, which were previously tails, will be tossed and land tails-up.


giving 2/3 tails and 1/3 heads again.

A: 
Will there be a convergence in distribution of heads vs. tails?

Yes, but not necessarily to $2/3$ tails and $1/3$ heads.
The visible distribution of heads and tails fluctuates, and does not converge.
The average of the distributions after the first $n$ moves by the robot converges (with probability $1$ and for large $n$) to a limiting distribution.
The limiting distribution for one coin that is moved infinitely often by the robot, is $2/3$ Tails and $1/3$ Heads.  This is the equilibrium distribution of the two state Markov chain that describes the robot operating forever on one coin.
The limiting distribution for the subset of coins that are moved infinitely often is $2/3$ T, $1/3$ H.
The limiting distribution for all the coins exists, but it can differ from $(2/3,1/3)$ if there is a positive probability for at least one coin to be visited only a finite number of times.  Whether that probability is positive depends on the programming of the robot and other factors.
A: 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.
