Markov chain problem

[Huan Dao]

Suppose you throw a large number of coins to the floor, assume that these coins are fair, and don't overlap each other.
A robot move randomly in the floor, and pick up coins, if it sees head, it flips the coins to tail and put it back to the original position. If it sees tail, it flips the coin and put it back. After a really long time, you pick up a coin from the floor, what is the probability that it shows head?

 

[koupparis]

The problem is symmetric in regards to heads vs tails, so it should be 1/2

 

[bob]

(p_H = \frac{1}{3})

(p_T = \frac{2}{3})

It's the eigenvector of the transition matrix with eigenvalue 1.

 

[pruse]

The problem is symmetric in regards to heads vs tails, so it should be 1/2

the problem is not symmetric in heads and tails. heads are flipped over to tails, but for tails the coin is tossed and takes the outcome.

 

[Huan Dao]

The problem is symmetric in regards to heads vs tails, so it should be 1/2

no it's not symmetric, if the coin is tail, the robot will flip the coin, and put back (head or tail).

bob's answer seems is correct, but I got a more simple solution:

let P(H)=A
P(T)=1-A
This is after a long time so P(H) and P(T) must be stable, that means if the robot continue to flipping coins, P(H) and P(T) will be unchanged, so if the robot picks up a coin, and flips it the probability of the flipped coin is head (F(H)) must be A or else P(H) and P(T) will be unstable.
So this is the math:

F(H)=(1-A)/2
F(H)=A
hence A=1/3
so P(H)=1/3
P(T)=2/3

 

[pruse]

no it's not symmetric, if the coin is tail, the robot will flip the coin, and put back (head or tail).

bob's answer seems is correct, but I got a more simple solution:

let P(H)=A
P(T)=1-A
This is after a long time so P(H) and P(T) must be stable, that means if the robot continue to flipping coins, P(H) and P(T) will be unchanged, so if the robot picks up a coin, and flips it the probability of the flipped coin is head (F(H)) must be A or else P(H) and P(T) will be unstable.
So this is the math:

F(H)=(1-A)/2
F(H)=A
hence A=1/3
so P(H)=1/3
P(T)=2/3

This is equivalent to what Bob did. An eigenvalue of 1 corresponds to "stability".

 

[Huan Dao]

This is equivalent to what Bob did. An eigenvalue of 1 corresponds to "stability".

is it cool that we need less math to solve this question?

 

[ThisGuy]

I'm guessing that koupparis was probably confused (as I was initially) by the dual uses of the word "flip". I think there's an simpler solution to this problem though: the only way to end with a head is for the previous state to be tails which then gets flipped and lands on heads. So (p_h = p_t/2).

 

[koupparis]

Misread the problem, assumed flipped meant the same as flips it over :D So yes it's not symmetric and Bob's solution is correct. Write down the transition matrix and find the eigenvector corressponding to eigenvalue of 1.

 

[pratikpoddar]

Thanks for the solution Bob