# Coin Flipping Robot

There is a large warehouse. Strewn all over the warehouse floor are thousands of coins. The coins are unbiased, each with a

*heads*side, and a*tails*side. Initially there is a random mix of heads and tails.Into this warehouse is placed a robot. The robot has been programmed with a very simple set of instructions:

- Move about and pick up a random coin.
- If the coin shows heads, gently turn it over and put it back down tails.
- If the coin shows tails, flip it high into the air, and let it randomly land (equal chance of landing heads or tails).
- Goto 1

If we leave the robot to do his thing for a long, long, time, what is will happen to the ratio of tails to heads for the coins in the room?

*After a long time, what will be the percentage of tails-up coins in the room?*

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## Solution

It’s possible to solve this problem using a Markov Chain, but it’s such a simple silly little puzzle that you can solve it using just a little common sense and come critical thinking.

The answer is that the percentage of tails-up coins in the room will trend towards 2/3rds.

The room will reach equilibrium when 2/3rds of the coins are tails (meaning one third are heads). When it is in this state:

- 1/3 of the coins (which were heads) will flip to tails.
- 1/2 × 2/3 of the coins will flip to tails (previously tails).
- 1/2 × 2/3 of the coins will flip to heads (previously tails).

(We started with 2/3rds tails, and the expected outcome is still 2/3rds tails).

Another way to think about this is that the room will reach equilibrium when the chance of a tail turning into a head is the same as a head turning into a tail. If

*h*is the proportion of heads, the chance of head to tail is just*h*. If*t*is the proportion of tails, then*t = 1 - h*. The chance of a tail turning to a head is*½(1 - h)*; It’s a coin flip! For these two to equal,*h = ½(1 - h)*, so*h = 1/3*.