1

An unfair coin turns up heads with probability 0.2. It is tossed repeatedly. What is the expected number of tosses until there are two heads in a row?

How can one use Markov chains to solve this problem?

kshah99
  • 19

1 Answers1

1

Since we know the probability of this coin turning up heads is 0.2, the odds of getting two heads in a row at any point in time would be 0.2 * 0.2 = 0.04.

Now let's ask the question: How many times would we need to flip the coin to reach an expected value of 1 for this outcome? Well if someone has a probability of 0.04, we know it would have an expected outcome of 1 time after 25 trials.

Therefore the expected number of tosses to get 2 heads in a row with this unfair coin would be 25 tosses.

Now for using Markov Chains, we would have two states. An H state and a T state for heads and tails respectively. Each state would have a recursive path on themselves, since you can of course flip the same result you just flipped, and you would have a path that goes to the other option because you could also not flip the same result.

On the H state, we would have the recursive path pointing back towards itself have a value of 0.2, because the odds of flipping another heads with this unfair coin is always 0.2, regardless of the previous outcome because flips are independent. The path going from H to T has a value of 0.8. On the H state, the recursive path would also have a value of 0.8, because that's always the probability of flipping a T with this coin, and lastly the path from T to H would have a value of 0.2.

So to visualize the Markov Chain diagram, it would look like this:

H -> H : 0.2

H -> T : 0.8

T -> T : 0.8

T -> H : 0.2

Drew Pesall
  • 167
  • Thanks for the great answer. How would one calculate the expected number of tosses after constructing the transitions probabilities using the Markov chain? – kshah99 Dec 11 '22 at 05:23