Question

In a series of coin flips, a run is a series of one or more consecutive coin flips that all have the same result. For example, in the sequence
\[TT \textcolor{red}{HHH} TTHHHTH,\]


the red letters form a run of length $3$. (A run of length $1$ is still considered a run.)


If a fair coin is flipped two times, what is the expected number of runs? (If you're confused about how to count the number of runs, the example sequence above has $6$ runs.)

Answer 1

To find the expected number of runs when flipping a fair coin two times, let's define a run. A run is a consecutive sequence of flips that yield the same outcome (either all heads or all tails).

We'll denote the results of the two flips as \( X_1 \) and \( X_2 \), where each \( X_i \) can be either heads (H) or tails (T).

We can outline the potential sequences from the two flips, which are:
1. HH
2. HT
3. TH
4. TT

Now, we will count the number of runs for each of these outcomes:

- For \( HH \):
- There is only 1 run (the "HH" itself).

- For \( HT \):
- There are 2 runs (the "H" and "T").

- For \( TH \):
- There are 2 runs (the "T" and "H").

- For \( TT \):
- There is only 1 run (the "TT" itself).

Next, let us summarize the number of runs for each outcome:
- Number of runs for \( HH \): 1
- Number of runs for \( HT \): 2
- Number of runs for \( TH \): 2
- Number of runs for \( TT \): 1

Now, we can calculate the expected number of runs \( E[R] \). Since the coin flips are fair, each outcome has a probability of \( \frac{1}{4} \):

\[
E[R] = \left(1 \cdot \frac{1}{4}\right) + \left(2 \cdot \frac{1}{4}\right) + \left(2 \cdot \frac{1}{4}\right) + \left(1 \cdot \frac{1}{4}\right)
\]

Calculating this gives:

\[
E[R] = \frac{1}{4} + \frac{2}{4} + \frac{2}{4} + \frac{1}{4} = \frac{1 + 2 + 2 + 1}{4} = \frac{6}{4} = \frac{3}{2}
\]

Thus, the expected number of runs in two flips of a fair coin is:

\[
\boxed{\frac{3}{2}}
\]