Tossing a coin

Problem

For a random experiment an ideal coin is tossed until it shows heads (\(H\)) for a second time or tails (\(T\)) for a second time. The event space is set to be: \(\{HH; TT; HTH; HTT; THH; THT\}\).

1. Argue that this random experiment is not a Laplace experiment.

2. The random variable \(X\) assigns to each event the number of coins tossed. Compute the expectation value of \(X\).

Solution of part a

For a Laplace experiment each outcome has the same probability. Here, this would mean that each outcome has a probability of \(\frac{1}{6}\). Because an ideal coin is used for the random experiment, we can determine the probabilities of each event as follows:

\[ \begin{align}\begin{aligned}P(HH) = P(TT) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\\P(HTH) = P(HTT) = P(THH) = P(THT) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\end{aligned}\end{align} \]

Because at least one of the probabilities now differs from \(\frac{1}{6}\), this random experiment is not a Laplace experiment.

The different probabilities also become apparent if one simulates the random experiment with Sage.

Solution of part b

The expectation value of \(X\) can be determined with the probabilities obtained in part a:

\[E(X)=2\cdot P(HH)+2\cdot P(TT) + 3\cdot P(HTH) + 3\cdot P(HTT) + 3\cdot P(THH) + 3\cdot P(THT) = 2.5.\]

A simulation with Sage yields: