# What's the event space of a single coin toss?

Take a probability triple, $$(\Omega,\mathscr{F},\mathbb{P})$$, representing a single coin toss.

Then \begin{align} & \Omega = \{H,T\}. \tag{Prop. 1} \\ \end{align}

Now, \begin{align} \text{A } \sigma \text{-algebra must be closed under countable union}, \tag{Prop. 2} \end{align}

so \begin{align} \text{if we have } \{H\},\{T\} \in \mathscr{F} \text{ then we must also have } \{H,T\} \in \mathscr{F}, \tag{Prop. 3} \end{align}

but \begin{align} \{H,T\} \text{ is not a possible event resulting from a single coin toss.} \tag{Prop. 4} \end{align}

What's gone wrong?

• You are mixing up events (collections of outcomes} and outcomes. $\{H,T\}=\Omega$ is an event since it is the set containing two outcomes, but it is not, by itself, an outcome. $H$ and $T$ are outcomes, $\{H\}$ and $\{T\}$ are the corresponding events consisting of single outcomes. – Dilip Sarwate Jun 3 '20 at 21:30

The event $$\{H,T\}$$ is that the result of the flip is either $$H$$ or $$T$$; this has probability $$1$$
The event $$\emptyset = \{\,\}$$ is that the result of the flip is neither $$H$$ nor $$T$$; this has probability $$0$$
So there is no problem; $$\mathscr{F}= \{\emptyset,\{H\},\{T\},\{H,T\}\}$$ as you might expect
• So proposition 4 is wrong. I guess that $\{H,T\}$ got interpreted as 'tails after heads' instead of 'tails or heads'. In that case it resembles an equivocation fallacy. – Sextus Empiricus Jun 3 '20 at 14:16