4
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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?

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  • 9
    $\begingroup$ 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. $\endgroup$ – Dilip Sarwate Jun 3 at 21:30
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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

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  • 8
    $\begingroup$ 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. $\endgroup$ – Sextus Empiricus Jun 3 at 14:16

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