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A wheel of fortune has outcomes $S=\left \{ 1000,100,50,20,0 \right \}$ as money prices. A consumer has the preferences

$$20\sim \left ( \frac{2}{100}\cdot1000 \oplus \frac{98}{100} \cdot 0 \right )$$ $$50\sim \left ( \frac{15}{100}\cdot1000 \oplus \frac{85}{100} \cdot 0 \right )$$ $$100\sim \left ( \frac{45}{100}\cdot1000 \oplus \frac{55}{100} \cdot 0 \right )$$

What does this notation mean? I figure that the direct sum is just showing that the percentage for winning $1000$ in the first equation is $2/100$ and $98/100$ for $0$. But what does the $20$ denote? Is this some utility and how would that be read with the indifference sign in front of it? Never seen the notation like this before.

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$$ \left(\frac{2}{100} \cdot 1000 \oplus \frac{98}{100}\cdot 0\right) $$ is the lottery where you get $1000$ with probability $2/100$ and $0$ with probability $98/100$.

The expression $$ 20 \sim \left(\frac{2}{100} \cdot 1000 \oplus \frac{98}{100}\cdot 0\right) $$ usually says that the decision maker is indifferent (in terms of preferences) between taking the lottery and having $20$ for sure.

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  • $\begingroup$ So say that the consumer face 2 wheels. the first wheel the consumer wins $1000$ with probability $1/4$, $50$ with probability $1/4$ and $20$ for the rest. The second he wins $1000$ with probability $1/5$, $100$ with probability $1/4$ and nothing for the rest. How do I determine from the preference which he chooses from the $3$ equations? $\endgroup$
    – mathstudent23
    May 15, 2021 at 14:28
  • $\begingroup$ (just to take a look at the usage of the preferences) $\endgroup$
    – mathstudent23
    May 15, 2021 at 15:00
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    $\begingroup$ This should be a separate question. $\endgroup$
    – VARulle
    May 15, 2021 at 21:29
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    $\begingroup$ @mathstudent23 $\left ( \frac{1}{5}\cdot1000 \oplus \frac{1}{4} \cdot 100 \oplus \frac{11}{20} \cdot 0\right ) \sim$ $\left ( \frac{1}{5}\cdot1000 \oplus \frac{1}{4} \cdot \frac{45}{100}\cdot1000 \oplus \frac{1}{4}\cdot\frac{55}{100} \cdot 0 \oplus \frac{11}{20} \cdot 0\right )$ or something like that, which you can simplify $\endgroup$
    – Henry
    May 16, 2021 at 17:13
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    $\begingroup$ @mathstudent23 It can be simplified, but your fractions do not add up to $1$. Instead $\left(\frac{5}{16}\cdot1000 \oplus \frac{11}{16} \cdot 0\right)$ $\endgroup$
    – Henry
    May 17, 2021 at 0:10

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