<p>Let's assume that the player is an expected utility maximizer, and that the utility they get from a payoff of $\$z$ is $u(z)$, for any $z$ in $\mathbb{R}$ such that $u$ is strictly increasing.</p>

<p>You are <em>essentially</em> asking about certainty equivalents. Call $\sigma$ the gamble that pays $y$ with probability $\frac{1}{2}$ and $0$ with probability $\frac{1}{2}$. Then, the utility of the $\sigma$ (under the EU hypothesis) is $U(\sigma) = \frac{1}{2}u(y) + \frac{1}{2}u(0)$. This means that the player is indifferent between this gamble and getting $u^{-1}(\frac{1}{2}u(y) + \frac{1}{2}u(0)) = u^{-1}(U(\sigma))$ for sure. This is called the <strong>certainty equivalent</strong> of $\sigma$. Notice, $U$ transforms a gamble over monetary payoffs into utilities, and $u^{-1}$ transforms the utilities back into monetary payoffs. If $x$ is a constant gamble then $u^{-1}(U(x)) = u^{-1}(u(x)) = x$.</p>

<p>If $x$ is larger than the certainty equivalent of $\sigma$, the player will choose $x$. You are asking the inverse (if $CE: \sigma \mapsto u^{-1}(U(\sigma))$ you are interested in $CE^{-1}$). Nonetheless the underlying mechanics are the same, once the utility function has been specified. The $y$ you are interested in satisfies</p>

<p>$$ y = u^{-1}(2u(x) - u(0)).$$</p>

<p>The deeper question here, however, is how is the certainty equivalent related to the gamble itself. In general (for a very ugly $u$), not much can be said. However, it is common practice to make some structural assumptions on $u$ to answer the questions you are asking. For example, what if we assume the utility index is differentiable and concave, for example $u(x) = ln(x)$. Then it is relatively straightforward to show that $y(x)$ is increasing and convex (loosely speaking, the relative difference between $x$ and $y$ is increasing, as suggested in your post, although it does not capture the complete reluctance to choose the risky prospect in the third example (although this can be captured by other, more complicated models)).</p>

<p>It is worth noting, all of this assumes EU, which while extensively used in applications, especially game theory, is not without its empirical failures. There are many other "psychological" models which might provide a different resolution to your inquiry.</p>

CC BY-SA 3.0
2016-07-05T16:13:42.217Z
2016-07-06T19:21:56.287Z
2016-07-06T19:21:56.287Z
201p
<p>I wonder how to analyze the following problem.</p>

<p>Suppose we play a one-shot game. We have two options. We can either get $x$ dollars or win $y$ dollars with probability 50% (or, probably, win nothing with the same chances). What is the minimal value that the game-designer should assign to $y$ to make you to choose the second option?</p>

<p>The game is one-shot. It means there is no way to play it several times. It works for me in the following way:</p>

<ol>
<li>if $x$ is 50 cents, I'm ready for \$1 at $y$;</li>
<li>if $x$ is \$100 dollars, \$300 will be enough.</li>
<li>if $x$ is \$100M, I won't choose the second option.</li>
</ol>

<p>I suppose the decision depends on your income, your tendency to risk and so on. So what factors are essential for rational decision? And how to compute $y(x)$?</p>

CC BY-SA 3.0
2016-07-05T15:26:54.910Z
Generally in economics you specify the utility (function) and then "rational" means maximising utility.
snoram
CC BY-SA 3.0
2016-07-05T15:42:55.973Z
Yes, I can: $u(t) = -|t - y(x)|$ :) The question is more about how to translate some kind of natural phenomenon into an optimization problem.
Vsevolod Oparin
CC BY-SA 3.0
2016-07-05T15:50:01.727Z
What is $t$ standing for?
snoram
CC BY-SA 3.0
2016-07-05T16:02:56.627Z
It was a joke with bad notions. Let $y_{right}(x)$ be a function that finds correct value of $y$ for each $x$. Then utility function is defined as $u_x(y) = -|y - y_{right}(x)|$. Actually, I think it doesn't matter what you specify: utility function or $y_{right}(x)$.
Vsevolod Oparin
CC BY-SA 3.0
2016-07-05T16:03:53.933Z
This basically means that $x$ is the 'certainty equivalent' of the lottery where with 50% probability you get $y$, and otherwise you get 0. If you know the Neumann-Morgenstern utility function of the person making the choice you can calculate this value from
$$
u(x) = 50\% \cdot u(0) + 50\% \cdot u(y).
$$
Giskard
<game-theory><decision-theory>
Trade-off between determenistic and random alternatives in a one-shot game
Vsevolod Oparin
# Trade-off between determenistic and random alternatives in a one-shot game

(Asked by: username_0 on 2016-07-05)

I wonder how to analyze the following problem.

Suppose we play a one-shot game. We have two options. We can either get $x$ dollars or win $y$ dollars with probability 50% (or, probably, win nothing with the same chances). What is the minimal value that the game-designer should assign to $y$ to make you to choose the second option?

The game is one-shot. It means there is no way to play it several times. It works for me in the following way:

1. if $x$ is 50 cents, I'm ready for \$1 at $y$;
2. if $x$ is \$100 dollars, \$300 will be enough.
3. if $x$ is \$100M, I won't choose the second option.

I suppose the decision depends on your income, your tendency to risk and so on. So what factors are essential for rational decision? And how to compute $y(x)$?

**Tags:** game-theory, decision-theory
> **Comments:**
> **username_1**: Generally in economics you specify the utility (function) and then "rational" means maximising utility.  
> **username_0**: Yes, I can: $u(t) = -|t - y(x)|$ :) The question is more about how to translate some kind of natural phenomenon into an optimization problem.  
> **username_1**: What is $t$ standing for?  
> **username_0**: It was a joke with bad notions. Let $y_{right}(x)$ be a function that finds correct value of $y$ for each $x$. Then utility function is defined as $u_x(y) = -|y - y_{right}(x)|$. Actually, I think it doesn't matter what you specify: utility function or $y_{right}(x)$.  
> **username_2**: This basically means that $x$ is the 'certainty equivalent' of the lottery where with 50% probability you get $y$, and otherwise you get 0. If you know the Neumann-Morgenstern utility function of the person making the choice you can calculate this value from $$ u(x) = 50\% \cdot u(0) + 50\% \cdot u(y). $$  

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## 1 answers

### Answer by username_3 on 2016-07-05. Score: 2 | Accepted answer

Let's assume that the player is an expected utility maximizer, and that the utility they get from a payoff of $\$z$ is $u(z)$, for any $z$ in $\mathbb{R}$ such that $u$ is strictly increasing.

You are *essentially* asking about certainty equivalents. Call $\sigma$ the gamble that pays $y$ with probability $\frac{1}{2}$ and $0$ with probability $\frac{1}{2}$. Then, the utility of the $\sigma$ (under the EU hypothesis) is $U(\sigma) = \frac{1}{2}u(y) + \frac{1}{2}u(0)$. This means that the player is indifferent between this gamble and getting $u^{-1}(\frac{1}{2}u(y) + \frac{1}{2}u(0)) = u^{-1}(U(\sigma))$ for sure. This is called the **certainty equivalent** of $\sigma$. Notice, $U$ transforms a gamble over monetary payoffs into utilities, and $u^{-1}$ transforms the utilities back into monetary payoffs. If $x$ is a constant gamble then $u^{-1}(U(x)) = u^{-1}(u(x)) = x$.

If $x$ is larger than the certainty equivalent of $\sigma$, the player will choose $x$. You are asking the inverse (if $CE: \sigma \mapsto u^{-1}(U(\sigma))$ you are interested in $CE^{-1}$). Nonetheless the underlying mechanics are the same, once the utility function has been specified. The $y$ you are interested in satisfies

$$ y = u^{-1}(2u(x) - u(0)).$$

The deeper question here, however, is how is the certainty equivalent related to the gamble itself. In general (for a very ugly $u$), not much can be said. However, it is common practice to make some structural assumptions on $u$ to answer the questions you are asking. For example, what if we assume the utility index is differentiable and concave, for example $u(x) = ln(x)$. Then it is relatively straightforward to show that $y(x)$ is increasing and convex (loosely speaking, the relative difference between $x$ and $y$ is increasing, as suggested in your post, although it does not capture the complete reluctance to choose the risky prospect in the third example (although this can be captured by other, more complicated models)).

It is worth noting, all of this assumes EU, which while extensively used in applications, especially game theory, is not without its empirical failures. There are many other "psychological" models which might provide a different resolution to your inquiry.


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