# Trade-off between determenistic and random alternatives in a one-shot game

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 $y(x)$?

• Generally in economics you specify the utility (function) and then "rational" means maximising utility. – snoram Jul 5 '16 at 15:26
• 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 Jul 5 '16 at 15:42
• What is $t$ standing for? – snoram Jul 5 '16 at 15:50
• 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 Jul 5 '16 at 16:02
• 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 Jul 5 '16 at 16:03

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)).