Pricing Repeated Gambles with Utility and Optimal Stopping

# Dice question - expected winnings of rolling dice $2$ times


# Dice question - expected winnings of rolling dice $2$ times












Typical trading interviews consider gambling problems such as rolling a dice and winning its face value. The expected winnings are $\\\$3.5$, $\\\$4.25$, $\\\$\frac{14}{3}$ for one throw, two throws, and three throws respectively. The expected winnings of $n$ throws may be calculated recusively knowing the value of $n-1$.

However, now consider a different game, in which one obtains $\\\$10,000,000$ for each face value of the dice. How much would you pay to enter this game?

Values of this big need us to be risk averse, in which the utility function is not linear but concave. A canonical choice is the logarithm, using which the money equivalent of the game is the geometric mean, $(\prod_{i=1}^6 i*10^8)^{1/6} \approx 2.9938* 10^8$. So a risk reverse investor will pay $30M$ to enter the game. However, what if now I can throw the dice two times? Obviously, the same strategy before cannot apply here. How should we price this game then? Using binomial tree?

## Answer by Kermittfrog (score 2, accepted)

https://quant.stackexchange.com/a/66626

Let's apply basic utility theory. Note that the expected utility is driven by the outcomes, their probabilities and the initial wealth $W_0$ of the gambler (see @Dimitri's comment to your question). Let $\mathrm{EU}$ denote expected utility

$$ \mathrm{EU}\equiv\sum\limits_{i=1}^np_iu(x_i+W_0) $$

and $\mathrm{CE}$ be the certainty equivalent (or money equivalent in your question) $$ \mathrm{CE}\equiv u^{-1}(\mathrm{EU})-W_0, $$ i.e. $u(\mathrm{CE+W_0})=\mathrm{EU}$. If the initial wealth is orders of magnitude above the gamble's outcomes, $W_0>>x$, then utility can be well approximated by a linear function, $u(W_0+x_i)\approx u(W_0)+cx_i$ with $c$ some constant depending on $W_0$ and $u$. This is reflected in the first part of your question.

Let's further introduce optimal decision making. With $n$ more rounds to go in the game, your gambler will always stop the game at some result $x_i+W_0$ whenever the utility of that result exceeds the continuation value,

$$ u(x_i+W_0)>\max{F_n} $$

where, somewhat sloppily, $\max F_n$ denotes the expected utility from optimally stopping the game at some future time. In your example, this problem can be solved recursively: With one more round to go, the gambler will stop at some outcome $x_j$ if $$u(W_0+x_j)>\sum\limits_{i=1}^np_iu(x_i+W_0)\Leftrightarrow x_j>u^{-1}(EU)-W_0\equiv \mathrm{CE}_1$$

The gambler will continue to play at any $x_j<\mathrm{CE}_1$. With two more rounds to go, they know their optimal decision in the next round and they can adjust the game's expected continuation value accordingly, i.e. now they stop if

$$ u(x_j+W_0)>\sum_{i:x_i < \mathrm{CE}_1}p_iu(W_0+\mathrm{CE}_1)+\sum_{i:x_i > \mathrm{CE}_1}p_iu(W_0+x_i) $$ And they will again find some certainty equivalent $\mathrm{CE}_2$ for this game, and so on.

At any step, the certainty equivalent (money value) of playing $n$ games can be calculated by calculating

$$ \begin{align} \mathrm{CE_n}&=u^{-1}(\mathrm{EU_n})-W_0\\ &=u^{-1}\left(\sum_{i:x_i < \mathrm{CE}_{n-1}}p_iu(W_0+\mathrm{CE}_{n-1})+\sum_{i:x_i > \mathrm{CE}_{n-1}}p_iu(W_0+x_i)\right)-W_0 \end{align} $$

N.B.: For completeness, the price of the gamble should, of course, also be reflected in the utility and outcome... Also: @Dimitri's comment is very insightful: You are only willing to pay some utility based reservation price for such a gamble, but the gambling house would, on average, loose money at such a ticket price.