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Suppose we are betting. We can either win or lose a certain amount of money at each bet. We can play two strategies, strategy $A$ with a winning ratio of $a$ and expected value $\alpha$, and strategy $B$ with winning ratio $b$ and expected value $\beta$, where:

\begin{equation} a = \frac{\text{bets won, while playing with strategy $a$}}{\text{total number of bets played}} \end{equation}

and

\begin{equation} \alpha = a\cdot(\text{gain from winning bet})+(1-a)\cdot(\text{loss from losing bet}) \end{equation}

(losses are negative values). $b$ and $\beta$ are defined in the same way. My question is: knowing the values of $a,b,\alpha$ and $\beta$, and knowing that

\begin{equation} a>b\qquad\qquad\text{and}\qquad\qquad\alpha<\beta \end{equation}

which strategy should we play (in order to maximize gains) if we can only play once?

My gut feeling says strategy $a$ even if the expected value is greater for the other strategy. But I don't know how to mathematically prove it (or disprove it).

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    $\begingroup$ If you want to maximise expected gain, then clearly you need to choose strategy B. You can probably justify strategy A by some other criterion (e.g. expected utility with some non-linear utility function), but you’ll need to specify what that criterion is for you to be able to determine if strategy A does in fact perform better. $\endgroup$ – Theoretical Economist Nov 9 at 13:23
  • $\begingroup$ Are you sure? The problem is that you can play only once. Strategy B could yield a huge gain just 0.01% of the time. Strategy A might be more stable and one might gain less but "more surely" $\endgroup$ – marco trevi Nov 9 at 13:25
  • $\begingroup$ To put it another way: what exactly do you mean when you say you want to “maximise gains”? $\endgroup$ – Theoretical Economist Nov 9 at 13:26
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    $\begingroup$ Yes. If you want to maximise expected gains, then the exact distribution does not matter, only the expected value. Nor does it matter how many times you play. This is just the definition of what it means to maximise expected gains. $\endgroup$ – Theoretical Economist Nov 9 at 13:27
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    $\begingroup$ Read your question again. That's not actually what you asked. If it was what you meant to ask, then perhaps you need to express it more clearly. In my first comment, I suggested one mathematical setup in my first comment that justifies choosing strategy A. However, it also uses expectations, so you may find it not to your liking. Bear in mind that using expectations (in some form) is still very common even when you only play once. You don't have resort to the law of large numbers, but that doesn't stop the expectation from being a good decision criterion. $\endgroup$ – Theoretical Economist Nov 9 at 13:45
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$\DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var}$ Suppose that you evaluate gambles according to the utility function $u(x) = \sqrt x$. More precisely, if the distribution of payoffs of the gamble is given by the random variable $X$, then your payoff from taking the gamble $X$ is $$ \E[u(X)]=\E\left[X^{1/2}\right]. $$ This means that between two gambles $X$ and $Y$, you prefer gamble $X$ if and only if $$ \E\left[X^{1/2}\right] \ge \E\left[ Y^{1/2} \right]. $$

Now, suppose that $$ X = \begin{cases} 100, & \text{with probability $\frac{1}{10}$} \\ 0, & \text{otherwise} \end{cases} $$ and $$ Y = \begin{cases} \left(\frac{101}{99}\right)^2, & \text{with probability $\frac{99}{100}$} \\ 0, & \text{otherwise} \end{cases}. $$

Then, $\E\left[X^{1/2}\right] = 1 < \E\left[Y^{1/2}\right] = 1.01 $.

Hence, if your preferences over lotteries are described by this utility function, then you will prefer gamble $Y$ to gamble $X$, even though gamble $X$ has expected value $10$, and $Y$ has expected value close to $1$.

This is a phenomenon called risk-aversion. It is often modelled via assuming preferences reflect the desire to maximise a concave utility function. (Notice that $u(x) = \sqrt x$ is concave.)

A decision-maker with a concave utility function demands a risk-premium for bearing risk. In particular, they will always prefer the expected value of a lottery for sure rather than the lottery itself. This is a simple consequence of Jensen's inequality. For any gamble $X$ and $u$ concave, we have that $$ u(\E[X]) \ge \E[u(X)], $$ which means that if the decision-maker were given the option of two gambles, one which paid $\E[X]$ for certain, and the other which pays the lottery $X$, then they would choose the former.

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Let's attach values to this game. Game A pays off $\$1$ with certainty, and game B pays off $\$100$ with probability $0.99$. I have a sense that virtually everyone would choose B.

When the numbers are closer, you're getting into psychological values like risk management and the non-linear utility of money. But without those factors, expectation is designed to be the number that you want to maximize.

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  • $\begingroup$ "I have a sense that virtually everyone would choose B"..OK but is there a way to prove that B is the best choice? $\endgroup$ – marco trevi Nov 9 at 13:34
  • $\begingroup$ What about winning really much 1% of the time with strategy B and not so much but most of the time with strategy A? Playing B could result in an almost-certain loss. $\endgroup$ – marco trevi Nov 9 at 13:36
  • $\begingroup$ @marcotrevi It depends on the criterion you’re using. How do you define “best”? $\endgroup$ – Theoretical Economist Nov 9 at 13:36
  • $\begingroup$ @marcotrevi Expectation lets us prove that A has a value of 1 and B has a value of $100(0.99)=99$. $\endgroup$ – Matthew Daly Nov 9 at 13:36
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    $\begingroup$ @marcotrevi You still need some way to make sense of the randomness (that is, turn the randomness into a numerical criterion). One way to do that is to use expected values. Just because you only play once and not multiple times does not rule this criterion out. $\endgroup$ – Theoretical Economist Nov 9 at 14:11

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