How to maximize the gain in a game? 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).
 A: 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.
A: 
$\DeclareMathOperator{\E}{E}
   \DeclareMathOperator{\Var}{Var}$
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.
