Defining a probability of success in a lottery

Question

Suppose that I have the following lottery:

$p(x)W-(1-p(x))L$

Where $p$ is the probability of winning W and x is a variable that influences the probability of success. The expected utility for the gambler is (omitting x for simplicity):

$E(U)=pW-(1-p)L$

My questions are:

1) it is possible to define $ p(x)$ such that as x goes to infinity p goes to 0,5 (not 1)?

2) if the answer is yes, what is an example of a function p(x)? Is it correct to say that $E(U)$ tends to $0.5W-0.5L$?

Answer 1

I'm not really sure what you mean by "$x$ influences $p$", but mathematically $p(x)$ with the desired properties are easy to come by. Basically you just need to shift by $1/2$ of whatever goes to zero, and scale it to make sure $p$ is always between $0$ and $1$.

And yes, $E(U) \to \frac12 (W - L)$ with such $p(x)$.

If you're interested only in the positive half $x \geq 0$, then

1. An example of winning chance grows monotonically from zero at $x=0$ to $1/2$ can be $\displaystyle p(x) = \frac12 \left(1 - e^{-x} \right)$.
2. Winning chance starting out 100% that oscillates can be $\displaystyle p(x) = \frac12\left( 1 +\frac{\sin x}x \right) $.

If somehow the whole $-\infty < x <\infty$ matters, then

1. The winning chance starts out to be $1/2$ at $-\infty$ that dips to zero then recovers can be $\displaystyle p(x) = \frac12 \frac{ x^2 }{1 + x^2 }$.
2. Coming from the left leveling at $1/2$, dip to zero, shoots to one, then back to $1/2$. It can be $\displaystyle p(x) = \frac12\left(1 +\sqrt{2}\, x\,\mathrm{Exp}\left(\frac12-x^2 \right) \right)$.
3. In from the left leveling at $1/6$ then turns to $1/2$. It can be $\displaystyle p(x) = \frac13 \left( 1 + \frac{\arctan x}{\pi} \right)$.