Probability of not losing money given the variances and equal expected value

Question

If you have two games A and B that have the same expected value $E$ when you play. The variance in the expected value of game $A$ is $vA$ and the variance in the expected value of game B is $vB$. What is the probability you don't lose money when you play each game? $(E>0)$

Thoughts

On average we expect to win E when we play either game. If we take the case $vA> vB$ then, intuitively, I think that playing game A has a higher likelihood of losing money. Is this reasoning correct? How do I use the variances to calculate the probability of $E<0$ in both cases?

Answer 1

To make the cases comparible let it be that there are equally distributed standardisations.

Let $\mu>0$ denote the common expected value, let $\sigma_A^2$ denote the variance in case A and let $\sigma_B^2$ denote the variance in case B.

If $\sigma_A>\sigma_B$ then: $$\frac{-\mu}{\sigma_A}>\frac{-\mu}{\sigma_B}$$ so that $$P(\mu+\sigma_AX<0)=P\left(X<\frac{-\mu}{\sigma_A}\right)>P\left(X<\frac{-\mu}{\sigma_B}\right)=P(\mu+\sigma_BX<0)$$ Here $X$ stands for the common standardisation.

---

This confirms your "reasoning" (or "thinking" actually, I cannot discern any reasoning in it) that in game $A$ there is a higher likelihood to loose money than in game $B$.

This however under an extra condition: equally distributed standardisations.

Answer 2

The variance alone doesn't give the answer. For example it's easy to construct examples of $A$ and $B$ with $vA>vB$ where one never loses playing $A$ but can lose playing $B$:

$A$: possible outcomes are $2,1$ and $0$ with equal probabilities, $B$: possible outcomes are $3,1,-1$ with $P(1)=0.98$, $P(3)=P(-1)=0.01$.

Answer 3

I was thinking more intuitive.

A coin flip $X$, where: $$P(Heads)=P(X=3)= \frac 12, P(Tails)=P(X=4)= \frac 12$$ And a dice roll $Y$, with possibilities $1,2,3,4,5,6$ where: $$P(Y=1)=\frac 16,P(Y=2)=\frac 16,P(Y=3)=\frac 16,$$ $$P(Y=4)=\frac 5{24},P(Y=5)=\frac 1{12},P(Y=6)=\frac 5{24}$$ Expectations are the same $E(X)=E(Y)=3.5$, variances are different; $\sigma_X^2= 0.25$ for the coin flip, $\sigma_Y^2=2.917$ for the dice. So $\sigma_Y>\sigma_X$, we have the same expectation, but clearly there are differences in the probability and distribution.

It is in fact not just that you have a higher likelihood of loosing/winning money, but also a higher likelihood of loosing more money than a game with same expectation.