Hedging bets to eliminate variance Let us say that your very generous uncle has decided to play a game with you. He is to roll a fair $6$-sided die and he will pay you $\$12$  for every pip that shows. For example, if he rolls a $5$ you get $\$60$ or if he rolls a $2$ you get $\$24$.
Your kind, but not so generous, aunty overhears this and decides to help you out. She will be your roulette, that is, she offers you the chance to stake any amount on any outcome (that pays out fairly) of your uncle's die roll.
For example, you may stake $\$5 $ on a $1$ or $2$ in which case if a $1$ or $2$ is rolled she pays you $\$15$ as this event has probability $\frac{1}{3}$. Or, you could have simply bet $\$5$ on a $1$ in which case you get paid out $\$30$ if it happens.
What bets should you make with your aunty to entirely eliminate your variance, i.e., to ensure you get paid out the same in net between your uncle and aunty for any roll?
 A: I managed to find an answer to this by plugging stuff in and then worked backwards to a solution. I thought I would post it on here incase anyone from the future wants an answer.
We actually only need to make $5$ bets due to the fact that placing/laying the same bet on each outcome does nothing, as seen below: (Consider just betting $\$1$ on each outcome and think about what this does to your payoff/variance)
Assuming some strategy,  $\bf{S}$ $ = (\theta_1 ,\theta_2 , \theta_3 , \theta_4 ,\theta_5, \theta_6) $  where $\theta_i$ denotes the bet placed on outcome $i$ achieves the zero variance we are after then so does $\bf{\bar{S}}$ := $(\theta_1- \theta_6,\theta_2- \theta_6 , \theta_3- \theta_6 , \theta_4- \theta_6 ,\theta_5- \theta_6,0)$
Hence we only consider a strategy with $5$ bets and nothing on the outcome of a  $6$
If we are in a position where we wish for total indifference between rolls that would mean that a $1$ and a $6$ payed out the same, thinking just in terms of "revenue" and not the net payoff we imagine we are in a position where all our bets have been made and our waiting for our uncle to roll the die:
When we roll a $6$ we get $\$72$ and so we want to make sure when we roll a $1$ and get paid only $\$12$ by our uncle we receive and additional $\$60$ from our aunty. So must stake $\frac{\$60}{6} = \$10$ with her.
Similarly we must ensure our "revenue" we we roll a $2$ is also $\$72$ and so must lay $\frac{72-24}{6} = \$8$ on a $2$
Continuing for the rest in this manor our strategy is $(10,8,6,4,2,0)$
