Hypothetical scenario:
You can bet on a
single dye roll (choosing a number between 1-6), but if you win you get paid x12 of your stake (instead of x6).
So the Expected Value is positive (see example below), but you'd still have 83% chance of losing.
1 - will you take that bet?
2 - if so, what % of your available funds would you put at stake (i.e. funds you're willing to gamble and afford to lose)?
Again, this is a single, non-repetitive bet.
-----------------------
Simulation for
BTC1 bet:
EV = (83% x -
BTC1) + (17% x
BTC11) = +
BTC1.04
Edit: calculation corrected, credit to Saint-loup
Edit2:
A somewhat related video on a subject on unreasonable risk aversion. Good watch and not very long:

I don't think this math checks out unless you have many rolls, not just the one. You cannot really calculate the statistical sum of the profits with a single dye roll. Just like you cannot calculate the average mean with only a single number as that would make the number the average mean itself and calculating it has no point. That just seems a bit silly, to be honest.
There is no need to calculate EV here. Its a one time thing, so either you win or you lose. The chances of winning is around 17%.
I would take the bet with twice the amount as regular dice game as it pays over double. That seems like a reasonable amount. But I would be wary of betting more, as the large winnings do not have any effect on the probability. Its still the same chance of winning as in a regular dice game.