In other words, even if you show that the probability of the ratio of variants within any given interval around the Median tends to 1, you'd still have to show that the resulting Payoff within that interval is zero.

Done that. Twice. If the ratios are all within the interval 0.1666<x<0.1667 then the resulting payoff is always less than 1, and it tends to zero as N tends to infinity. An upper bound for the payoff is: Payoff < (0.7^{0.1666}*0.8^{0.1666}*0.9^{0.1666}*1.1^{0.1667}*1.2^{0.1667}*1.5^{0.1667})^{N} < 0.9998^{N} < 1. In plain words: over time, one hundred percent of the variations will be close enough to the median variation, that their payoff is zero.

To understand **why** this matters, do not concern yourself with the nominal values of the stakes and payoffs, but rather with their effect on your bankroll.

If your bankroll is $1, and you bet $1 and hit 0.7, your bankroll is reduced from 1 to 0.7, a change by the factor 0.7. Here the nominal change equals the bankroll change.

However, if your bankroll had been $2 instead, and you bet $1 and hit 0.7, your bankroll would be reduced from 2 to 1.7, a change by the factor 0.85. Because you only bet half your bankroll, the change is not as dramatic. If you bet half your bankroll instead of your entire bankroll, the payoffs would result in a change of 0.85, 0.9, 0.95, 1.05, 1.1, or 1.25. The geometrical mean of these changes is (0.85*0.9*0.95*1.05*1.1*1.25)^{1/6} ~ 1.008, which means you now have a positive expected effect on your bankroll. If you bet half your bankroll every time, you can expect it to grow exponentially (1.008^{N}).

Now we might ask, when is the inflection point between a good game and a bad game? How much of your bankroll can you bet before you expect to lose money?

Well, if you bet the fraction x of your bankroll, the payoffs would be 0.7x, 0.8x, 0.9x, 1.1x, 1.2x, and 1.5x. Adding the 1-x you have remaining on the side, the change in bankroll would be 1-0.3x, 1-0.2x, 1-0.1x, 1+0.1x, 1+0.2x, or 1+0.5x. The geometrical mean would be GM = ((1-0.3x)(1-0.2x)(1-0.1x)(1+0.1x)(1+0.2x)(1+0.5x))^{1/6}, which can be solved for x, at GM=1. The solution in the interval 0<x<1 is a bit over 0.989, which means the game is good for you as long as you bet between 0% and 98,9% of your bankroll.

But why am I not concerned with the nominal values of the winnings, but rather their effect on my bankroll? Consider this: if you had a $1,000,000 total fortune (this includes your house and valuables), would you risk it all for a 50% chance to triple it? Would you risk it all for an 80% chance to double it? Would you risk $999,000 of it for a 60% chance to double it? Very few people would be silly enough to take those risks, even though the EV is very good. The true difference between having one million and being dead broke is a lot more dramatic than the difference between having two millions and having one million, but the nominal difference is the same.