The probability distribution of the outcomes is positively skewed. You are looking at the median outcome which is below the average outcome.
If I may propose looking at it in this way: Certainly this is not the case in an automated trading environment where the exit is either a stop at a predetermined risk level or a profit at a predetermined profit objective. Also, over the long run, even a discretionary method that involves all sorts of variability - breakeven stops, trailing stops, scaling in, scaling out, etc. should yield you with a set of data that will determine an average win rate and an average payout over the long run. Even in poker and blackjack, payouts vary from hand to hand as do the probabilities of winning or losing any particular hand (even using perfect basic strategy with counting, one will lose hands he "should" have won and will win hands he "should" have lost - just as each trade is independent from each one before and after, so too is each hand dealt - even a "high" count does not assure one will win each hand until the shoe is completely dealt). Each hand is, in a sense, like a discretionary trade. If you look at the OP's example, the scenario he has proposed, based on just six trials, is an even money bet with a 50% probability of winning. In this case there is no advantage, i.e. he has NO edge, and over the long run, (the very long run) his equity should be unchanged minus transaction costs/vig. Over the short run, however, such even money bets with a 50/50 probability will still experience some equity volatility (deviation, variance - what some identify using the word "randomness"). Given the assumptions of the OP's six trials, he has no advantage, and therefore he should not bet. Such is the folly of the Van Tharp 2% bet mantra: The optimal bet size is edge-dependent, and a "one size fits all" money management strategy may sell books and course, but it won't make you money except by accident. Without knowing one's edge, one cannot accurately figure out the proper money management protocol.
Mostly correct. Yes his scenario is breakeven and zero-edge. But he is not betting even money, he is betting a fixed fraction, which makes him a net loser. If his fraction was 50% instead of 2%, he would need a gain of 100% after losing 50% just to get back to even. This is why he shouldn't bet at all: betting a fixed dollar amount gains him nothing and betting a fixed fraction greater than zero actually hurts him. Agreed. 2% may look safe but if the edge disappears for whatever reason, suddenly 2% becomes very unsafe.
Even money, as I used it, refers to a 50/50proposition with a 1:1 risk to reward - not his 2% bet scheme.
Drawing on what other posters have said, you have no edge so the Kelly optimum is not to bet. Since you're betting more than that zero you'll end up losing money (you'll never hit zero, since in this stylised example that's impossible - you always get out at your "stops"). The average terminal wealth would be less than you started with; the loss growing over time. For example if I simulate bets of 10,000 runs with an equal chance of winning or losing 2%, then the average run sees you losing $5,000. With runs of 100,000 you lose on average $98,000! If you simulated this with a small positive edge, such that 2% was less or equal to kelly optimum, then your average terminal wealth would be greater than $100K. So for example with a 50.1% chance of winning 2%, and a 49.9% chance of losing 2%, with runs of 10,000 I end up with an average profit of $58,000.