Let a fair coin decide at the start of the year whether the share runs at 15 or 35 percent, price calls in that world, then read the volatilities back out with the constant-volatility formula: 28.43, 25.91, 24.97, 25.68 and 27.16 percent across five strikes. The floor sits at the money and below the 25 percent average of the two regimes, which one second derivative settles without any numerics. The usual explanation for the wings is refuted here, because the coin-flip world is less likely to clear 130 than a flat 25 percent and its option is still worth 27 percent more.
Two weights that add to 0.98 give the variance forecast a half-life of 34.31 days; delete the second one and the half-life is 0.2744 days, gone before the next open. The same recursion turns strictly normal daily draws into a year with kurtosis exactly 297/67, and one shuffle of those same numbers separates the fat tail from the clustering. It also has a condition nobody quotes: stationarity is not enough for that kurtosis to be finite.
Four settlements in five come back below the 1.50 outlay, and the average payoff is still 1.80, an edge of 0.30 a contract or twenty percent of the money at risk. The reflex is not bad arithmetic, it is the mode standing in for the mean. The article carries the tally over one full cycle, the threshold saying you need the large outcome more often than one time in eight, and the reason waiting longer can leave you less likely to be ahead.
Both games pay 3.5 million dollars on average, so "they match" is true and the inference that it is a wash is not. Shrinking the ticket by a million divides the spread by a million while adding a million independent rolls multiplies it back by only a thousand, so the ratio of standard deviations is exactly root of a million: 1,707,825 against 1,707.83. The whole argument rests on independence, and at a correlation of 1 the diversified game becomes the single roll exactly.
The standard deviation of 1, 2, 3, 4, 5 is either 1.4142 or 1.5811, and offering one of them without asking which question you are answering is the only wrong move. The sum of squared deviations is 10 either way, so everything turns on whether you divide it by 5 or by 4. Bessel's correction makes the variance unbiased and leaves the standard deviation biased low by about six percent at this sample size, and a third divisor beats both of them if you optimise for mean squared error instead.