Lets put it in perspective; the chance of one person out of a million getting a 100% streak is 9.332 × 10^-296 %. That's significantly less than 0.000000000000000000000000001%.
If there are people who have a 100% record of being correct on every guess, the probability of them being geniuses is way higher than the probability of them being lucky.
That being said, I get your point. We don't have enough metrics from the research to determine whether or not these super forecasters are just lucky or geniuses.
For anyone interested, the full set of steps (that produces a numerically identical result):
Prob[1 or more in 1,000,000 right]
= 1 - Prob[all 1,000,000 wrong]
= 1 - Prob[person 1 is wrong AND person 2 wrong AND ... person 1,000,000 wrong]
= 1 - Prob[person 1 is wrong]^1,000,000
= 1 - (1 - 0.5^1000)^1,000,000
= 1 - exp(1,000,000 * log(1 - 0.5^1000))
= 1 - exp(1,000,000 * log1p(-0.5^1000))
≈ 1 - exp(1,000,000 * -9.33 × 10^-302)
= 1 - exp(-9.33 × 10^-296)
= -expm1(-9.33 × 10^-296)
= 9.33 × 10^-296
log1p(x) = log(1 + x) but is more accurate when x is near zero.expm1(x) = exp(x) - 1 but again is more accurate when x is near zero.
Both are necessary here to get a result other than "0".
Here's a quote from a New York Times article about the project -
> In the second year of the tournament, Tetlock and collaborators skimmed off the top 2 percent of forecasters across experimental conditions, identifying 60 top performers and randomly assigning them into five teams of 12 each. These “super forecasters” also delivered a far-above-average performance in Year 2. Apparently, forecasting skill cannot only be taught, it can be replicated.
So the answer to the question "What is the probability any one member of the group correctly guesses the result of the next coin toss?" appears to be "reasonably high".