It turns out not to matter (since the numbers are so small) but your math implies that 10^305 people guessing would give a probability of about 9,300 that at le

It turns out not to matter (since the numbers are so small) but your math implies that 10^305 people guessing would give a probability of about 9,300 that at least one of them is right on all 1000 coin flips. Since probabilities can't be larger than 1, that's a bit of a problem. 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".