[corecirculator]

Other commenters are wrong in saying that the payout is different for an adversarial choice. The crux of the payout derivation is: we can only cover 1 number in step 1, 2 in step 2, 4 in step 3, 8 in step 4, and so on. You can choose your initial number in binary search randomly, and as long as you meet the above condition is met (# of possible numbers covered in each step), payout should be same as 0.2

[dagw]

If I 'know' that my opponent is adversarial, then I might assume that he's not picking from the set of 100 possible numbers, but actually from a smaller set of 'adversarial' numbers, like the set that will always take 6 or 7 guesses using the naive binary search approach, and I can adjust my strategy accordingly.

[LudwigNagasena]

You should assume that your opponent is adversarial to your specific strategy.

[alexey-salmin]

Your calculation assumes that probability of each number is the same which is not true for adversarial choice.