Think of a number, answer each 'greater than?' question honestly, and watch the grid of candidates halve every time; the counter of questions asked equals the bits gathered.
The claim above is what this widget demonstrates. Disagree with it?
A secret number from 1 to 64 is always identified in exactly 6 halving yes/no questions because log₂ 64 = 6, so each question resolves one bit of the starting uncertainty.
Builds on Surprise in bits. Taught in Bits & Surprise.
Think of a number, answer each 'greater than?' question honestly, and watch the grid of candidates halve every time; the counter of questions asked equals the bits gathered.
The claim above is what this widget demonstrates. Disagree with it?