...or just add up the answers mod 10.
This has the property that if even a single person answers uniformly at random, then the final number you compute will be uniformly random, regardless of how everyone else answers.
This has the property that if even a single person answers uniformly at random, then the final number you compute will be uniformly random, regardless of how everyone else answers.
In a room of 8500 people, it is likely that there is someone in the room smarter than you and knows a better way to pick random numbers. Come into the room and tell everyone that your objective is to pick a random number from 1-10. Maybe even go as far as to tell them your mod 10 idea. Wait a bit to let them to think about it, then start asking for the random numbers. It is likely that someone will have come up with a better way to pick a random number than your mod 10 solution, and thus their answer will be more uniformly random that what your solution can produce. And thus your solution becomes at least as uniformly random as theirs.
This is how my board game groups pick the first player.
Has any human ever proven they're capable of this? Generating truly random sequences is more or less impossible for humans AFAIK. (E.g., see https://news.ycombinator.com/item?id=19336754 which challenges you to do just that. Spoiler: you will probably fail miserably.)
It's an interesting idea, but in practice I think relying on the assumption that "even" one person is truly answering randomly (let alone uniformly at random) is a non-starter. But perhaps if there are enough people, the resulting sequence blends enough entropy together to get something that looks almost like a uniform random variable anyway? It would be interesting to test empirically.
Philosophically, no it is not possible, though neither can any natural phenomenon for which we reasonably rely on for randomness.
Practically I suppose your goal is just to generate numbers such that the next number cannot be predicted given only the previous numbers but also without considering any outside knowledge. Potentially possible. Potentially impossible to test. If you can beat the test, it probably just means your method beats that specific test
But if you do have independence, the proof is easy! Let S be the sum of everyone else and X be the discrete uniform random in [0, 9]. Then:
Pr(X + S mod 10 = i)
= \sum_j Pr((X + j) mod 10 = i | S = j) Pr(S = j)
= \sum_j Pr(X = i | S = j) Pr(S = j)
= \sum_j Pr(X = i) Pr(S = j)
= Pr(X = i) \sum_j Pr(S = j)
= Pr(X = i)