[hanoz]

> Wouldn't putting "everyone in a numbered sequence unknown to them" require a random number, which we don't have?

Yes. The above answer just obfuscates the issue. Consider ten people overwhelmingly biased towards 7. The suggested approach in all likelihood gets you 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Then what? You're clearly not much closer to getting a single random number without a random way of picking one of them.

[simonh]

>You're clearly not much closer to getting a single random number without a random way of picking one of them.

That’s a characteristic of the original formulation of the question, not specifically in my solution. It’s implicit in the article that you can choose an arbitrary person in the room without bias.

[hanoz]

It's a shortcoming specifically introduced by your proposal. If you've got a random way of picking one of ten people, then you've already got a random number between one and ten. Your answer essentially reduces to: ask one person to think of a number then add an actual random number and mod 10.

[simonh]

From the article:

>The easy thing to do is to ask someone “Hey, pick a random number from 1 to 10!”. The person replies “7!”. Great! Now you have a number. However, you start to wonder, is the number uniformly random?

So the problem reduces to transforming one form of randomness (picking a person) to another form (a uniform numeric distribution. That's implicit in the whole rest of the article.

>If you've got a random way of picking one of ten people, then you've already got a random number between one and ten.

You've got a source of randomness, but you still need to turn that into a number. The article gives one method and I've given another. You could also assign the people numbers from a stack of shuffled numbered tiles, then pick a person although that requires you have the tiles which is not really part of the setup. The conceit is that you have to use people's ability to pick numbers pseudo-randomly. There are probably numerous ways.