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Comment by idlewords | Hacker News Reader

"Here's your winning strategy. Before you see A, choose a random number R"

Isn't it true that there's no uniform distribution over the reals? That seems to be what the whole apparent paradox hinges on.

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You're right, but it doesn't have to be uniform, and of course it can't be. It only needs to be everywhere positive.

I wrote it up in more detail here: http://math.stackexchange.com/questions/655972/help-rules-of...

One way to resolve this problem is to use a continuous, increasing map f from the reals to say, (-1, 1). Something like arctan(x)/pi. You can then pick a number randomly from (-1, 1), then compare that with f(A), rather than a random real with A.

But then it's not really uniform, is it? That innocent instruction to 'pick a random number' is carrying a lot of conceptual baggage.

Right, though I don't think uniform is the necessary condition here. We only need a positive chance of picking an R so that f(A) and f(B) are on different sides of R, which isn't as much of a problem if we're picking R from (-1, 1).

There's also the question of how the first two numbers were chosen; in other words, what it means that 'the numbers are completely arbitrary'.

I can't math real good, but this is the part that smells fishy to me.

So long as you have a non-zero chance of your number "R" being between A and B, you have an advantage.

It doesn't have to be a random number from any particular distribution - you can roll 1d20, or use your age, or anything else you like.

The point of the paradox is that picking a value and trading anything below it is always a non-losing strategy, and may also be a winning strategy, depending on factors that are not defined in the puzzle (i.e. whether your opponent's choices are anywhere close to the number you chose).

Rolling 1d20 doesn't work, because if the first player always chooses numbers bigger than 20 or smaller than 1, the second player wins with only probability ½.

To get a winning probability greater than ½ the second player needs to choose from a distribution that is everywhere positive, which means that a discrete distribution won't do.

Ah, I see what you mean - I meant that any old number will suffice for having a non-losing strategy, assuming that the opponent's strategy is unknown and undefined.

There is no paradox, because under the conditions of the problem, the advantageous solution cannot be shown to occur with a probability greater than 0.

Whether a solution is actually advantageous depends on whether your chosen number lies between your opponents choices, which are not defined. The author notes that if you select from an unbounded distribution, you can in principle have a nonzero (though in practice very small) advantage regardless of the opponent's choice.

True that there's no uniform distribution over the reals (or even over the integers or the natural numbers). But just pick whatever distribution you like.

Per my understanding, it doesn't take the weirdness involved in the reals... there's not even a uniform distribution over the integers.

True. If somebody asks you to choose a random number, you have to ask: From what distribution?

Any distribution suffices, as long as it is everywhere positive.

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