5 ms·
"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
by idlewords 11y ago
"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.
- sidedishes 11y agoOne 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.
- idlewords 11y agoBut then it's not really uniform, is it? That innocent instruction to 'pick a random number' is carrying a lot of conceptual baggage.
- sidedishes 11y agoRight, 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).
- idlewords 11y agoThere'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.
- aidenn0 11y agoSo long as you have a non-zero chance of your number "R" being between A and B, you have an advantage.
- dllthomas 11y agoPer my understanding, it doesn't take the weirdness involved in the reals... there's not even a uniform distribution over the integers.
- analog31 11y agoTrue. If somebody asks you to choose a random number, you have to ask: From what distribution?
- mjd 11y agoAny distribution suffices, as long as it is everywhere positive.
- monochromatic 11y agoTrue 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.
- fenomas 11y agoIt 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).
- JackFr 11y agoThere is no paradox, because under the conditions of the problem, the advantageous solution cannot be shown to occur with a probability greater than 0.
- fenomas 11y agoWhether 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.
- mjd 11y agoRolling 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.
- deleted 11y ago[deleted]
- fenomas 11y agoAh, 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.
- mjd 11y agoYou'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-a-game-whose-details-i-dont-remember/656426#656426 http://math.stackexchange.com/questions/655972/help-rules-of...