You don't need a uniform distribution to get guaranteed better than even odds. And many nonuniform distributions work just fine.
There are no hidden conditions. It is just a shocking result that we don't expect.
There are no hidden conditions. It is just a shocking result that we don't expect.
If your solution is the same as this article's, it's plain wrong. Even the natural number case is plain strong.
It's very easy to demostrate as well: consider a trivia case where the distribution is just {P(1)=1/3, P(2)=1/3, P(3)=1/3} and you see 2 in the first envelope. There is no strategy to get a better chance than 50%. Therefore, any strategy that gives a better chance than 50% must implicitly make an assumption over the initial distribution (and therefore excludes a distribution like {P(1)=1/3, P(2)=1/3, P(3)=1/3})
Actually the article is even "wronger" than this, because "started A" and "switched" aren't independent and one can't simply use the product of their probability. The above example is a quick way to demonstrate it's not a general strategy without assumption to get >50% winning chance. Similarily, one can just use {P(1)=1/3, P(2)=1/3, P(3)=1/3} (this is a valid distribution over real numbers!) to demonstrate the real number strategy isn't general.
Again, for both natural number and real number case, the discussion over strategies is only meaningful is we know something about the distribution.
Interestingly, this article is wrong more or less in the same way as believing switching does give you more expected value in the original "twice money in another envelope" variation.
Edit: For people who are interested in the switching strategy, check Randomized Switching in the Two-Envelope Problem (2009). Spoiler: full of discussion over the initial distribution.
I don't actually care how you convince yourself. But the explanation is right. If your random number is outside of the range, you've got even odds. If it is inside of the range, you've got 100% odds. As long as there is a positive probability of being between, you've got strictly better than even, by half of the probability of being between.
Many, many distributions guarantee positive odds of being in between. The one I chose for my program was:
(log(rand) * (flip_coin() ? 10 : -10 ))
The strategy is straightforward and bulletproof (if you allow a random generator of real numbers, otherwise you may keep tossing coins indefinitely): keep tossing coins until you get tails. If the number you saw is less than the number of heads you got, you don't switch.
For the simplest case assume that one envelope always contains 1 and another always contains 2. You choose one envelope randomly, so in 50% of cases you get 1, which you switch in 50% of cases. And in 50% of cases you get 2, which you switch in 25% of cases. Hence, you pick the higher number in 62.5% of cases. The same works with any numbers N, M; or any complex distributions; or even real numbers with a bit more complicated strategy. You don't have to know whether you are between two values in advance, you just have to guess.
In other words, I'm merely trying to be informative.
I thought you meant the strategy can make the winning chance always >50% even after the player opens the first envelope, which isn't possible.
However you actually meant the strategy can make the expected winning chance >50% before the player opens the first envelope, for any well-defined distribution of real number, even the distribution is not known to the player, which now I realize is true.
(I haven't thought through some edge case like Cantor distribution, but now I incline to it's true not just for "many distributions". Of course for a discrete distributions, we need to specifiy the two envelopes can't have the same number. Besides that, it seems to hold true for any distribution?)
But before you pick, your odds were still bigger than 50%. Just not by much.
Seeing 2 is only one of the many possible cases. You haven’t calculated the total probability.
It's not saying that after the player see the number in the first envelope, the strategy guarantees a >50% outcome.
It's saying that give any distribution, over all possible outcomes, >50% times the strategy will end up pick the larger number. You can say this >50% is the expected winning chance before the player see the number in the first envelope.
I'd say this is "intuitve" because, if your strategy can guarantee "when the player see a large number in the first envelope, he's less likely to switch than if he saw a small number", it would be better than blindly switching by coin toss. So intuitively such a strategy exists.
The only "trick" here is that since the player doesn't know the initial distribution, they can't tell "how large counts as large?" therefore they needs something that preserves some property over the whole real number line. That's why the strategy involves sampling from a another distribution whose PDF is non-zero everywhere.