I've been thinking a bit more about this, and I don't think what I've quoted below has been fully resolved. And I wonder if this is some sort of philosophical paradox rather than a purely mathematical question. Like e.g. "This statement is false" but with more baggage around it so that it's difficult to distil it to its purest form to isolate and understand it.

But to revisit a couple of things: You can pick any distribution that fits the parameters of the question and show that the solution given seems to basically work. I.e. pick the distribution, and then pick your starting number, and you can work out mathematically that you will win over 50% of the time if you play multiple times.

E.g. you could pick a normal distribution with a mean of 2 and standard deviation of 2 and pick 5 as your starting number. Play that out hundreds of times and you'll win more often than not (that's been discussed elsewhere but I can go through it again if needed).

But it works for any number that you pick. And despite what was said previously in this thread, it has nothing to do with picking the number beforehand, because the above paragraph is true regardless of when a number is picked, and that's where it starts to get paradoxical.

For example, I have 2 as my number. The first card is revealed as a 4. I'm about to say "lower" when I think to myself that whatever underlying distribution is being used here, it also works for 6, so I could just as easily go for higher.

Alternatively, if you don't like that, someone could walk into the room and say that he's playing too and that 6 is his number. Paul brushed over this by saying that playing long term, both numbers would win in the long run, but that's not the point. We have a one-off event. One strategy is saying higher and the other is saying lower and there's no way to distinguish between them. Do I have a greater than 50/50 chance of winning? Does the other guy? We can't both have.

Also if you ask someone who wasn't playing the game what they know about the distribution if they see a 4 come out as the first number, they know exactly the same amount about the distribution as someone who is playing and picked 2 beforehand. Picking your number and seeing a card turned over gives you no more knowledge about the underlying distribution than simply seeing a card turned over.

But as I said previously, it would seem very strange if the strategy given as the solution would work over a long run, but not give you a greater than 50/50 chance on the first attempt. After all, the first go is exactly the same as the rest of them.

But the bottom line is that solving this problem doesn't mean doubling down on either 1 or 2 in the quote below. It means resolving the tension between the two of them. So if someone comes back to this thread now just to double down on 2 and spell it out in minute detail, they would be missing the point.

Gavin Chipper wrote: ↑Sat Mar 18, 2023 6:45 pm
I thought I might give this a revisit, for a laugh.

Basically we have two competing things that both seem correct, but are also seemingly in conflict with each other.

1. We just have a single random number from a distribution we don't really know anything about and no other relevant information about it, so can have no way of knowing if another number taken from that distribution is likely to be higher or lower, so it must be 50/50.

2. It is true that for any distribution (one that's relevant for this anyway), you can pick any number (c) and if you pick a number (a) from the distribution, then more often than not, another number (b) from the distribution will be c-side of a.

What those setting the puzzle do is double down on (2). They say it's correct, so (1) must be incorrect. What those being set the puzzle do is double down on (1). It's correct so (2) must be incorrect. But there's little interaction between (1) and (2) in the discussions, so both sides go away thinking that they must still be right. (It reminds me of the aeroplane on a treadmill in that respect, though I can't remember any of the details about that.)

Edit - One other thing that could be said is that it's been said many times on here that you can't have a uniform distribution across all real numbers. Because of this, any underlying distribution will be in some way "biased". Specifically, the median of the distribution must be "approximately" zero. That is, I will be able to pick a number n where the median of the distribution is zero to the nearest n.

Also the number that the player picks will also come with exactly the same bias, so the distributions they come from are therefore related in some manner. So the player using their own pick as a proxy for a sample of the dealer's distribution makes more sense from that perspective.

But that still doesn't change the fact that the guy standing behind you is saying the opposite of what you're saying...