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A game for two players: you each pick a real number between in the interval [0,1]. Whoever picks the highest number wins. If you both pick the same, you go again.

Play three times. You can't pick the same number more than once.

Is there a strategy?

@mur2501 @christianp That wouldn't be allowed. The square brackets means non-inclusive.

This problem is identical to the game where you have to pick a finite number, and the one who picks the greatest number wins. All you have to do to convert from one to the other is to say that my number is 1-1/hugenumber.

Now, how to describe a huge number in a finite number of characters is an interesting exercise. https://en.wikipedia.org/wiki/Rayo%27s_number

@loke I think you are mistaken ... square brackets means inclusive, it's round brackets that means the limit points are not included. I can find references if you like.

And if you pick "1" for the first game you can't pick it again, so there are strategies there. It's non-trivial.

CC: @mur2501 @christianp

@loke Not at all. I suggest you work through the scenarios. Remember, you're playing three times, you get to see what your opponent is doing.

If you pick "1" on the first move and your opponent doesn't then on a subsequent turn, if your opponent chooses, they have a guaranteed win.

Etc.

CC: @mur2501 @christianp

Anyway, if you pick 1 first, there's nothing to stop your opponent doing the same, and then you'd play that round again until you pick different numbers

@mur2501 For your first move, OK. But you can't pick the same number twice ... what do you pick for the second number?

CC: @christianp

@ColinTheMathmo @christianp

For second 0.999999999......

@mur2501 In standard mathematics that's equal to 1, so it's the same number, and it wouldn't be permitted under the rules.

CC: @christianp

@ColinTheMathmo @christianp

It's very close to 1 but it's not exactly equal to one

@mur2501 I've had this discussion too many times, and I'm not going to have it again.

Read this:

https://en.wikipedia.org/wiki/0.999...

If you don't understand something then maybe I'll take time to explain it, but I've wasted a lot of time in the past trying to help people see it.

All of modern mathematics says that zero point nine recurring is another representation of 1, just as 6/8 is another representation of 9/12.

CC: @christianp

Well yes I have seen various proofs of that and they are not wrong

Just that the meaning of infinity in mathematics is still very much dynamic and flexible and I think this property also depends on what we think infinity to be

@christianp

@mur2501 If they are proofs then the thing they are proving is correct. That means that point 9 recurring equals one.

If you want to move into non-standard analysis then you're on your own.

My time doing graduate work and then research in maths gives me a fairly definite view on this. I've tried to share this in the past, to help people understand why maths is the way it is, and people just keep waving their hands and going "infinity is complicated, and you can ..."

Well ...

CC: @christianp

@mur2501 ... these things have been studied and researched for centuries, and in standard maths the conclusion is that point nine recurring equals one.

If you want to develop your own theory for why it isn't, I can point you at non-standard analysis, hyper-reals, and the surreals, and bid you good luck, and fair winds.

CC: @christianp

@ColinTheMathmo

I now dive into the unknown territory

This would be fun

@christianp

@mur2501 @ColinTheMathmo @christianp Note that @ColinTheMathmo was very clear when he said "standard mathematics".

We can have a completely separate discussion about hyperreals or even surreal numbers. That's a really interesting discussion on its own, but as much as I am a fan of hyperreal numbers, that was not the topic at hand.

Also, even when discussing hyperreals, you have to somehow argue that 0.999... is equal to 1-ε. And while I've seen that assertion made, I don't think I've seen a convincing argument why that should be the case.

And again, all of that it outside of standard mathematics, so again not in the scope of the original question.

@christianp pick 1. you will never lose!

@kat what if the other player picks 1 too? And what about rounds 2 and 3?

@christianp It was a slightly joky response (hope that's ok)

does a round not just repeat infinitely until a round winner is declared?

I don't actually think it's a winning strategy, but thought it was a non losing strategy

@christianp lol! I have to think harder now! Thanks!

@christianp Same problem as "pick the higher natural number", no? Take a number n, calculate your guess: 1-(1/10^n), hope your n is larger than theirs. On repeats or in follow-up games, always increase n.

@patrick yes, I think that's right

𝓓𝓸𝓷 𝓟𝓲𝓪𝓷𝓸@mur2501@qoto.org@christianp

I would just pick 1