Gaming philosophy conundra

[whswhs]

Originally Posted by malloyd

Surely whoever rolls second always wins. The probability space between the first roll X and infinity is always infinitely larger than the space between 0 and X, right?

I didn't think of that one, but it sounds right, at least if you're rolling a die with countably infinite sides.

I'm not sure what happens if your die has continuous sides. If you give the south pole a value of 0 and the north pole a value of 1, and every point in between has some fractional value, then for example if you land on the point whose fractional value is exactly 0.5, your opponent has a 50% chance of beating you, right? There are no more points between 0.5 and 1 than there are between 0 and 0.5.

Can you treat the interval [0, 1] as isomorphic to the interval from 0 to infinity? I'm not sure you can. "Infinity" here probably means aleph sub null, the infinity of natural numbers, which can be visualized as a line; but I don't think the infinity of the continuum is on that line. So maybe there's a conceptual difference between a sphere and an infinite-sided polyhedron, or an apeirohedron (I just looked on Wiktionary and they say that word already exists, so I'm not the first to make it up)?

It seems as if you have said that when you throw an apeirohedral die, the number of sides with larger numbers must always exceed the number of sides with smaller numbers. Is that the same as saying that it must always land on the "bottom"?

[Anthony]

Originally Posted by malloyd

Surely whoever rolls second always wins. The probability space between the first roll X and infinity is always infinitely larger than the space between 0 and X, right?

Unless your d∞ is infinite because it can produce non-integer values, in which case a range from 0 to 1 is still infinite.

[Lovewyrm]

Originally Posted by The Colonel

This assumes, does it not, that amongst an infinity of numbers, one of them must be "sword" ... or am I misreading?

No, it's not a sword.
But consider this number
0.034749592304805347502348023489574509683045803570 24803457045382304702357606035739045203485704596804 80345870345830674056

vs
0.555555554

First one's longer because it has more digits.
Second is larger, numerically, but that's boring.

In other words, I was proposing it from the perspective of "The die must be able to accomodate all the digits of ∞ on its surface so the roller can read the results."
A result with sufficient length would possibly be suitable to bonk one another with.

[whswhs]

Originally Posted by Lovewyrm

No, it's not a sword.
But consider this number
0.034749592304805347502348023489574509683045803570 24803457045382304702357606035739045203485704596804 80345870345830674056

vs
0.555555554

First one's longer because it has more digits.
Second is larger, numerically, but that's boring.

In other words, I was proposing it from the perspective of "The die must be able to accomodate all the digits of ∞ on its surface so the roller can read the results."
A result with sufficient length would possibly be suitable to bonk one another with.

The number 1/9, in decimal representation, is 0.11111 ..., with an infinite repetition of the digit one. So it seems it would be infinitely long. That might be a challenge to wield as a sword.

[zoncxs]

Originally Posted by Lovewyrm

No, it's not a sword.
But consider this number
0.034749592304805347502348023489574509683045803570 24803457045382304702357606035739045203485704596804 80345870345830674056

vs
0.555555554

First one's longer because it has more digits.
Second is larger, numerically, but that's boring.

In other words, I was proposing it from the perspective of "The die must be able to accomodate all the digits of ∞ on its surface so the roller can read the results."
A result with sufficient length would possibly be suitable to bonk one another with.

If it is only whole numbers (1,2,3,4,etc.), then it can, but if it also includes all the real numbers(0.1,0.01,0.001,0.0....001, etc), then it is impossible.

There are more real numbers between 0 and 1 than there are whole numbers from 0 to positive infinity.

[whswhs]

Originally Posted by zoncxs

There are more real numbers between 0 and 1 than there are whole numbers from 0 to positive infinity.

On one hand, there are as many rational numbers between 0 and 1 as there are whole numbers between 0 and infinity. But on the other hand, there are more real numbers between zero and one than there are rational numbers between zero and one. (See Cantor's diagonal proof.)

Long ago, when I was taking real analysis, I asked my instructor why Cantor's argument didn't apply equally to whole numbers. He answered that rational numbers, in decimal representation, could have an infinite number of digits, but whole numbers could not. So using an apeirohedral die would give you many numbers with infinite digits; in fact I think that statistically that would be the expected result, and perhaps getting a number with finite digits might be so unlikely as to have effective probability 0. A sword of infinite length would be unwieldly. But that problem doesn't need real numbers; you can get it already just with rational numbers.

[Anthony]

Originally Posted by whswhs

Long ago, when I was taking real analysis, I asked my instructor why Cantor's argument didn't apply equally to whole numbers. He answered that rational numbers, in decimal representation, could have an infinite number of digits, but whole numbers could not.

The issue isn't precisely infinite digits, the issue is infinite information content. If you limit yourself to real numbers that can be expressed by a finite equation, you wind up with a set that is no larger than the set of whole numbers.

[whswhs]

Originally Posted by Anthony

The issue isn't precisely infinite digits, the issue is infinite information content. If you limit yourself to real numbers that can be expressed by a finite equation, you wind up with a set that is no larger than the set of whole numbers.

I don't think that's relevant to the diagonal proof. The diagonal proof turns precisely on constructing a real number whose first digit differs from the first digit of the first rational (or algebraic) number in a list, whose second digit differs from the second digit of the second rational number in the list, and so on. That results in a number with infinitely many digits that is not in the (by hypothesis) complete list of rational (or algebraic) numbers, but that IS a real number. If real numbers could not have infinitely many digits, then your proof would be showing only that C was larger than the largest finite number of digits, which wouldn't be very interesting.

But whole numbers cannot possibly have infinitely many digits. The number 0.111 ... is a rational number but the number ... 111. is not a whole number.

[Anthony]

Originally Posted by whswhs

I don't think that's relevant to the diagonal proof. The diagonal proof turns precisely on constructing a real number whose first digit differs from the first digit of the first rational (or algebraic) number in a list, whose second digit differs from the second digit of the second rational number in the list, and so on. That results in a number with infinitely many digits that is not in the (by hypothesis) complete list of rational (or algebraic) numbers, but that IS a real number.

I am aware of that, but it turns out that the resulting number is not in the set of numbers that can be defined by finite algorithms (with finite inputs), because it's provably possible to map that set of numbers to the integers -- just pick an encoding and the integer map is the encoded form of the algorithm.

[whswhs]

Originally Posted by Anthony

I am aware of that, but it turns out that the resulting number is not in the set of numbers that can be defined by finite algorithms (with finite inputs), because it's provably possible to map that set of numbers to the integers -- just pick an encoding and the integer map is the encoded form of the algorithm.

It seems as if it still must be the case that if you represent every number on the apeirohedron as a decimal (which is how Lovewyrm chose to represent them), there must be many numbers whose representation has infinitely many digits and therefore counts as a "sword" of infinite length.