Gambling in higher dimensions

Hi guys, long time no see. It seems I am in desperate need of a Vegas trip, since I came up with these pointless problems:

In four-dimensional craps, played with hypercubes of course, what is the equivalent of the number “seven”?

On the roulette wheel, all the areas where the ball might land are of the exact same shape and size, with no space in between. Spherical roulette must satisfy the same condition. How many numbers (max) are there in spherical roulette? (edit: all the edges between the spaces must also be identical)

If that number seems too low for your gambling pleasure, what is the maximum number in four dimensional 4-sphere roulette?

Some slot machines with a classic two-dimensional playing fields are advertised as having 243 way to win. How many ways to win do their three-dimensional counterparts have, with the playing field going three layers deep. (You still need to make a combination from left to right.)

To avoid spoilers, just post the total of all your four results added together.

There's no way to answer the roulette problems as, just as the number of spaces in "two-dimensional" roulette is arbitrary, the number in 3-dimensional roulette is also arbitrary - divide the sphere into two hemispheres, then divide each hemisphere into as many "sectors" as you want.

To make this a little easier to describe, use Earth as an example; each sector has a "length" of 90 degrees of latitude, running from the equator to either pole, and a "width" of 3 degrees of longitude, so there are 240 numbers. Change the width to 1 degree, and there are now 720 numbers.

As for the sum of the craps and slot problems, I get 59,073.

I'm picturing a casino on the Starship Enterprise.

Quote: ThatDonGuy

There's no way to answer the roulette problems as, just as the number of spaces in "two-dimensional" roulette is arbitrary, the number in 3-dimensional roulette is also arbitrary - divide the sphere into two hemispheres, then divide each hemisphere into as many "sectors" as you want.

To make this a little easier to describe, use Earth as an example; each sector has a "length" of 90 degrees of latitude, running from the equator to either pole, and a "width" of 3 degrees of longitude, so there are 240 numbers. Change the width to 1 degree, and there are now 720 numbers.

As for the sum of the craps and slot problems, I get 59,073.

Thank you for that comment, you are right. I failed to mention that that each space's edges connecting to adjacent spaces must also be identical. (to ensure randomness :) ) (will edit)

As for the other problems, you my have a different interpretation of the hypercube craps problem than I intended. We agree on the slots problem.

Edit: did you mean to post 59,074? In that case, I understand where you are coming from. In that case however, I submit to you that in regular craps, the number of possible outcomes on a single die is 6, not 12.
Interestingly though, a craps table could easily be constructed that allows for twelve outcomes per die.

"Variable gravity gaming"

you mean like in space??

https://patents.google.com/patent/US20060217169

Quote: Canyonero

Edit: did you mean to post 59,074? In that case, I understand where you are coming from. In that case however, I submit to you that in regular craps, the number of possible outcomes on a single die is 6, not 12.
Interestingly though, a craps table could easily be constructed that allows for twelve outcomes per die.

You are correct; I should have put 59,074 (9⁵ = 59,049 "lines" on a 3 x 3 x 5 slot machine; I count 24 sides on a hypercube, so the equivalent of "seven" is 25).

And even if the edges of each of the spaces on a "roulette sphere" have to be the same, I think you can still have as many spaces as you want, if each space consists of four edges - two from, say, the north pole to 30 degrees north, and two from 30 degrees north that meet at 30 degrees south. For example, you can have 20 identical spaces by using the vertices of the inscribed regular icosahedron as the vertices of the sides.

Quote: Deucekies

I'm picturing a casino on the Starship Enterprise.

I'm thinking Spock or Data needs to streamline the rules since none of these games are explainable in 30 seconds or less.

Quote: ThatDonGuy

For example, you can have 20 identical spaces by using the vertices of the inscribed regular icosahedron as the vertices of the sides.

Your example is valid. However, you will find that there is no other pattern of all identical edges filling a sphere with a higher number of spaces. Which answers this problem.

168:-)

Quote: Canyonero

Your example is valid. However, you will find that there is no other pattern of all identical edges filling a sphere with a higher number of spaces. Which answers this problem.

Identical "edges"? I'm pretty sure the "diamond" layout I mentioned earlier does this (e.g. to have 720 spaces, have 360 spaces with a vertex at the north pole, one at latitude 30 north and longitude (X + 1/2), one at latitude 30 north and longitude (X - 1/2), and the fourth at latitude 30 south and longitude X; the other 360 have vertices at the south pole, latitude 30 south and longitude X, latitude 30 south and longitude (X + 1), and the fourth at latitude 30 north and longitude (X + 1/2).

If you want identical angles as well, then you are right in saying that 20 is the limit.

Quote: ThatDonGuy

Identical "edges"? I'm pretty sure the "diamond" layout I mentioned earlier does this (e.g. to have 720 spaces, have 360 spaces with a vertex at the north pole, one at latitude 30 north and longitude (X + 1/2), one at latitude 30 north and longitude (X - 1/2), and the fourth at latitude 30 south and longitude X; the other 360 have vertices at the south pole, latitude 30 south and longitude X, latitude 30 south and longitude (X + 1), and the fourth at latitude 30 north and longitude (X + 1/2).

If you want identical angles as well, then you are right in saying that 20 is the limit.

I think I understand your diamond layout better now. It teaches me that I still haven't managed to verbalize the conditions for my roulette sphere very well. There seems to be this flaw in your design: What happens when the ball lands on top of a "pole" while the sphere stops with a pole at the very bottom. The ball would just lie there without going into any numbered space. Also the third dimension is kinda useless in this layout, since all the spaces are equally available on the equator, so we might as well stick with a disc. Unless I still don't fully understand your layout.

It seems to me we have to stick with circle based roulette for now. (and we haven't even gotten into the devastating effect of the gyroscopic effect on randomness when spinning roulette spheres)