Canyonero
Canyonero
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January 21st, 2019 at 3:12:31 PM permalink
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.
Last edited by: Canyonero on Jan 21, 2019
ThatDonGuy
ThatDonGuy
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Thanks for this post from:
Canyonero
January 21st, 2019 at 4:11:14 PM permalink
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.
Deucekies
Deucekies
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January 21st, 2019 at 4:19:30 PM permalink
I'm picturing a casino on the Starship Enterprise.
Casinos are not your friends, they want your money. But so does Disneyland. And there is no chance in hell that you will go to Disneyland and come back with more money than you went with. - AxelWolf and Mickeycrimm
Canyonero
Canyonero
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January 21st, 2019 at 4:24:20 PM permalink
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.
heatmap
heatmap
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January 21st, 2019 at 6:11:44 PM permalink
"Variable gravity gaming"

you mean like in space??

https://patents.google.com/patent/US20060217169
ThatDonGuy
ThatDonGuy
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January 21st, 2019 at 6:17:51 PM permalink
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.
Gialmere
Gialmere
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January 21st, 2019 at 7:30:53 PM permalink
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.
Have you tried 22 tonight? I said 22.
Canyonero
Canyonero
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January 22nd, 2019 at 12:13:47 PM permalink
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.
rdw4potus
rdw4potus
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January 22nd, 2019 at 12:26:21 PM permalink
168:-)
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
ThatDonGuy
ThatDonGuy
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January 22nd, 2019 at 4:31:16 PM permalink
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.

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