My thoughts on looking for the proportions of a fair d3: Can we smoothen a d6? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Looking for a probability distributionLooking for Anthony?What should be the proportions of a three sided coin?Altitude to the Hypotenuse ProportionsConfidence Interval for ProportionsProbability Question, can you confirm my thoughtsWhat proportions make a regular right prism a fair dice?A Geometric Problem: Looking for the Correct Intervalhypothesis testing for population proportionsCan a fair coin have dependent tosses?
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My thoughts on looking for the proportions of a fair d3: Can we smoothen a d6?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Looking for a probability distributionLooking for Anthony?What should be the proportions of a three sided coin?Altitude to the Hypotenuse ProportionsConfidence Interval for ProportionsProbability Question, can you confirm my thoughtsWhat proportions make a regular right prism a fair dice?A Geometric Problem: Looking for the Correct Intervalhypothesis testing for population proportionsCan a fair coin have dependent tosses?
$begingroup$
A year or so ago, there was a video by Matt Parker talking about a fair d3, using a cylinder. Matt and his friend then set out to make cylinders with their calculations, and then did a fairly large sample of dice throws.
The stats professor in the video said that Matt and his friend were wrong, but they did find a lower and upper bound.
When I posted this question in a similar group on other social media, it is said that creating a d3 at this point is just using a d6 then mod 3 the result.
Me being a person of results, I wish for a more satisfying conclusion to this problem.
Would it be possible to smoothen every other edge of a d6 so as to have a polyhedron with 3 curved "sides"?
I suck at draeing so I can't draw what I mean, but just imagine a cube with every other edge being a smooth continuation from one face to the other, as opposed to a hard edge.
probability geometry probability-theory probability-distributions euclidean-geometry
asked Mar 31 at 13:37
juventas_polluxjuventas_pollux
83
$endgroup$
add a comment |
$begingroup$
A year or so ago, there was a video by Matt Parker talking about a fair d3, using a cylinder. Matt and his friend then set out to make cylinders with their calculations, and then did a fairly large sample of dice throws.
The stats professor in the video said that Matt and his friend were wrong, but they did find a lower and upper bound.
When I posted this question in a similar group on other social media, it is said that creating a d3 at this point is just using a d6 then mod 3 the result.
Me being a person of results, I wish for a more satisfying conclusion to this problem.
Would it be possible to smoothen every other edge of a d6 so as to have a polyhedron with 3 curved "sides"?
I suck at draeing so I can't draw what I mean, but just imagine a cube with every other edge being a smooth continuation from one face to the other, as opposed to a hard edge.
probability geometry probability-theory probability-distributions euclidean-geometry
asked Mar 31 at 13:37
juventas_polluxjuventas_pollux
83
$endgroup$
1
$begingroup$
img02.deviantart.net/175b/i/2010/223/3/7/…
$endgroup$
– Aretino
Mar 31 at 15:06
2
$begingroup$
thingiverse.com/thing:635946
$endgroup$
– Aretino
Mar 31 at 15:15
$begingroup$
Thanks to the people who sent images!
$endgroup$
– juventas_pollux
Apr 1 at 3:55
add a comment |
$begingroup$
A year or so ago, there was a video by Matt Parker talking about a fair d3, using a cylinder. Matt and his friend then set out to make cylinders with their calculations, and then did a fairly large sample of dice throws.
The stats professor in the video said that Matt and his friend were wrong, but they did find a lower and upper bound.
When I posted this question in a similar group on other social media, it is said that creating a d3 at this point is just using a d6 then mod 3 the result.
Me being a person of results, I wish for a more satisfying conclusion to this problem.
Would it be possible to smoothen every other edge of a d6 so as to have a polyhedron with 3 curved "sides"?
I suck at draeing so I can't draw what I mean, but just imagine a cube with every other edge being a smooth continuation from one face to the other, as opposed to a hard edge.
probability geometry probability-theory probability-distributions euclidean-geometry
asked Mar 31 at 13:37
juventas_polluxjuventas_pollux
83
$endgroup$
A year or so ago, there was a video by Matt Parker talking about a fair d3, using a cylinder. Matt and his friend then set out to make cylinders with their calculations, and then did a fairly large sample of dice throws.
The stats professor in the video said that Matt and his friend were wrong, but they did find a lower and upper bound.
When I posted this question in a similar group on other social media, it is said that creating a d3 at this point is just using a d6 then mod 3 the result.
Me being a person of results, I wish for a more satisfying conclusion to this problem.
Would it be possible to smoothen every other edge of a d6 so as to have a polyhedron with 3 curved "sides"?
I suck at draeing so I can't draw what I mean, but just imagine a cube with every other edge being a smooth continuation from one face to the other, as opposed to a hard edge.
probability geometry probability-theory probability-distributions euclidean-geometry
probability geometry probability-theory probability-distributions euclidean-geometry
asked Mar 31 at 13:37
juventas_polluxjuventas_pollux
83
asked Mar 31 at 13:37
juventas_polluxjuventas_pollux
83
asked Mar 31 at 13:37
juventas_polluxjuventas_pollux
83
asked Mar 31 at 13:37
juventas_polluxjuventas_pollux
83
asked Mar 31 at 13:37
juventas_polluxjuventas_pollux
83
83
1
$begingroup$
img02.deviantart.net/175b/i/2010/223/3/7/…
$endgroup$
– Aretino
Mar 31 at 15:06
2
$begingroup$
thingiverse.com/thing:635946
$endgroup$
– Aretino
Mar 31 at 15:15
$begingroup$
Thanks to the people who sent images!
$endgroup$
– juventas_pollux
Apr 1 at 3:55
add a comment |
1
$begingroup$
img02.deviantart.net/175b/i/2010/223/3/7/…
$endgroup$
– Aretino
Mar 31 at 15:06
2
$begingroup$
thingiverse.com/thing:635946
$endgroup$
– Aretino
Mar 31 at 15:15
$begingroup$
Thanks to the people who sent images!
$endgroup$
– juventas_pollux
Apr 1 at 3:55
1
1
$begingroup$
img02.deviantart.net/175b/i/2010/223/3/7/…
$endgroup$
– Aretino
Mar 31 at 15:06
$begingroup$
img02.deviantart.net/175b/i/2010/223/3/7/…
$endgroup$
– Aretino
Mar 31 at 15:06
2
2
$begingroup$
thingiverse.com/thing:635946
$endgroup$
– Aretino
Mar 31 at 15:15
$begingroup$
thingiverse.com/thing:635946
$endgroup$
– Aretino
Mar 31 at 15:15
$begingroup$
Thanks to the people who sent images!
$endgroup$
– juventas_pollux
Apr 1 at 3:55
$begingroup$
Thanks to the people who sent images!
$endgroup$
– juventas_pollux
Apr 1 at 3:55
add a comment |
1 Answer
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$begingroup$
You have the right idea, but not with "every other edge". You need to round only three edges out of the 12.
Imagine looking down one body diagonal of the cube. You see three edges aimed towards you, there are three opposing hidden edges meeting at the other end of the diagonal, and the remaining six form a zig-zag band around the "middle". Select either set of three alternating edges in this band and break out your carpenter's tools. The two possible sets of three edges correspond to mirror image configurations.
From a symmetry point of view, the cube starts out with $O_h$ symmetry, having 48 point-group symmetry elements. Your modified die ends up $D_3$, with six elements. This implies you have eight choices for your modification. We just saw two of them given one body diagonal along which to view the cube. You can choose any of the four body diagonal for your vantage point, properly giving $4×2=48/6=8$ possible orientations.
answered Mar 31 at 13:53
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
add a comment |
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1 Answer
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$begingroup$
You have the right idea, but not with "every other edge". You need to round only three edges out of the 12.
Imagine looking down one body diagonal of the cube. You see three edges aimed towards you, there are three opposing hidden edges meeting at the other end of the diagonal, and the remaining six form a zig-zag band around the "middle". Select either set of three alternating edges in this band and break out your carpenter's tools. The two possible sets of three edges correspond to mirror image configurations.
From a symmetry point of view, the cube starts out with $O_h$ symmetry, having 48 point-group symmetry elements. Your modified die ends up $D_3$, with six elements. This implies you have eight choices for your modification. We just saw two of them given one body diagonal along which to view the cube. You can choose any of the four body diagonal for your vantage point, properly giving $4×2=48/6=8$ possible orientations.
answered Mar 31 at 13:53
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
add a comment |
$begingroup$
You have the right idea, but not with "every other edge". You need to round only three edges out of the 12.
Imagine looking down one body diagonal of the cube. You see three edges aimed towards you, there are three opposing hidden edges meeting at the other end of the diagonal, and the remaining six form a zig-zag band around the "middle". Select either set of three alternating edges in this band and break out your carpenter's tools. The two possible sets of three edges correspond to mirror image configurations.
From a symmetry point of view, the cube starts out with $O_h$ symmetry, having 48 point-group symmetry elements. Your modified die ends up $D_3$, with six elements. This implies you have eight choices for your modification. We just saw two of them given one body diagonal along which to view the cube. You can choose any of the four body diagonal for your vantage point, properly giving $4×2=48/6=8$ possible orientations.
answered Mar 31 at 13:53
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
add a comment |
$begingroup$
You have the right idea, but not with "every other edge". You need to round only three edges out of the 12.
Imagine looking down one body diagonal of the cube. You see three edges aimed towards you, there are three opposing hidden edges meeting at the other end of the diagonal, and the remaining six form a zig-zag band around the "middle". Select either set of three alternating edges in this band and break out your carpenter's tools. The two possible sets of three edges correspond to mirror image configurations.
From a symmetry point of view, the cube starts out with $O_h$ symmetry, having 48 point-group symmetry elements. Your modified die ends up $D_3$, with six elements. This implies you have eight choices for your modification. We just saw two of them given one body diagonal along which to view the cube. You can choose any of the four body diagonal for your vantage point, properly giving $4×2=48/6=8$ possible orientations.
answered Mar 31 at 13:53
Oscar LanziOscar Lanzi
13.6k12136
$endgroup$
You have the right idea, but not with "every other edge". You need to round only three edges out of the 12.
Imagine looking down one body diagonal of the cube. You see three edges aimed towards you, there are three opposing hidden edges meeting at the other end of the diagonal, and the remaining six form a zig-zag band around the "middle". Select either set of three alternating edges in this band and break out your carpenter's tools. The two possible sets of three edges correspond to mirror image configurations.
From a symmetry point of view, the cube starts out with $O_h$ symmetry, having 48 point-group symmetry elements. Your modified die ends up $D_3$, with six elements. This implies you have eight choices for your modification. We just saw two of them given one body diagonal along which to view the cube. You can choose any of the four body diagonal for your vantage point, properly giving $4×2=48/6=8$ possible orientations.
answered Mar 31 at 13:53
Oscar LanziOscar Lanzi
13.6k12136
edited Mar 31 at 16:25
answered Mar 31 at 13:53
Oscar LanziOscar Lanzi
13.6k12136
answered Mar 31 at 13:53
Oscar LanziOscar Lanzi
13.6k12136
answered Mar 31 at 13:53
Oscar LanziOscar Lanzi
13.6k12136
13.6k12136
add a comment |
add a comment |
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$begingroup$
img02.deviantart.net/175b/i/2010/223/3/7/…
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– Aretino
Mar 31 at 15:06
2
$begingroup$
thingiverse.com/thing:635946
$endgroup$
– Aretino
Mar 31 at 15:15
$begingroup$
Thanks to the people who sent images!
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– juventas_pollux
Apr 1 at 3:55