What are the probabilities of a die with an offset center of gravity?

[Ad VanderVen]

TL;DR

What about the odds with a die where the center of gravity is off-center?

With a pure die, all odds are equal. With a pure die, the center of gravity is exactly in the middle of the die. But what if the center of gravity is not in the center? How are the odds then. For example, how do the odds become if the center of gravity is exactly on the line that runs through the center of the 1 plane and the center of the 6 plane and then exactly in the middle between the center of the die and the center of the 6 plane. The probabilities of a 2, 3, 4 and 5 remain the same and the probability of a 6 will be greater than 1/6, but how big exactly?

 

[.Scott]

And we should also stipulate that the die is a perfect cube with no rounded corners or edges.
But still, we don't know how elastic the collision will be between the die and the surface; how much "spin" used to cast the die; and how slippery the surface is.
If we assume that the collision will be completely inelastic, the surface is perfectly slippery, and the throw never includes any angular momentum, then we might be able to estimate the result by considering all initial orientations of the die and tracing the center or gravity directly down. Our estimate would be that whichever face is directly below the center of gravity will be the one that ends up facing down.

Using that estimate (and estimating the arithmetic), a COG on the 1 dot would select 6 50% of the time, and 1, 2, 3, 4, and 5 about 10% each.

Edit:
Actually, there's an easy way to calculate the odds for the case I described. If the die lands on 2, 3, 4, or 5, it is at best astable - and unless it lands perfectly on that side, it will immediately roll to show the six. On the other hand, if it lands with 1 up, it will always stay that way. So the odds become:
1: 1/6 (16.67%)
2 to 5: 0
6: 5/6 (83.33%)

 

[Ad VanderVen]

Regardless of the physical properties of an actual die, you can still speak in probability theory of the different probabilities of the number of pips with a pure die, where the center of gravity is exactly in the middle. These probabilities are

P (X = k) = 1/6, k = 1, 2, 3, 4, 5 ,6.

Likewise, I think you should be able to calculate the probability for a pure die where the center of gravity is off-center.

 

[Stephen Tashi]

Regardless of the physical properties of an actual die, you can still speak in probability theory of the different probabilities of the number of pips with a pure die, where the center of gravity is exactly in the middle.

This is not a conclusion required by probability theory. If the conclusion can be reached, it would fall under the study of physics (a symmetry argument).

It is correct to say the terminology "fair die" is used in probability texts to mean a die where the probability of each face is 1/6. That is a convention of language.

Likewise, I think you should be able to calculate the probability for a pure die where the center of gravity is off-center.

That would also be a topic for the physics sections of the forum.

 

[hutchphd]

As described and with no bounce or spin one can calculate the solid angle subtended by each face relative to the new center of mass: this will be proportional to the probability that that face ends up "down". Clearly 2,3,4,5 will be equal and less than 1/6. "1" will be even less and "6" will be the big winner.
The exercise. is left to the OP.