If the center of gravity changes on a die, how do the odds change?

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SUMMARY

Changing the center of gravity (COG) on a die directly affects the probabilities of landing on each face. When the COG is displaced, the die exhibits different leverage during rolls, leading to a higher likelihood of settling on certain faces over others. The exact equations governing these new probabilities are complex and not universally defined, as they depend on the specific displacement of the COG. Further exploration into the dynamics of rolling unfair dice can yield insights into optimizing outcomes.

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Ad VanderVen
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TL;DR
If the center of gravity changes on a die, how do the odds on each of the eyes change?
If the center of gravity changes on a die, how do the odds on each of the eyes change?
 
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When you say "how do", are you asking why the probabilities change, or are you asking for a formula for the new probabilities?
Why the probabilities change is because there is different leverage when they roll and they are more likely to settle into some positions than into others.
The exact equations for the probabilities are far beyond my knowledge and I will leave that to others.
 
Ad VanderVen said:
Summary:: If the center of gravity changes on a die, how do the odds on each of the eyes change?

If the center of gravity changes on a die, how do the odds on each of the eyes change?
A Google search on math of unfair die returns lots of good hits. Maybe have a look through those search results to see if you can find what you are looking for. :wink:
 
Ad VanderVen said:
If the center of gravity changes on a die, how do the odds on each of the eyes change?
Uh ... you think maybe that answer to that might depend on HOW the center of gravity changes? I could posit a die that has its COG very close to the middle of the face opposite the one. What would the odds be of rolling anything but a one?
 
Ad VanderVen said:
If the center of gravity changes on a die, how do the odds on each of the eyes change?

Interesting question! I've been playing around with a variation on it for a little while. Consider an ##n##-sided polygon in the plane of side length ##s##. Label each side with a number between ##1## and ##n##. Let the centre of mass be displaced from the centre of the polygon, toward the ##i^{\mathrm{th}}## edge, by a distance ##a##.

To analyse the dynamics, consider the change in energy & angular momentum which occur on each subsequent roll (pivoting from one edge to the next). Given an initial angular velocity ##\omega##, how many rolls does it take for the polygon to come to rest? Assuming that the edge labelled ##1## is initially in contact with the table, you can work out the relationship between the initial angular velocity and the edge on which the polygon comes to rest.
 
On a related note, I did a Google search trying to see if there is a technique to rolling unfair dice, like to you want to minimize or maximize the number of bounces and disatance travelled, etc. So far no luck with my searching. It does seem like there would be a best way to "roll" the unfair dice to get the weighting to do the best job of landing a favorable face up...
 

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