Suppose I bias a cubic plastic die by incorporating a lead pellet hidden just behind the '1' face, so that the probability $P_6$ of rolling a 6 is greater than the symmetric 1/6. Its exact behaviour when I throw it will be complicated, the nondiagonal inertia tensor will affect the spin, and the kinetics of bouncing and toppling on the table (assuming it falls on a table) will also be changed. So let's not ask for a full solution and instead ask a simple question: does the outcome depend on gravity? Will $P_6$ be the same on the earth and on the moon?

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    $\begingroup$ Why would the inertia tensor be "non-diagonal"? $\endgroup$ Commented May 10, 2023 at 10:58
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    $\begingroup$ @JánLalinský: I suppose it would be non-diagonal in some set of coordinates. But "asymmetric" is probably a better word for what the question is describing. $\endgroup$ Commented May 10, 2023 at 12:59
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    $\begingroup$ @MichaelSeifert that's not a good word either, since the tensor is symmetric. Better word would be anisotropic. $\endgroup$ Commented May 10, 2023 at 13:37
  • $\begingroup$ The question formulated as "does the outcome depend on gravity?" in the text should be the headline as well, since the word outcome is contextually crucial. The headline as it reads now, asks a trivial question (without gravity you cannot define the outcome of any die - loaded or not). $\endgroup$
    – Steeven
    Commented May 10, 2023 at 20:26
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    $\begingroup$ My reason for asking was not to run a shady lunar casino, but because for an unloaded, symmetric die $P_6$ is a hard number (1/6), and classical real probability is based on this. Once the symmetry is lost then $P_6$ depends on other stuff, not just the die, and is not such a 'property',and one has to use Frequentist or Bayesian probability. More details would probabky get this question shunted to the philosophy stack... $\endgroup$ Commented May 13, 2023 at 9:12

7 Answers 7


The experimental research paper "Cuboidal Dice and Gibbs Distributions" by Wolfgang Riemer, Dietrich Stoyan, and Danail Obreschkow, (link here) indicates that when symmetry is broken (in their case, by the shape of the die being perturbed from exactly cubical), even the qualities of the surface on which the die lands affect the probability distribution of the outcomes. Therefore, it seems highly likely that in any practical situation, the force of gravity would also be a factor since it would affect the strength of interaction between the die and the surface.


Edited to acknowledge the realities that make it likely that $P_6$ depends on $g$ to a limited extent. With the new answer and discussion in the comments, this answer has morphed into more of a discussion on some reasons one might think $P_6$ is weakly invariant, and some exceptions to those arguments.

Argument 1

In certain models of the environment, the motion of the dice will be the same between the two planets upon rescaling time. There are two exceptions to this fact.

  • First, you need to drop the dice in a particular way, like from a random height and angle at rest, otherwise the initial conditions don't rescale in the same way. But the idea that $P_6$ might depend on how you throw the dice is kind of antithetical to the whole concept of dice. You obviously need to drop the dice with sufficient energy such that chaos erases the effect of the initial conditions.
  • Second, and more imporantly, when the dice bounces on the ground, there are more forces than gravity. Most notably friction. "Bounces" will remain mostly invariant if you assume a model where a fixed fraction of energy is lost in a bounce (a well known empirical model for bouncing balls). Friction is another surface interaction that will rescale properly (the forces scale $\propto g$ just like the force of gravity). But there are probably many kinds of surface interactions that do not rescale properly. Another answer points out (with a reference) that changing the bouncing surface does indeed change $P_6$.

Argument 2

Consider a 4 sided (square) dice in a 2d universe, and think of "rolling a dice" as a statistical mechanical system in the angle of the dice and the height the dice is off the ground. enter image description here

As the dice is rolled, the temperature drops over time. The potential of a biased dice as a function of angle at fixed $h$ looks like this:

enter image description here

So when the temperature drops below $T_0=v_0/k_B$ the dice chooses which potential well to permanently fall into, and the ratio of probabilities $P_4/P_1$ (four sided dice biased to land on $4$) is $\sim \exp(\delta/k_BT_0)=\exp(\delta/v_0)$. When we change the force of gravity, both $\delta$ and $v_0$ scale the same way (they are both proportional to $g$), so this probability ratio should remain invariant.

Obviously, this argument isn't rigorous - it's not clear that this system can be analyzed with statistical mechanics, nor that this way of treating "which potential well we fall into as temperature lowers" is really valid. But I think for the derivation of how $P_6/P_1-1$ scales with the potential difference between sides and $g$, it's valid. So I think this argument is a valid to argue that $P_6$ is at least weakly independent of $g$.

There's an obvious exception to this argument, which again argues that $g$ probably doesn't matter but the bouncing surface probably does. When you change the bouncing surface, it may change the rate at which the temperature drops. Consider for example a less bouncy surface that makes the temperature drop faster. This may lower the effective temperature at which the dice's fate is decided, increasing $P_6$.

Argument 3

This is a joke... or is it? $P_6$ is unitless and $g$ has units distance per time squared. The dice can be described by many different quantities: total mass, a density field $\rho(\mathbf{x})$, the moment of inertia tensor $I_{ij}$. But there is no quantity to describe the dice that includes any dimensions of time. So no unitless quantity can be written in terms of $g$ and properties of the dice, so $P_6$ cannot depend on $g$. So if $P_6$ is an intrinsic property of the dice, and it isn't affected by the bouncing surface (which it is), it will not change. Adding the bouncing surface, for example, will give you the speed of sound in the bouncing surface, which is independent of $g$ and has units of distance/time. Other small corrections to the dice's motion like air resistance also provide several quantities with units involving time. To some extent this explains why argument 1 and 2 got the answer they did - they couldn't have gotten any other answer until you start to consider things that affect the dice roll quite subtly.

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    $\begingroup$ In argument 3, I think you forgot something: Under your assumptions, the die would never stop jumping around (for the unit-analysis reason you explain, there can be no time when it stops jumping!), so for it to stop jumping, it needs something like air drag or plastic deformation - and those things have time in their units, which can explain when the die will stop jumping around, but can also theoretically make P6 depend on time. So I think argument 3 isn't useful. $\endgroup$ Commented May 10, 2023 at 16:21
  • $\begingroup$ @NadavHar'El Good point. But that implies something similar to argument 1 - that $P_6$ can only depend on $g$ if it also depends on things like the bounciness of the floor or the air density. $\endgroup$
    – AXensen
    Commented May 10, 2023 at 19:44
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    $\begingroup$ For argument 3, remember that the atmosphere viscosity is expressed in pascal-second. Changing the gravity changes the effect of viscosity. Viscosity is what slows the die down and allows it to settle on the loaded face. $\endgroup$
    – Jeffrey
    Commented May 10, 2023 at 21:22
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    $\begingroup$ The air drag, initial velocity, die's plastic/elastic parameters, and other things, as well as the strength of gravity, obviously effect the outcome of a single throw. And in theory, if you throw the die with exactly same initial conditions, it will fall the same every time. But the question posed was whether result averaged over all slightly different initial conditions (with fixed gravity) will change if we change the fixed gravity. $\endgroup$ Commented May 11, 2023 at 8:44
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    $\begingroup$ The currently most upvoted response (with a useful source!) makes it clear that the bouncing surface changes the way a dice is loaded. This argument clearly refutes the assumptions of arguments 1 and 3. I'm leaving this answer up because I think it has valid arguments for why changing gravity - and nothing else - under certain assumptions - should either change the result very little or not at all. But I think the caveats pointed out in the comments are an important part of understanding under what circumstances all three arguments may be wrong. $\endgroup$
    – AXensen
    Commented May 11, 2023 at 13:20

Let's model the throw where the initial orientation is random (other things are random too, but this one I'll need)

Let's start with the situation where the atmosphere around the die has high viscosity and we have low gravity.

In the case where viscosity tends towards infinity, due to friction the die will stop rotating, during its fall, with the '6' on top, always.

However, if we raise the gravity (and don't have infinite viscosity), from this setup, we get to a situation where the die doesn't have time to settle with the '6' on top, because the die will quickly hit the ground. (ok, it will bounce, put it will still settle faster)

So, at high viscosity, higher gravity means less bias towards '6'.

At regular viscosity, the effect will be less, but I hypothesize there will still be an effect where higher gravity weakens the bias.


Probabilities of die faces are not a property of the die and gravity strength only, but also depend on the initial height, velocity and angular velocity of the throw. Initial die angular orientation is assumed to be unknown with uniform distribution.

Let's assume only throws with zero initial angular velocity, and zero vertical velocity.

When throwing from a great height and with a velocity close to zero, the motion will be mostly vertical. When throwing near the table with great horizontal velocity, the motion will be mostly horizontal. These two classes of throw lead to different motions and thus can have different probabilities of results. Changing any single one of these parameters of throw can change probabilities of the die faces.

However, if we change the throw specification not only in gravity but also in other parameters in such a way that the die trajectory in any single throw is the same, then we can expect both kinds of throw to have the same probabilities.

If we restrict ourselves to throwing horizontally from the same height, this means that we can achieve the same trajectories and thus the same probabilities if with changing the gravity by a factor $k$, we also change the initial horizontal velocity by a factor of $\sqrt{k}$. This way, the die moves along the same trajectory, only with $k$-times greater acceleration of the center of mass and angle of orientation. After a long enough time, the angle of orientation in any single throw in the increased gravity should be the same as in the original gravity. Thus the probabilities should be the same.


I like Jeffrey's answer, but I think we can simplify it even more. Suppose we imagine a rod of uniform thickness/density. We spin it about its center and drop it, and ask: "Which side will touch the ground first?" If the rod is "fair", $P_0$ should be 0.5.

But suppose we instead attach a weight to one end (the $P_0$ end) so that it's biased. What will the rod do as it is spinning? For simplicity, let's make the weight 100x the mass of the original rod. Now it's basically spinning around the weighted end (because that's where the COM is, more or less), and the rod itself becomes a "tail".

In the limit of infinite fall time, and without friction, $P_0$ should still be 0.5. If we introduce air resistance, in the limit of infinite fall time, the rod will land on the weighted side with probability 1.0. That's because the drag on the rod will lower the angular velocity to 0. But because the air resistance is equivalent to a wind blowing up from the ground, the final orientation will be with the rod pointing up.

Some time between $t_0$ and $\infty$, the probability of $P_0$ goes from 0.5 to 1.0, as air resistance converts the uncertainty of initial angular position + momentum into a certain outcome. Let us call $t_{win}$ the time it takes for $P_0$ to be > 0.99.

Since the air resistance determines how quickly angular velocity goes to 0, we could say that the outcome mostly depends on the air density/viscosity and the $c_{drag}$ of the rod. So where does gravity come in?

Well, $t_{win}$ alone doesn't tell us everything we want to know. For instance, it doesn't tell us how high we need to drop from to get the desired outcome. To get that, we need to know local $g$.

And thus, we see that $P_0$ depends on drag, the initial angular velocity, and the height of the drop. If you increase $g$, you must increase $h$ to achieve the same $P_0$. Or, if you increase $g$ but hold $h$ fixed, then $P_0$ decreases.

However, we started talking about a rod. let's go back to the die. It's much less obvious that air resistance is so significant in this case. But the basic principle is the same: the outcome is decided only when the angular velocity goes to zero. When the die bounces on the surface, if the surface (and die) are perfectly elastic, and there's no air resistance, then the die will bounce forever, which is equivalent to the infinite free-fall case. If there is only air resistance, then $c_{drag}$ will determine the relaxation time along with the height (because drag will make each bounce smaller). A non-elastic surface should be equivalent to a much higher drag, in that it will decrease the relaxation time.

Thus, the argument for $t_{win}$ vs. $P_0$ vs. $g$ should transfer directly to the die scenario. $P_0$ depends on $g$ because $g$ helps determine the height at which $P_0$ exceeds whatever threshold you like.

In the limit where $g$ goes to $\infty$, the die will land on a side immediately, and the outcome will depend only on its orientation when it is released. If the toss is fair, then $P_0$ will be $1/6$. For non-pathological $g$, you can make $P_0$ get arbitrarily close to 1 simply by raising the height of the toss, and $g$ itself will help determine that height.


In the limit of arbitrarily strong gravity, the die drops straight to the floor in arbitrarily little time, landing on whatever face was already pointing up when it was released. There won't be any bounce to speak of, since we've defined gravity to the dominant force. In the limit of zero gravity, the die may not come to rest at all. With moderate gravity, we know that a 6 is most likely. Clearly, there must be some dependence of the outcome of the roll on the strength of gravity.


I would posit that it likely depends on gravity. The strength of gravity determines how many times it gets to spin before hitting the ground. On an extreme case like Jupiter, it may barely tilt from its initial position before landing heavy side down. Also, any bounces are much less likely to cause a change in the orientation in heavy gravity. E.g. imagine on earth dropping a gaming die vs a cube of steel or lead of equal size (you can do this experiment yourself).

My hypothesis would be that higher gravity equals more bias in favor of the heavy side. In essence changing gravity is the same as changing the severity of the bias. It might be that below a certain threshold value where the die spends a significant time spinning in the air between bounces, the probability levels off to a fairly constant value. I.e. tossing it on Mercury may be no different from tossing on the Moon or an asteroid.


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