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# Stand and topple

## Question

A fair die was rounded on its four corners
just tangent to top and bottom edges.
On a fair throw ..will you bet for "stands" or "toppled" ?

## Recommended Posts

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If the die is dropped, the probability of it landing on either side of its tipping point is 50%...

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What did you mean when you said the "die was rounded on its four corners just tangent to top and bottom edges"?

Does that mean that the base of the resulting cylinder is the same size as the largest circle that can be drawn within a face of the original die?

toppled

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It must mean that the now-circular 3 and 4 faces are the inscribed circles of the original square faces.

I also vote for topple. Once 3 and 4 become side faces, the die can roll without need to the center of mass to elevate. So it will tend to stay in that orientation. For 3 or 4 to become the top face, the center of mass will have to elevate.
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It must mean that the now-circular 3 and 4 faces are the inscribed circles of the original square faces.

I also vote for topple. Once 3 and 4 become side faces, the die can roll without need to the center of mass to elevate. So it will tend to stay in that orientation. For 3 or 4 to become the top face, the center of mass will have to elevate.

Why would you say that?

the center of mass will be equal height from the ground, whether the dice "stands" or is "toppled".

I do bet for toppled, however

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Why do I say that?

Let's say that the 1-6, 2-5 and 3-4 positions are initially equally likely.
Of course 1-6 and 2-5 are now imaginary, but you could put dots on the curved surfaces.
That transition does not raise cm.
But transitioning to or from 3-4 does raise cm.
So if that's the only influence in play, odds are 2:1 for toppled.
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Consider the weight of the white paints of dots compensates the cavities of dots so that the center of mass is 1/2 height and along the circle center.

Edited by TimeSpaceLightForce
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This explanation might not be the greatest but...

Let's assume that the die is rolled. The die has two rotational degrees of freedom, but since it is symmetrical around its center, we can just consider one of them. If the die is rolled along its curved surface, it is easy to see that it is guaranteed to stay toppled. If the die is rolled after being rotated 90 degrees through the rotational axis orthogonal to the axis along which it is being rolled, such that the die alternates between toppled and standing until it stops, it is easy to see that the probability of it being topped is 50%. Here's where I take a speculative leap... The two aforementioned orientations are clearly the two extremes. For all the intermediate orientations in the 90 degrees interval between the two extremes, the probability of a toppled outcome should take all the intermediate values between 50% and 100%. Obviously, that would mean that toppled is the more probable outcome overall.

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This explanation might not be the greatest but...

Let's assume that the die is rolled. The die has two rotational degrees of freedom, but since it is symmetrical around its center, we can just consider one of them. If the die is rolled along its curved surface, it is easy to see that it is guaranteed to stay toppled. If the die is rolled after being rotated 90 degrees through the rotational axis orthogonal to the axis along which it is being rolled, such that the die alternates between toppled and standing until it stops, it is easy to see that the probability of it being topped is 50%. Here's where I take a speculative leap... The two aforementioned orientations are clearly the two extremes. For all the intermediate orientations in the 90 degrees interval between the two extremes, the probability of a toppled outcome should take all the intermediate values between 50% and 100%. Obviously, that would mean that toppled is the more probable outcome overall.

If the die is dropped, the probability of it landing on either side of its tipping point is 50%...

..Most likely correct.Nice analysis!

I posted a similar problem on a science forum before..it is about the thick coin that has a head- tail-side probability of 1/3,i got a good reply-
Edited by TimeSpaceLightForce
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This explanation might not be the greatest but...

Let's assume that the die is rolled. The die has two rotational degrees of freedom, but since it is symmetrical around its center, we can just consider one of them. If the die is rolled along its curved surface, it is easy to see that it is guaranteed to stay toppled. If the die is rolled after being rotated 90 degrees through the rotational axis orthogonal to the axis along which it is being rolled, such that the die alternates between toppled and standing until it stops, it is easy to see that the probability of it being topped is 50%. Here's where I take a speculative leap... The two aforementioned orientations are clearly the two extremes. For all the intermediate orientations in the 90 degrees interval between the two extremes, the probability of a toppled outcome should take all the intermediate values between 50% and 100%. Obviously, that would mean that toppled is the more probable outcome overall.

If the die is dropped, the probability of it landing on either side of its tipping point is 50%...

..Most likely correct.Nice analysis!

I posted a similar problem on a science forum before..it is about the thick coin that has a head- tail-side probability of 1/3,i got a good reply-

That's a small paper!

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Oh... Haha. :S

That makes a lot more sense.

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