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Using a blank cube and a bunch of circular stickers, an average person constructs a d6. A d6 is also known as a six-sided die, or sometimes, a dice.

For the purpose of this puzzle, the average person

  • knows the shape of the correct pip pattern for each number (as in "a 3 is three pips on a diagonal line" and "a 5 is like a 4 but with a pip in the middle")
  • knows that the sum of two opposing sides is always 7
  • doesn't know any more specifics on dice manufacturers' conventions (and doesn't have a reference handy)
  • given options, chooses one randomly, as long as the things he knows will apply in the resulting die.

What is the probability for the resulting pip pattern being the exact same one that official casino dice always use? (Seen in both of the dice in the image below)

enter image description here

(There's no need to account for any minor inaccuracies in the pip positions, we are only interested in the overall pattern being the same.)

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    $\begingroup$ Should this be puzzling, or math.stackexchange.com? :-) $\endgroup$ – T.J. Crowder Feb 15 '18 at 15:58
  • $\begingroup$ @T.J.Crowder that's a good question. Since this is small, simple, and bit silly problem with no other interesting features than the fun of figuring out all the little twists you can do to the sides, I'd say puzzling. On the other hand, this is pretty close to a math textbook problem, which are officially off topic here. As always, it's up to the community, and at least so far, looks like the PSE has liked it. $\endgroup$ – Bass Feb 15 '18 at 16:23
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I make it

1 in 16.
There are two possible orientations of the Six and the One - Six horizontal and Six vertical - the One is symmetrical about all axes.
There are also two possible orientations of the Five and the Two - Two Up left to right, and Two down left to right - the Five is symmetrical about both axes.
There are also two possible orientations of the Four and the Three - Three Up left to right, and Three down left to right - the Four is symmetrical about both axes.
Finally there are two possible positions for the numbers around the middle faces of the cube - 5 4 2 3 or 5 3 2 4 (its mirror).
2 x 2 x 2 x 2 = 16.

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  • $\begingroup$ Well put, this is exactly how I thought about it myself. For extra clarification, I put a picture of the dice in the self-answer below. $\endgroup$ – Bass Feb 17 '18 at 19:10
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I decided to solve this in the second hardest possible way myself, and created a 3D model of all the possibilities. Here it is:

enter image description here How to read the image:
* The green dice are right-handed (if 6 is up, 2 is to the right of 3), the red(dish) ones are left-handed
* In the red row nearer the camera, and the green row next to it, the 3 leans away from the 2. In the rear 2 rows, the 3 leans towards the 2.
* In the transparent dice the 6 has three pips towards the 2. In the opaque ones, the 6 has three pips towards the 3.
* In the dice with white pips, the 2 leans towards the 3. In the ones with colour in their pips, the 2 leans away from the 3

The sides that are not visible are fully determined by the visible ones, and their patterns are symmetric.

The die with the pips in the "official casino dice" pattern is the one on the far left in the image.

The nice thing about 3D models is that you can actually roll them:

enter image description here

Hmph. No Yahtzee.

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No pair of 1, 2 and 3 can be on opposite sides, as they wouldn't make the required sum of 7. Hence they must neighbor to each other. Then they either make a clockwise or counter-clockwise configuration. This accounts to 2 possibilities.

In each of those cases,

the two dots of digit 2 may lay either on the diagonal that coincides with the common 1-2-3 corner, or on the other one. These are two possibilities, making 2×2=4.

Now,

similary the 3 may be in one of two configurations: 4×2=8.

Now the location for other digits is determined, and the only possible change is in orientation of the patterns. However,

4 and 5 are symmetric, so only 6 may be in one of two distinct positions.
Which makes a final number of possible configurations 8×2 = 16.

Hence a probability is

1/16

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The odds are

100%

Because

After placing 1 and 6 there are only two possibilities to place 3 and 4. However whichever one you choose, you can rotate the die to make it the same as the other, leaving one spot left for 2 and 5. There is only one orientation for the pips on a D6.

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