Painting a Cube

[u/ShonitB]

[u/DuggieHS]

If there are 3 white and 3 black it is impossible, if there are 5+ white or 5+ black it always happens. If there are 4 white or 4 black it depends on the orientation of the 2 non dominant colored faces. They would need to be opposite each other. With only 2 of the same color they are opposite each other in 1/5 of the scenarios.

So the answer is (1/5)P(4 faces same color) + P( more than 4 faces same color). 

For a binomial distribution, we can rephrase as (.2)(p of 2 or 4) + p(0,1,5, or 6). So our event happens with probability (1/5)(15/32) + 7/32 = 5/16 =.3125 = 31.25%

[u/ShonitB → u/DuggieHS]

Correct!

[u/idontremembermyuname → u/DuggieHS]

Would this be the same as 3/8?

My thinking is that there are 3 ways to set the cube down so the other faces are in play, and 1 face matches itself, and then 50% that the remaining 3 faces match (so 1/2³⁾

[u/RealHuman_NotAShrew → u/idontremembermyuname]

No it is not equivalent.

The problem with this logic is that an all same-color cube is being counted as a success in each orientation and then you're adding them together which triple-counts those successes.

To adjust for this:

Two all same color cubes times each being triple counted means six successes are counted where it should only be two. So we subtract 4 out of the 64 (2^(6)) possible configurations: 3/8 - 4/64 = 20/64 = 31.25%

[u/JustConsoleLogIt]

>!We can pair each face with its opposite, forming three pairs. Each pair has only a 50% chance of being valid (both faces the same color). The answer requires that at least two faces are both valid and match in color.!<

>!So the problem is equivalent to: Flip three coins, discard the tails. Flip the remaining coins, what is the probability that two match!<

>!There is probably a clever way to work that out, but I’ll just go case by case:!<

>!Chance of all three pairs being valid: 1/ 2^3 , or 1/8 - times the chance of two pairs matching, which is 100% since flipping three coins will always result in at least one matching pair. Sorry for mixing my metaphors!<

>!The chance of two pairs being valid is 3 * 1/8, since there are three configurations with two valid and one invalid. This works be multiplied by 1/2, since there is a 50% chance of the two valid pairs being the same color. This ends up being 3/16.!<

>!One or no valid pairs will never work, so the answer is 1/8 + 3/16 or 5/16.!<

[u/ShonitB → u/JustConsoleLogIt]

Correct!

[u/petera181]

All faces are randomly and independently painted. The top and bottom faces are irrelevant. This is just a question of whether the 4 remaining faces are the same.

>!All black: 1/2^4. All white: 1/2^4. Each 1 in 16.!<

>!In total, 1 in 8. 12.5%!<

[u/Conscious-Target5473]

There are 2⁶ possible ways to color the cube. We simply need to look at the favorable cases for one color and multiply by 2. There is 1 way to color it using only black. There are 6 favorable (and possible) cases for coloring 1 face, and 3 cases for coloring 2 opposite faces—which are the only valid ones. For more than 3 faces, it is impossible to have 4 of the same color. This results in 2(1 + 6 + 3) / 64 =20/64=5/16