Timeline for Hard combinatorics and probability question.
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Feb 22, 2015 at 19:27 | history | suggested | psmears | CC BY-SA 3.0 |
Improve grammar and wording
|
| Feb 22, 2015 at 19:26 | review | Suggested edits | |||
| S Feb 22, 2015 at 19:27 | |||||
| Feb 22, 2015 at 19:02 | comment | added | Alex R. | @barakmanos: I think it might be easier to view the cubes as number and the possible cube slots numbered as well. This fixes the cube orientation and now you don't need to worry about symmetries because you don't care about which numbers each placed cube has, just its type. | |
| Feb 22, 2015 at 11:31 | comment | added | Marc van Leeuwen | @Tryss: Yes, I know, that's what I would do too. But I was trying to explain to barakmanos that even using the full set of symmetries leads to the same answer. | |
| Feb 22, 2015 at 11:29 | comment | added | Marc van Leeuwen | @barakmanos: I don't follow. I don't see what it means to have more than one identical (not just indistinguishable) configuration, nor what reason there could be to count the same configuration multiple times (and if so, how many times?), nor what kind of counting of symmetries of a cube could have $144$ as answer. Maybe you could start explaining that last point. | |
| Feb 22, 2015 at 11:22 | comment | added | Tryss | @MarcvanLeeuwen: I used a simpler way to compute the probability to get the small cube in the correct orientation : for the face cube, you can only interest you in face, hence 1/6. The rotations doesn't change the facts that only one face will be outside, and that each face is equiprobable. Same with the edge : there is 12 edges, and they are all equiprobable to be on the edge of the cube. And for the corner, there is 8 corners and all are equiprobable to be on the corner off the cube. Hence the result | |
| Feb 22, 2015 at 11:21 | comment | added | barak manos | @MarcvanLeeuwen: For counting different variations of the cube you are correct. But for calculating probability, I think that one needs to account for identical (symmetric) versions of the cube as if they were different. | |
| Feb 22, 2015 at 11:19 | comment | added | Marc van Leeuwen | @barakmanos: No, you are wrong and this answer is right. To begin with, there are only $24$ orientation-preserving symmetries of a cube, not $144$. And for a corner cube there are $3$ of them that will produce the correct colouring (rotation about the axis passing through the corner is OK). And so forth. | |
| Feb 22, 2015 at 11:15 | comment | added | barak manos | Although some of these rotations yield "so called" identical cubes, in terms of probability, the blind man can still reassemble each one of these identical versions, and so they need to be accounted for when computing the probability. | |
| Feb 22, 2015 at 11:12 | comment | added | barak manos | I think you're forgetting the fact that each sub-cube can be rotated and placed in $144$ different ways. In order to place them correctly, the ones in the corners have only $1$ legal rotation. The ones on the sides have $2$ legal rotations. The ones on the center of each face have $6$ legal rotations. The one in the center has $144$ legal rotations. | |
| Feb 22, 2015 at 10:31 | comment | added | user26486 | @BrianM.Scott Close. It's one chance in $\approx 5.465\times 10^{36}$ (Wolfram Alpha). | |
| Feb 22, 2015 at 10:21 | comment | added | Brian M. Scott | That's a little less than one chance in $2\times 10^{35}$, if my calculator was behaving. | |
| Feb 22, 2015 at 10:20 | history | edited | Tryss | CC BY-SA 3.0 |
added 24 characters in body
|
| Feb 22, 2015 at 10:19 | history | edited | Brian M. Scott | CC BY-SA 3.0 |
added 7 characters in body
|
| Feb 22, 2015 at 10:16 | vote | accept | Aditya Kumar | ||
| Feb 22, 2015 at 10:14 | history | edited | Tryss | CC BY-SA 3.0 |
added 918 characters in body
|
| Feb 22, 2015 at 9:56 | history | answered | Tryss | CC BY-SA 3.0 |