dgfgrafgdfge111

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Benutzernamedgfgrafgdfge111
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dgfgrafgdfge111  05.08.2019
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Here's what I tried:

 

Let's call the faces of the cube as follows:

 

Front - 1

Back - 4

Right - 3

Left - 5

Up - 6

Down - 2

 

I tried to upload an image of it, but Nginx said it was too big for some reason. You can see the image here though: https://ibb.co/GtHW21P

 

The starting will be face 1. By side, I mean any of the faces 3,6,5,2.

 

So, as Melody said,

If the second last position (3rd) is front or back then it is impossible. So the 3rd position must be a side.

So, the 3rd position, as in my diagram, must be on faces 3,6,5,2.

 

We have 2 cases:

  1. front - side - other side - other side (maybe back to the first side) - front
  2. front - side - back - side - front

Melody did the same thing (so credit goes to Melody). Now we have to calculate the number of possibilities for each case:

  1. 1 x 4 x 2 x 2 x 1 = 16
  2. 1 x 4 x 1 x 4 x 1 = 16

But Wait!!! I just realized that we can go back to the front face before the 4 minutes are over!!!

 

So now we have more cases:

  1. front - side - front - side - front

And our possibilities for this case are:

  1. 1 x 4 x 1 x 4 x 1 = 16

So, our total is 48/256 (Melody found out the denominator) = 3/16.

 

Thanks Melody. It just occured to me that we could go back to the front face. I got the same answer as you previously, but something struck me when I saw the 4 x 4 x 4 x 4 = 256. I realized that then we were going back faces, so we could go back to the front face. Thanks again laugh

14.03.2019