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Conditional Cubelet Conundrum

July 17, 2008 pm31 7:07 pm

Over at Math Notations, Dave recently posed a variation on a cube dissection problem. Here’s another:

A cube is painted on the outside and then cut into 27 equal cubes by dividing each edge in three with six planes which are parallel to the faces of the original cube.

The cubelets are placed in a bag. One is removed at random and tossed. The face on top is unpainted.

If that same cubelet is tossed again, what is the probability that an unpainted face will appear on top again?

12 Comments leave one →
  1. July 17, 2008 pm31 8:07 pm 8:07 pm

    spoiler alert. i’m gonna try to work this out
    right here in the comment editor.

    eight “corners” with three faces painted;
    twelve “edges” with two;
    six “faces” with one;
    one “center” with zero
    (total of 27; it checks).
    all equally likely to be the chosen cubie.

    imagine the appropriate “tree” as needs be
    to see that one should multiply each number pair,
    sum the results, and divide by 162 (=27*6):
    (8*3 + 12*2 + 6*1+ 1*0)/162
    (this is the hard part)

    hmm. i’ll bet there’s some “conceptual” way to see this …

  2. July 17, 2008 pm31 8:22 pm 8:22 pm

    Since the most-painted cubelet has 3/6 faces unpainted, the probability of getting an unpainted face cannot be less than 3/6.

    The 8, 12, 6, 1 part is correct.

  3. July 17, 2008 pm31 9:09 pm 9:09 pm

    Ooh, good one. I’m going to have to work on this.

  4. July 17, 2008 pm31 9:42 pm 9:42 pm

    oh. sorry. i did *painted* faces.
    mumble, mumble … principle of complementation …
    you know the rest …

    you corrected me in fine math-teacher style,
    btw (don’t just flat-out give the right answer;
    point out the mistake and invite the solver
    to work it out from there) … you seem to’ve
    played this game before.

  5. July 17, 2008 pm31 10:02 pm 10:02 pm

    oh, no, wait. dagnab. there’s a lot more to it than this.
    this (or rather, 2/3) is just the probability of the condition
    that the first face is unpainted. gotta now do an entire
    “conditional probability” thing. P(A|B) = P(A&B)/P(B).
    i’ve got P(B) = 2/3.
    so what the devil is P(A&B)?
    (where A is the event “the second throw is unpainted”?)
    hmm. better go back to the 8, 12, 6, 1 tree.
    i might need a piece of paper here. rats.
    here it is: muliply the prob for each initial “branch”
    (8/27, 12/27, 6/27, 1/27) by the chance
    of throwing *two* unpainted faces on that cubie
    (i.e. (3/6)^2 for the “corner”, … (6/6)^2 for the “center”):
    (8*3^2 + 12*4^2 + 6*5^2 + 1*6^2)/(27*6^2) = 25/54.
    dividing this by 2/3 (per the “conditional prob” discussion above)
    gives, okay, i’m touching the calculator again, 25/36.

    how’s that?

  6. July 17, 2008 pm31 11:17 pm 11:17 pm

    “This game” is one of the few skills I possess.

    Not an easy little puzzle, in concept, or in execution.

    I’m thinking of an extension (generalize for any number of cuts) or working in another directions. Cubes are too much fun just to drop.

  7. July 18, 2008 am31 2:02 am 2:02 am

    My preferred approach: the unpainted face is equally likely to be any unpainted face. There are 24 on the corners, 48 on the edges, 30 on the faces, and 6 on the centre (total 108). So it’s (24×1/2 + 48×2/3 + 30×5/5 + 6×1)/108, though of course you can cancel through a 6 and improve life.

    It’s the same calculation as Vlorbik’s, but I prefer the thought process :).

  8. July 18, 2008 am31 2:04 am 2:04 am

    “A cube is painted on the outside and then cut into 27 equal cubes by dividing each edge in three with four planes which are parallel to the faces of the original cube.”

    It seems to me that you need six planes (even if you stack pieces after each slice, you must make six cuts to expose the six unpainted faces of the central piece).

  9. July 18, 2008 am31 4:10 am 4:10 am

    Fixed, thanks Rick.

    Also, the arithmetic all three of us match on, but I come closer to your thought process (calculate probability that the cube is an edge, mult by 4/6 to get that piece of the expected value, etc.)

  10. July 18, 2008 pm31 5:02 pm 5:02 pm

    This is a nice one that I heard recently, btw – the first part is old, the second part was new-to-me.

    A friend mentions that you have new neighbours who have two kids, and he knows one’s a girl but doesn’t know about the others. What’s the probability both are girls?

    You go to introduce yourself and a girl answers the door. How does this change the situation, if at all?

  11. July 18, 2008 pm31 5:08 pm 5:08 pm

    A different version was featured in last week’s Carnival of Mathematics: Chance of having two boys.

    It was Math-Play’s “2 boys” and Dave’s “cube dissection” that led me to dig out this old puzzle to post. I wrote it, and I think from scratch, so it is likely to appear as vaguely novel, which can be good.


  1. Little cube question « JD2718

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