[Guest]

You have a black box. In this black box is 1 penny inside a box, 1 shoe in another box, 1 shoe in another box (the shoes are otherwise identical), 1 safety pin in one box, 1 safety pin in another box, and 1 safety pin in another box (the safety pins are otherwise identical), and 1 button in another. Peculiarly, the boxed items have been boxed in such a way that 6 boxes are contained by the seventh box. Think of Russian Dolls as an example. Unfortunately, the order of the boxes (inmost to outmost) has been "randomized" by a malicious 4-dimensional creature. After opening the first four boxes, determine what the probability is to the nearest millionth that the penny, one shoe, and one safety pin remain.

If my description isn't clear, I can supply a diagram illustrating the idea.

Good luck!

Edited March 31, 2011 by omnihistor

[Guest]

In anticipation, I've decided to put up a diagram anyway.

That is the initial stage of the problem, where each letter represents one of the items described. The "randomization" will rotate the order of the boxes, e.g., A goes to G, G to F, F to E, E to D, D to C, C to B, B to A (or backwards). The task is to determine the probability of the three remaining boxes (highlighted in red) containing the penny, a shoe, and a safety pin (their order being irrelevant).

Hope this prevents any confusion and increases enthusiasm for the problem.

[curr3nt]

Either 100% or 0% chance. After opening 4 boxes you would know what the other three have to be.

[Guest]

Either 100% or 0% chance. After opening 4 boxes you would know what the other three have to be.

Apologies, but I should have mentioned that the first four boxes have already been opened and their contents weren't given to you. The brunt of the problem lays in determining the probability of those three items together, given the stated conditions in addition to not knowing what the contents of the previous items were.

In other words, what is are the chances that the last three boxes will have the penny, a shoe, and a safety pin?

Edited March 31, 2011 by omnihistor

[plainglazed]

.171429 (or a pretty simple fraction). gonna give this one some more thought and most likely take another stab later.

fun puzzles, omnihistor. thanks

[Guest]

.171429 (or a pretty simple fraction). gonna give this one some more thought and most likely take another stab later.

fun puzzles, omnihistor. thanks

You're right. Care to show your method?