[Guest]

Here is yet another matryoshka probability problem similar to the one I posted last time:

However, there are a few differences that make this one slightly more difficult (or so one hopes).

1 box contains 1 apple. 2 boxes contain 1 orange each. 3 boxes contain one banana each. Before each box has been opened the order has been randomized: A goes to B, B to C, C to D, E to F, F to A, or backwards. (In the last one, there was only one randomization before opening the boxes.) After the first three boxes have been opened, determine the probability of there being exactly 1 apple, 1 orange, and 1 banana remaining in the last three boxes (highlighted in red). Be sure to give the answer in decimal to the nearest millionth.

Diagram:

Good luck!

[plainglazed]

am not quite sure I understand how the boxes are randomized. would you please elaborate on how?

[Guest]

am not quite sure I understand how the boxes are randomized. would you please elaborate on how?

Certainly. Imagine that the boxes don't contain each other but that they must be opened in a specific order (the outside box being the "first" box). The randomization basically rotates the order of the boxes, first to second, second to third, fourth to fifth, sixth to first, or backwards. You will not know where the randomization will stop, but the basic operation of the randomization should be straightforward enough; just like tossing a die is a randomization procedure reliant on gravity and other physical processes to yield a seemingly random result for each throw.

[Guest]

Certainly. Imagine that the boxes don't contain each other but that they must be opened in a specific order (the outside box being the "first" box). The randomization basically rotates the order of the boxes, first to second, second to third, fourth to fifth, sixth to first, or backwards. You will not know where the randomization will stop, but the basic operation of the randomization should be straightforward enough; just like tossing a die is a randomization procedure reliant on gravity and other physical processes to yield a seemingly random result for each throw.

Okay, let's assume even that wasn't clear enough. In that case, imagine that the boxes are set in a row and are labeled (each) A, B, C, D, E, and F. You have a six-sided standard die that has its faces labeled (each) A, B, C, D, E, and F. In order to determine which box to open, you throw the die, and whichever face shows upward determines which box you will open. For each box, you throw the die and do this for the first three boxes. Hence, what are the chances that the last three boxes will have 1 apple, 1 orange, and 1 banana?

That scenario is probabilistically equivalent to this problem. Perhaps that helps understand it a bit.

Edited April 1, 2011 by omnihistor

[superprismatic]

.300000

or 3/10.

[Guest]

.300000

or 3/10.

...you do realize that that is the same answer you would get if the boxes had been randomized only once? What about two more randomizations for the next two boxes? Are you sure that's your final answer?

[superprismatic]

...you do realize that that is the same answer you would get if the boxes had been randomized only once? What about two more randomizations for the next two boxes? Are you sure that's your final answer?

that's my final answer.

[Guest]

that's my final answer.

Good show! I guess I'll have to come back some other time with more difficult puzzles.