Fun Seemingly Impossible Riddle

[Mint]

One of my favorites

[JimB..]

Very cool

[stubshaft]

My brain hurts now...

[Shanghai Jack]

Well obviously - not.

[BLAHUT]

Dang YOU made me think not Kool

[bbogue1]

Very cool. I never knew it would be so simple. My brain is kind of mushy I guess. Thanks.

[Mint]

Dang YOU made me think not Kool

This made me laugh out loud

[abunaitoo]

I don't get it.

[Mint]

I don't get it.

Is there a part that doesn't make sense, I could explain it better.

Imagine it this way... you have 4 envelopes on your desk. You put them on your desk left to right. Number them 1,2,3,4.

Now, take 4 slips of paper labeled 1,2,3,4 and randomize them. Place them face down in the 4 envelopes.

Now, open the [envelopes] in order and look at the papers inside them.

[1] = paper with a 2
[2] = paper with a 3
[3] = paper with a 1
[4] = paper with a 4

Pay attention to the "loops" that are created in this example. There are 2 loops total, one loop is a chain of 3 (1->2->3->back to 1) and a chain of 1 (envelope 4 contains its own number).

To see what I mean by "loop"... pick a random envelope. Let's say envelope #1. Open it, and there is a 2 inside. Now go to envelope #2 with a 3 inside. Now go to envelope #3 and theres a 1 inside. Now you can stop, because you've already opened envelope #1 (it makes a loop). Envelope 4 is a loop of a single item because it points to itself.

There can be any number of loops, of different lengths, but it's always guaranteed they will add up to the total number. For example here examples...

2 loops of 2 items each:
[1] => 2
[2] => 1
[3] => 4
[4] => 3

1 loop of 4 items:
[1] => 2
[2] => 3
[3] => 4
[4] => 1

etc, etc. It's all random how the loops are. In the very first example of this post, its 2 loops total... 1 loop of 3 items and 1 loop of a single item.

So this is the core principle behind the riddle are these loops. There are 100 boxes and each prisoner gets 50 guesses. The 100 boxes form one or more loops just like the envelope examples.

Now the second important point of this... WHY does a prisoner start with their own number?

If a prisoner were to walk into the room and pick ANY box, the only thing that would be guaranteed is that they are in ONE OF the different loops, but its NOT guaranteed that that loop contains their number. In the envelope example, pretend you are trying to find the number 4 and you pick box 1. It goes 1->2->3->1 again, never finding your own slip of paper for #4).

So, by starting with the box with your number, you are guaranteeing that the box BEFORE the box you opened is going to point to your number. Think of the envelope example, and pretend you are #2. So, you open envelope 2 and it contains a 3. You go to 3 and you get a 1. and now box one points to your number so you win. This is based on the guarantee that the loops are CLOSED, there is no such thing as an "open" loop. If you try to imagine any random combination for the envelopes where one of the loops doesn't meet back at the start, you cant.

The last part of the riddle which is less interesting is the 33% part. Because they each get 50 guesses, this only works for loops LESS THAN or equal to 50 long. If there is a loop thats 51 long, the last guy will not ever reach his number. So the 33% is basically saying, thats the probability that 100 randomly generated boxes will contain ONLY loops with 50 or less items.

[Wag]

What's fascinating to me is that I understood every step of the explanation and Mint's explanation BUT I could never have figured it out on my own and I would never be able to explain it to anyone else. Chronic problem of my life through school was that in any math or calculus class, I could understand every explanation the teacher gave and even understand the book but come homework time or test time and I was done. Could rarely get any of it to work on paper.

But I still find it all very cool to read or hear!

--Wag--