Dicing with Death

[Tahnan]

This may be something of a chestnut; it's probably not that hard anyway. Enjoy it if you haven't seen it before. Part II is courtesy of, I believe, Victor Serebriakoff; I'll confirm that once I get around to digging up the answer.

And this one isn't a lateral thinking puzzle, so the answer doesn't involve eating any grapes (or cubes), or twisting the language, or otherwise re-interpreting. It's just math. :-)

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You've been taken prisoner by the Evil Dictator of Puzzledonia. (I know it has a president now; this is an historical puzzle.) One afternoon you're dragged from your cell to the throne room, where the Dictator cackles manaically and produces a largish wooden cube, painted red. (You hope it's only paint. He's got a pretty evil reputation.)

"This cube," says the EDP, "is of an ordinary brownish wood; only the outside is painted red. My Royal Torturer here, who is so good at cutting things, will saw it into a three by three by three stack of cubes, in pretty much the normal way you'd expect, nothing tricky there." This doesn't sound so bad so far.

"Then we'll give you the cubes, and these two bags here." Sure enough, he has two bags. "Divide the cubes between the two bags as you like. Then I'll pick a bag at random (by flipping a one-puzmid coin), and draw a cube at random from the bag." Still doesn't sound so bad. "We'll roll that cube; if it comes up on an unpainted, wooden side, we'll let you sit in the dungeon for another few years or so, and if it comes up red, you die in the morning."

Oh. Now it does sound pretty bad.

"You can put all the cubes into one bag, if you like, in which case we'll skip the coin-flip and just use the one bag. Or you can divide them. Ready?"

No, not really, but you don't have much choice, do you?

Part I. If you put all the cubes in one bag, what's your chance of survival?

(Part I-a. If the Dictator had had the large cube cut into an n-by-n-by-n stack instead of a three-by-three-by-three stack, for some integer n greater than 0, what would be your chance of survival if you put all n-cubed cubes into one bag? Why?)

Part II. What is your best strategy for survival?

[Zealot]

Part 1: 2/3?

I might as well try grab the easy part quick before the people who are actually good at math get here.

[Icarus]

I'm thinking part one is 1/3 chance RED

And of course, someone beat me to it.



[This message has been edited by Icarus (edited 03-06-2002 07:12 PM).]

[Lucky Wizard]

Part I: 2/3
Part Ia: (n-1)/n
Part II: My guess would be that it involves having very few cubes (possibly only the one in question) in the bag that contains the middle cube, the one with no red sides. This is based on my memory of the Prisoner's Balls. If you put only that one cube in one of the bags, you get a 43/52 chance of survival.

[Icarus]

For Part 2 - my strategy would be put the 26 cubes with at least one red side in one bag, and put the one completely brown cube in the second bag

[JToomey]

I'd suggest that I'll happily go off to be executed if the cube cutter can do so with less than 6 cuts, rearranging the wood after each cut however he likes.



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The Big, Stupid Puzzle:
http://www.yark.org/puzzle

[GH]

I'm interested in Part II-a. If the large cube had been cut into n^3 smaller cubes, what's the best strategy for survival? Is it still to put 1 of the no-red-sides cubes into one bag, and put everything else into the other bag? What if n=2?

[dead mith chap]

If n=2, all the cubes are identical, so it doesnt matter what you do.

[Daft Dog Fred]

Eeh, lad. Ah’d ask ‘im ter cut ‘is cube into a 39 x 39 x 39 stack. He’d ‘ave ter ‘ave a bloomin’ good eye ter make all ‘em 114 cuts jus’ right, so’s yer got 59,319 equal cubes that roll right.

Ah’d think yer best idea is to get ‘im to agree to the most cuts that ye can. Then ye jus’ bung all yer unpainted bits in one bag and cross yer paws.

Ah’d widdle on ‘is throne an all, if I thought he weren’t lookin’.

[mathgrant]

No, just put ONE no-red in the first bag. The rest will increase the chances of you winning if he happens to get the other.

That's the answer to any problem. Put a single cube with no red in a bag, and everything else in the other. So for a 4x4x4 you have:

8 with 3 red
24 with 2 red
24 with 1 red
8 with 0 red

If you do this:

Bag A:
1 no red

Bag B:
8 3 reds
24 2 reds
24 1 reds
7 no reds

You only have an 16/63=0.2540- of dying.

[dead mith chap]

Random side puzzle:

With the 3x3x3 (this wouldn't work with a bigger cube), you have to choose one cube, then put all the rest of the cubes in one of two bags, A and B. You then roll the cube you chose, and if it comes up red, you pick a cube from bag A to roll, while if it comes up non-red, you pick a cube from bag B to roll. What's your best strategy now?

[mikegoo]

I think this is the best strategy for the random: Take one of the cubes with one side painted as the first rolling cube. Put the blank cube in bag B and the rest in bag A. I think this is the best you can do.

have a Doris Day