# Talk:Eight options with six sides

## This seems too easy

Determining the least number of rolls he will have to make is so trivial I wouldn't even call it a puzzle. Determining the greatest number of rolls he may have to make is a slightly less trivial, but still very simple. One can optimise either of these values. Perhaps one for Category:Easy puzzles?

As far as I can tell, I'm not missing anything. Zerrakhi 03:43, 22 October 2010 (EDT)

Hmmm. I'm not seeing this as easy. Perhaps I am missing something. That or I worded the question in an unclear way. Could you tell me what you think the answer is? That would help me understand what is going on here. Oscarlevin 15:03, 18 November 2010 (EST)
To optimise the LEAST number of rolls he WILL have to make:
Let N be the least integer such that 6^N >= D where D is the number of dishes.
Of the 6^N possible permutations of rolled numbers, allocate a certain number to each dish and leave the rest blank.
The number of permutations allocated to each dish is the greatest integer A such that A <= (6^N)/D.
Roll N dice. If the permutation that's rolled is allocated a dish, select that dish. Otherwise roll again.
With this method, an infinite number of rolls MAY be needed, but the least number of rolls that WILL be needed is simply N.
If D = 8 then N = 2 and A = 4.
To optimise the MOST number of rolls he MAY have to make:
Let N be the least integer such that 6^N is a multiple of D.
Allocate permutations arbitrarily to dishes, as above except that now every permutation is allocated a dish (there are no blank spaces).
With this method, exactly N rolls are needed, no more no less.
If D = 8 then N = 3 and A = 27 (because 6 ^ 3 = 8 * 27).
The least easy bit is characterising the values of D such that N exists, but that's not part of the main puzzle.
Zerrakhi 21:40, 23 December 2010 (EST)
Very good. I was definitely hoping for the second sort of answer - I've updated the puzzle to hopefully make that clearer. I always have trouble phrasing puzzles like this: the goal is to find methods using N rolls such that at the completion of the algorithm the task will be complete (here, a fair choice is made) and then minimize N over all such methods. That's just not all that poetic however.
As for your solutions, your method (in the second case) is sound, however there are ways to get the number of rolls down lower than 3. It is possible to do with only one roll, although I admit the puzzle becomes more of a lateral thinking puzzle then. Think about the other sides of the die when you roll.
Oscar Levin 08:56, 29 December 2010 (EST)