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)