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Little math problems October 2019
10-06-2019, 10:41 PM (This post was last modified: 10-06-2019 11:09 PM by jpcuzzourt.)
Post: #6
RE: Little math problems October 2019
(10-06-2019 09:40 PM)Thomas Okken Wrote:  I don't know how to prove it rigorously, but it seems to me like you can't do better than four dice rolls. Because of the way the probabilities are distributed, whatever scheme you come up with must have a multiple of 16 possible outcomes, and since a die roll has 6 = 2 * 3 possible outcomes, getting the required factor of 2^4 requires 4 dice rolls.

That is, unless you allow re-rolls; then you could do something like this: roll two dice and add the numbers, and then say 3 -> 0; 2, 4, 5 -> 1; 6, 7, 12 -> 2; 8, 10 -> 3; 11 -> 4; 9 -> try again. That means you end up needing, on average, 2 1/4 dice rolls, but of course there is no upper limit to the number of rolls you might end up needing on any given turn.

Well, since we require a space of at least 16 possibilities, 2 6-sided dice yields 36 possible states, if taken in order. I've worked out a mapping that allows us to discard 4 of those as invalid, requiring a re-roll if they happen (11.11% probability), and all the others are mapped such that the order doesn't matter, condensing the remaining 32 states to 16. Despite my best efforts at making it logical, it would not be easily explained or learned, I think. Here is one possible mapping. BTW, lets redesignate the '6' as a '0' so that each die displays 0-5.

03,30,24,25 are disallowed = rethrow

01 or 10 = '0'
45 or 54 = '4'
12,21,14,41,15,51,05,50 = '1'
13,31,23,32,34,43,35,53 = '3'
00,11,22,33,44,55,02,20,04,40,05,50 = '2'

You can see that you get a '2' if you roll doubles or a 0 with anything but a 1 (or 3, since 03, 30 are invalid.)
stating the other rules is a bit trickier. Let me simplify the table:

01 = '0'
05,12,14,15 = '1'
13,23,34,35 = '3'
45 = '4'
doubles and 02,04,05 = '2'

That's the best I can do for 'minimal', but the rules are pretty ugly. Someone might devise prettier rules, but as long as I could scrounge 4 dice, I think I'd do it that way. Or throw 1 die 4 times, or 2 dice, twice!
I'll keep my eye on this space for better solutions though.

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Little math problems October 2019 - pier4r - 10-06-2019, 07:07 PM
RE: Little math problems October 2019 - jpcuzzourt - 10-06-2019 10:41 PM



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