
#1
Feb2411, 08:33 AM

P: 519

Originally posted on another forum with the title "are some things impossible?" The question was more of a philosophical one, but it got me thinking mathematically.
I thought I had a solution like this: Put die #1 in a base 5 numbering system, you get the following values for the 6 rolls for die 1. Roll Value (in base 5) 1 1 2 2 3 3 4 4 5 10 6 11 If you keep the second die in base 10 then, the rolls (5,3) and (6,2) give you 13. But that's like trying to add 1 (binary) to decimal 2 and getting 3. I think I'm onto something but I don't know how to express it yet. Any ideas? Dave KA 



#2
Feb2411, 08:45 AM

Math
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PF Gold
P: 38,877

No, [itex]11_5+ 3_{10}\ne 13[/itex] in any base! If you are determined to get a roll of 13 with two dice, replace 1, 2, 3, 4, 5, 6 on one die with 2, 4, 6, 8, 10, 12. then "13" would correspond to a roll of (1, 12).




#3
Feb2411, 12:39 PM

P: 70

You could label one die with 0,6,12,18,24,30.
Then with the other, normal die, each result in the range 136 will be equally likely. 



#4
Feb2411, 04:12 PM

P: 153

Roll a 13 using 2 ordinary dice
If one desired to make 13, in essence, the new 7, then you could label the dice as a straight run from 1 > 6, and then 7 > 12, in which case the least likely rolls would be 8 and 18.
6, 7 > 13 5, 8 > 13 4, 9 > 13 3, 10 > 13 2, 11 > 13 1, 12 > 13 6, 12 > 18 1, 7 > 8 



#5
Feb2411, 05:45 PM

P: 519

I think when the OP (on the other forum) said "ordinary dice" he meant those labeled 16, meaning one dot up to six dots. So I was trying to come up with a way to do it without changing the dots themselves. That's why I thought of a different base numbering system. So the symbols could stay the same but represent something else.




#6
Feb2411, 06:34 PM

P: 153

10 > b4, b5, b6, b7, b8, b9  02 > 02, 02, 02, 02, 02, 02 03 > 03, 03, 03, 03, 03, 03 04 > 10, 04, 04, 04, 04, 04 05 > 11, 10, 05, 05, 05, 05 06 > 12, 11, 10, 06, 06, 06 07 > 13, 12, 11, 10, 07, 07 08 > 20, 13, 12, 11, 10, 08 09 > 21, 14, 13, 12, 11, 10 10 > 22, 20, 14, 13, 12, 11 11 > 23, 21, 15, 14, 13, 12 12 > 30, 22, 20, 15, 14, 13 If, as per my previous post, one wanted to make 13 the new 7 (i.e. the most likely roll), base 4 would be the way to go. 



#7
Feb2411, 11:06 PM

P: 105





#8
Feb2511, 04:31 AM

P: 688

Perhaps one easy way is to interpret the dice result as in base 8; thus 6 + 5 = 13 (base 8).
It sounds like cheating to me, anyway. :) 



#9
Feb2511, 01:23 PM

P: 519

M 



#10
Mar111, 12:23 PM

P: 891




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