Roll a 13 using 2 ordinary dice

[dkotschessaa]

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

 

[HallsofIvy]

No, $11_5+ 3_{10}\ne 13$ 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).

 

[Goongyae]

You could label one die with 0,6,12,18,24,30.
Then with the other, normal die, each result in the range 1-36 will be equally likely.

 

[Raphie]

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

 

[dkotschessaa]

I think when the OP (on the other forum) said "ordinary dice" he meant those labeled 1-6, 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.

 

[Raphie]

I was trying to come up with a way to do it without changing the dots themselves.

Not sure if this is what you were getting at, but here are some equivalencies to base 10 dice rolls in other bases...

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.

 

[Mensanator]

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

Use dice of different colors. Choose one color and add 6 to each face.

 

[dodo]

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. :)

 

[dkotschessaa]

Perhaps one easy way is to interpret the dice result as in base 8; thus 6 + 5 = 13 (base 8).

I think that's what I'm after. Simple! I'm not good with different bases.

It sounds like cheating to me, anyway. :)

Totally. It was just a mathematical answer to a non-mathematical question on a board of sci-fi geeks.

-M

 

[ramsey2879]

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. :)

Consider the base as 12 and roll a 1 and 1. If one objects that you can't tell which die is the 12's digit one can roll one die at a time so you know which is the first of the ordered pair. No one said that there can't be gaps in the result rolled. But what I think is the best answer (in line with the op's original post ) is to consider the result in base 5; then, if you roll a 6+2, it would be 13 in base 5. Dies don't have bases, you just count the number of dots, 8 is 13 in base 5. You could also roll a 6+4 to get a 13 in base 7