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You need to design a pair of dice where each number one 1-12 has an equal chance of being rolled.  You have been given the task of designing the new dice. You are to use 2 standard cubes each with six faces, and number them in such a way that the probability of rolling the numbers (the sums of the two up faces) one through1-12 are the same.

What's the answer, and how would I find it.

Guest Feb 15, 2015

Best Answer 

 #7
avatar+81070 
+10

At last, Nauseated

The die is cast

As Sisyphus hath said

And found

Such problems do not lend themselves

To Numbers

"Round"

 

CPhill  Feb 16, 2015
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7+0 Answers

 #1
avatar+17711 
+10

Try this:  

Label one die with the numbers:          2, 4, 6, 8, 10, and 12.

Label the other die with the numbers: -1, -1, -1, 0, 0, and 0.

When the second die shows a -1, the sum of the two dice will be 1, 3, 5, 7, 9, or 11.

When the second die shows a 0, the sum of the two dice will be 2, 4, 6, 8, 10, or 12.

Since there is an equal chance for the first second die to show a -1 or a 0,

there will be equal chances for the two dice to sum to 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, or 12.

geno3141  Feb 16, 2015
 #2
avatar+81070 
0

Very nice, geno....I hadn't considered negative numbers.....!!!!

 

CPhill  Feb 16, 2015
 #3
avatar+91479 
0

That's brilliant Geno, I never would have thought of that.  

Melody  Feb 16, 2015
 #4
avatar+26406 
+5

Could also have one die with faces numbered 1 to 6 as normal, and the other die with half the faces numbered 6 and half numbered zero.

 

Or one die with faces numbered 7 to 12, and the other die with half numbered -6 and half zero.

 

Or one die with faces numbered 13 to 18 and the other with half numbered -12 and half numbered -6.

 

Or one die with ...!!!

.

Alan  Feb 16, 2015
 #5
avatar+81070 
+5

So.....by Alan's reasoning ....the general solution would be for one die to have six faces labeled  n, n+1, n+ 2, n+3, n + 4  and n+ 5  for n ≥ 1. And the other die would have three faces labeled  -n + 1  and three labeled -n + 7

 

CPhill  Feb 16, 2015
 #6
avatar+1037 
+10

A long with being true that is quite funny, CPhill. You should get a professional troll’s license. It’s to die for!

 

Now, for a very important question: How could we do it with roman numerals?

Nauseated  Feb 16, 2015
 #7
avatar+81070 
+10
Best Answer

At last, Nauseated

The die is cast

As Sisyphus hath said

And found

Such problems do not lend themselves

To Numbers

"Round"

 

CPhill  Feb 16, 2015

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