The Diciest Challenge
We know that when rolling two standard 6-sided dice, each numbered 1 to 6, we can roll a sum of 2 with a chance of 3 6 1 , roll a sum of 3 with a chance of 3 6 2 , ... , and roll a sum of 12 with a chance of 3 6 1 .
Now you are given two blank 6-sided fair dices; prove that there is one other way to label the two dice with not necessarily distinct positive integers of your choice in such a way that we get the exact same probability for each sum as when rolling the two standard 6-sided dice shown above.
Let the numbers on the faces of the first dice be denoted as a 1 , a 2 , a 3 , a 4 , a 5 , a 6 , where a 1 ≤ a 2 ≤ a 3 ≤ a 4 ≤ a 5 ≤ a 6 .
Similarly, let the numbers on the faces of the second dice be denoted as b 1 , b 2 , b 3 , b 4 , b 5 , b 6 , where b 1 ≤ b 2 ≤ b 3 ≤ b 4 ≤ b 5 ≤ b 6 and a 6 < b 6 .
Submit your answer the concatenation of these 12 digits, a 1 a 2 a 3 a 4 a 5 a 6 b 1 b 2 b 3 b 4 b 5 b 6 .
For example, if the respective faces of the two dices that you've found are ( 2 , 3 , 3 , 6 , 6 , 7 ) and ( 1 , 4 , 5 , 6 , 6 , 9 ) , submit your answer as 233667145669.
The answer is 122334134568.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
5 solutions
+1 Your solution is a great read:
Using the generating function approach shows us what the various terms we should be looking at are. In particular, we want
f
(
x
)
g
(
x
)
=
(
x
+
x
2
+
x
3
+
x
4
+
x
5
+
x
6
)
2
such that
1.
f
(
x
,
)
,
g
(
x
)
have non-negative coefficients, since each number appears non-negative times
2.
f
(
1
)
=
g
(
1
)
=
6
, since the dice has 6 faces. (Technically, because we could have a face of 0, we want
f
(
1
)
,
g
(
1
)
≤
6
. But since we also need
f
(
1
)
g
(
1
)
=
6
2
, hence we must have
f
(
1
)
=
g
(
1
)
=
6
.
In this way, we can eliminate various partitions of the individual factors and conclude that there is only 1 other way to obtain { f ( x ) , g ( x ) } .
Thanks for contributing and helping other members aspire to be like you!
Why not labelling the dice A with all sides as ones and dice B with twos?
- To get Sum of 2 in one way, its obvious both the dice have label 1
Dice A 1
Dice B 1
- To get sum of 3 in 2 ways... the only different way to achieve that is to have 2 twice on Dice A
Dice A 1, 2, 2
Dice B 1
(2, 1)(2, 1) being the 2 ways
- We need to get sum of 4 in 3 ways... now we have options
We cannot have 2 on Dice B... so if we put 3 on Dice B and 3 twice on dice A, we get our 3 ways to sum up to 4
Dice A 1, 2, 2, 3, 3
Dice B 1, 3
(1, 3) (3, 1) (3, 1) being the 3 ways
- Sum of 5 in 4 ways... till now we have (2, 3), (2, 3) so if we put a 4 on both the Dice, we get (1, 4), (4, 1) so
Dice A 1, 2, 2, 3, 3, 4
Dice B 1, 3, 4
- Sum of 6 in 5 ways... we have 4 ways already (2, 4), (2, 4), (3, 3), (3, 3) so if we put a 5 on Dice B, we get our 5th way (1, 5)...
Dice A 1, 2, 2, 3, 3, 4
Dice B 1, 3, 4, 5
- Sum of 7 in 6 ways ... we have 5 ways (2, 5), (2, 5), (3, 4), (3, 4), (4, 3) so if we put a 6 on Dice B we get (6, 1) as our 6th way
Dice A 1, 2, 2, 3, 3, 4
Dice B 1, 3, 4, 5, 6
- Of necessity to sum up to 12, we must have 8 as the last number on Dice B
Dice A 1, 2, 2, 3, 3, 4
Dice B 1, 3, 4, 5, 6, 8
-
Sum of 8 we have our 5 ways (2, 6), (2, 6), (3, 5), (3, 5), (4, 4)
-
Sum of 9 we have our 4 ways ... (1, 8), (3, 6), (3, 6), (4, 5)
-
Sum of 10 we have our 3 ways... (2, 8), (2, 8), (4, 6)
-
Sum of 11, we have our 2 ways (3, 8), (3, 8)
-
Sum of 12, we have our 1 way... (4, 8)
Let D n stand for "a uniformly distributed number in { 0 , 1 , … , n − 1 } ". Then the situation of two dice can be written as 2 + D 6 + D 6 .
Because six is a composite number, we can decompose its outcomes as D 6 = D 2 + 2 D 3 or D 6 = D 3 + 3 D 2 In other words, random choice of { 0 , 1 , 2 , 3 , 4 , 5 } = random choice of { 0 , 1 } + random choice of { 0 , 2 , 4 } ; random choice of { 0 , 1 , 2 , 3 , 4 , 5 } = random choice of { 0 , 1 , 2 } + random choice of { 0 , 3 } . In particular, 2 + D 6 + D 6 = 2 + ( D 2 + 2 D 3 ) + ( D 3 + 3 D 2 ) . We are free to rearrange these terms. However, if we are to end up with six-sided dice, we need to combine each D 2 with a D 3 term. The only other way in which this can be done is ⋯ = 2 + ( 2 D 3 + 3 D 2 ) + ( D 2 + D 3 ) , and because we are not allowed to put zeroes on the dice, ⋯ = ( 1 + 2 D 3 + 3 D 2 ) + ( 1 + D 2 + D 3 ) . Working out the details: 1 + { 0 , 2 , 4 } + { 0 , 3 } = { 1 , 3 , 5 , 4 , 6 , 8 } ; 1 + { 0 , 1 } + { 0 , 1 , 2 } = { 1 , 2 , 2 , 3 , 3 , 4 } ; in the format required the answer becomes 1 2 2 3 3 4 1 3 4 5 6 8 .
The solution above says "if we are to end up with six-sided dice"... But what if we aren't? See this problem I wrote to explore the possibilities for a four-sided die and a nine-sided die.
There are also solutions for a three-sided die plus a twelve-sided die: { 1 , 2 , 3 } + { 1 , 2 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 , 8 , 9 } ; or { 1 , 3 , 5 } + { 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , 6 , 7 } .
These are also known as the "Sicherman dice"! :0)
Why you don't accept the simplest solution which is (0,1,2,3,4,5) and (2,3,4,5,6,7) 012345234567
while it may generate the correct probabilities, the problem specifies "not necessarily distinct positive integers".
So a generating function for the dice is just ( x + x 2 + x 3 + x 4 + x 5 + x 6 ) 2 . Factorising gives ( x − 1 ) x 7 − x = x ( x + 1 ) ( x − 1 ) ( x 2 + x + 1 ) ( x 2 − x + 1 )
With two copies of each of these factors, we need to partition them. After a quick search, this should yield x ( x + 1 ) ( x 2 + x + 1 ) = x + 2 x 2 + 2 x 3 + x 4 and x ( x + 1 ) ( x 2 + x + 1 ) ( x 2 − x + 1 ) 2 = x + x 3 + x 4 + x 5 + x 6 + x 8
From which the solution follows.