Question

Show all of your work and explain your steps.

Suppose you can replace one of the number cubes
with a nonstandard number cube, where any of the
numbers 1 through 6 can appear on multiple faces.

2. How can you arrange the numbers on the nonstandard cube so that the mean of the rolls is the same as that of two standard number cubes, but that the standard deviation is as
large as possible? What is this largest standard deviation?
value?

Answer 1

@phi

Answer 2

The new die (what's with "number cubes"??) still has 6 faces.
they want an assignment of 6 numbers (which can be repeated multiple times)
so that:
1) the mean is the same as the original die
2) the standard deviation is as large as possible.

First step: what is the mean of a normal die?

Answer 3

a normal die or 2 normal dices

Answer 4

if its two normal ones then the mean is 3.5

Answer 5

Re-reading it:
the mean of the rolls is the same as that of two standard number cubes

so that sounds like you roll two die. (and I assume we add the value of each die to get the value of a roll)

also, it sounds like we replace one of the dice with a new version, and roll a normal die along with the new version. does this sound like what they are doing ?

Answer 6

2,2,2,4,5,6 (new die) 1,2,3,4,5,6 (standard die) = 42/12=3.5

Answer 7

for me I chose to repeat "2" 3 times in order to get the total amount of 42 which is what the 2 standard dices equal to.

Answer 8

ok, but before working the details, I need to get clear what we are doing.

the problem says
Suppose you can replace one of the number cubes

I am thinking replace one of the dice means we have 2 dice, and we replace one of them with a new version. And then we roll 2 dice: a normal one along with a new version.

I have not read the entire question, so it may mean that (or not?)

Answer 9

We have 2 dices but one of them is different from the normal die, meaning that new die may contain the same number on multiple sides, but we need to add the new die and a regular die to get 42 as the total

Answer 10

ok. for 2 dice:
we could list all 36 pairs that show up, add up the values and divide by 36
or we can say
expected value( die1 + die2) = E(die1)+E(die2)
because the individual dies are independent.

the Expected value of 1 die (i.e. the mean value after many rolls) is
1/6(1+2+3+4+5+6) = 1/6 * 21 = 3.5

and the E(die1+die2) = 3.5 + 3.5 = 7

Answer 11

for the pair with a new die and a normal die we want
E(funny) + E(normal) = 7
but E(normal)= 3.5
so E(funny) = 3.5

Answer 12

E(funny)= 1/6 (sum of faces) = 3.5
sum of faces = 6*3.5 = 21

we need an assignment that adds up to 21

Answer 13

you found one such assignment 2,2,2,4,5,6

but I think to make the standard deviation as large as possible, we want the biggest spread in the numbers:
1,1,1,6,6,6 (I think? is the biggest spread we can get)

Answer 14

we need to find the variance (and then the standard deviation)
the variance for 1 normal die is
1/6 sum (# - 3.5)^2
which is 1/6( (1-3.5)^2 + (2-3.5)^2 + (3-3.5)^2 + (4-3.5)^2 + (5-3.5)^2 + (6-3.5)^2 )= 2.9167

the variance for the 1,1,1,6,6,6 die is
1/6( 3*(1-3.5)^2 + 3*(6-3.5)^2)= 6.25

the variance for the sum of two normal dice is the sum of the variances
2.9167+2.9167= 5.8334
the standard deviation for two normal dice is sqr(5.8334)= 2.415

the variance of the sum of a funny die and a normal die is
2.9167+6.25= 9.1667
the standard deviation of the sum of a funny die and normal die = sqr(9.1667)=3.028

Answer 15

so technically the last one is my answer?

Answer 16

that is one way to do it (assuming I understood the problem)