Question

what is the expected value for throwing m dice, if they have n sides?
All m dice are thrown at once, m and n are constants and positive integers

THE DICE ARE FAIR!

Answer 1

use the QH @dan815

Answer 2

@TuringTest solve his weird question thingy

Answer 3

lel

Answer 4

how many times will you be throwing it?

Answer 5

m dice are thrown at once

Answer 6

Ok is n a constant value ??

Answer 7

Why do you and ganeshie always ask pointless no answer questions? ._.

Answer 8

it is not pointless!

Answer 9

bahahaha I do have a point to this question i am working on something

Answer 10

And one more question is n give one side per throwing ??

Answer 11

n sides, so the numbers on each side is 1,2,3... to n

Answer 12

Is it unbiased?

Answer 13

@khanacademy

Answer 14

His username is actually @saifoo.khan

Answer 15

And hey lol dan expected value ? I don't get what is that , what else expect numbers ?

Answer 16

like for exampel 2 dice with 6 sides
have an expected value of 7

Answer 17

(1*2+2*3+3*4+4*5+5*6+6*7+5*8+4*9+3*10+2*11+1*12 )/36=7

Answer 18

The average

Answer 19

would it always just be the median i guess?

Answer 20

mean=median when its a normal distribution

Answer 21

if there are m dice with n sides

m*n/2 or if odd floor( m*n/2) +1

Answer 22

LEts see then
for 2 dice and 6 sides
floor(6*2/2) +1
=7
for 2 dice and 4 sides
4*2/2 +1
=5

1 2 3 4
1 5
2 5
3 5
45

Answer 23

3 dice and 4 sides
4*3/2 +1 = 7
sums
1 1 1 -- 3
1 1 2 ---4
1 2 1
2 1 1
1 1 3 -----5
1 3 1
3 1 1
1 2 2
2 1 2
2 2 1
1 1 4 ------6
1 4 1
4 1 1
.
1 2 3 = 3! ways of this
...
2 2 2
--------7 sum
1 2 4 = 3! ways of this
1 3 3 = 3!/2
2 2 1 = 3!/2
-----------8 ssum
1 3 4 ---- 3! ways
2 3 3 ----3 ways
so symmetry along 7, it is probably the expected value

Answer 24

I think the main questions to answer here is, how many distinct possible outcomes do we have in the general scenario?

By definition of expected value, we have
\[\large \mathbb{E}(x)=\sum_{x\in X}xp(x)\]It's clear that \(p(x)=\dfrac{1}{n^m}\) for any given outcome. Now it's a matter of determining the cardinality of the sample space \(X\).

Answer 25

Right someone was telling me that we would just consider the median because, we know that after an infinite number of throws we must have a binomial distribution

Answer 26

hey what does cardinality mean

Answer 27

Size of the set, or number of elements. Basically the total number of outcomes.

Answer 28

It comes down to finding a proper closed form for the upper limit of the sum.

Answer 29

okay but, how come you are saying p(x) = 1/n^m for any outcome should it be like
p(x)= a/n^m, a <n^m

Answer 30

for example the number 3, from throwing 2 dice and 6 sides that has a prob of
2/6^2

Answer 31

I'm interpreting each outcome to be distinct. Let's say \(m=3\) and \(n=2\). We have \(9\) total outcomes, including repreats:
\[\begin{matrix}111&\color{red}{112}&122&222\\
&\color{red}{121}&212\\
&\color{red}{211}&221\end{matrix}\]We have \(2^3=8\) total outcomes, each with \(\dfrac{1}{8}\) probability of occurring, but this doesn't take into account that all the red ones are the same.

Answer 32

ahh i gotcha!

Answer 33

Whoops, meant \(8\) not \(9\), but yeah

Answer 34

:)