1. ## Dice Probability

A die is rolled 1000 times. Show that the probability that the sum of the numbers shown is 1100 is the same as the probability that the sum of the numbers shown is 5900.

Not really sure how to go about doing this.
Some hints would be nice.

I would appreciate it if I wasn't given the answer all written out. I just want an idea of where to start.

2. Hello,

Hmm I'll give it some thought...

I think the main idea for this is :
$\mathbb{P}(X_n=i)=\mathbb{P}(X_n=7-i) \quad \forall i\in\{1,2,3,4,5,6\}$, and where $X_n$ is the score of the n-th dice.
But I haven't tried to use it for a formal proof yet.

3. Okay, let me rephrase that. (note : I edited my first version, where I put $6-X_n$ instead of ${\color{red}7}-X_n$) :

$X_n$ and $7-X_n$ follow the same distribution.
That is $\forall i\in\{1,2,3,4,5,6\} ~,~ \mathbb{P}(X_n=i)=\mathbb{P}(7-X_n=i)$

And here is my try for a formal proof... I'll ask my teacher for a more beautiful one, if you're interested.

Spoiler:

Let's study the n-tuple $X=(X_1,\dots,X_n)$
This follows the same distribution as $X'=(7-X_1,\dots,7-X_n)$

This means that for any measurable application h, we have $\mathbb{E}(h(X))=\mathbb{E}(h(X')) \quad {\color{red}\star}$

Let :
$h\equiv h_a~:~ \mathbb{R}^n\rightarrow \mathbb{R}$
$h_a((x_1,\dots,x_n))=\begin{cases} 1 \quad \text{if }x_1+\dots+x_n=a \\ 0 \quad \text{otherwise} \end{cases}$

It is obvious that :
$\mathbb{E}(h_a(X))=\mathbb{P}(X_1+\dots+X_n=a)$
and
$\mathbb{E}(h_a(X'))=\mathbb{P}(7-X_1+\dots+7-X_n=a)=\mathbb{P}(X_1+\dots+X_n=7n-a)$

By $\color{red} \star$, we get $\boxed{\mathbb{P}(X_1+\dots+X_n=a)=\mathbb{P}(X_1+ \dots+X_n=7n-a)}$

Which is a generalization of your situation (n=1000, a=1100)

4. That was a beautiful proof Moo! Thank you.

I like the fact that you labeled it a "spoiler".

5. It's a symmetric distribution with a mean of 3500.