# Independence of 2 variables derived from dice rolls

A while ago, I was told that if you roll a d%, by rolling 2 ten sided dice and treating one as the tens digit and the other as the ones, then you can gain 2 independently distributed values by flipping which is 10s digit and which is ones.

I want to prove or disprove this, It feels like it shouldn't be true.

Ie: for X, Y independent and Uniformly discretely distributed, on 0,1,2..,9 Show, that for random variables A=10X+Y B=10Y+X A and B are Independent (or show the converse)

So it is clear that:

$P_X (x) = P_Y (y)=1/10$

$P_XY (x,y) = 1/100$ (from independence)

$P_A (a) = P_B(b) = 1/100$

Now I must show that $P_AB(a,b)=1/10000=P_A(a)\times P_B(b)$ to show independence. Just not to sure where to start with this, I tried to use the convolution rule for sums of random variables, but I must have made a mistake since i ened up with $P_AB(a,b)=1$

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Two events $A,B$ are independent if and only if $P(A|B) = P(A)$.

Clearly for a given number $x$, $P(A = x | B ) = 1$ if and only if $B$ is number that has the same (but reversed) digits as $x$. Otherwise, $P(A = x | B ) = 0$.

So, $A$ and $B$ cannot be independent.

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Ah, I was over thinking it. I should have been looking for a explicit counter example. Rather than trying to calculate a distribution – oxinabox Oct 10 '12 at 6:11
@oxinabox : Happens to us all the time. Glad I could help. Cheers~ :) – Legendre Oct 10 '12 at 6:16

I don't understand your notation, so I'll use my own.

You can calculate the probability that $A=42$, right?

You can also calculate the probability that $A=42$, given that $B=17$, right?

Doesn't that tell you what you need to know about independence of $A$ and $B$?

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