# Thread: Conditional distribution for two dice values

1. ## Conditional distribution for two dice values

Hi!

I have two dice values X, Y and should calculate the conditional distributions
X+Y given Y = 3 or 5
Y given X>Y

I don't know how to start with that. Can anyone give me a hint?

2. Originally Posted by mork
I have two dice values X, Y and should calculate the conditional distributions
X+Y given
1) Y = 3 or 5
2) Y given X>Y
For #1, your given subspace is $\displaystyle B=\{(1,3),(2,3),(3,3),(4,3),(5,3)(6,3),(1,5),(2,5) ,(3,5),(4,5),(5,5)(6,5)\}$.
Now $\displaystyle P(X+Y=5|B)=\frac{1}{12}$, WHY?

But $\displaystyle P(X+Y=7|B)=\frac{2}{12}$, WHY?.

Now you finish.

3. Originally Posted by Plato
For #1, your given subspace is $\displaystyle B=\{(1,3),(2,3),(3,3),(4,3),(5,3)(6,3),(1,5),(2,5) ,(3,5),(4,5),(5,5)(6,5)\}$.
Now $\displaystyle P(X+Y=5|B)=\frac{1}{12}$, WHY?

But $\displaystyle P(X+Y=7|B)=\frac{2}{12}$, WHY?.

Now you finish.
OK, I have your B already and I know how to calculate $\displaystyle P(X+Y=n|B)$ for every n, but how should the conditional distribution look like? Is it formula which can be derived from this or do I have to "see" what it should look like?

The only thing I can see is that P is $\displaystyle 2\cdot\frac{1}{12}, 4\cdot\frac{2}{12},2\cdot\frac{1}{12}$ between 4 and 11. I cannot come up with a formula that returns this behaviour.

4. Originally Posted by mork
I cannot come up with a formula that returns this behaviour.
There is no formula.
It is just a matter of counting. There is only one way $\displaystyle X+Y=4$.
There are two ways $\displaystyle X+Y=7$.
Count all the other cases.

5. Originally Posted by Plato
There is no formula.
It is just a matter of counting. There is only one way $\displaystyle X+Y=4$.
There are two ways $\displaystyle X+Y=7$.
Count all the other cases.
So I was looking for something which isn't there.
How do I write the result?
$\displaystyle P(X+Y|Y \in \{3,5\}) = \{\frac{1}{12},\frac{1}{12},\frac{2}{12},\frac{2}{ 12},\frac{2}{12},\frac{2}{12},\frac{1}{12},\frac{1 }{12}$\}

By the way: thank you so far for your patience with me

6. Originally Posted by mork
So I was looking for something which isn't there.
How do I write the result?
$\displaystyle P(X+Y|Y \in \{3,5\}) = \{\frac{1}{12},\frac{1}{12},\frac{2}{12},\frac{2}{ 12},\frac{2}{12},\frac{2}{12},\frac{1}{12},\frac{1 }{12}$\}

By the way: thank you so far for your patience with me
This makes no sense because

1. probability is a number. Your right hand side is not a number.

2. What is the condition X + Y has to satisfy?

And what is $\displaystyle Y \in \{3,5\}$ meant to mean? Does it mean that Y = 3 or Y = 5?

Plato's replies say exactly what to do. I'm not sure what the trouble can stiil be ....? The result (that is, I assume, the answer) is given in your very first post!