Vtyjkyuk

By definition,

$$ P( X = x mid maxX,Y = z ) = fracP(X = x, maxX,Y = z )P(max X,Y = z).$$ (am I right?)

I've found that

$$P(maxX,Y = z) = frac2z - 136,$$

but I can't find the other expression. I've tried to use Bayes' theorem
and it has lead me to nothing.

Assuming X and Y are independent and have a uniform distribution $mathbf X, mathbf Y sim mathcalU1,6$ (which makes sense for two six-sided dices) we should start with the most immediate conclusions we can make:

$$eqalign
P(mathbf X = x, mathbf Y = y) &= P(mathbf X = x) P(mathbf Y = y) \
&= left( 1 over 6right)^2 \
&= 1 over 36$$

for all $x, y in 1, 2, ldots, 6$.

And all possible values for $max mathbf X, mathbf Y $ given $mathbf X$ and $mathbf Y$ are:

XY || 1 2 3 4 5 6
====================================
 1 || 1 2 3 4 5 6
 2 || 2 2 3 4 5 6
 3 || 3 3 3 4 5 6
 4 || 4 4 4 4 5 6
 5 || 5 5 5 5 5 6
 6 || 6 6 6 6 6 6

And we know that by definition
$$P( mathbf X = x mid max mathbf X, mathbf Y = z) = P(mathbf X = x, max mathbf X ,mathbf Y = z) over P(max mathbf X , mathbf Y ) = z$$

I'll show a solution with intuition and a purely analytical one.

Solution with intuition

From the table we can calculate the probability of an event $A$ by counting all the possible occurences of $(mathbf X, mathbf Y)$ in which the event $A$ occurs and divide this by 36 since each $(x, y)$ has a probability of $1 over 36$ to occur.

If $max mathbf X , mathbf Y = z$ we can see that if $mathbf X = z$ then $mathbf Y$ can have any of the $z$ values in $ 1, ldots, z $ and if $mathbf X$ has any of the $z-1$ values in $1, ldots, z-1 $ then $mathbf Y = z$. So these are $z + (z - 1) = 2z - 1$ combinations of $( mathbf X , mathbf Y )$ such that $max mathbf X , mathbf Y = z$ and therefore $P(max mathbf X , mathbf Y = z) = 2z-1 over 36$.

Now lets check $P( mathbf X = x, max mathbf X , mathbf Y = z )$.

Clearly if $x geq z$ then $P( mathbf X = x, max mathbf X , mathbf Y = z) = 0$.

From the table we can see that whenever $x lt z$, then we will have just one combination where $max mathbf X , mathbf Y = z$ because $mathbf Y = z$ is the only possible value of $mathbf Y$. So $P( mathbf X = x, max mathbf X , mathbf Y = z) = 1 over 36$ for $x lt z$.

And when $x = z$ then $mathbf Y$ can take any of the $z$ values where $y leq z$. So $P(mathbf X = x, max mathbf X , mathbf Y = z) = z over 36$ for $x = z$.

In conclusion:
$$P(max mathbf X , mathbf Y = z) = 2z - 1 over 36$$
and
$$P( mathbf X = x, max mathbf X , mathbf Y = z) =
begincases
1 over 36 & x lt z \
z over 36 & x = z
endcases
$$

therefore
$$P( mathbf X = x mid max mathbf X , mathbf Y = z) =
begincases
1 over 2z - 1 & x lt z \
z over 2z - 1 & x = z
endcases
$$

Purely analytical solution

Since $mathbf X$ and $mathbf Y$ are independent
$$P(max mathbf X , mathbf Y = z) = P( mathbf X = z, mathbf Y leq z) + P( mathbf X lt z, mathbf Y = z)$$
and calculating each summand separately we can see that
$$eqalign
P( mathbf X = z, mathbf Y leq z) &= P(mathbf X = z, mathbf Y = 1) + cdots + P( mathbf X = z, mathbf Y = z) \
&= z over 36
$$
and
$$eqalign
P( mathbf X lt z, mathbf Y = z) &= P( mathbf X = 1, mathbf Y = z) + cdots + P( mathbf X = z-1, mathbf Y = z) \
&= z-1 over 36
$$
So
$$eqalign
P(max mathbf X , mathbf Y = z) &= P( mathbf X = z, mathbf Y leq z) + P( mathbf lt z, mathbf Y = z) \
&= z over 36 + z-1 over 36 \
&= 2z - 1 over 36
$$
For $P( mathbf X = x, max mathbf X, mathbf Y = z)$ we can see different cases.

If $x lt z$ then
$$eqalign
P( mathbf X = x, max mathbf X , mathbf Y = z) &= P( mathbf X = x, mathbf Y = z) \
&= 1 over 36
$$
If $x = z$ then
$$eqalign
P( mathbf X = x, max mathbf X , mathbf Y = z) &= P( mathbf X = z, mathbf Y = 1) + cdots + P(mathbf X = z, mathbf Y = z) \
&= z over 36
$$
So
$$ P( mathbf X = x, max mathbf X , mathbf Y = z) =
begincases
1 over 36 & x lt z \
z over 36 & x = z
endcases
$$
therefore
$$P( mathbf X = x mid max mathbf X , mathbf Y = z) =
begincases
1 over 2z - 1 & x lt z \
z over 2z - 1 & x = z
endcases
$$

For $xgt z, P=0$.

For $xlt z, P=frac12z-1$.

For $x=z, P=fracz2z-1$.

Here $P=P(X=z,max(X,Y)=z)$.

This is a result of noting that for each $xlt z$, there is exactly one possibility, while for $x=z$, there are exactly $z$ possibilities leading to $2z-1$ possibilities for $max(X,Y)=z$.

You need to condition the numerator over $Y$, then enumerate. For example if $y=6$, then $z=6$, and $Xsim U(0,6)$. If $y=1$ and $z=1$, then $P(X=1)=1$. The basic formula you want is

$$
P(X=xmidmax(X,Y)=z) = sum_y P(X=xmidmax(X,Y)=z,Y=y)P(Y=y).
$$

Conceptually, if $y<z$ you already know exactly what $x$ is.