A tough dice probability conundrum There are 2 fair die. You randomly roll them at the same time. To win is to have the sum of the die or some combination of the sum or the value of each die cover the numbers 1-9 inclusive. 
You can roll something like 2&3 and  count it as both 2&3 or count it as the sum which is  5. If you have already counted one of the outcomes once and you roll it again, you MUST count the sum. If the sum is one you have already counted for, you lose. Thus, what is the chance of winning this game (counting all numbers 1-9 without repeats in consecutive rolls)?
 A: Using the following Matlab program, I found the following probabilities: In the current 
version, double(n) must count as 2n.  A sum of 10,11 or 12 can not be counted, so 4-6 and up must be treated as separate numbers.
$859736710700/36^9 = 0.008465311909368$ if double$(n)$ can count as $n$
$14398449280/36^8 = 0.005103824293391$ if double$(n)$ must count as $2n$
Sorry, I don't know how to include a program   
[a,b,c,d,e,f,g,h,i]=ndgrid(0:1);
abc=[a(:) b(:) c(:) d(:) e(:) f(:) g(:) h(:) i(:)];
Win=zeros(1,512);Win(512)=1;
for CountLeft=1:9,
    for State=1:512,
        Vector=[abc(State,:) 1 1 1];
        if sum(Vector)==12-CountLeft,
            R=zeros(6);
            for Die1=1:6,
                for Die2=1:6,
                    BothOK=(~Vector(Die1))&(~Vector(Die2))&(Die1~=Die2);
                    SumOK=~Vector(Die1+Die2);
                    if BothOK,
                        VectorBoth=Vector;
                        VectorBoth([Die1 Die2])=1;
                        BothState=1+VectorBoth(1:9)*2.^[0:8]';
                        BothScore=Win(BothState);
                    else BothScore=0;
                    end;
                    if SumOK,
                        VectorSum=Vector;
                        VectorSum(Die1+Die2)=1;
                        SumState=1+VectorSum(1:9)*2.^[0:8]';
                        SumScore=Win(SumState);
                    else SumScore=0;
                    end;
                    R(Die1,Die2)=max(SumScore,BothScore);
                end;  end;
            Win(State)=mean(mean(R));
        end;  end;  end;
disp(Win(1));
A: * This is not a full answer. *
It considers only the case where you must win in 6 rolls. Maybe it will help in some way, or be a starting point for a more robust answer.
The game can be won in 6 rolls, at the least. For example, here is one of those ways:

So, lets first figure out how many ways we can roll 6 times. We can roll 2 dice 36 different ways, so 6 rolls we can roll 36^6 = 2,176,782,336 different ways.
Now let's figure out how many ways we can win in 6 rolls. Three of the rolls need to be added, one to obtain 7, one to obtain 8, and one to obtain 9. There are 4 ways to roll a 9 (3,6;6,3;4,5;5,4), 5 ways to roll an 8 (4,4;3,5;5,3;2,6;6,2) and 6 ways to roll a 7 (3,4;4,3;2,5;5,2;1,6;6,1). For the individual numbers (not summed), there are 6! ways to roll them.
Therefore, we have 4*5*6*6! = 86,400 possible ways to win in 6 rolls.
Chance of winning in 6 rolls:
86,400 / 2,176,782,336 = 0.00397% (rounded)
