Solving for HR and FAR, given ACC and C Lets say I have an equal variance SDT framework with an equal number of target present and target absent trials:

$d'=Z^{-1}(HR)-Z^{-1}(FAR)$

$C=\frac{Z^{-1}(HR)+Z^{-1}(FAR)}{-2}$

$ACC=\frac{HR+(1-FAR)}{2}$

where $HR=$ hit rate, $FAR=$ false alarm rate, $d'=$ sensitivity, $C=$ bias, and $ACC=$ accuracy.

If I know the value of $C$ and $ACC$, how do I solve for $HR$ and $FAR$?

To put it another way. In ROC space, I know my iso-bias curve and I know my iso-accuracy curve, and I want to know the coordinates where they intersect in terms of $HR$ and $FAR$.

It seems that I should take my equations for $C$ and $ACC$ in terms of $HR$ and $FAR$, and work them to give me $HR$ and $FAR$ in terms of $C$ and $ACC$, but I haven't managed to successfully untangle the inverse CDFs in order to do that.

Edit:

I can rearrange the equation for $ACC$, to express $FAR$ in terms of $HR$ and $ACC$:

$FAR=HR+1-2ACC$

And then I can plug that into the equation for $C$:

$C=\frac{Z^{-1}(HR)+Z^{-1}(HR+1-2ACC)}{-2}$

And rearrange that a bit:

$Z^{-1}(HR)=-2C-Z^{-1}(HR+1-2ACC)$

And then I could do this:

$HR=Z(-2C-Z^{-1}(HR+1-2ACC))$

And then I get stuck, because I'm not sure how to "free" the $HR$ buried on the right-hand side.
I tried to do the math and got stuck. I asked for help at math.se and didn't get any further. My thinking is that it is not possible to write down a nice simple answer even though it is two equations and two unknowns, since system is not linear. When faced with math I cannot do, sometimes it is easier to just bring out the MATLAB hammer:

function [HR, FAR] = findHRFAR(c, acc)
    ACC = @(HR, FAR)(HR+1-FAR)./2;
    C = @(HR, FAR)(norminv(HR)+norminv(FAR))./2;

    HR = fminbnd(@(HR)temp(HR, c, acc), 0, 1);
    [SSE, FAR] = temp(HR, c, acc);

    [HR, FAR, C(HR, FAR), ACC(HR, FAR), SSE]

    function [SSE, FAR] = temp(HR, c, acc)
        [FAR, SSE] = fminbnd(@(FAR)((c-C(HR, FAR)).^2)+...
            ((acc-ACC(HR, FAR)).^2), 0, 1);
    end
end


It runs pretty fast, but ... the nested bounded optimization is ugly, and there may be a more efficient way to find the answer. I also don't see any reason this could not be done in Octave, Python, or any similar type program of your choice.