Your intuition is better than mine! I crunched it out (just now edited my answer to show how) but still have no intuitive sense of why it should be the golden ratio. Would love to hear some intuition for that!
But generally...
The process is something like:
If you choose 0.5, then 0.625 beats you because that is the expected average of someone redrawing below 0.5. If you choose 0.625, then 0.617[2] beats you because that is the expected average of someone redrawing below 0.625. If you choose 0.618, then 0.6182 beats you...
Lets ignore those numbers for a moment and focus on the process: "Given a number to redraw below, redraw below the expected value of redrawing below that number".
Trying to find a fixed point for this kind of process, the golden ratio has some kind of... mythological fit? Like, I've heard so many stories like this that ended up with a formula involving the golden ratio...
I wouldn't necessarily expect the answer to be phi, but I would so not be surprised to find it to be some formula involving phi, a rational factor, and maybe pi (just because I have absolutely no intuition for which series happen to sum to rational factors of pi).
[2] I'm getting these numbers by simulation, it's too late for me here in Israel to do math
Jr'er ybbxvat sbe n svkrq cbvag bs "vs gur bccbarag pubbfrf gb erqenj orybj k, jr erqenj orybj uvf rkcrpgrq inyhr".
V gevrq gb qrevir gur sbezhyn sbe uvf rkcrpgrq inyhr n srj gvzrf naq nyjnlf tbg pbashfrq ba jurer V fubhyq hfr k, jurer (1-k) naq jurer 0.5, fb riraghnyyl V gevrq qrevivat gur fcrpvny pnfr jurer k=0.625 naq gung jnf rnfvre:
(1-0.625) [bqqf bs abg qenjvat ntnva] * (1 + 0.625) / 2 [rkcrpgrq inyhr va guvf pnfr] + 0.625 [bqqf bs erqenjvat] * 1 / 2 [rkcrpgrq inyhr va guvf pnfr]
Fb va trareny vg jbhyq or:
s(k) = (1 - k)(1 + k) / 2 + k / 2
Gur svkrq cbvag vf jura s(k) = k, be
(1 - k)(1 + k) / 2 + k / 2 = k
naq...
uggc://jjj.gvtre-nytroen.pbz/qevyy/(1-k)(1_k)/2_k/2=k/
Guvf tvirf n pybfr rabhtu nafjre? Fb V cebonoyl unir fbzr fznyy zvfgnxr ohg unir gur evtug vaghvgvba?
V arrq gb tb gb fyrrc, fb V'yy yrnir vg sbe gur vagrearg gb svaq zl fznyy zvfgnxr, nffhzvat gung'f jung vg vf, naq V pbafvqre zl rkcynangvba orggre orpnhfr vg erdhverf bayl uvtufpubby-yriry zngu :-)
Now, obviously a smart opponent would not choose a number k where k != s(k), because s(k) would be a better choice. So the only numbers it makes sense for him to choose would be fixed points of s(k), points where k = s(k). And the smartest thing for me to do in that case would be to choose s(k), which just so happens to be equal to k.
[1] which may be more or less than s(k), e.g. if my opponent chooses to always redraw, i.e. k=1, it should be the same as if he chooses to never redraw, k=0, and we expect for both cases s(k)=0.5 < 1 (and it indeed is)