3 ms·

OK I almost got it (edit: rot13 isn't very good at scrambling one-variable equations, so read at your peril, spoilers ahead): Jr'er ybbxvat sbe n svkrq cbvag b

OK I almost got it (edit: rot13 isn't very good at scrambling one-variable equations, so read at your peril, spoilers ahead): 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 :-)

boromi 10y ago
why do you seek the fixed point and not the k value that maximizes s(k) = (1 - k)(1 + k) / 2 + k / 2 ?
- SonOfLilit 10y ago
  s(k) is the function for "The number I should choose if my opponent chooses k" [1]. We arrived at the formula by asking "what is the best number for us to choose assuming our opponent chooses k?". 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)
  - murbard2 10y ago
    Why do you bring up expected value in the calculation. You may end up with the right result, but there's no good justification for why expected value should come into play.

boromi 10y ago

why do you seek the fixed point and not the k value that maximizes s(k) = (1 - k)(1 + k) / 2 + k / 2 ?

SonOfLilit 10y ago
s(k) is the function for "The number I should choose if my opponent chooses k" [1]. We arrived at the formula by asking "what is the best number for us to choose assuming our opponent chooses k?". 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)
- murbard2 10y ago
  Why do you bring up expected value in the calculation. You may end up with the right result, but there's no good justification for why expected value should come into play.