This may also end up failing as well, but I though I'd share.
B's plan will be to always pick and even card (from 2 to 48) based only on the even cards that have been picked. C=code number. B=B's card
Case 1. All cards are odd (O1,O2,O3,O4). C=B. A will see a single even card and know C=B
Case 2. One even (O1,O2,O3,E1). B=E1+C mod 100. A will see two even cards, E1 and B. E1B is too big, but BE1=C.
Case 3. Two Even (O1,O2,E1,E2). if E1>E2 then B=(E1E2)+C if E2>E1 then B=(E2E1)+C. A will see three even cards, E1, E2, and B and will try to decode to find C. There are three choices:
B(E1E2) = BE1+E2 = C
E1(BE2) = B+E1+E2 = not C
E2(BE1) = B+E1+E2 = not C
The unique solution is the correct one, the two incorrect ones come out the same. [Is there an example that doesn't work?]
Case 4. Three even (O1,E1,E2,E3). I haven't tried this yet but I'm thinking if E1>E2>E3 then B=E1E2+E3+C may work but I haven't checked it yet.
Case 5. Four even (O1,O2,O3,O4). Again I haven't tried this yet but I'm thinking if E1>E2>E3>E4 then B = E1E2+E3E4+C may work but I haven't tried it yet.

Posted by Jer
on 20070514 12:59:05 