2 ms·
Using this on only 3 questions, you get 2/3 success rate: 000,001,010,100,000 -> buddy signals 0, I write 000 111,110,101,011,111 -> buddy signals 1, I write
by szemet 10y ago
Using this on only 3 questions, you get 2/3 success rate:
000,001,010,100,000 -> buddy signals 0, I write 000
111,110,101,011,111 -> buddy signals 1, I write 111
with 7 bit you definitely win 14 questions from 21
- SomeStupidPoint 10y agoYou can get at least 17 by using the 1 bit to express preference for just the first 12 questions, and the remaining 11 bits the answers to the last 11 questions exactly. I've been trying to think if there's a clever scheme to spread the knowledge around to beat 17 with statistical tricks. (As a fun aside: 17/23 is about 73%, which typically gets you a C -- so you'd pass the exam.)
- szemet 10y ago6 bit to win 12 question from the first 18 (group by 3) 5 bit for the remaining 5 17 answer and one bit remained
- SomeStupidPoint 10y agoWhich answer do you use the extra bit on that's guaranteed to give you one more correct? For any you choose, there are answer patterns where you already counted that answer being correct in your guaranteed number as part of the 2/3rds correct. So I don't think you can (trivially) use that bit to guarantee another correct answer.
- jcranmer 10y agoThe last bit for the extra answer doesn't guarantee you an extra point in all cases.
- jcranmer 10y agoI wrote a simple program to test all possible majority-of-group breakdowns for generic number of questions and bitlengths, and found that the optimal result in all cases for this class was to use all but one bit for individual questions and then use that one bit for majority of the rest. You can probably get better bit usage by doing some sort of more complex error code correction scheme--imagine having overlapping groups that you know majority vote on. It's been a decade since I've touched that stuff, so I don't know any concrete numbers here.