In a true/false question that is true with probability p, if you assign probability q, your probability of losing is p(1−q)2+(1−p)q2. (The probabily the answer is true and you spin false twice plus the probability the answer is false and you spin true twice.)
This probability is minimized when its derivative with respect to q is 0, or at the boundary. This derivative is −2p(1−q)+2(1−p)q, whis is 0 when q=p. We now know the minimum is achieved when q is 0, 1, or p. The probability of losing when q=0 is p. The probability of losing when q=1 is 1−p. The probability of losing when q=p is p(1−p), which is the lowest of the three options.
In a true/false question that is true with probability p, if you assign probability q, your probability of losing is p(1−q)^2+(1−p)q^2. (The probabily the answer is true and you spin false twice plus the probability the answer is false and you spin true twice.)
This probability is minimized when its derivative with respect to q is 0, or at the boundary. This derivative is −2p(1−q)+2(1−p)q, whis is 0 when q=p. We now know the minimum is achieved when q is 0, 1, or p. The probability of losing when q=0 is p. The probability of losing when q=1 is 1−p. The probability of losing when q=p is p(1−p), which is the lowest of the three options.
(Why does the two-spin work?)
In a true/false question that is true with probability p, if you assign probability q, your probability of losing is p(1−q)2+(1−p)q2. (The probabily the answer is true and you spin false twice plus the probability the answer is false and you spin true twice.)
This probability is minimized when its derivative with respect to q is 0, or at the boundary. This derivative is −2p(1−q)+2(1−p)q, whis is 0 when q=p. We now know the minimum is achieved when q is 0, 1, or p. The probability of losing when q=0 is p. The probability of losing when q=1 is 1−p. The probability of losing when q=p is p(1−p), which is the lowest of the three options.
Copied without LaTeX:
In a true/false question that is true with probability p, if you assign probability q, your probability of losing is p(1−q)^2+(1−p)q^2. (The probabily the answer is true and you spin false twice plus the probability the answer is false and you spin true twice.)
This probability is minimized when its derivative with respect to q is 0, or at the boundary. This derivative is −2p(1−q)+2(1−p)q, whis is 0 when q=p. We now know the minimum is achieved when q is 0, 1, or p. The probability of losing when q=0 is p. The probability of losing when q=1 is 1−p. The probability of losing when q=p is p(1−p), which is the lowest of the three options.
This is called either Brier or quadratic scoring, not sure which.
Not exactly. Its expected value is the same as the expected value of the Brier score, but the score itself is either 0 or 1.
For some reason, the latex is not rendering for me. I can see it when I edit the comment, but not otherwise.
The comment has just started rendering for me.
Edit: Oh wait no, you just added another comment without LaTex.
Huh, that’s really weird. The server must somehow be choking on the specific LaTeX you posted. Will check it out.
Ok, I found the bug. I will fix it in the morning.
And you did! Cheers for your hard work. :)