I teach a large enrollment course, and as such I give multiple choice quizzes and exams. Last semester I started using a method to award partial credit that allows students to select two answers (usually there are 5 choices), essentially allowing them to hedge their bet. There is only ever one correct option for a question, so this doubles their chance that they choose the right answer (and then earn partial credit).
I use the following grading scheme:
Correct answer:
1 answer selected: 1 point
2 answers selected: 0.7 points
Incorrect answer:
1 answer selected: 0 points
2 answers selected: -0.25 points
As you can see, this encourages students to guess when they can narrow down the choices. But it discourages students from just choosing two answers for every question, because of the -0.25 cost if neither option they pick is correct.
My question here: is there a way to evaluate my point breakdown and determine if it is accomplishing the goal I've set out to do? Ideally, someone that selected two options at random for each question, should end up with a negative score (lets say 20 question on a exam).
Happy to provide further information/details if they are needed.
To answer your last question, we can compute the expected score on each question, given the strategy is choosing two answers at random. There's probability $0.4$ that this strategy will pick the correct answer, and probability $0.6$ that this strategy will pick the wrong answer, so the expected value is $$0.4\cdot 0.7 + 0.6 \cdot (-0.25) = 0.28 - 0.15 = 0.13$$ So this strategy won't, on average, give a negative score; it will give on average $0.13$ per question.
On the other hand, a strategy of guessing one answer at random will give $$ 0.2 \cdot 1 + 0.8 \cdot 0 = 0.2 + 0 = 0.2$$ So it is better for the student, if the student is guessing randomly, to guess one answer than it is to guess two.
Let's say the student is (correctly) confident one choice is wrong and guesses randomly among the other 4. Then the probabilities come out to the following: 2-choice strategy: $$ 0.5 \cdot 0.7 + 0.5 \cdot (-0.25) = 0.225$$ 1-choice strategy: $$0.25 \cdot 1 + 0.75 \cdot 0 = 0.25$$ So it's still a little better to choose 1 answer.
If the student is (correctly) confident two choices are wrong and guesses randomly among the other 3, the 2-choice strategy gives: $$ \frac23 \cdot 0.7 + \frac13 \cdot (-0.25) \approx 0.38 $$ and the 1-choice strategy: $$ \frac13 \cdot 1 + \frac23 \cdot 0 \approx 0.33 $$ So if the student can narrow down to three choices, it becomes advantageous to choose two of those three at random.