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I want to create a multiple-choice exam consisting of $N$ problems. Each problem has $M$ mutually-exclusive options. If you don't choose any option, you will get $0$. If you wrongly choose, you will get $-1$. Otherwise you will get $1$.

To pass the exam, at least you need to have a score of $x$ where $0<x\leq N$. When an uneducated toddler plays with the answer sheet and behaves randomly, what is his/her probability to pass? The toddler knows that she/he cannot choose 2 or more options for each problem.

If you think my question is not clear enough, you can make some answers, each for the possible case you think.

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For those who give answers, be patient. I am waiting for more answers and comments to make a final decision. This comment will be removed later. – kiss my armpit Nov 4 '12 at 9:48
The more interesting question is how many questions the student should guess to maximize their pass probability. – Phira Nov 4 '12 at 10:13
You mean, skip with probability 1/2 and choose any given option with probability 1/(2M)? – Did Nov 4 '12 at 10:44
@did: You have the same chance to answer or skip. When you answer, you have the same chance to select one option among the others. I don't know how to rephrase this mathematically. – kiss my armpit Nov 4 '12 at 10:47
I don't know how to rephrase this mathematically... I just did. – Did Nov 4 '12 at 10:53

The score on any answer seems (finally!) to be $0$ with probability $\frac12$, $+1$ with probability $\frac1{2M}$, and $-1$ with probability $\frac{M-1}{2M}$. The mean is $\mu=-\frac{M-2}{2M}$. To compute the probability that the sum $S_N$ of $N$ i.i.d. scores distributed like this is such that $S_N\geqslant x$ for some $1\leqslant x\leqslant N$ is tedious (and boring) but one can describe some asymptotics:

  • If $M=2$, $\mu=0$ hence $S_N$ is asymptotically gaussian in the sense that $\mathbb P(S_N\geqslant\sqrt{N}z)\to1-\Phi(2z)$ when $N\to\infty$.

  • If $M\geqslant3$, $\mu\ne0$ hence $S_N\sim\mu N$ in the sense that $\mathbb P(S_N\geqslant Nz)$ converges to $1$ for every $z\lt\mu$ and to $0$ for every $z\gt\mu$. Since $\mu\lt0$, one sees in particular that $\mathbb P(S_N\geqslant0)\to0$ (and this convergence is exponentially fast) hence any nonnegative threshold $x$ is basically irrelevant.

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