How probable is this trick effective? I know my question is too general but:
Suppose there is a 4 choice exam, a student gets a score, but its not a good score, how possible is it for him to get more score by saying to his teacher that he chose choices for all questions seperatly in reverse order meaning if he chose 1 for a question, the trick makes it 4 or of he chose 3 it would become 2.
Can we find how probable this will improve his score with rescpect to number of questions and the score he got?
An example would be:
There was an exam with 40 questions and somebody got 17.5% true, is it worth for him to do the trick?
 A: Assume out of n questions he answered r correctly and (n-r) incorrectly.  After the re-scoring r will be incorrect (since will be switched from a correct answer to an incorrect one), whereas the (n-r) questions will have a $\frac13$ chance of being correct.  So, his expected final score is $\frac{n-r}3$.  To benefit from re-scoring her initial grade should be below 25%.
For example if the initial grade was 10/100, re-scoring will increase the expected score to (100-10)/3 = 30.
A: The question is how probable it is to do better this way.
Very short answer: If they got 25%, then the probability they do any better by switching answers is 50%. If they got less than 25% it's likely that they will do better by switching. If they got more than 25% it's unlikely.
Short answer: The probability of the student doing better by switching the answers is about $\Phi(-\tfrac{12s-3}{2-2s})$, where $s$ is the proportion of answers they got correct in the first place and $\Phi(x)$ is the cumulative distribution function of the standard normal distribution.
In particular that if $s=0.25$ (i.e. the student got 25% correct initially) then the probability that they do better by switching is 50%.
If $s=0.3$ (they got 30% right) then the probability they do better is about 35%. If $s=0.4$ (40%) the probability drops to 7%.
Long answer: If there are $n$ questions, and $r$ were answered correctly, then as karakfa's answer says, if we swap the answers then $r$ will definitely be incorrect and the other $n-r$ will each be correct with probability $\tfrac13$. Hence the number of correct answers $X$ is distributed binomially as $X \sim \operatorname{Bin}(n-r,\tfrac13)$. The answer is then the probability that $X$ is more than $r$: $\mathbb{P}(X > r)$.
If we assume there are quite a lot of answers, so $n-r$ is large, then we can approximate this distribution as $X \sim N(\tfrac13(n-r),\tfrac13\tfrac23(n-r))$ and in particular $X > r$ if and only if $Z > \frac{12s-3}{2-2s}$ where $s=\tfrac rn$ is the proportion of initially correct answers and $Z \sim N(0,1)$ is standard normal.
Hence the probability of doing better is $\mathbb{P}(X > r) \approx \Phi(-\tfrac{12s-3}{2-2s})$.
A: Edit: @karakfa 's answer is correct. This one is not, but it's a natural place to start and happens to lead to a similar answer. I will leave it rather than deleting it so that others can learn from my mistake.
If I read this correctly the student is essentially asking to switch all his answers to a random answer. That would give him a score of $25\%$.
If his original score was leas than that, this strategy would probably improve his exam score, but probably not help his final grade too much.
He might get lucky. The probabilities there depend on the number of questions on the exam and the number he got right to begin with.
