What is the probability of guessing the correct answer to a multiple-choice question with four choices?

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If you have carried out an assessment where someone makes a response by choosing from a set of possible responses (e.g. by pointing at a picture), you can use this to work out how likely they could have scored what they got on the test by chance. More information.

In a multiple choice question there are 4 alternative answers of which 1, 2, 3, or all may be correct. A candidate decides to tick answers at random. If he is allowed upto 5 chances to answer the question, the probability that he will get the marks in the question is?

It is equally likely for 1, 2, 3, or all to be correct so probability is 1/4.

Case1 when only 1 option is correct. Since the candidate is allowed 5 chances, probability of getting correct answer is 1.

Case2 when 2 options are correct. Toltal ways in which 2 options can be correct is $\left(^4_2\right)$, which is 6. Out of these only 1 is correct. So probability of selecting correct answer is 1/6. Since he has 5 chances the probability of getting marks is $1 - \left(\frac{5}{6}\right)^5$

Case3 when 3 options are correct.
Total ways in which 3 options can be correct is $\left(^4_3\right)$, which is 4. Since he has 5 chances, probability of getting correct answer is 1.

case4 when all options are correct. Only one way in which all can be correct. Probality of getting marks is 1.

So answer should be $\left(\frac{1}{4}\times1\right) + \frac{1}{4}\times\left(1 - \left(\frac{5}{6}\right)^5\right) + \left(\frac{1}{4}\times1\right) + \left(\frac{1}{4}\times1\right)$. Which is indeed wrong.

A friend of mine did this question as follows. Total options: $\left(^4_1\right) + \left(^4_2\right) + \left(^4_3\right) + \left(^4_4\right)$, i.e 15. Since he has 5 chances to answer, probability would be 5/15. I know this is wrong (or not?) but beacause my textbook says answer is 1/3 I couldn’t argue.

Please help. Thanks in advance.