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Suppose a school test has 100 multiple-choice questions, each with four options $A, B, C, D$. Each question carries 1 mark. To prevent students from scoring just by pure luck, all the options carry equal weight, i.e. each of the options is the answer to 25 of the questions.

A student who has not studied anything attempts the test. Which of these two strategies will be better?:

1.Choose any option and answer all the 100 questions as that option.

2.Answer randomly chosen 25 of the questions as $A$, 25 as $B$, 25 as $C$ and 25 as $D$.

Clearly, the first strategy ensures a score of 25. I think, the second strategy can result in any score from 0 to 100. Let's say the second strategy is better than the first strategy if the probability of getting 25 marks or more by using that strategy is more than 50%. So, is this strategy better than the first?

EDIT: I think my question is a bit different from Bernoulli trails. In Bernoulli trails, the student answers each question independently, so the total number of ways the student can answer is $4^{100}$. While in the second strategy, the answers are not independent. For example, if the student has already answered first $25$ questions as $A$ then this prevents them from answering any other question as $A$.

The total number number of ways to answer 25 questions as A, 25 as B, 25 as C and rest as D is: $\frac{100!4!}{(25!)^4}=3.869298019717862276844777519113e+58$ which is less than $4^{100}=1.6069380442589902755419620923412e+60$

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  • $\begingroup$ I can't see how the condition prevents students from scoring by sheer luck. Now, substracting 0.5 points from every wrong answer would achieve this more effectively, imo...but this is a little beyond mathematics. $\endgroup$ – DonAntonio Nov 21 '17 at 10:11
  • $\begingroup$ Suggestion: See if you can find the probability that the answer to the first question is correct. Then see if you can use linearity of expectation to find the expected number of correct answers. $\endgroup$ – awkward Nov 21 '17 at 13:11
  • $\begingroup$ @awkward: that shows the expected number of right answers is $25$ but does not say whether the chance of $25$ or more is at least $50\%$ $\endgroup$ – Ross Millikan Nov 21 '17 at 14:56
  • $\begingroup$ @RossMillikan You are right, I overlooked that part of the question. Oops. $\endgroup$ – awkward Nov 22 '17 at 13:43
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P(n Correct) = ${100\choose n}(\frac{1}{4})^{n}(\frac{3}{4})^{100-n}$

P(maximum 25 Correct) = $\sum_{n=0}^{25}{100\choose n}(\frac{1}{4})^{n}(\frac{3}{4})^{100-n} = 55.34\%$

Thus P(26 or more correct) = $44.66\%$

Thus it is unfavorable to randomly answer the question

I obtained the first formula using Bernoulli Trials

Aside from the maths, if the students know each option occurs exactly 25 times, the test gets a bit more easier :P

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  • $\begingroup$ For shorter tests it's actually better to randomly guess because of the border condition. I'm curious as to what the minimum value of n is such that it no longer becomes a better strategy to randomly guess n problems. $\endgroup$ – Zhuli Nov 21 '17 at 16:44
  • $\begingroup$ My results seem to tell the opposite.. unless i fucked up somewhere $\sum_{n=0}^{a/4}{a\choose n}(\frac{1}{4})^{n}(\frac{3}{4})^{a-n}$ This gives me a result of almost 70% for lower values(4 and 8 to be exact) and it never reaches below 50% (checked as far as 1200) so it is always better to pick just one option (assuming equal weigthage) $\endgroup$ – Anvit Nov 21 '17 at 16:49
  • $\begingroup$ If f(n) is defined as the probability of scoring at least the average, I've confirmed the following values: f(4)=17/24, f(8)=341/504, f(12)=238751/369600 $\endgroup$ – Zhuli Nov 21 '17 at 16:57
  • $\begingroup$ Yeah, my answer includes the 25th in maximum and yours in minimum $\endgroup$ – Anvit Nov 21 '17 at 17:07
  • $\begingroup$ I feel like my question is a bit different. In Bernoulli trials, the student would give answer to each question independently. It is equivalent to the student randomly guessing each answer. So, one of the possible combinations the student can give answer is: first 50 questions as A and the rest as B. But this is against the second strategy because in the second strategy the student must ensure that they answer 25 questoins as A, 25 as B, 25 as C rest as D. $\endgroup$ – Ryder Rude Nov 22 '17 at 2:12

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