In an exam with $12$ yes/no questions with $8$ correct needed to pass, is it better to answer randomly or answer exactly $6$ times yes and 6 times no, given that the answer 'yes' is correct for exactly $6$ questions?

I have calculated the probability of passing by guessing randomly and it is

$$\sum_{k=8}^{12} {{12}\choose{k}}0.5^k0.5^{n-k}=0.194$$

Now given that the answer 'yes' is right exactly $6$ times, is it better to guess 'yes' and 'no' $6$ times each?

My idea is that it can be modelled by drawing balls without replacement. The balls we draw are the correct answers to the questions.

Looking at the first question, we still know that there are $6$ yes and no's that are correct. The chance that a yes is right is $\frac{6}{12}$ and the chance that a no is right is also $\frac{6}{12}$.

Of course the probability in the next question depends on what the first right answer was. If yes was right, yes will be right with a probability of $5/11$ and a no is right with the chance $6/11$. If no was right, the probabilities would change places.

Now that we have to make the choice $12$ times and make the distinction which one was right, we get $2^{12}$ paths total. We cannot know what the correct answers to the previous questions were. So we are drawing $12$ balls at once, but from what urn? It cannot contain $24$ balls with $12$ yes and $12$ no's. Is this model even correct?

Is there a more elegant way to approach that?

I am asking for hints, not solutions, as I'm feeling stuck. Thank you.

Edit: After giving @David K's answer more thought, I noticed that the question can be described by the hypergeometric distribution, which yields the desired result.

  • 5
    $\begingroup$ In terms of balls and urns: Maybe it helps to think about it as: You have a blue urn and a red urn, and you have $6$ red balls and $6$ blue balls. You randomly put $6$ of the twelve balls in the blue urn, and the other $6$ in the red urn. Now: what it the chance that at least $8$ balls are in the 'right' (i.e. same colored) urn? Looking at it that way, you can actually just focus on the red balls, and you find that you need $4$, $5$, or all $6$ red balls in the red urn. $\endgroup$
    – Bram28
    May 6, 2018 at 16:27
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    $\begingroup$ Wait, do you know that there are 6 "yes" and 6 "no" answers?? If so, responding "yes" to every question is a sure way of failing. $\endgroup$
    – Darkhogg
    May 6, 2018 at 19:44
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    $\begingroup$ @Darkhogg you are correct $\endgroup$
    – B.Swan
    May 6, 2018 at 19:48
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    $\begingroup$ I've written a program to check all of the $\binom{12}{6} = 924$ possible answer keys, and exactly $262$ of them are passing grades. The odds are approximately $0.28355$, on the assumption that each answer key is equally likely, and independent of your guessing. $\endgroup$
    – user14972
    May 7, 2018 at 0:38
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    $\begingroup$ How is it that no one suggested one should try to actually solve the problems? Wait... is this one of the problems on the test? $\endgroup$
    – Kimball
    May 7, 2018 at 2:51

6 Answers 6


We are given the fact that there are $12$ questions, that $6$ have the correct answer "yes" and $6$ have the correct answer "no."

There are $\binom{12}{6} = 924$ different sequences of $6$ "yes" answers and $6$ "no" answers. If we know nothing that will give us a better chance of answering any question correctly than sheer luck, the most reasonable assumption is that every possible sequence of answers is equally likely, that is, each one has $\frac{1}{924}$ chance to occur.

So guess "yes" $6$ times and "no" $6$ times. I do not care how you do that: you may guess "yes" for the first $6$, or flip a coin and answer "yes" for heads and "no" for tails until you have used up either the $6$ "yeses" or the $6$ "noes" and the rest of your answers are forced, or you can put $6$ balls labeled "yes" and $6$ labeled "no" in an urn, draw them one at a time, and answer the questions in that sequence.

No matter what you do, you end up with some sequence of "yes" $6$ times and "no" $6$ times. You get $12$ correct if and only if the sequence of correct answers is exactly the same as your sequence. That probability is $\frac{1}{924}.$

There is no way for you to get $11$ correct. You get $10$ correct if and only if the correct answers are "yes" on $5$ of your "yes" answers and "no" on your other "yes" answers. The number of ways this can happen is the number of ways to choose $5$ correct answers from your $6$ "yes" answers, times the number of ways to choose $5$ correct answers from your $6$ "no" answers: $\binom 65 \times \binom 65 = 36.$

There is no way for you to get $9$ correct. You get $8$ correct if and only if the correct answers are "yes" on $4$ of your "yes" answers and "no" on your other "yes" answers. The number of ways this can happen is the number of ways to choose $4$ correct answers from your $6$ "yes" answers, times the number of ways to choose $4$ correct answers from your $6$ "no" answers: $\binom 64 \times \binom 64 = 225.$

In any other case you fail. So the chance to pass is $$ \frac{1 + 36 + 225}{924} = \frac{131}{462} \approx 0.283550, $$ which is much better than the chance of passing if you simply toss a coin for each individual question but not nearly as good as getting $4$ or more heads in $6$ coin tosses.

Just to check, we can compute the chance of failing in the same way: $6$ answers correct ($3$ "yes" and $3$ "no"), $4$ answers correct, $2$ correct, $0$ correct. This probability comes to $$ \frac{\binom 63^2 + \binom 62^2 + \binom 61^2 + 1}{924} = \frac{400 + 225 + 36 + 1}{924} = \frac{331}{462} \approx 0.716450, $$ which is the value needed to confirm the answer above.

  • $\begingroup$ Assuming it's a maths exam (specifically probabilities), I think you're going to breeze it. $\endgroup$
    – Strawberry
    May 7, 2018 at 12:01
  • $\begingroup$ This is one of those cool (or annoying) problems where you're forced to hedge your bets. You could "go for broke" by answering 6 each Y and N, thus possibly getting 100% but every wrong answer means two wrong answers; or you can choose to give, e.g., 8 and 4, thus increasing the chance of passing but guaranteeing you can't get 100%. $\endgroup$ May 7, 2018 at 16:13
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    $\begingroup$ @CarlWitthoft Suppose you answer yes 8 times. You can get 6 correct "yeses" and 4 correct "noes" in $\binom 86$ ways; 5 correct "yeses" and 3 correct "noes" in $\binom 85 \binom 43$ ways; if 4 or more of your "yes" guesses are wrong then at least 2 of your "no" guesses also are wrong. $\endgroup$
    – David K
    May 7, 2018 at 20:10
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    $\begingroup$ @CarlWitthoft The probability that if you choose 8 yeses and 4 noes you get 8 or more correct is $\left[\binom{6}{6}\binom{6}{2}+\binom{6}{5}\binom{6}{3}\right]\div\binom{12}{8}\approx0.272727$, which is less than 0.283550. $\endgroup$ May 8, 2018 at 11:04
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    $\begingroup$ @CarlWitthoft No matter what number of times you choose to answer "yes," you commit to some minimum number of wrong answers and all additional mistakes occur in pairs. But you're not forced to hedge your bets in this particular exam, because the "go for broke" strategy turns out to be maximal in that case. (As noted in a comment, if you know there are actually exactly 7 questions whose correct answer is "yes," you should answer more than 7 questions "yes"; that is, "go for broke" is not a winning strategy for all variations of this problem.) $\endgroup$
    – David K
    May 8, 2018 at 12:43

In terms of balls and urns: Maybe it helps to think about it as follows:

You have a red urn and a blue urn, and you have $6$ red balls and $6$ blue balls. You randomly put $6$ of the twelve balls in the red urn, and the other $6$ in the blue urn. Now: what it the chance that at least $8$ balls are in the 'right' (i.e. same colored) urn?

Well, to get $8$ correct, you either need to get all $6$ red balls in the red urn ($1$ possibility), or $5$ red ones and $1$ blue in the red urn (${6 \choose 5} \cdot {6 \choose 1} = 6 \cdot 6 = 36$ possibilities), or $4$ red ones and $2$ blue ones (${6 \choose 4} \cdot {6 \choose 2} = 15 \cdot 15 = 225$ possibilities). This is out of a total of ${12 \choose 6} = 924$ possibilities, and so the probability is $\frac{1+36+225}{924}$

NOTE: Thanks to @DavidK for pointing out my initial answer was wrong! Everyone please upvote his answer!

  • $\begingroup$ In this model, the fact that we know that 6 answers are in fact 'yes' is represented by the capacity of the 'yes' urn, right? $\endgroup$
    – B.Swan
    May 6, 2018 at 16:56
  • $\begingroup$ @B.Swan No, it's rather that you have $6$ 'yes' balls and $6$ 'no' balls. $\endgroup$
    – Bram28
    May 6, 2018 at 16:58
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    $\begingroup$ @B.Swan Yes, that struck me like a large increase as well .. not something I would have guessed ... in fact, I was reminded of those 'matching problems' where you have to match A,B,C,D with 1,2,3,4 ... frustratingly, with those problems, if you do one wrong you automatically do a second wrong, so I have often played with the idea of maybe repeating a letter so getting 3 right becomes an option. Never done the math, but thanks to your problem I'm thinking that might actually not be a good idea after all! $\endgroup$
    – Bram28
    May 6, 2018 at 17:20
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    $\begingroup$ There are 924 possible sequences of answers in which 6 are yes and 6 are no. I do not see how you can have a better than 1/924 chance to guess the sequence exactly. But this answer is based on the premise that the chance is the same as flipping 6 heads in a row, which is 1/64. I do not think this answer is correct. $\endgroup$
    – David K
    May 6, 2018 at 22:55
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    $\begingroup$ Anyway, whatever else happened, we now have a correct accepted answer (this one, after it was edited), so I'm happy! $\endgroup$
    – David K
    May 7, 2018 at 0:53

Try the following approach.

First, assume that the answerer puts "yes" for the first six and "no" for the last six. Since the order of his answers clearly cannot change his probability of passing, this is an okay assumption.

Note that if there are $x$ questions correct on the first half of the test, there are $x$ questions correct on the second half as well (think about why). So it's sufficient to get the probability that $4$ of the first $6$ answers are yes.

Now, try to calculate the probability that a randomly ordered row of $12$ balls -- $6$ "yes" balls and $6$ "no" balls -- has $4$ "yes" balls among the first $6$ balls in the row.


A quick way to find the correct answer, which IMHO also gives insight into why this works better than 12 independent coin flips:

  • Let $X =$ no. of correct answers if you randomly guess exactly 6 Yes and 6 No.

  • As others have observed, $X \in \{0, 2, 4, 6, 8, 10, 12\}$.

  • $P(X=6) = {{6 \choose 3}{6 \choose 3} \over {12 \choose 6}} = {400 \over 924} \approx 0.4329$.

  • By symmetry, $P(X \ge 8) = {1 - P(X=6) \over 2} = {262 \over 924} \approx 0.2835$.

IMHO, this is ultimately why this method works so much better than 12 independent coin flips. If the passing grade were 7 correct answers, flipping 12 independent coins would have been better:

  • Let $Y =$ no. of correct answers if you flip 12 independent coins.

  • $P(Y = 6) = {12 \choose 6} ({1\over 2})^{12} \approx 0.2256.$

  • By symmetry, $P(Y \ge 7) = {1 - P(Y = 6) \over 2} \approx 0.3872.$

  • Meanwhile, of course $P(X \ge 7) = P(X \ge 8) \approx 0.2835$.

However, the passing grade is 8 correct answers, and you have invented a method which is guaranteed not to have 7 correct answers thereby eliminating the (useless-to-you) 7 correct answers situation.

Bonus: if the passing grade is 7 correct answers, then the correct strategy is to actually randomly answer 5 (or 7) of them as Yes!

  • Let $Z =$ no. of correct answers if you randomly guess 5 Yes and 7 No. (Math is same if you guess 7 Yes and 5 No.)

  • The only possible values are $Z \in \{1, 3, 5, 7, 9, 11\}$

  • By symmetry, $P(Z \ge 7) = {1 \over 2} > P(Y \ge 7) \approx 0.3872 > P(X \ge 7 ) \approx 0.2835$.

I bet this observation generalizes: i.e., if the passing grade is odd, guess 5 (or 7) as Yes, and if the passing grade is even, guess 6 as Yes. What a way to game an exam!

  • $\begingroup$ Can you explain why you divided the term for $P(Y\geq 7)$ by 2? And does your $Y$ take into consideration that you know 6 answers are correct and 6 are not? $\endgroup$ May 8, 2018 at 13:48
  • $\begingroup$ @ChristianIvicevic - i have edited my answer to (hopefully) improve the explanation. my $Y$ did NOT take advantage of knowing there are 6 Yes and 6 No. i wanted to show that, if the passing grade is 8, then $X$ outperforms $Y$, but if the passing grade is 7, then $Y$ outperforms $X$ (even though $Z$ outperforms both $X,Y$). $\endgroup$
    – antkam
    May 8, 2018 at 14:22
  • $\begingroup$ @ChristianIvicevic - in other words, $Y$ is never the best, but if you want to take advantage of knowing there are exactly 6 Yes and 6 No, then your best strategy ($X$ or $Z$) still depends on what passing grade (8 or 7) you are aiming for. $\endgroup$
    – antkam
    May 8, 2018 at 14:29

Brute force:

Suppose the answer has a set position of yes and no answers

Suppose you guess 6 yesses, then the rest of them you answer no.

There are ${6 \choose k}$ ways for you to have guessed $k$ of the yeses correct. If you guess $k$ yesses correctly then you have guess $6-k$ yesses incorrectly. There are $6$ nos and $6-k$ of the must be in spots where you guessed them yes incorrectly. The remaining $k$ must be in spots you guessed no. So you will have gotten $2k$ correct.

So there is ${6\choose k} $ ways to get get $2k$ out of $12$ correctly and it is impossible to get an odd number correct.

So the total ways to guess is $\sum_{k=0}^6 {6 \choose k}$ and the total ways of passing are $\sum_{k=4}^6{6\choose k}$.

So the probability of passing is $\frac {\sum_{k=4}^6{6\choose k}}{\sum_{k=0}^6 {6 \choose k}}$. As ${6 \choose k} = {6\choose 6-k}$ we know ${\sum_{k=0}^6 {6 \choose k}}= {6\choose 3} + 2\sum_{k=4}^6 {6\choose k}$

So probability of passing is $\frac {{6\choose 4}+{6\choose 5} + {6\choose 6}}{2({6\choose 4}+{6\choose 5} + {6\choose 6}) + {6\choose 3}}=\frac {15 + 6 + 1}{2(15+6+1) + 20} = \frac {22}{64} = \frac {11}{32}$ more than $1$ in $3$.

If you guess randomly:

The probability that you guess them exactly correct is $\frac 1{2^{12}}$.

The probability that you guess the first one wrong and the $2-12$ correctly is the same $\frac 1{2^{12}}$, but if you guess exactly one wrong there are $12$ possible questions that could be wrong. So the probabiblity that exactly one of your guess is wrong is $12*\frac 1{2^{12}}$.

If you guess exactly $k$ wrong there are ${12\choose k}$ ways to do this so the probability of guessing exactly $k$ wrong is ${12\choose k}\frac 1{2^{12}}$.

So to guess $0, 1,2,3$ or $4$ wrong is $\frac 1{2^{12}}({12\choose 0}+{12\choose 1}+{12\choose 2}+{12\choose 3}+{12\choose 4})=\frac 1{2^{12}}(1+12+66+220+ 495)= \frac {794}{4096} $ almost $1$ in $5$.

So first is better.

  • $\begingroup$ Your answers have both different result than those calculated with other models, could it be that you are forgetting to account for the different combinations out of the $12$ questions? Something is (almost certainly) off. I am pretty sure the calculation of totally random answering with the binominal distribution yields the correct result. $\endgroup$
    – B.Swan
    May 6, 2018 at 17:50
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    $\begingroup$ Yeah... I was just noticing that. You and I both did $\sum_{k=8}^12 {12\choose k}\frac {1}{2^12}$ so ..... Isn't ${12\choose 12}={12\choose 0}=1$ and ${12\choose 1} = 12$ and ${12\choose 2} = \frac {12*11}2=66$ and .... Oh.... I only did 4 of the five terms .... DOH!!!!! $\endgroup$
    – fleablood
    May 6, 2018 at 18:24
  • $\begingroup$ But may first way is way off... $\endgroup$
    – fleablood
    May 6, 2018 at 18:37
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    $\begingroup$ Rookie mistake. There are ${12 \choose 6}$ possible answer keys but one we fix one it doesn't matter which and that has nothing to do with the sample space. $\endgroup$
    – fleablood
    May 6, 2018 at 18:50
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    $\begingroup$ Since $32$ does not divide $924,$ the probability $11/32$ cannot be correct. And we now have an exhaustive count in software that comes up with $262/924.$ The source of error seems to be taking $\binom 6k$ as the number of ways to get $2k$ correct out of $12$, when the correct number is actually $\binom 6k^2$. $\endgroup$
    – David K
    May 7, 2018 at 0:50

Suppose we reframe it as the following: we have some probability $p_r$ of passing if youwe choose randomly. We can also define $p_n$ as the probability of passing if we restrict ourselves to random assignments with exactly $n$ "yes". So your specific question is: is $p_6>p_r$? More generally, we ask "How does $n$ affect $p_n$?" If we were to assign all the questions "no", there is no way we could pass; since only 6 questions are actually "no", we can get at most 6 right. To pass, we have to assign at least 2 questions "yes". So we have $p_n = 0$ for $|n-6|>4$. We have non-zero $p_n$ for only $n$ in $\left\{2,3,4,5,6,7,8,9,10 \right\}$. Since $p_r$ is a weighted average of $p_n$ for all $n$ (including values for which $p_n =0$), the only way $p_r$ could greater than $p_6$ is if $p_6$ is less than the average $p_n$ for $|n-6|\leq4$=. Intuitively, however, $p_6$ should be at least as large as $p_n$ for any n.


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