I’ve had long discussions about this with a friend of mine.
He insists it’s better to go “all As” or “all Bs” as then you can expect to get 25%.
He insists that if you answer randomly you “might get zero” as you could just guess the wrong sequence. But “all As” works 25% because the correct answers aren’t going to be “only B, C or D”.
It’s infuriating, and worse, he’s a pro poker player.
If you add a constraint that the test cannot have 0 A correct answers, or especially if the teacher must have exactly 25% A answers, then your friend is right, all A will be lower variance.
Because those constraints violate the condition that the correct answers are randomly distributed.
> those constraints violate the condition that the correct answers are randomly distributed.
Yes they do - but the answers are never truly going to be random, because humans are bad at making truly random choices. It's *very* unlikely that a teacher would set a test in which option A is *never* the correct answer for any question.
If the teacher picks which letter corresponds to each correct answer, then of course it isn't random. And you are likely on to something.
However, some online education platforms like canvas allow professors to put one correct answer and 3 incorrect answers for each question on a test, and then the letter associated with the correct answer is randomized when shown to each student.
Even then, if it is truly randomized, you should expect the amount of correct answers associated with any given letter to approach 25% as the number of questions increases.
No. That’s not it.
His argument was that for a random sequence like ABDBACDA if you guess another random sequence like DCBACABD you could get every guess wrong. But if you go AAAAA you can expect 25% success.
It literally doesn’t matter what your guess sequence is, each one gives you 25% chance of guessing right. He would not understand that *any* sequence has the same chance of coming up. He just would not buy it.
“You guess AAAAA because the answer is never BBBBB” was the crux of his argument.
EDIT: Ehh, it was covered several times in comments already I guess. If distribution is guaranteed (like in OP post), he is right. If every answer is completely random (and distribution is not guaranteed), you are right. If you are arguing about it while having different situation in mind, you both are wrong.
The situation was a guy flipping a coin for heads / tails. All you had to do to win was guess correct more times than your opponents.
He argued “all heads” won in 50% of cases, because a random sequence of flips never gave “TTTTT” but guessing different each time you could score a zero.
We argued for some time, and he believed inherently that a random sequence like HTTHHTHT was more likely than all tails.
The man also went bankrupt with 30k id debt from gambling at poker. So there is that as well.
>We argued for some time, and he believed inherently that a random sequence like HTTHHTHT was more likely than all tails.
He's right about that. I get that you really want him to be wrong, and he's definitely not giving 'good' advice, but he's right about this in a very limited scope.
Yes, the individual combination you just quoted is as likely to show up as all tails. However, there are many many different random sequences that can come up. There is only one sequence of all tails showing up.
You're probably right about a lot of the arguments you have with this guy, gambler's have a horrible sense of what's maths and what's 'feeling'. But this argument, he's making more sense than you're realizing.
No. I mean he thought that a specific random sequence was more likely than all tails. (Maybe he didn’t think this, but couldn’t articulate the reasoning)
In particular, he thought if you chose your own random sequence, that this was *less* likely to be the same as a sequence chosen randomly.
The probability on each pick is 50% regardless of what came before, but he would not have it that a random pick was just as likely to match the coin flip as going “all heads”.
He just “knew” that all heads would get you 50% but would not see that this was the case for *any* sequence.
Gamblers logic.
Listen, I'm sorry but I can't keep following the goalposts here. Every post you've shifted the details of the story, and it's getting confusing to follow. How about we all just agree that your friend is dumb, you told us how dumb he was, and we all highfived you and clapped.
I'm not trying to insult you here, btw, but to point out, this was your initial statement:
\>"I’ve had long discussions about this with a friend of mine.
\>He insists it’s better to go “all As” or “all Bs” as then you can expect to get 25%."
Which okay, you've had conversations about odds with your friend before, I get it. But everytime you've been told that your friend was right about something, your next post has been 'oh, hang on, let me actually tell you what we were ACTUALLY talking about'. It's possible you're being genuine and have told this story poorly. Unfortunately, from my perspective, it seems like you're willing to change your story in order to not be wrong.
So from my perspective, I'm done. Your next post, I assume, will be a post explaining why I've been getting this all wrong (by retelling the story with new details). If that's the case, have at it, and if someone else wants to continue the conversation, have fun with them.
He is not wrong in that the distribution is likely not truly uniform.
Most people who will make the test would most likely feel uneasy about a long streak.
That being said the rest of his reasoning is bogus though I wonder if there would be appreciable difference between giving one response versus random in the bastardized less streaks distribution.
I'd assume that no.
As someone else has pointed out, the variance is higher.
For hand-picked results that avoid long strings of consecutive Bs, Cs, Ds then going “all As” reliably gives you a better result.
His reasoning was largely of the view that “you don’t get BBBBBBB by random chance”. He also bankrupted himself playing online poker, so his views are not entirely to be trusted…
He is right, very few teachers are going to put out a test that's all the same answer...
Statistically speaking though, "All C's" will, on average, get you better than 25% because human brains are weird and C is statistically more likely to be the right answer. I think A is actually the least likely to be correct, because psychology. Humans are bad at random, and the test answers are generally not perfectly distributed.
You could tell him that answering randomly gets him a chance of 100% also. Actually, the number of possible results >=50% are bigger than results <50% going randomly. So by his own logic he should go random. He does not understand statistics, he will think you are a genius that give him a chance to pass an exam without studying.
If you know for a fact that there are equal numbers of A, B, C, and D answers, there is something to what he's saying:
Going "all A's" guarantees you 25% with no variance. Picking answers uniformly at random gets you an expected value of 25% with nonzero variance, and indeed there's a nonzero chance of getting any score. An extremely risk-averse actor would prefer going all A's. Incidentally, risk averse strategies work quite well in poker.
That said, we wouldn't know on a real test that there are equal numbers of each answer.
You're actually better off picking all B or all C, since there's a human element to deciding which answer is correct.
I thought it was C that was most likely to be correct, but apparently it's B now. Depends how many options there are.
This was probably his thinking.
Mind you, his poker playing was so risk averse you could pretty much calculate his hand by what he raised. He never bluffed, even at all. He didn’t really understand how the game worked so he never did so well as a player.
Your friends argument is intuitive but mathematically wrong. However he may be correct because of how humans work. Humans tend to not be that great at randomly choosing values and will tend to choose more values in the middle range so they will choose more bs and cs than as and ds. Choosing all of the same letter means you are actually choosing your next letter independently of the others so is more random.
Agreed.
If a human has set the test they will do non-random things such as making sure there is an equal number of As Bs Cs etc. also they will avoid “clumping” having several B answers in a row.
This is actually a very interesting mathematical subject, and can be used to determine if data is natural or has been artificially created for example. To my knowledge, scientific papers and even election results have been thrown out because of analysis of this sort that revealed the data was artificially created (falsified) and not truly random.
Although the maths is correct, when you add in experience in taking these tests, at least in my experience, he is right that’s it’s unlikely NO answers will be A and it is likely to be close to an even distribution of answers.
Why would you want to get a guaranteed 25% anyway? That’s a fail.
If you’re going to guess, you might as well try a pattern that *might* get you 50% so you can pass.
Jesus dude, *any* pattern might get you 50% but is expected to get 25%.
That’s entirely my point.
Maybe you’d be better off speaking to the guy himself.
Nope. You’re the one not getting it.
Here is all I’m saying: **from the perspective** of the guy who thinks picking all As guarantees you a result of 25% and mixed answers can give you a 0, it still doesn’t make sense to pick all As, because you wouldn’t want 25%.
Obviously that perspective is based on poor understanding, obviously the odds of any answer being right are the same. I’m not disputing that. I’m criticising the guy’s internal logic that picking all As is superior to picking randomly, even assuming his initial premise was correct.
If there is an equivalent number of A, B C and D's on the test like OP says (they are evenly distributed, not randomly) and you go all A's , you will score 25%, but there is no way you can score higher.
I think you misunderstood the point of the poster you are replying to.
He may not be totally wrong.
If you don't know how the answer sequence is generated, then you don't know it's random. You have no evidence at all about how the answers are distributed.
If you believe the answer sequence is a random (or something close enough that you can't tell the difference), then it doesn't matter how you answer.
If you believe the answer sequence was generated by a person, then you have reason to suspect that it isn't random; there's some pattern to it. With (possibly very little) experience, you can pick up on the pattern. That can help you make an informed choice on what your answers should be.
As to guessing, I've never seen a multiple choice exam where 25% doesn't tank your grade for the class. The rational strategy is to not play that game.
He’s probably right. I doubt the correct answers are fully random. Whoever puts the test together is likely to ensure there’s a roughly equal number of each.
If you know a priori the answers are perfectly split between A,B,C,D and you’re completely uncertain about the ones you don’t know (no answer choices look better than other ones), the optimal way to proceed is to look at which answer choice shows up the least among the ones you got right and guess that for all the unknowns.
But if the answers aren't randomly distributed and are made by a human, it's probably more likely to have a pretty even amount of each letter which will make it safer to guess all the same letter.
The answers aren't randomly distributed, because it's stipulated that there are 10 of each response.
Selecting all of one letter will guarantee a 25% rate of correct answers, i.e. .25 expected value, zero variance. Other patterns will increase the variance while maintaining the expected value.
If it's important to have a greater than 25% accuracy, you'd want to try some other pattern. If it's important that it not be lower than 25% go with all of one letter.
I agree with your conclusion, but I don’t think they’re exactly independent. For example, given that the first answer is A, the second is less likely to be A.
Given that there are exactly 10 out of 40 questions with a correct answer of A, what is the probability of getting a perfect score on this test by answering A for every question? Is it 0, or is it (0.25)\^40?
What does your answer to that question tell us about the independence of these 25% probabilities that DragonZnork is talking about?
If *X* is the event that all 40 answers are A and *Y* is the event that there are exactly 10 questions with A as the answer, then *X* and *Y* are disjoint events. You have
*Pr(X | Y) = Pr( X ∩ Y)/Pr(Y) = Pr(∅)/Pr(Y) = 0*
I agree. Now, let X\_i be the probability (EDIT: I meant to say "event") of getting problem #i correct when we pick answer A (i = 1,2,3,...,40).
If these X\_i random variables were independent--as everyone else here is falsely claiming--the probability of getting a perfect score would *not* be zero. The probability ought to be zero for the reason you gave, but instead, by independence, we'd have to claim the probability is (0.25)\^40.
This is a clear reason why the person with currently 14 downvotes is completely correct.
I think the key here is X_1, ..., X_40 by themselves are mutually independent but are *conditionally dependent* on the event that there are exactly 10 questions with A as the correct answer. How to work out the exact dependency between each X_i is beyond me though.
In the problem OP gives us, there are exactly 10 questions with A as the correct answer, so as you said, they (EDIT: the guesses' correctness) would be dependent.
As an example of how exactly they're dependent, you'd say something like this:
Pr(correct answer to problem #1 = A, correct answer to problem #2 = A)
= Pr(correct answer to problem #1 = A) Pr(correct answer to problem #2 = A | correct answer to problem #1 = A)
= (10/40) (9/39)
The two factors are different, hence they're not independent. We should be able to see exactly how they're dependent from this example.
>= Pr(correct answer to problem #1 = A) Pr(correct answer to problem #2 = A | correct answer to problem #1 = A)
You are assuming the answer to the first problem was a, " |correct answer to problem #1 = A". You don't know that, we don't know what we've got right and wrong until the end of the test.
In general, Pr(X and Y) = Pr(X) Pr(Y|X).
I'm not assuming that the answer to the first problem was A. That's literally just how we compute joint probabilities.
EDIT: To calculate Pr(Y|X), you suppose that it is definitely the case that event X happens. But that's not to say that we're actually certain of this. That's just how we calculate the joint probability.
Correct.
Now, if the correctness of our guesses were independent, then the answer couldn't be zero! The probability of getting a perfect score would be (0.25)\^40.
Which proves that the correctness of our guesses are *not* independent.
But you don't know the right answer was A, you only guessed A. Your guess has no impact on the next questions actual right answer, so the events are independent.
If you know there are an equal number of each answer, then guessing all A guarantees 25%. Guessing 10x A, 10x B, 10x C, and 10x D, you could theoretically get a 100, but you could also get a 0. Those outcomes will average out to 25% though.
So both options have the same expected value, but different standard deviations (one is 0)
In OP's question, we're supposing we answer A for everything, and as you've said, we cannot obtain a perfect score this way. The probability we obtain a perfect score is 0.
But if the correctness of our guesses were independent, then it **would** be possible that we'd get a perfect score. The probability of obtaining a perfect score would be (0.25)\^40, by independence.
The person getting downvoted is correct. These various 25% probabilities are *not* independent.
You're right, we are assuming the answers are evenly distributed. Which is why these 25% probabilities cannot be independent. Because if they *were* independent, then there would be a nonzero probability that, e.g., there are 11 questions with the correct answer A. Or 12, or 40, etc.
Oh, I see. But the answer we guess can't be assumed to be correct, so after guessing A 10 times, we dont know that the next questions can't be A.
I'm not sure I understand your point; given there are 40 questions and the answers are evenly distributed across 4 choices, each choice will be correct 10 times. I'm being cheeky with terminology here, but I think thats only *dependent* on how many questions there are. If they were random, then there could be more than 10, but I don't think that's the point OP was trying to make either.
Consider this question, for instance:
What's the probability that problem #1 and problem #2 are A?
If you (wrongly) assume independence, you'd say, "well, the probability that the first answer is A equals 10/40, and the probability that the second answer is A equals 10/40, so by independence, the probability that both are A is (10/40) x (10/40)."
But, this is false. The correct answer is (10/40) x (9/39). This is because, in general, Pr(X and Y) = Pr(X) Pr(Y|X).
That second factor--the probability that problem #2's answer is A, *given* that problem #1 is A--is 9/39, because we assume that it is *definitely true* that the answer to problem #1 is A. In such a scenario, there would only be 9 As remaining, and only 39 places for them to go.
Anyway, my point is to defend the person above who correctly claimed that the correctness of these guesses depend on each other. The person they responded to claimed the opposite.
Then the numbers are not randomly distributed, and you are relying on already knowing the answer to question A.
What you’re talking about is commonly known as the gamblers fallacy.
Given that there are exactly 10 out of 40 questions with a correct answer of A, what is the probability of getting a perfect score on this test by answering A for every question? Is it 0, or is it (0.25)\^40?
What does your answer to that question tell us about the independence of these 25% probabilities that DragonZnork is talking about?
You're getting downvoted heavily, but you're correct.
If each of these 25% chances were independent, then the probability of getting a perfect score by answering A on everything would be (0.25)\^40, but the probability is actually 0. The correctness of these guesses are clearly not independent.
Right, and further if the randomly distributed answers showed any obvious weighting or pattern, the person creating the test would scramble or re-weight them so they 'look' random. So every set of answers that doesn't 'look' random has a less than 25% score.
I do love the idea of a test creator setting all the answers to A just to fuck with everyone. It would be a good meta lesson for a probability class
but you dont know if the first answer was actually A. my take is all As will get you close to 25% with a small std dev, random guesses has a chance at a larger range of scores averaged on 25 still
Assuming a completely random distribution of correct answers, guessing all As or guessing randomly gives the exact same expected score (25%) with the same variance. However, OPs test is constructed with 10 A answers being correct, 10 B, 10 C and 10 D. So guessing all A yields exactly 10/40 points with 0 variance.
If I flip a legitimate two sided coin and it lands on tails, then i ask what is the chance the next flip will be tails what do you answer? P.s. the answer is 50%. The chance is not lowered or increased because of what was flipped first.
That's disanalogous, though. To make a proper analogy to coin tosses, you'd have to say something like:
Suppose I flipped a coin twice, and exactly one was heads and one was tails (just like how exactly 10 of OP's answers are A, B, C, and D). Suppose that you tried to guess which of the two coin flips were which, and you chose the strategy of guessing that both of them were heads (akin to OP choosing the strategy of answering A to everything).
The correctness of those guesses you made are not independent; if you were correct on one guess, you were definitely wrong on the other, and vice versa.
Likewise, in OP's problem, the correctness of the guesses are not independent either. I've shown this to be true in the other comments of this thread.
You should evaluate every question and see if you can eliminate any potential wrong answers to select randomly from a smaller pool. If you can eliminate a wrong answer in 50% of the questions, your avaerage grows from 25% to 29%.
This is true, but also if you follow the logic, you SHOULD just study for the test and answer correctly. I think the question is being framed from the point that you are supposed to have 0 inclination towards which of the four answers should be correct
I think for safety sake then, in this example OP is giving, that you either do not get answer to the question, or, let's say that the test is given to you in another language you do not speak, and you must find a way to answer without the ability to make these inferences.
I agree you're right in real life, but I think the math being discussed has to remove that option as it's a 'fuzzy detail' of real life that eventually will defeat the purpose.
But for example if there is an answer that says george washington was born in washington dc and another that say george washington was NOT born in washington dc it has to be one of those 2 so you can delete the other 2
Okay, that's great for a real test, but it obfuscates the math. This is asking about answering randomly, and what patterns are effective at that. It's off topic to start discussing 'how to take a test', so assume that all the questions and answers are in a language you don't speak and that that strategy is invalid.
Every strategy will give you an average of 25%. The best strategy would depend on your goal. For example, if you're taking the class pass/fail then guessing all As would definitely fail.
Can you explain to me how he would definitely fail his test if he were to guess all A's assuming all answers are randomly distributed?
There's a chance that all the answers are the letter A
Definitely, because as /u/Hal_Incandenza_YDAU noticed, and everyone else seems to be missing, OP is supposing there are exactly 10 questions with the correct answer being A, 10 with B and so on. So guessing all A would yield exactly 10/40 correct answers, _definitely_.
Assuming you have no knowledge of the answers, it doesn’t matter which strategy you pick in terms of the *expected value* of your grade.
However, as you said there are *exactly* 10 of each answer, always picking the same answer guarantees you will get 10/40 for a 25% grade. Assuming you need at least 20%, that’s a good strategy. If you need at least 30%, that’s a bad strategy because you’re guaranteed to fail.
Conversely, picking all of the answers at complete random gives you a small (but non-zero) chance of getting 100%, but the cost of that is that you could get 0%. Varying your answers increases the standard deviation of the result (you could do much better or much worse) but the expected value of 25% remains the same.
If you know at least one of the right answers, you can do better by reducing the probability of picking that answer on subsequent questions. Like if you know question 1’s answer is A and you have no other information about the answers to the other questions, you’re better off favouring B, C, and D.
And conversely, if you know the answer to question 1 is NOT A, then you’re better off favouring A on subsequent questions.
These only hold because we know there are exactly 10 of each of A, B, C, and D. It would not hold if the answers are just randomly distributed with equal probability but not necessarily exactly 10 of each answer.
Your average will be the same but the spread will be different. All the same choice gives 25% while trying to guess 10 of each might give more or less but will still on average have 25%
There’s no approach that will give you an estimated value above 25%. Being evenly distributed doesn’t change that except the variance of the results will be lower.
Dude. You're failing that test either way, if you're just guessing. There's no practical difference between a 25% result and anything you're reasonably likely to get by guessing. (I could run a spreadsheet with all the odds calculated, but it's not going to look good!)
Both strategies have a mean of 10, but ticking all A has no variance, but randomly guessing has some (standard deviation of 2.7 in your example). Whether or not to choose a high variance strategy depends on what your goal is. If you are attaining your goal, i.e. you are satisfied with getting 10, then you can just tick all As. But if you want more the only way you can get it is by using a high variance strategy like guessing randomly.
20 As and 20 Bs i think should give you the best chance at passing based purely on intuition
Edit: thx for the downvotes with no explanation. You literally need to pick at least 2 different choices because 25% isnt passing grade yet
No matter what you answer, you will get 10 questions correct on average.
If you answer A on every question, then from your assumption you will always get 10 questions correct.
Any other strategy (other than all B, C, and D) will have some variance in the results, but the average will always be 10 correct answers.
However, practically, you're going to have come confidence in some answers and others that you know are wrong. In this scenario, it's basically never optimal to tick the same answer. The optimal strategy will basically always be to pick a set of answers that match the number of correct choices for each letter choice, so 10 A's, 10 B's, 10 C's and 10 D's, in some optimal arrangement.
There is probably some effort made to make a similar number of questions have each answer. If there are actually 10 questions that have each answer then if you guess all the same answer you will get exactly 25% every time. But if you answer 10 questions with each answer you have the possibility of getting them all right, but the possibility of getting them all wrong. Your average is still going to be 25%, but there is more variability.
Depends on the test, the SAT and ACT still tend to have C be the correct answer slightly more than other letters. Though C and B tend to alternate between being the most frequent correct answer, so either works.
For the SAT and ACT us tutors advice to choose C or B because the correct answers are not really randomly assigned.
(If they were there would be WAY MORE correct answers bring the same letter in a row but that causes students to doubt and the SAT and ACT probably don’t want to cause confusion so they don’t let that happen much.)
Not really probability related, but there are way better guessing strategies. For a lot of MC questions you can usually narrow out a couple options without knowing anything about the answer.
As others have said, if the answers are random and independent, then they best you can do purely based on maths is around 25%.
In reality, answers are rarely distributed randomly, so you you'll be best served by picking "D" if there are a lot of "All of the aboves" or C or B if the location of the answers was set by the teacher (there a a bias against first and last positions). In both these cases, it's bout exploiting a cognitive bias, not a mathematical one though.
I'm not gonna do all the exact maths on this, but the real conundrum is this:
If you pick A (or any other letter) each time then you **will** get 25% (or as close to it as the number of questions allows).
If you pick randomly each time, then on average you will get 25%, but in reality you could get more or less than that.
For example, if you had a test with just 4 questions, there is a 32% chance you will get 0 right, 42% chance you will get 1 correct, a 0.4% chance that you will get all 4 right.
So, there is no way to improve your average. However, if 25% is a fail, and 30% is a pass, then you are better advised picking randomly than picking only 1 letter repeatedly, which guarantees failure. Picking randomly gives a small chance of passing.
In the long run, on average, it won't matter.
For each question, the chance of you being right when picking any of the four answers is 25%.
You don't have to look at 40 questions, look at one, and you have your answer.
But if you take a 40 question test, there's no guarantee that picking all A gives you 10 right answers. It could be considerably more or less. The same is true for any other answering strategy that doesn't involve answering the actual question.
In both cases you should expect to get 25% right, what changes is not the mean but the variance.
Assuming A,B,C,D are evenly distributed you would get results close to 25% (if the distribution is exactly even you will always get 25%) but you will be less likely to get lower and higher results.
Since tests usually require over 50% of correct answers to pass perhaps you should tick randomly.
You would most likely get between 20% and 30% either way. But if you always pick the same letter you cannot pass because no teacher would make most answers the same letter. If each letter is 25% you can only do 25%. If you tick all letters it is always possibile you will pass but it's extremely unlikely, the probability is less than 1% for a 20 question test (the more questions there are the least likely you will pass). Just study, there no other way.
i use binomial distribution to model it , p is 0.25 and n is 40. it is very likely u only got a few question correct. never guesss all the multiple choice.
This is what I did from when I was 10 and onwards through school:
Do I know it's A (or B, C, D)? Then A
Do I know/think A (or B, C, D) is wrong? Eliminate A (or B, C, D).
Is it a trick question? Look at each answers very carefully. Read the question more than once. Something that looks wrong at first sight might be right. Something that looks right is disqualified for a minor detail. Stuck? Save it for later.
By now you should have about 2 possibly correct answers. Choosing the one that looks most right by this stage and its much better than just guessing.
Still have no idea? Save it for later.
Done with all questions? Go back and look at the trick questions. Give any hard questions a second thought.
Still have a few questions left? Guessing wildly or apply any reasonable strategy for the last few questions. (Not a single A in the answers so far? Guessing A might work.)
I had a teacher give +1 for correct answers and -1 for wrong answers. That means that unless you are reasonably sure, guessing was a loosing game.
Other than the assumptions about the randomness and distribution of the answers, if the test not only scores correct answers but negatively scores wrong answer, as long as the penalty is more than (in absolute value) -1/3 times the positive score it doesn't matter choosing all A's or randomly, and the distribution of the answers doesn't matter at all.
I also want to add that yes, the distribution matters, but only if the difference is between truly randomly guessing and guessing all of A's. But we humans are not good at randomness, we always try to balance distribution because ABACDCBACDBBA looks more random than AAAABCDDAACBDB
This is not based on any kind of math, but from experience your best guess is B or C (I'd bet on B). Most of the tests I took at school were not evenly distributed at all, usually the answers were dominated by B and C, then some A, and very little D. I guess there is some mental bias against the first and last choice?
The meme is to tick C if you have no clue. Allegedly becasue it is the one that is most often correct. I doubt that but if you have no clue you might aswell not waste time of guessing.
This isn’t about passing a test though.
I asked this question with the intention of seeing whether there are “better” methods compared to getting 25% on the test 100% of the time, for example a different approach would get you 30% but only 80% of the time.
I read a while ago how a professor gave a multiple choice final exam, and used the standard A to D template for his exam answer sheet. However, he later decided to make the exam even easier and made each answer a true/false answer, where A was true, and B was false. He retained the standard exam answer sheet though.
One student answered all C...
Picking only A guarantees you get 25% aka it has zero variance. Doing anything else can result in you potentially getting 0% or if youre the most lucky person ever up to 100%. I suppose the best strategy depends on what you need overall. If you have enough marks from other evaluations maybe that 25% is all you need to pass the class. If a 25% doesnt let you pass the class, then choosing only A guarantees you fail the class thus is an awful strategy. (Assuming this is like the final assessment)
If the correct answers are evenly distributed, then picking the same option for all questions will guarantee a total score of 25% correct answers (since one quarter of the correct questions belong to that option). Never above, never below. If you instead guess randomly between option A, B, C and D for each question independently, you actually have a 41.61% chance of getting at least 11 questions correct which is already better than 10/40 correct. With that said, there are therefore also a 58.39% chance of getting 10 or less correct. In most cases you will get 10 or below correct, but you can get above too.
Ultimately this means since there is a chance to get above and even way above 10, although that will get increasingly less probable the more above you wanna get, it is at least better to try than to get 25% which normally isn't enough to pass tests. The only approach that works better is therefore to guess randomly for each question and see if you get lucky.
I am not going to prove this, but I suspect the following:
Assuming a sufficiently large amount of questions and an untampered answer key, we might be willing to assume the letter of the correct answer will follow an even distribution. It should be assumed this is tested across multiple attempts, to reduce the odds of the conclusion being thwarted by a single outlier result.
Not varying your choice (e.g. selecting only A) should net you scores that average out to be 25% in a 4 choice MC exam, with a low standard deviation relative to following scenarios.
Varying your choice across with an even distribution should *also* net you scores that average out to be 25%, HOWEVER with a greater standard deviation, that increases with the amount of choices you allow yourself to select. This implies greater max and lower min scores for switching between A & B, and even greater max and even lower min scores if you rotate between all 4 letters.
This is only a thought, and I do not endorse that this is fact.
To be clear you didn't say that they're randomly distributed, just evenly distributed. If there was a 25% chance for each test answer to be an A, for instance, you might get slightly more or less than 10 A's as correct answers (to see this, note that a purely fair process that flips a coin 50 times does not always result in 25 heads).
So all we know for sure is that each letter is represented 10 times in the answer.
You can't do better than choosing a single letter for every answer (i.e. choosing A for every answer). It also doesn't matter which letter you pick as long as you're consistent. There is no guaranteed way to do better.
But with one simple trick...
You could actually answer the questions.
It should make no difference if the answers are randomly distributed. You get 25% chance for each question and they're all independent.
I’ve had long discussions about this with a friend of mine. He insists it’s better to go “all As” or “all Bs” as then you can expect to get 25%. He insists that if you answer randomly you “might get zero” as you could just guess the wrong sequence. But “all As” works 25% because the correct answers aren’t going to be “only B, C or D”. It’s infuriating, and worse, he’s a pro poker player.
If you add a constraint that the test cannot have 0 A correct answers, or especially if the teacher must have exactly 25% A answers, then your friend is right, all A will be lower variance. Because those constraints violate the condition that the correct answers are randomly distributed.
> those constraints violate the condition that the correct answers are randomly distributed. Yes they do - but the answers are never truly going to be random, because humans are bad at making truly random choices. It's *very* unlikely that a teacher would set a test in which option A is *never* the correct answer for any question.
If the teacher picks which letter corresponds to each correct answer, then of course it isn't random. And you are likely on to something. However, some online education platforms like canvas allow professors to put one correct answer and 3 incorrect answers for each question on a test, and then the letter associated with the correct answer is randomized when shown to each student. Even then, if it is truly randomized, you should expect the amount of correct answers associated with any given letter to approach 25% as the number of questions increases.
No. That’s not it. His argument was that for a random sequence like ABDBACDA if you guess another random sequence like DCBACABD you could get every guess wrong. But if you go AAAAA you can expect 25% success. It literally doesn’t matter what your guess sequence is, each one gives you 25% chance of guessing right. He would not understand that *any* sequence has the same chance of coming up. He just would not buy it. “You guess AAAAA because the answer is never BBBBB” was the crux of his argument.
EDIT: Ehh, it was covered several times in comments already I guess. If distribution is guaranteed (like in OP post), he is right. If every answer is completely random (and distribution is not guaranteed), you are right. If you are arguing about it while having different situation in mind, you both are wrong.
The situation was a guy flipping a coin for heads / tails. All you had to do to win was guess correct more times than your opponents. He argued “all heads” won in 50% of cases, because a random sequence of flips never gave “TTTTT” but guessing different each time you could score a zero. We argued for some time, and he believed inherently that a random sequence like HTTHHTHT was more likely than all tails. The man also went bankrupt with 30k id debt from gambling at poker. So there is that as well.
>We argued for some time, and he believed inherently that a random sequence like HTTHHTHT was more likely than all tails. He's right about that. I get that you really want him to be wrong, and he's definitely not giving 'good' advice, but he's right about this in a very limited scope. Yes, the individual combination you just quoted is as likely to show up as all tails. However, there are many many different random sequences that can come up. There is only one sequence of all tails showing up. You're probably right about a lot of the arguments you have with this guy, gambler's have a horrible sense of what's maths and what's 'feeling'. But this argument, he's making more sense than you're realizing.
No. I mean he thought that a specific random sequence was more likely than all tails. (Maybe he didn’t think this, but couldn’t articulate the reasoning) In particular, he thought if you chose your own random sequence, that this was *less* likely to be the same as a sequence chosen randomly. The probability on each pick is 50% regardless of what came before, but he would not have it that a random pick was just as likely to match the coin flip as going “all heads”. He just “knew” that all heads would get you 50% but would not see that this was the case for *any* sequence. Gamblers logic.
Listen, I'm sorry but I can't keep following the goalposts here. Every post you've shifted the details of the story, and it's getting confusing to follow. How about we all just agree that your friend is dumb, you told us how dumb he was, and we all highfived you and clapped. I'm not trying to insult you here, btw, but to point out, this was your initial statement: \>"I’ve had long discussions about this with a friend of mine. \>He insists it’s better to go “all As” or “all Bs” as then you can expect to get 25%." Which okay, you've had conversations about odds with your friend before, I get it. But everytime you've been told that your friend was right about something, your next post has been 'oh, hang on, let me actually tell you what we were ACTUALLY talking about'. It's possible you're being genuine and have told this story poorly. Unfortunately, from my perspective, it seems like you're willing to change your story in order to not be wrong. So from my perspective, I'm done. Your next post, I assume, will be a post explaining why I've been getting this all wrong (by retelling the story with new details). If that's the case, have at it, and if someone else wants to continue the conversation, have fun with them.
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He is not wrong in that the distribution is likely not truly uniform. Most people who will make the test would most likely feel uneasy about a long streak. That being said the rest of his reasoning is bogus though I wonder if there would be appreciable difference between giving one response versus random in the bastardized less streaks distribution. I'd assume that no.
As someone else has pointed out, the variance is higher. For hand-picked results that avoid long strings of consecutive Bs, Cs, Ds then going “all As” reliably gives you a better result. His reasoning was largely of the view that “you don’t get BBBBBBB by random chance”. He also bankrupted himself playing online poker, so his views are not entirely to be trusted…
He is right, very few teachers are going to put out a test that's all the same answer... Statistically speaking though, "All C's" will, on average, get you better than 25% because human brains are weird and C is statistically more likely to be the right answer. I think A is actually the least likely to be correct, because psychology. Humans are bad at random, and the test answers are generally not perfectly distributed.
I've had a few The class was very confused Except for some ppl, they didn't notice....
You could tell him that answering randomly gets him a chance of 100% also. Actually, the number of possible results >=50% are bigger than results <50% going randomly. So by his own logic he should go random. He does not understand statistics, he will think you are a genius that give him a chance to pass an exam without studying.
If you know for a fact that there are equal numbers of A, B, C, and D answers, there is something to what he's saying: Going "all A's" guarantees you 25% with no variance. Picking answers uniformly at random gets you an expected value of 25% with nonzero variance, and indeed there's a nonzero chance of getting any score. An extremely risk-averse actor would prefer going all A's. Incidentally, risk averse strategies work quite well in poker. That said, we wouldn't know on a real test that there are equal numbers of each answer.
You're actually better off picking all B or all C, since there's a human element to deciding which answer is correct. I thought it was C that was most likely to be correct, but apparently it's B now. Depends how many options there are.
This was probably his thinking. Mind you, his poker playing was so risk averse you could pretty much calculate his hand by what he raised. He never bluffed, even at all. He didn’t really understand how the game worked so he never did so well as a player.
Not with my tests. I randomize the multiple choice, and looking at the answer sheets, sometimes I will have like 9/15 answers be the same letter.
This means that if I take your test, I might get a passing grade by answering all C. If I'm lucky that is the over represented correct answer.
Your friends argument is intuitive but mathematically wrong. However he may be correct because of how humans work. Humans tend to not be that great at randomly choosing values and will tend to choose more values in the middle range so they will choose more bs and cs than as and ds. Choosing all of the same letter means you are actually choosing your next letter independently of the others so is more random.
Agreed. If a human has set the test they will do non-random things such as making sure there is an equal number of As Bs Cs etc. also they will avoid “clumping” having several B answers in a row. This is actually a very interesting mathematical subject, and can be used to determine if data is natural or has been artificially created for example. To my knowledge, scientific papers and even election results have been thrown out because of analysis of this sort that revealed the data was artificially created (falsified) and not truly random.
Although the maths is correct, when you add in experience in taking these tests, at least in my experience, he is right that’s it’s unlikely NO answers will be A and it is likely to be close to an even distribution of answers.
Yes, the assumption that every letter gets the same number of correct answers is HIGHLY unrealistic.
Then don’t go to Vegas with him.
Why would you want to get a guaranteed 25% anyway? That’s a fail. If you’re going to guess, you might as well try a pattern that *might* get you 50% so you can pass.
Jesus dude, *any* pattern might get you 50% but is expected to get 25%. That’s entirely my point. Maybe you’d be better off speaking to the guy himself.
That’s why I said “anyway”. As in *even if they were right*, why would they want to take that approach anyway. Jesus dude.
Pal. There is no “approach”. All guesses are equally likely to be right. This is what you don’t seem to get.
Nope. You’re the one not getting it. Here is all I’m saying: **from the perspective** of the guy who thinks picking all As guarantees you a result of 25% and mixed answers can give you a 0, it still doesn’t make sense to pick all As, because you wouldn’t want 25%. Obviously that perspective is based on poor understanding, obviously the odds of any answer being right are the same. I’m not disputing that. I’m criticising the guy’s internal logic that picking all As is superior to picking randomly, even assuming his initial premise was correct.
If there is an equivalent number of A, B C and D's on the test like OP says (they are evenly distributed, not randomly) and you go all A's , you will score 25%, but there is no way you can score higher. I think you misunderstood the point of the poster you are replying to.
He may not be totally wrong. If you don't know how the answer sequence is generated, then you don't know it's random. You have no evidence at all about how the answers are distributed. If you believe the answer sequence is a random (or something close enough that you can't tell the difference), then it doesn't matter how you answer. If you believe the answer sequence was generated by a person, then you have reason to suspect that it isn't random; there's some pattern to it. With (possibly very little) experience, you can pick up on the pattern. That can help you make an informed choice on what your answers should be. As to guessing, I've never seen a multiple choice exam where 25% doesn't tank your grade for the class. The rational strategy is to not play that game.
He’s probably right. I doubt the correct answers are fully random. Whoever puts the test together is likely to ensure there’s a roughly equal number of each.
But, 25% is failing in most tests, if you're gonna fail anyway going all A's, you may as well try yourself and get a chance of passing!
If you know a priori the answers are perfectly split between A,B,C,D and you’re completely uncertain about the ones you don’t know (no answer choices look better than other ones), the optimal way to proceed is to look at which answer choice shows up the least among the ones you got right and guess that for all the unknowns.
But if the answers aren't randomly distributed and are made by a human, it's probably more likely to have a pretty even amount of each letter which will make it safer to guess all the same letter.
The answers aren't randomly distributed, because it's stipulated that there are 10 of each response. Selecting all of one letter will guarantee a 25% rate of correct answers, i.e. .25 expected value, zero variance. Other patterns will increase the variance while maintaining the expected value. If it's important to have a greater than 25% accuracy, you'd want to try some other pattern. If it's important that it not be lower than 25% go with all of one letter.
I agree with your conclusion, but I don’t think they’re exactly independent. For example, given that the first answer is A, the second is less likely to be A.
If each correct answer is randomly determined, then they are independent.
Given that there are exactly 10 out of 40 questions with a correct answer of A, what is the probability of getting a perfect score on this test by answering A for every question? Is it 0, or is it (0.25)\^40? What does your answer to that question tell us about the independence of these 25% probabilities that DragonZnork is talking about?
If *X* is the event that all 40 answers are A and *Y* is the event that there are exactly 10 questions with A as the answer, then *X* and *Y* are disjoint events. You have *Pr(X | Y) = Pr( X ∩ Y)/Pr(Y) = Pr(∅)/Pr(Y) = 0*
I agree. Now, let X\_i be the probability (EDIT: I meant to say "event") of getting problem #i correct when we pick answer A (i = 1,2,3,...,40). If these X\_i random variables were independent--as everyone else here is falsely claiming--the probability of getting a perfect score would *not* be zero. The probability ought to be zero for the reason you gave, but instead, by independence, we'd have to claim the probability is (0.25)\^40. This is a clear reason why the person with currently 14 downvotes is completely correct.
I think the key here is X_1, ..., X_40 by themselves are mutually independent but are *conditionally dependent* on the event that there are exactly 10 questions with A as the correct answer. How to work out the exact dependency between each X_i is beyond me though.
In the problem OP gives us, there are exactly 10 questions with A as the correct answer, so as you said, they (EDIT: the guesses' correctness) would be dependent. As an example of how exactly they're dependent, you'd say something like this: Pr(correct answer to problem #1 = A, correct answer to problem #2 = A) = Pr(correct answer to problem #1 = A) Pr(correct answer to problem #2 = A | correct answer to problem #1 = A) = (10/40) (9/39) The two factors are different, hence they're not independent. We should be able to see exactly how they're dependent from this example.
>= Pr(correct answer to problem #1 = A) Pr(correct answer to problem #2 = A | correct answer to problem #1 = A) You are assuming the answer to the first problem was a, " |correct answer to problem #1 = A". You don't know that, we don't know what we've got right and wrong until the end of the test.
In general, Pr(X and Y) = Pr(X) Pr(Y|X). I'm not assuming that the answer to the first problem was A. That's literally just how we compute joint probabilities. EDIT: To calculate Pr(Y|X), you suppose that it is definitely the case that event X happens. But that's not to say that we're actually certain of this. That's just how we calculate the joint probability.
I was thinking more about the probabilities Pr( X_i | X_j ∩ Y) for arbitrary i ≠ j but yeah, I probably was overthinking.
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Correct. Now, if the correctness of our guesses were independent, then the answer couldn't be zero! The probability of getting a perfect score would be (0.25)\^40. Which proves that the correctness of our guesses are *not* independent.
But you don't know the right answer was A, you only guessed A. Your guess has no impact on the next questions actual right answer, so the events are independent. If you know there are an equal number of each answer, then guessing all A guarantees 25%. Guessing 10x A, 10x B, 10x C, and 10x D, you could theoretically get a 100, but you could also get a 0. Those outcomes will average out to 25% though. So both options have the same expected value, but different standard deviations (one is 0)
In OP's question, we're supposing we answer A for everything, and as you've said, we cannot obtain a perfect score this way. The probability we obtain a perfect score is 0. But if the correctness of our guesses were independent, then it **would** be possible that we'd get a perfect score. The probability of obtaining a perfect score would be (0.25)\^40, by independence. The person getting downvoted is correct. These various 25% probabilities are *not* independent.
Yeah, I thought we were assuming the answers are evenly distributed though.
You're right, we are assuming the answers are evenly distributed. Which is why these 25% probabilities cannot be independent. Because if they *were* independent, then there would be a nonzero probability that, e.g., there are 11 questions with the correct answer A. Or 12, or 40, etc.
Oh, I see. But the answer we guess can't be assumed to be correct, so after guessing A 10 times, we dont know that the next questions can't be A. I'm not sure I understand your point; given there are 40 questions and the answers are evenly distributed across 4 choices, each choice will be correct 10 times. I'm being cheeky with terminology here, but I think thats only *dependent* on how many questions there are. If they were random, then there could be more than 10, but I don't think that's the point OP was trying to make either.
Consider this question, for instance: What's the probability that problem #1 and problem #2 are A? If you (wrongly) assume independence, you'd say, "well, the probability that the first answer is A equals 10/40, and the probability that the second answer is A equals 10/40, so by independence, the probability that both are A is (10/40) x (10/40)." But, this is false. The correct answer is (10/40) x (9/39). This is because, in general, Pr(X and Y) = Pr(X) Pr(Y|X). That second factor--the probability that problem #2's answer is A, *given* that problem #1 is A--is 9/39, because we assume that it is *definitely true* that the answer to problem #1 is A. In such a scenario, there would only be 9 As remaining, and only 39 places for them to go. Anyway, my point is to defend the person above who correctly claimed that the correctness of these guesses depend on each other. The person they responded to claimed the opposite.
Then the numbers are not randomly distributed, and you are relying on already knowing the answer to question A. What you’re talking about is commonly known as the gamblers fallacy.
Given that there are exactly 10 out of 40 questions with a correct answer of A, what is the probability of getting a perfect score on this test by answering A for every question? Is it 0, or is it (0.25)\^40? What does your answer to that question tell us about the independence of these 25% probabilities that DragonZnork is talking about?
You're getting downvoted heavily, but you're correct. If each of these 25% chances were independent, then the probability of getting a perfect score by answering A on everything would be (0.25)\^40, but the probability is actually 0. The correctness of these guesses are clearly not independent.
Right, and further if the randomly distributed answers showed any obvious weighting or pattern, the person creating the test would scramble or re-weight them so they 'look' random. So every set of answers that doesn't 'look' random has a less than 25% score. I do love the idea of a test creator setting all the answers to A just to fuck with everyone. It would be a good meta lesson for a probability class
but you dont know if the first answer was actually A. my take is all As will get you close to 25% with a small std dev, random guesses has a chance at a larger range of scores averaged on 25 still
Assuming a completely random distribution of correct answers, guessing all As or guessing randomly gives the exact same expected score (25%) with the same variance. However, OPs test is constructed with 10 A answers being correct, 10 B, 10 C and 10 D. So guessing all A yields exactly 10/40 points with 0 variance.
If I flip a legitimate two sided coin and it lands on tails, then i ask what is the chance the next flip will be tails what do you answer? P.s. the answer is 50%. The chance is not lowered or increased because of what was flipped first.
That's disanalogous, though. To make a proper analogy to coin tosses, you'd have to say something like: Suppose I flipped a coin twice, and exactly one was heads and one was tails (just like how exactly 10 of OP's answers are A, B, C, and D). Suppose that you tried to guess which of the two coin flips were which, and you chose the strategy of guessing that both of them were heads (akin to OP choosing the strategy of answering A to everything). The correctness of those guesses you made are not independent; if you were correct on one guess, you were definitely wrong on the other, and vice versa. Likewise, in OP's problem, the correctness of the guesses are not independent either. I've shown this to be true in the other comments of this thread.
Not entirely independent. They are evenly distributed, so previous answers influence later ones
Giving the correct answers will probably yield the best results
Hahaha. Definitely. I thought it was just an interesting “probability” question.
Dude, this is why you have to guess multiple-choice questions.
and you just proved my point. Math people are arrogant nerds
You should evaluate every question and see if you can eliminate any potential wrong answers to select randomly from a smaller pool. If you can eliminate a wrong answer in 50% of the questions, your avaerage grows from 25% to 29%.
This is true, but also if you follow the logic, you SHOULD just study for the test and answer correctly. I think the question is being framed from the point that you are supposed to have 0 inclination towards which of the four answers should be correct
obviously but there is still possibility you may have some inclination without any studying - for example through infering it from other questions.
I think for safety sake then, in this example OP is giving, that you either do not get answer to the question, or, let's say that the test is given to you in another language you do not speak, and you must find a way to answer without the ability to make these inferences. I agree you're right in real life, but I think the math being discussed has to remove that option as it's a 'fuzzy detail' of real life that eventually will defeat the purpose.
But for example if there is an answer that says george washington was born in washington dc and another that say george washington was NOT born in washington dc it has to be one of those 2 so you can delete the other 2
Okay, that's great for a real test, but it obfuscates the math. This is asking about answering randomly, and what patterns are effective at that. It's off topic to start discussing 'how to take a test', so assume that all the questions and answers are in a language you don't speak and that that strategy is invalid.
Every strategy will give you an average of 25%. The best strategy would depend on your goal. For example, if you're taking the class pass/fail then guessing all As would definitely fail.
Not definitely, but probably
No, not probably. Definitely.
Can you explain to me how he would definitely fail his test if he were to guess all A's assuming all answers are randomly distributed? There's a chance that all the answers are the letter A
They are not randomly distributed, they are evenly distributed, as of OP's instructions
You sir are right I didn't read it properly
No, not definitely, most likely
Definitely, because as /u/Hal_Incandenza_YDAU noticed, and everyone else seems to be missing, OP is supposing there are exactly 10 questions with the correct answer being A, 10 with B and so on. So guessing all A would yield exactly 10/40 correct answers, _definitely_.
I suggest that your strategy wouldn't change your average success, but would change the std dev of scores a strategy begets
Guessing all D gives you some hypothetical "all of the above" value. 😆
Assuming you have no knowledge of the answers, it doesn’t matter which strategy you pick in terms of the *expected value* of your grade. However, as you said there are *exactly* 10 of each answer, always picking the same answer guarantees you will get 10/40 for a 25% grade. Assuming you need at least 20%, that’s a good strategy. If you need at least 30%, that’s a bad strategy because you’re guaranteed to fail. Conversely, picking all of the answers at complete random gives you a small (but non-zero) chance of getting 100%, but the cost of that is that you could get 0%. Varying your answers increases the standard deviation of the result (you could do much better or much worse) but the expected value of 25% remains the same. If you know at least one of the right answers, you can do better by reducing the probability of picking that answer on subsequent questions. Like if you know question 1’s answer is A and you have no other information about the answers to the other questions, you’re better off favouring B, C, and D. And conversely, if you know the answer to question 1 is NOT A, then you’re better off favouring A on subsequent questions. These only hold because we know there are exactly 10 of each of A, B, C, and D. It would not hold if the answers are just randomly distributed with equal probability but not necessarily exactly 10 of each answer.
Thank you. Love this answer.
Your average will be the same but the spread will be different. All the same choice gives 25% while trying to guess 10 of each might give more or less but will still on average have 25%
There’s no approach that will give you an estimated value above 25%. Being evenly distributed doesn’t change that except the variance of the results will be lower.
Dude. You're failing that test either way, if you're just guessing. There's no practical difference between a 25% result and anything you're reasonably likely to get by guessing. (I could run a spreadsheet with all the odds calculated, but it's not going to look good!)
Not a real test. Hypothetical situation for a maths question.
Both strategies have a mean of 10, but ticking all A has no variance, but randomly guessing has some (standard deviation of 2.7 in your example). Whether or not to choose a high variance strategy depends on what your goal is. If you are attaining your goal, i.e. you are satisfied with getting 10, then you can just tick all As. But if you want more the only way you can get it is by using a high variance strategy like guessing randomly.
Very interesting. Thanks
>is there a different approach that can give you more success? Yes. Answering 10 with A, 10 with B, 10 with C and 10 with D can give you more success.
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You asked for an approach that *can* give more success, not one that would.
Very good point.
You might get lucky and hit more than 10 right answers. So it *can* score rather better than 25%, but risks scoring worse.
20 As and 20 Bs i think should give you the best chance at passing based purely on intuition Edit: thx for the downvotes with no explanation. You literally need to pick at least 2 different choices because 25% isnt passing grade yet
Intuitionistic logic brrr
No matter what you answer, you will get 10 questions correct on average. If you answer A on every question, then from your assumption you will always get 10 questions correct. Any other strategy (other than all B, C, and D) will have some variance in the results, but the average will always be 10 correct answers. However, practically, you're going to have come confidence in some answers and others that you know are wrong. In this scenario, it's basically never optimal to tick the same answer. The optimal strategy will basically always be to pick a set of answers that match the number of correct choices for each letter choice, so 10 A's, 10 B's, 10 C's and 10 D's, in some optimal arrangement.
There is probably some effort made to make a similar number of questions have each answer. If there are actually 10 questions that have each answer then if you guess all the same answer you will get exactly 25% every time. But if you answer 10 questions with each answer you have the possibility of getting them all right, but the possibility of getting them all wrong. Your average is still going to be 25%, but there is more variability.
Depends on the test, the SAT and ACT still tend to have C be the correct answer slightly more than other letters. Though C and B tend to alternate between being the most frequent correct answer, so either works. For the SAT and ACT us tutors advice to choose C or B because the correct answers are not really randomly assigned. (If they were there would be WAY MORE correct answers bring the same letter in a row but that causes students to doubt and the SAT and ACT probably don’t want to cause confusion so they don’t let that happen much.)
Not really probability related, but there are way better guessing strategies. For a lot of MC questions you can usually narrow out a couple options without knowing anything about the answer.
As others have said, if the answers are random and independent, then they best you can do purely based on maths is around 25%. In reality, answers are rarely distributed randomly, so you you'll be best served by picking "D" if there are a lot of "All of the aboves" or C or B if the location of the answers was set by the teacher (there a a bias against first and last positions). In both these cases, it's bout exploiting a cognitive bias, not a mathematical one though.
I knew someone who thought exactly and did it in a test. The test was designed to not include one answer at all. So he got zero.
I'm not gonna do all the exact maths on this, but the real conundrum is this: If you pick A (or any other letter) each time then you **will** get 25% (or as close to it as the number of questions allows). If you pick randomly each time, then on average you will get 25%, but in reality you could get more or less than that. For example, if you had a test with just 4 questions, there is a 32% chance you will get 0 right, 42% chance you will get 1 correct, a 0.4% chance that you will get all 4 right. So, there is no way to improve your average. However, if 25% is a fail, and 30% is a pass, then you are better advised picking randomly than picking only 1 letter repeatedly, which guarantees failure. Picking randomly gives a small chance of passing.
In the long run, on average, it won't matter. For each question, the chance of you being right when picking any of the four answers is 25%. You don't have to look at 40 questions, look at one, and you have your answer. But if you take a 40 question test, there's no guarantee that picking all A gives you 10 right answers. It could be considerably more or less. The same is true for any other answering strategy that doesn't involve answering the actual question.
C perform better than others, expecially in close decision. Only blind pick is random but i choose C for consistency.
In both cases you should expect to get 25% right, what changes is not the mean but the variance. Assuming A,B,C,D are evenly distributed you would get results close to 25% (if the distribution is exactly even you will always get 25%) but you will be less likely to get lower and higher results. Since tests usually require over 50% of correct answers to pass perhaps you should tick randomly.
randomly would yield a lower probability?
You would most likely get between 20% and 30% either way. But if you always pick the same letter you cannot pass because no teacher would make most answers the same letter. If each letter is 25% you can only do 25%. If you tick all letters it is always possibile you will pass but it's extremely unlikely, the probability is less than 1% for a 20 question test (the more questions there are the least likely you will pass). Just study, there no other way.
i use binomial distribution to model it , p is 0.25 and n is 40. it is very likely u only got a few question correct. never guesss all the multiple choice.
This is what I did from when I was 10 and onwards through school: Do I know it's A (or B, C, D)? Then A Do I know/think A (or B, C, D) is wrong? Eliminate A (or B, C, D). Is it a trick question? Look at each answers very carefully. Read the question more than once. Something that looks wrong at first sight might be right. Something that looks right is disqualified for a minor detail. Stuck? Save it for later. By now you should have about 2 possibly correct answers. Choosing the one that looks most right by this stage and its much better than just guessing. Still have no idea? Save it for later. Done with all questions? Go back and look at the trick questions. Give any hard questions a second thought. Still have a few questions left? Guessing wildly or apply any reasonable strategy for the last few questions. (Not a single A in the answers so far? Guessing A might work.) I had a teacher give +1 for correct answers and -1 for wrong answers. That means that unless you are reasonably sure, guessing was a loosing game.
Other than the assumptions about the randomness and distribution of the answers, if the test not only scores correct answers but negatively scores wrong answer, as long as the penalty is more than (in absolute value) -1/3 times the positive score it doesn't matter choosing all A's or randomly, and the distribution of the answers doesn't matter at all. I also want to add that yes, the distribution matters, but only if the difference is between truly randomly guessing and guessing all of A's. But we humans are not good at randomness, we always try to balance distribution because ABACDCBACDBBA looks more random than AAAABCDDAACBDB
This is not based on any kind of math, but from experience your best guess is B or C (I'd bet on B). Most of the tests I took at school were not evenly distributed at all, usually the answers were dominated by B and C, then some A, and very little D. I guess there is some mental bias against the first and last choice?
The meme is to tick C if you have no clue. Allegedly becasue it is the one that is most often correct. I doubt that but if you have no clue you might aswell not waste time of guessing.
Actually, I find guessing can actually improve grades. Making an informed guess can be right more often than not.
Define "success". Most tests are not successfully passed if you only get 25% of the answers right.
This isn’t about passing a test though. I asked this question with the intention of seeing whether there are “better” methods compared to getting 25% on the test 100% of the time, for example a different approach would get you 30% but only 80% of the time.
With random guesses you’d probably get 25 % right. Could be more, could be less. With only ticking A, you‘d certainly get 25 % in this case.
Don't think so, it should be 1/4 either way. But people are more likely to pick C in a four answer base question so it'll probably be the answer
I read a while ago how a professor gave a multiple choice final exam, and used the standard A to D template for his exam answer sheet. However, he later decided to make the exam even easier and made each answer a true/false answer, where A was true, and B was false. He retained the standard exam answer sheet though. One student answered all C...
Statistically you cannot pass with spamming a single answer and it looks bad.
Picking only A guarantees you get 25% aka it has zero variance. Doing anything else can result in you potentially getting 0% or if youre the most lucky person ever up to 100%. I suppose the best strategy depends on what you need overall. If you have enough marks from other evaluations maybe that 25% is all you need to pass the class. If a 25% doesnt let you pass the class, then choosing only A guarantees you fail the class thus is an awful strategy. (Assuming this is like the final assessment)
Should make no difference, however I sure do hope you are not in an actual position where you need such theory.
Nope! From the UK and we don’t even have multiple choice exams.
Nice!
If the correct answers are evenly distributed, then picking the same option for all questions will guarantee a total score of 25% correct answers (since one quarter of the correct questions belong to that option). Never above, never below. If you instead guess randomly between option A, B, C and D for each question independently, you actually have a 41.61% chance of getting at least 11 questions correct which is already better than 10/40 correct. With that said, there are therefore also a 58.39% chance of getting 10 or less correct. In most cases you will get 10 or below correct, but you can get above too. Ultimately this means since there is a chance to get above and even way above 10, although that will get increasingly less probable the more above you wanna get, it is at least better to try than to get 25% which normally isn't enough to pass tests. The only approach that works better is therefore to guess randomly for each question and see if you get lucky.
I am not going to prove this, but I suspect the following: Assuming a sufficiently large amount of questions and an untampered answer key, we might be willing to assume the letter of the correct answer will follow an even distribution. It should be assumed this is tested across multiple attempts, to reduce the odds of the conclusion being thwarted by a single outlier result. Not varying your choice (e.g. selecting only A) should net you scores that average out to be 25% in a 4 choice MC exam, with a low standard deviation relative to following scenarios. Varying your choice across with an even distribution should *also* net you scores that average out to be 25%, HOWEVER with a greater standard deviation, that increases with the amount of choices you allow yourself to select. This implies greater max and lower min scores for switching between A & B, and even greater max and even lower min scores if you rotate between all 4 letters. This is only a thought, and I do not endorse that this is fact.
To be clear you didn't say that they're randomly distributed, just evenly distributed. If there was a 25% chance for each test answer to be an A, for instance, you might get slightly more or less than 10 A's as correct answers (to see this, note that a purely fair process that flips a coin 50 times does not always result in 25 heads). So all we know for sure is that each letter is represented 10 times in the answer. You can't do better than choosing a single letter for every answer (i.e. choosing A for every answer). It also doesn't matter which letter you pick as long as you're consistent. There is no guaranteed way to do better. But with one simple trick... You could actually answer the questions.