Althouse: "If you choose an answer to this question at random..."

August 21, 2014

"If you choose an answer to this question at random..."

"... what is the chance you will be correct?"

70 comments:

Rae said...: Humans don't make good random number generators, as any computer scientist or security expert will attest.; 8/21/14, 9:15 AM
Ignorance is Bliss said...: Could someone post the full text of the question? The link is blocked for me.; 8/21/14, 9:17 AM
SteveBrooklineMA said...: Zero?; 8/21/14, 9:18 AM
William said...: In real life we are constantly faced with multiple choice decisions, and the answer is always 100% or 0%. There are some people who can do no wrong, and others who can do no right. I will leave vague and unspecified who are the righteous and who are the incorrigibly wrong, because we all know who they are.; 8/21/14, 9:22 AM
Ann Althouse said...: I changed the link. It should work for everyone now.; 8/21/14, 9:24 AM
Ann Althouse said...: You need to go to the link. Or I'll put the picture up.

This is only a good post if you see the multiple choice!; 8/21/14, 9:24 AM
Ignorance is Bliss said...: Thanks for the picture. Based on this set of choices, Steve is correct, the answer is zero. ( assuming that choosing at random implies a uniform distribution of the choices. It is actually possible to assign different probabilities to the different choices and come up with correct results. ); 8/21/14, 9:30 AM
Ignorance is Bliss said...: The question gets more difficult if you replace option C with 0.; 8/21/14, 9:31 AM
David said...: One in three I suppose. Though the phrasing of the question is so strange that you could say "unknowable."; 8/21/14, 9:39 AM
JPS said...: Strange one. A bit off-topic, it reminds me of a professor emeritus who back in the day gave multiple-choice exams. His class averages were consistently below zero.

He did the usual normalization: point for each correct answer, fractional point off for incorrect answers, so that if you guessed randomly throughout, your score should be zero.

But every question had a little twist such that unless you had an excellent understanding, you were more likely to arrive at the wrong answer than the right one. Thus a few people scored well into the positive range. Most did not.

I don't approve, but on some level I admire it.; 8/21/14, 9:48 AM
traditionalguy said...: This is a good example of where knowing more about the subject than the question maker will ruin you.; 8/21/14, 9:52 AM
Big Mike said...: Given numbers 0.25, 0.25, 0.5 and 0.6, the probability of guessing correctly is 0.5. However, assuming this is your standard “fill in the bubble” multiple choice test, only one answer can be correct, meaning A and D count as different answers. Therefore the answer would be A or D, not A and D, because the probability of picking the right answer from four possible answers truly at random is 0.25.

But! If the test was designed properly, then there can be only one correct answer, and consequently the duplicate number must be wrong. So if we are allowed an “educated” random answer, we can immediately eliminate A and D as possibilities. Between B and C the probability of guessing correctly returns to 0.5.

Which, of course, a true mathematician would pick with probability 1.0.; 8/21/14, 9:58 AM
Birkel said...: I'm waiting around until either The Crack Emcee or "garage mahal" swings by to enlighten us.; 8/21/14, 10:05 AM
Tank said...: I can't answer this because I don't understand the question.; 8/21/14, 10:05 AM
Bill, Republic of Texas said...: Here's an uneducated redneck's guess at the answer.

Zero chance of correct answer is not listed.

Now I'll go and read the comments and the article.; 8/21/14, 10:08 AM
Edmund said...: Perhaps we could conjure up the ghosts of Pascal and Gödel and have them debate it. Schrödinger's cat can moderate.; 8/21/14, 10:10 AM
Bill, Republic of Texas said...: OK. I still don't understand.; 8/21/14, 10:11 AM
Ignorance is Bliss said...: Schrödinger's cat can moderate.

Or not.; 8/21/14, 10:12 AM
traditionalguy said...: Ocam's razor says to chose the first answer, but be careful not to cut yourself.; 8/21/14, 10:19 AM
I'm Full of Soup said...: That's good.; 8/21/14, 10:20 AM
David said...: Damn well better give us the supposed answer at some point, professor. We can grade ourselves. It's pass-fail, with probably a lot more failures than usual.

I still say one in three.; 8/21/14, 10:26 AM
Unknown said...: If you write the choices on ping pong balls and draw them out of a hat you'll get 25% half the time. The other two will each show up a quarter of the time.

25% can then say "I won" and the other two have to do whatever 25% says or they're racists.; 8/21/14, 10:34 AM
Anonymous said...: No correct answer to this question.; 8/21/14, 10:54 AM
Mary Beth said...: It was my understanding that there would be no math.; 8/21/14, 11:03 AM
clint said...: 0%.

None of the listed answers are correct, so choosing from them at random cannot produce the correct answer.

(And Ignorance is Bliss is right. If C were "0%", then that wouldn't be the correct answer anymore.)

Aren't recursive questions fun!; 8/21/14, 11:15 AM
Fernandinande said...: SteveBrooklineMA said...
Zero?

The "problem" is to choose an answer "at random", which doesn't necessarily imply a rigorous generation of a random number, so 100%.; 8/21/14, 11:23 AM
Jonathan Card said...: Oh, this would be a great question to follow with "Defend:" and grade the essay.; 8/21/14, 11:26 AM
Fernandinande said...: If you choose an answer to this question at random, what is the chance you will be correct?
A - One Fish
B - Two Fish
C - Red Fish
D - A; 8/21/14, 11:30 AM
Tank said...: Fernandinande said...
SteveBrooklineMA said...

Zero?

The "problem" is to choose an answer "at random", which doesn't necessarily imply a rigorous generation of a random number, so 100%.

I like this answer.; 8/21/14, 11:30 AM
Sigivald said...: JPS said: But every question had a little twist such that unless you had an excellent understanding, you were more likely to arrive at the wrong answer than the right one. Thus a few people scored well into the positive range. Most did not.

This is horrible in the real world - because it's hard, as a test taker, especially the first time, to tell it from poor and ambiguous wording.

Back in my college days I occasionally wrote somewhat longer than normal answers to such questions explaining my answers to the various ways the poorly-described question could be taken, or at least explaining how I'd been stuck interpreting it.

Properly designed questions on an exam are absolutely unambiguous.; 8/21/14, 11:40 AM
Anonymous said...: The question has the same logical structure as the old "If God is omnipotent, can He make a rock so heavy even He can't lift it?" We learn nothing about omnipotence, God, rocks, "at random," or probability from these questions. We're merely reminded that the ability to form a grammatically correct utterance is insufficient to imbue it with meaning.

Incidentally, people who enjoy this sort of logical-impossibility puzzle will enjoy much of Raymond Smullyan's work.; 8/21/14, 11:48 AM
rehajm said...: When this thing first appeared it drove a fairly large group of us nuts. Brain hurts, heated arguments, stopped speaking to each other, the works. I'm now at peace with the correct answer...

Goes along with the best graph of all time; 8/21/14, 11:53 AM
MaxedOutMama said...: This comment has been removed by the author.; 8/21/14, 12:05 PM
Paul said...: B.; 8/21/14, 12:10 PM
FullMoon said...: 1 out of 4 is 25%. 50% of the possible answers are"25%".

Close your eyes and point. 50% chance of pointing at correct answer.; 8/21/14, 12:14 PM
sane_voter said...: the correct answer is 1/3 or 33.3%. So none of the above.; 8/21/14, 12:17 PM
Fernandinande said...: Here's a math teaser rather than a semantics teaser:

"I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"

The answer is not 50%.; 8/21/14, 12:49 PM
Anonymous said...: That's a clever version of the Spanish Barber that I haven't seen before!; 8/21/14, 1:44 PM
Left Bank of the Charles said...: It's fun to read as people try to reason their way through this.

Here's the truth. Both the question and the answer have been poisoned with iocane powder.; 8/21/14, 1:50 PM
Xmas said...: I think that's a question in "Godel, Escher, Bach".; 8/21/14, 2:54 PM
Yancey Ward said...: If there were only 3 selections 25%, 50%, and 60%, you would select "correctly" 1/3 of the time. While the addition of answer (d), which is the same as (a), seems to make it more likely you would choose 25%, I would argue the odds are still 1/3 unless you couldn't see the answers themselves, but then does it matter? Maybe not. In that last case, you will select (a) and (d) half the time and select (b) and (c) the other half of the time. In that case....

33% * 50% = 16.6 and
66% * 50% = 33.3

Or, it is still twice as likely the answer is b or c than it is a and d.

Of course, I am totally ignoring the semantic problems caused by the question's structure- I am assuming the answer to the question isn't described by any of the answers given below, but that those answers are to some other question.; 8/21/14, 3:35 PM
BJK said...: "I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"

The answer is not 50%.

The answer at Fernandinande's link is inherently wrong, because it assigns a value to the order of events, eliminating the double counting.

If you analyze it as pairs, without respect to order, there are fourteen couplings which satisfy the terms of the given information:
{(B-T, B-Su), (B-T, B-M), (B-T, B-T), (B-T, B-W), (B-T, B-R), (B-T, B-F), (B-T, B-Sa), (B-T, G-Su), (B-T, G-M), (B-T, G-T), (B-T, G-W), (B-T, G-R), (B-T, G-F), (B-T, G-Sa)}

All pairs are of equal probability. 7 of the 14 pairs involve a boy as the other child, and 7 of the 14 pairs involve a girl. The answer is 50%.

The author's problem is that his model fails to account for double counting, since there are two equally probable events which would produce (B-T, B-T).
- If the parent providing the information referenced the first child; and
- if the parent providing the information referenced the second child.

---

As to the original question, the problem is the question is ambiguous. If there are 4 possible answers, and we assume only one of them is right, you have a 25% chance of being correct. Since 2 of the possible answers are the 'correct' answer, your probability increases to 2 in 4 (50%)...but if 50% becomes the 'correct' answer, only one of the 4 (25%) is 'correct,' as only one of the choices reads 50%.

As written, the question is unanswerable; if you assume there can be only one 'correct' answer, then the answer is "B) 50%." (...as then the answer cannot be 'A' or 'D' individually, so your choice is between 'B' or 'C'). The question does not provide enough information to make that assumption, however.; 8/21/14, 3:46 PM
wildswan said...: Thank you, BJK. My mind was beginning to vibrate and make a strange high pitched whine.; 8/21/14, 4:48 PM
Ann Althouse said...: The correct answer is 50% but at the instant that becomes true, it is untrue, and the answer is 25%, but there are two answers for that, and we're back to 50% being correct again, and suddenly, it's wrong again. Repeat forever.; 8/21/14, 5:17 PM
Ann Althouse said...: Do I get the math prize or the literature prize?; 8/21/14, 5:18 PM
Fritz said...: Just because a question can be asked doesn't mean it has an answer.; 8/21/14, 5:48 PM
Pat Moffitt said...: We are never given the question so we can't answer; 8/21/14, 5:50 PM
The Godfather said...: "Examinations are formidable, because the greatest fool can ask what the wisest man cannot answer."; 8/21/14, 6:13 PM
Unknown said...: This sentense has three erors.; 8/21/14, 7:14 PM
traditionalguy said...: The Professor should get 50% of the Math prize and 50% of the Literature prize, but she is disqualified for using excessive IQ. Unlike El Rushbo, she will not tie half of her IQ behind her back.; 8/21/14, 9:02 PM
SeanF said...: Ann Althouse: The correct answer is 50% but at the instant that becomes true, it is untrue, and the answer is 25%, but there are two answers for that, and we're back to 50% being correct again, and suddenly, it's wrong again. Repeat forever.

The correct answer, as others have already said, is 0%. Change option C from 60% to 0% and then there is no answer.; 8/21/14, 9:22 PM
SeanF said...: BJK: The answer at Fernandinande's link is inherently wrong, because it assigns a value to the order of events, eliminating the double counting.

The answer at Fernandinande's link is inherently correct because it assigns a value to the order of events.

(We discussed this earlier on this blog, on a post regarding the probability of consecutive births at a hospital being all the same gender.)

If you have two kids, you are twice as likely to have a boy and a girl as you are to have two boys. When calculating probability involving children, you must take that into account.

If the question were simply, "I have two children and at least one of them is a boy - what are the odds the other is also a boy?" The answer is 1/3, because the possibilities are B-B, B-G, and G-B. If you avoid the order of events, to eliminate the "double-counting," you get the wrong answer.; 8/21/14, 9:36 PM
Gabriel Hanna said...: The correct answer is 0%.

It works like this:

You can choose 1 of 4 answers, but two correspond to the same number, 25%.

Your probability of picking 25% is 50%, your probability of picking 50% is 25%, and your probability of picking 60% is 25%.

So if we can identify which of these numbers is equal to its probability, we know the probability of picking the correct answer.

25% is not equal to 50%
50% is not equal to 25%
60% is not equal to 25%

That means we have a 0% probability of choosing the correct answer, which is 0%.

0% = 0%. QED. We have consistently applied the same standard throughout.; 8/21/14, 10:25 PM
Gabriel said...: Let's suppose that the 25% answer was not duplicated. If we apply the same logic I did we should get the correct answer.

If 25% appears only once, then we have a 25% probability of picking it. 25 = 25.

If 25% does not appear in the answers at all, we have a 0% chance of picking it, and since 25 is not equal to 0 25% cannot be the correct answer.

The logic is the same in these two obvious cases as it is in this counterintuitive one where 25% appears more than once.; 8/21/14, 10:31 PM
Smilin' Jack said...: Ann Althouse said...
Do I get the math prize or the literature prize?

How about a dunce cap? As others have pointed out at length, the correct answer is simple: 0.; 8/22/14, 12:55 AM
Anonymous said...: "How about a dunce cap? As others have pointed out at length, the correct answer is simple: 0"

And since "0" is not one of the choices therefore it is not the correct answer.
Dunce cap for you jack (I'm not surprised).

To repeat; "No correct answer to this question"; 8/22/14, 1:39 AM
SeanF said...: Phil D: And since "0" is not one of the choices therefore it is not the correct answer.

Consider the following:

"What is 2+2?

A) 5
B) 1
C) 7
D) 3"

The correct answer is not listed as a choice, but the correct answer is still 4, is it not?; 8/22/14, 6:06 AM
Anonymous said...: "If you choose an answer ..."
The question as posed would imply that the answer must be a member of the collection {A, B, C , D}.
Since the answer "0" isn't part of that collection therefore it is not the answer.

Mind, if you you disagree with my "implied" than we can start going around in circles. The answer is "0", but did you "choose" that answer "randomly"?
This sort of thing makes my head hurts.

So, in other words, the question is nonsense. that stops the merry-go-round.; 8/22/14, 7:29 AM
SeanF said...: Phil: "If you choose an answer ..."
The question as posed would imply that the answer must be a member of the collection {A, B, C , D}.
Since the answer "0" isn't part of that collection therefore it is not the answer.

Some answers are correct answers, and some answers are incorrect answers, but they are all answers.; 8/22/14, 8:12 AM
Gabriel said...: @PhilD

"If you choose an answer at random" does not imply that one answer must be correct, just like "If you owned a unicorn, would you keep it in the backyard?" doesn't imply that unicorns exist.

If you choose A, B, C, or D the probability that you have chosen the correct answer to the question is 0%. There's no contradiction. The question does not assert that one of A, B, C, or D is the correct answer, and the question is not a multiple choice question. It is a short-answer question about probability.; 8/22/14, 8:13 AM
Anonymous said...: "This question"
"What is the chance that you will be correct".

So a question about a question. But if you split this in 2 questions then "this question" is no longer defined. So if I say "papperdelap" then I have a more than 0% probability that I'm correct.

Or, I can indeed choose any answer, but if I choose "0" and "0" is correct, than the probability of me answering correctly was more than "0", so foiled again.

Like I said, these things give me a headache.; 8/22/14, 8:38 AM
Anonymous said...: Godel;

"The basic idea at the heart of the incompleteness theorem is rather simple. Gödel essentially constructed a formula that claims that it is unprovable in a given formal system. If it were provable, it would be false, which contradicts the idea that in a consistent system, provable statements are always true"

My contention is that this question is not consistent. But maybe I'm overthinking (or underthinking) this.; 8/22/14, 8:44 AM
BJK said...: @SeanF

If you have two kids, you are twice as likely to have a boy and a girl as you are to have two boys. When calculating probability involving children, you must take that into account.

Having the knowledge that only one of the children is identified removes a variety of data sets from the potential outcomes. In attempting to identify the possible pairings which could produce this outcome, the author disregards that there are two possible outcomes which produce the same result. Look at it this way, with the child identified by the parent in bold:

{(B-T, B-Su), (B-T, B-M), (B-T, B-T), (B-T, B-W), (B-T, B-R), (B-T, B-F), (B-T, B-Sa), (B-T, G-Su), (B-T, G-M), (B-T, G-T), (B-T, G-W), (B-T, G-R), (B-T, G-F), (B-T, G-Sa), (G-Su, B-T), (G-M, B-T), (G-T, B-T), (G-W, B-T), (G-R, B-T), (G-F, B-T), (G-Sa, B-T), (B-Su, B-T), (B-M, B-T), (B-T, B-T), (B-W, B-T), (B-R, B-T), (B-F, B-T), (B-Sa, B-T)}

There is your full 28 pairs, doubling the number from the previous example to take order into account. Because we do not know which child the parent described, it is equally probable that the parent referred to one boy rather than the other; they are separate solutions which match the given information.

Out of that 28, there are 14 possible solutions in which the non-identified child is a boy, and another 14 where the non-identified child is a girl.

The odds involving consecutive children of the same gender are different, because there is chance on the first one, and chance on the second. There is no chance with respect to the gender of one of the children identified here: one of the children is always going to be a boy born on a Tuesday, 100% of the time.; 8/22/14, 9:31 AM
Ignorance is Bliss said...: BJK-

I think you are making a mistake by counting the possible ways a situation can be described, rather that the possible ways it can occur.

Consider this:

Say the first child is a B-Tu, what are the odds that the second child will be a G-Su? I get 1 in 14. However, if I count up the odds based on the possibilities that you list, it appears to be 1 in 15. Likewise, the odds of the second child being a B-Tu should be 1 in 14, however, if I count up the cases that you list, I see that possibility twice in the 15 cases.

You are listing the B-Tu, B-Tu case twice, giving it twice the probability, even though it is no more likely than any other pairing.; 8/22/14, 12:09 PM
BJK said...: "Not so fast, says probabilist Yuval Peres of Microsoft Research. That naïve answer of 1/2? In real life, he says, that will usually be the most reasonable one.

Everything depends, he points out, on why I decided to tell you about the Tuesday-birthday-boy. If I specifically selected him because he was a boy born on Tuesday (and if I would have kept quiet had neither of my children qualified), then the 13/27 probability is correct. But if I randomly chose one of my two children to describe and then reported the child’s sex and birthday, and he just happened to be a boy born on Tuesday, then intuition prevails: The probability that the other child will be a boy will indeed be 1/2. The child’s sex and birthday are just information offered after the selection is made, which doesn’t affect the probability in the slightest."

https://www.sciencenews.org/article/when-intuition-and-math-probably-look-wrong

(As cited in the original link.); 8/22/14, 2:30 PM
SeanF said...: BJK: Out of that 28, there are 14 possible solutions in which the non-identified child is a boy, and another 14 where the non-identified child is a girl.

As IisB has already pointed out (and the original link made clear), there are only 27 possible solutions, and only thirteen of them have two boys. B-Tu/B-Tu is one solution, not two.

BJK: "Not so fast, says probabilist Yuval Peres of Microsoft Research. That naïve answer of 1/2? In real life, he says, that will usually be the most reasonable one.

Everything depends, he points out, on why I decided to tell you about the Tuesday-birthday-boy.

I don't believe this is correct either. For two children, there are a total of 196 possible gender/weekday combinations, all of which are equally likely. For any given specific single combination (Boy born on Tuesday, girl born on Friday, etc.), there will be 27 combinations that match it, of which 13 will have the other child of the same gender and 14 will have the other child of the opposite gender.; 8/22/14, 2:56 PM
ken in tx said...: My Ed Psych class was given a series of 20 question tests, back in the 60s, in which each question had possible answers A, B, C, or D. We were not told any questions. I routinely made 80% on these tests. I don't remember what the point that exercise was.; 8/22/14, 4:18 PM
Smilin' Jack said...: "I have two children. One is a boy born on a mmmff-day. What is the probability I have two boys?"

"Sorry, I didn't quite catch the day?"

But before your interlocutor can respond, he's struck dead by a bolt of lightning. What's the probability he had two boys?; 8/23/14, 12:15 AM
Gabriel said...: The more irrelevant information you include in the "two children" problem, the closer the probabilities get to 1/2.

In real life, we have great deal of information about a person we meet who has two children, even if we just met them--just on what we can see and hear--and so the odds are going to be effectively 1/2 even if he tells us about one of them being born on Tuesday or whatever it is. By "effectively" I mean that failing to account for something in your explicit analysis is not going to give you a meaningfully different answer from 1/2.; 8/23/14, 5:16 PM
SeanF said...: Smilin' Jack: What's the probability he had two boys?

Are you going to be surprised that it's not the same answer when you know the day (about 48%), but it's not 50% either? :)

Your hypothetical is no different than merely knowing that there's at least one boy, so the odds of two boys are 33.33333~% (exactly 1/3). There are 196 total possibilities (when we look at both gender and birth weekday). Of those total possibilities, 98 are one boy and one girl, 49 are two boys, and 49 are two girls. "One of each" outnumbers "two boys" by 2-to-1.

Here's the thing - there are 13 cases of "Two boys, with at least one boy born on a Sunday" (1 where both boys were born on Sunday and 12 where the other boy was born on a different day). That is true for all 7 days. Quick math will tell you that 7*13 is 91 - but there are not 91 total two-boy possibilities, there are only 49. This is because all of the "different day" possibilities are included in two groups. If you have one boy born on a Sunday and one boy born on a Monday, you would be included in the "at least one boy born on a Sunday" group and the "at least one boy born on a Monday" group. So going from "At least one boy" to "At least one boy born on a Sunday" does not reduce your "Two boys" possibilities to 1/7th.

There are 14 cases of "One boy and one girl, with the boy born on a Sunday." That is true for all seven days. 14*7 is 98, which is the total number of "One boy and one girl" possibilities. Each of these possibilities only gets included in one group, because there's only one boy in it. So going from "At least one boy" to "At least one boy born on a Sunday" does reduce your "one of each" possibilities to 1/7th.

That's why the odds of two boys when it's simply "At least one boy" (33.33333~%) are different than the odds of two boys when it's "At least one boy born on a Sunday" (about 48%).

The odds are not 50% in the second case because the "two boy" possibilities number 13 while the "one boy and one girl" possibilities number 14.; 8/23/14, 11:45 PM