# Differentiate between the numbers from 1 to 5 with one single yes/no question

This question requires a bit of logical distortion to get the information required to answer the question.

However, I've never seen it asked before what would happen if more modes of question unanswerability were introduced.

I'm thinking of a number from 1 to 5, and you're allowed to ask one yes-no question to which I can answer one of the following things:

• Yes
• No
• Sometimes (Both yes and no are known to be true with a nonzero frequency.)
• I don't know (There is a definite answer, but it is not known.)
• Not applicable (This question does not have an answer for my number.)

What question can you ask to determine what number I'm thinking of?

There may be many correct answers to this question, but try to make it as elegant as possible.

• Can you give an example of a question that would be answered with "not applicable" (that could plausibly be useful)? For example, "Is your number an orange?" would be responded to with "not applicable", but isn't applicable to one number but not another (which is required to make it useful). – WendiKidd May 22 '14 at 22:25
• "Is the square root of (your number - 2) greater than 1.5?" would be "not applicable" to 1 since negative numbers don't have square roots, at least not ones that can be compared in any order. – Joe Z. May 22 '14 at 22:29
• Also that works for negative numbers, but the question is referring to positive 1, 2, 3, 4, and 5, right? – WendiKidd May 22 '14 at 22:58
• "Sometimes" and "I don't know" can be interpreted to mean somewhat the same thing, no? If not, how do you differentiate between them? (Other than this point, I have an answer.) – Aza May 23 '14 at 1:18
• @arshajii A clever idea, might make a good puzzle answer, but I don't think it works here, because he said the question must be a yes/no question, which is normally understood to mean a question in the form "is it ...?" or "does it ...?" – Jay May 29 '15 at 15:44

A slightly simpler possibility:

I have a number in mind: it’s either 14 or 15, but I’m not saying which. If you multiply together your number and its smallest prime factor, then roll an ordinary 6-sided die and add the result, will the total be at least as big as than the number I’m thinking of?

1. n/a (since 1 has no prime factors)
2. No. (2*2=4; the maximum achievable total is 10.)
3. Sometimes. (3*3=9; so if you roll a six, it’s a “yes”, if you roll a four, it’ll be “no”).
4. Don’t know. (4*2=8; so if I’m thinking of 15, it’ll be “no”, but if I’m thinking of 14, there’s a chance of “yes”.)
5. Yes. (5*5=25.)
• This is very nice +1, but isn't the question about discerning between 5 numbers using 1 question with 5 responses? – d'alar'cop Oct 26 '14 at 4:55
• That's exactly what it does, though. – Joe Z. Oct 28 '14 at 1:03

I'm thinking of a number, $s$, which equals either $0$ or $1$ with equal probability. If $s=1$, then $p$ is either $0$ or $1$ with an unknown probability distribution - otherwise $p=0$.

If your number is $n$, where $n\neq 4$, is the $\displaystyle\lim_{x\to n}\frac{1}{x-2-s-p}\geq0$?

• If $x=1$, then the denominator is either $-1$, $-2$, or $-3$, for which the answer is no.
• If $x=2$, then the denominator is either $0$, $-1$, or $-2$, for which the answer is sometimes. The $0$ case - the only one that matters - occurs with known probability of $50\%$.
• If $x=3$, then the denominator is $1$ if $s=0$ (known probability), $0$ if $s=1$ and $p=0$, or $-1$ if $s=1$ and $p=1$. Since the last two cases where $p$ changes with unknown probability produce a different result, the answer is I don't know.
• If $x=4$, then question does not apply, so the answer is not applicable.
• If $x=5$, then the denominator may be $1$, $2$, or $3$, for which the answer is yes.
• You can rephrase the "unknown probability distribution" part to "I'm thinking of another integer p that's either 0 or 1." – Joe Z. May 23 '14 at 2:16
• A logician would reply yes if thinking of 4. – Peter Taylor May 23 '14 at 8:42
• @Emrakul I can explain, the logical implication $p\rightarrow q$ is true iff $\lnot p \lor q$. If your number is $4$, that makes $n\neq 4$ not true, which makes the combined statement true. Saying "$\text{where} \; n\neq 4$" may remove this problem. – SQB May 23 '14 at 14:05
• (There was a chat record here about this, which I can't seem to find anymore. In short, I am surprised but accept the explanation) – Aza May 26 '14 at 15:56
• You realize a lot of people are gong to answer "I don't know", regardless, yes? – imallett Feb 8 '15 at 20:13

This isn't a complete solution and probably may not even be completely right but I'm posting anyway since I found the partial solution quite simple and the Author stressed on elegance. Also since an answer has long been accepted. I'm hoping others could add to this answer to make it complete.

The Question can be

Is One subtracted from the Number you are thinking of prime?

1 - 1 = 0 (Not Applicable, 0 is not a Natural Number)

2 - 1 = 1 (I Don't know, Controversial Answer! But neither prime nor composite!)

3 - 1 = 2 (Yes)

4 - 1 = 3 (Yes)

5 - 1 = 4 (No)

Now, the only problem is for 2 & 3. "Sometimes" means the question has to do with probability or a random physical event.

Will edit my answer if I can complete it!