You have one question to tell whether the number I'm thinking of is 1, 2, or 3

Question

This is an interesting puzzle which was passed to me by a friend some time ago. I do know the answer, but will refrain from self-answering on this to see where it goes.

I'm thinking of a number: 1, 2, or 3. You may ask me one question, which I will answer to the best of my ability. I may not, however, tell you my number or any codified version of my number. I can only answer yes, no, maybe, I don't know, etc.

How do you tell the number I'm thinking of?

Answer 1

If you flip that many coins, will you get two the same?

Because

As Doorknob put it... "1 => no, because it's only one, 2 => maybe, because it could be one of TH, HT, TT, or HH, 3 => yes, because a coin flip has only two possible outcomes"

Answer 2

Would it work if I ask:

"I'm thinking of a number: either 0 or 1. Is the sum of our numbers greater than 2?" I could figure out the answer if that's a question I'm allowed to ask.

If your number is 3, then you would say "yes" because no matter what you add, it would be true.

If your number is 2, you would say "I don't know". If my number was 0, then the answer would be no, but if my number is 1 the answer is yes. So you don't know.

If your number is 1, then you would respond "no". No matter what number you add, the number will never be greater than 2.

Answer 3

Here is my answer:

I am thinking of an odd number , does your number divide the number I am thinking of?

Because

1 -> yes
2 -> no
3 -> maybe

Answer 4

Here's my thought process on this:

• The three answer possibilities must be "yes," "no," and... something else.
• Perhaps there is a way to get an answer of "maybe" or "sometimes?"
• Therefore, this would have to involve some kind of randomness.
• The question should be in the form of "Either X or Y is true. Is your number {some condition}?"
• A simple condition I can think of is "even or odd."
• Is there a way to get a known even and known odd result for two of the numbers, and an unknown one for the other?

• Let's make a little chart:

  1 2 3
  O E O n^2
  E O O floor(n/2)
  O O E ceil(n/2)
  E E O floor(n/3)
  O O O ceil(n/3)
  

• Hey, look! For floor(n/2) and floor(n/3), our conditions are satisfied!

• So, my final question is:

  I will divide your number by either 2 or 3 and round it down. Is the result even or odd?

  If the answer is "even," the number was 1. If it's "odd," the number was 3. If it's "I don't know," the number was 2.

• Note that an alternate question could use n^2 and floor(n/3) instead (I will either square your number or divide it by three and round down).

Answer 5

Slightly tongue-in-cheek, but a variation on the "maybe" option provided the other player interprets it literally enough.

Keep evaluating powers of your number until you reach one that exceeds 6. Is it an odd power? Answer yes = 2, no = 3, infinitely long pause = 1 …

Answer 6

Here's my simple answer:

What word has that many letters? Please answer with yes or no.

Example:

- Yes -> 3
- No -> 2
- Er... I can't. -> 1

Answer 7

As is the case with all information puzzles, we have to find away to map each outcome (yes, no, I don't know) to a different number (1, 2, 3). To do this, we need some property that's unknown for exactly one of these three numbers.

There are multiple ways to do this. The most basic (and most reliable) way would be to introduce a number that only you know so I wouldn't know:

I'm thinking of either 1.5 or 2.5. Is your number greater than mine?

1 is smaller than either of these numbers, so the answer will always be no. 3 is greater, so yes. And 2 could be either greater or smaller, so I don't know.

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Another way, which is less reliable, is to introduce an unknown is to use some unknown property in mathematics:

Is TREE(your number) divisible by your number?

The Kruskal tree theorem produces a sequence of numbers such that TREE(1) = 1 and TREE(2) = 3, but TREE(3) is so astronomically large that it makes Graham's number look like epsilon in comparison. Moreover, it's not even known how to calculate it, so we cannot even tell whether it's even or odd.

In this case, if I'm thinking of 1, the answer is yes, if I'm thinking of 2, the answer is no, and if I'm thinking of 3, I'm certainly not going to know whether TREE(3) is divisible by 3. The reason this one is less reliable is because if we ever do find a way to calculate TREE(3), the uncertainty will no longer be there.

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Yet a third way, the least reliable of all, is to use some form of contingent unknown, like a form of obscure trivia or some event that has a relatively unpredictable or unknown occurrence.

Is the current world population greater than (your number + 5.3) billion?

Will Easter Sunday ten years from now happen less than (your number * 32) days after February 1?

Is the last bit of the SHA-256 hash of "entanglement" strictly smaller than (your number - 1)?

The current world population is currently between 7.2 and 7.3 billion. It reached 6.3 billion ten years ago, and won't reach 8.3 billion for another ten years, but somebody who didn't actually look up the population tables wouldn't know this fact for sure.

Easter Sunday is defined as the Sunday after the first full moon after the vernal equinox, so it can't happen before March 20 in any year (which is the earliest date the vernal equinox can happen), and it can't happen after April 25 (36 days after the vernal equinox) either. So it's not going to happen before March 3 or after May 8 either. But the date in the middle, April 5/6, is right in the middle of possible dates for Easter, so unless you've memorized the Computus table, it's not generally feasible to predict what day Easter is going to fall on.

The SHA-256 hash of "entanglement" is generally impossible to calculate in a few seconds in your head, unless you looked it up ahead of time, but you do know that its last bit (which is either 0 or 1) will always be less than (3 - 1), and never less than (1 - 1).

Although this method isn't completely reliable in that they might know that piece of trivia, the key is to provide two completely ridiculous values on each side and one that can be argued either way in the middle that the average person would answer "I don't know" to.

For the record:

The current world population at the time of posting is about 7.23 billion, which is less than 7.3 billion. This makes the answer "no" when my number is 2.

Easter Sunday will happen on March 31 in 2024, which makes the answer "no".

The SHA-256 hash of "entanglement" is 961B164F23EB33F8FDA12C95E8BD93F6 32A08A8B8A0A18B3DDE1CFE8926875FF, which has a last bit of 1. So the answer is "no".

Answer 8

The question I would ask is this (where $n$ is the number you are thinking of):

Is $\lim_{x \to n} \frac{1}{x-2} > 0$ (where we are approaching $n$ from either the left or the right)?

The possible answers are:

• If $n = 1$: $\frac{1}{1-2} = -$, so the answer is "No".
• If $n = 2$: $\lim_{x \to 2^+} \frac{1}{x-2} = +\infty$ but $\lim_{x \to 2^-} \frac{1}{x-2}$ = $-\infty$ so the answer is "I don't know".
• If $n = 3$: $\frac{1}{3-2} = 1$, so the answer is "Yes".

Answer 9

I would ask

"Are there at least $N + 27$ days in February, where $N$ is your number?"

Because

• if $(N = 1)$ then $(N + 27 = 28)$ --> Yes

• if $(N = 2)$ then $(N + 27 = 29)$ --> Maybe

• if $(N = 3)$ then $(N + 27 = 30)$ --> No

Answer 10

I wanted a solution that avoids uncertainty. So here's a good logical contradiction:

Is exactly one of these statements true?

(a) Your number is not 2.

(b) Your number is 3 and your answer to this question is Yes.

Analysis

If the number is 1, then (a) is true and (b) is false, so the answer is Yes.

If the number is 2, then (a) is false and (b) is false, so the answer is No.

If the number is 3, then (a) is true. Now:

If the answer is Yes, then (b) is true and therefore the answer must be No.

If the answer is No then (b) is false and therefore the answer must be Yes.

By contradiction, neither Yes nor No can be answered, and some other response must be used (e.g. Unanswerable).

Summary

If Yes, then the number is 1.

If No, then the number is 2.

If Unanswerable, then the number is 3.

Answer 11

I would ask

"Is there infinitely many n such as n and n + 'your number' -1 are primes?"

Because

1. Yes (Euclid)

2. No, obviously

3. I don't know, still working on twin prime conjecture

Answer 12

As you can see in this topic, there are a lot of possible solutions of course.
But there is general solution, which includes all of them. It is:

Are you thinking of 1 OR you are thinking of 2 AND [statement with unknown result]?

The interpretation of the answer is clear:

1 -> yes
2 -> I do not know
3 -> no

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My most preferable solution is

Are you thinking of 1 OR you are thinking of 2 AND tomorrow will be rain?

Just because it is simple and it is clear how it works.

But any other solution will be equivalent to given general solution (up to a interchange of numbers).

For example accepted answer of Ice-9:

"I'm thinking of a number: either 0 or 1. Is the sum of our numbers greater than 2?"

is equivalent to:

Are you thinking of 3 OR you are thinking of 2 AND I thinking of 1?

Or another solution of Doorknob:

"I will divide your number by either 2 or 3 and round it down. Is the result even or odd?"

is equivalent to:

Are you thinking of 1 OR you are thinking of 2 AND I thinking of 3?

Even such sophisticated solution, which arshajii gave:

"Is $\lim_{x \to n} \frac{1}{x-2} > 0$ (where we are approaching $n$ from either the left or the right)?"

is equivalent to:

Are you thinking of 3 OR you are thinking of 2 AND $\lim_{x \to 0} \frac{1}{x} > 0$?

Answer 13

I would say:

Considering that I am also thinking of a number between 1 and 2, is you number higher than mine?

1-> No
2-> Maybe
3-> Yes

Answer 14

You should ask:

We have 1 ball on the table and 0 or 1 balls under a concealed cup. Now I ask you is the amount of balls on the table is larger than the number on your mind?

1. No, because 1 is not larger than the 1 or 2 balls on the table

2. Maybe, because maybe it's larger than 1, but it's not larger than 2

3. Yes, because 3 is larger than the 1 or 2 balls on the table.

Answer 15

I would

write 2.XX on a piece of paper, cover its fractional part with my hand

and ask,

"Is your number greater than mine?"

Because if

1. Definitely no

2. Maybe (my number would have been $2.00$, so giving no for an answer would not be correct)

3. Definitely yes

Answer 16

Divide 2 by one less than the number you were thinking of. Is the result an even number?

• Yes: You were thinking of 2 (because 2/(2-1) = 2)

• No: You were thinking of 3 (because 2/(3-1) = 1)

• Unanswerable: You were thinking of 1 (because 2/(1-1) = ∞)

Answer 17

Soccer World Cup is in progress, so that warrants a themed question (though just a bit late, as group stage is over now - oh well):

Is your number both valid and not less than a number of points that a team got after finishing a group stage match in which they scored at least the same number of goals as their opponent?

A soccer team receives 1 point for a draw and 3 points for a win, 2 is not a valid number of points, so:

1 => yes
2 => no
3 => maybe

I feel this doesn't bode well with the spirit of the posted question as some of the other answers do, though. ;)

Answer 18

Does the number of letters in your answer match the number you are thinking of?

So if thinking of 1, answer is no.
If thinking of 3, answer is yes.
If thinking of 2, can't truthfully answer yes or no

Answer 19

If I pull your number of socks from a drawer with two different pairs of socks in it, will I have a matching pair? 1 - No, 2 - Maybe, 3 - Yes.

Answer 20

Wimbledon is in progress, so let's put that to good use:

Is your number less than a number of sets played in a straight-set win in a Wimbledon tennis match?

Women win in 2 sets, men in 3 sets.

1 -> yes
2 -> maybe
3 -> no

Answer 21

I would

consider a prime number $p\neq 2$

and ask

"Does your number divide $p$?"

Because

• If he answers Yes, then the number is $1$.

• If he answers NO, then the number is $2$.

• If he answers MAYBE, then the number is $3$.

Answer 22

I am thinking of a multiple of 5, but not of 15 - does your number divide mine?

If the number the person is thinking of is 3, it cannot divide my number since I blocked factors of 3, so we get a "no".

If 1, it will always divide my number, since 1 is a factor of every number, so we get a "yes".

If 2, we get a "maybe", because I could just as easily be thinking of 20 as 25, so there is not enough information.

Answer 23

I would ask:

If you are driving and take your number's worth of turns, (right or left, no off-ramps or U-turns), will you get multiple turns to the same direction? This will work because:

If their number is 1

They will say "no", for obvious reasons.

If their number is 2

They will say "maybe" or "I don't know", because with two turns, you could (where R = a right turn and L = a left turn) do: RL, LR, RR, or LL.

If their number is 3

They will say "yes." Possible turns are: RLL, RLR, RRR, >!RRL, LRR, LRL, LLR, LLL. All of these include multiples of the same direction.

Answer 24

Yet another answer:

I would ask: If $N$ is your number, is $N-1$ a natural number which is also an identity element in $\mathbb{Q}$ ($\mathbb{R}$ or $\mathbb{C}$ would also work)?

Answers:

3: No (2 is neither a multiplicative identity nor an additive one).
2: Yes (1 is multiplicative identity).
1: Maybe (also "I don't know", "It depends" etc.) (0 is additive identity, but some authors don't consider it a natural number).

Answer 25

Does any user of puzzling.stackexchange have that many arms?

• If your number is 3, the answer will be "no" (unless we have any aliens on the site)
• If your number is 2, the answer will be "yes"
• If your number is 1, the answer will most probably be "maybe".

Answer 26

I would ask

"Supposing that the number you are thinking of is $n$, if I was to get $n$ parallel one-sided mirrors and face the glass to each other, would I get infinite reflections on each mirror, theoretically?"

Because

1. Nope. There is only one mirror.

2. Yes. Two mirrors can create an infinity mirror if parallel and facing each other.

3. Maybe. It depends on the size. Imagine these lines being one-sided mirrors: $\large\color{blue}{\mid}\:\color{red}{\mid}\:\color{green}{\mid}$.
   If the glass side of each blue and red were facing each other, then since they are one sided, there would not be infinite reflections on the glass of green. The same would apply for blue if I turned around red to face green instead, which is facing in the direction of blue and red.
   We would need red to be shorter with blue and green facing each other; i.e., ${\large\color{blue}{\mid}}\color{red}{\mid}{\large\color{green}{\mid}}$.
   Red can face whichever other mirror (blue or green)... but the question never mentioned the size of the mirrors, so it all depends on that, particularly the size of the red mirror.

Answer 27

would this work?

she is thinking of a number; 1, 2, or 3, and only able to answer yes no or I don't know. I'm given only one question to know her number. so...

"using only 1,2, and 3, if you subtract the number preceding the number you are thinking, and add the number following the one you are thinking, then divide that answer by your number, is the resulting number even?"

the following would happen:
if 1: it would be (1+2)/1 becoming 3/1 simplified into 3 odd: NO

if 2: it would be (2-1+3)/2 becoming 4/2 simplified into 2 even: YES

if 3: it would be (3-2)/3 becoming 1/3 which is a fraction being neither odd nor even so: I DON'T KNOW

this has been in my head for years since my friend told me this and this is the answer I came up with that makes the most sense to me so far.

it's the "I don't Know" part that has me flustered, because that can always be swept under as "NO."

Answer 28

Does your answer have the same number of letters as is your number numeric value?

A little tricky, question seems not to ask for enough information, however:

answer: NO(2) >> 1 ; answer: YES(3) >> 3 ; answer MAYBE(5)/DON'T KNOW(8) >> 2 (two)

If the number is 2, answers YES (3 letters) and NO (2 letters) are both wrong!