Guessing right number from 1 to 27, with 3 Yes/No/Maybe questions.

[jasen]

Your best friend (Bob) comes to you with this problem.

The Problem

His teacher (Mr. Sam) will ask Bob to work out a number in a child (Joan) head from 1 and 27.
 Bob has to find the correct number, by asking Joan three Yes/No questions about the number.
 Joan can respond with 3 different answers: "Yes", "No", or "I do not know".

Joan is a smart boy, but he doesn't know about another base number except 10. So by asking that kind of question he will answer "I do not know"

Example:
If the number is 12

Is it an even number? Yes.
 Is it divisible by 9? No.
 If I take a random number between 10 to 15, will I get the right number? I do not know.

------------------

Another Example:
If the number is 9

Is it an even number? No.
 Is it divisible by 9? Yes.
 If I take a random number between 10 to 15, will I get the right number? No.

------------------
 

Find the strategy, so every number from 1 to 27 can be guessed correctly.
 

[bonanova]

Spoiler

Let's say the correct number is x

Divide the numbers into three equal-size groups

A: { 1  2  3  4  5  6  7  8  9}
B: {10 11 12 13 14 15 16 17 18}
C: {19 20 21 22 23 24 25 26 27}

Ask Q1: If I pick a number at random from B, (the middle group), will it be greater than x?
Answers (Yes / IDK / No) indicate x is in (A/B/C) respectively.
Suppose Joan answered Yes. Then x is in A. Divide A into three equal-size groups:

D: {1 2 3}
E: {4 5 6}
F: {7 8 9)

Ask Q2: if I pick a number at random from E, (the middle group) will it be greater than x?
Answers (Yes / IDK / No) indicate x is in (D/E/F) respectively.
Suppose Joan answers IDK. Then x is in E, and therefore is 4, 5, or 6.

Now we need a question different in form from the first two. In order for IDK to be a possible response, randomness has to be present. Since there is no randomness in a number picked at random from a single-element set, we need to get creative.

Ask Q3: if I were to pick a number at random from E , (the group identified by Q2),that turned out not to be x, would it be greater than x?
Answers (Yes / IDK / No) indicate x is (4 / 5 / 6) respectively.

So we can identify the correct number using three questions. The first two ask about a number picked at random from the  middle of three multiple-number groups. The third asks about a number picked at random, from the group of three numbers that survives Q2, that turns out not to be x..

Summary:

1. Divide the numbers into three groups, as above. Ask: If I were to select a number at random from the middle group, would it be greater than x? Select the (first / second / third) group if the answer is (Yes / IDK / No).
    
2. Divide the selected group into three groups, as above. Ask (again): If I were to select at random a number from the middle group, would it be greater than x? Select the (first / second / third) group if the answer is (Yes / IDK / No).
    
3. Finally, ask: If I were to select at random a number from (the last selected group) that turned out not to be x, would it be greater than x? Select the (first / second / third) number from the group if the answer is (Yes / IDK / No).

 

[bonanova]

First thoughts:

Spoiler

It sounds like the weighing problem where one of 27 coins is counterfeit, differing in weight from all the others.

The difference is that a set of balances can give 3 different outcomes, whereas a normal yes/no question gives only two. But IDK has been added to the mix. So perhaps the solution hinges on finding a way to use {yes, no, IDK} responses in an aalogous way to using {left-heavier, equal, right-heavier} for the weighing problem.

If with a single question we can deduce which of {1 2 3} was chosen, then we should with three questions be able to determine which of {1 2 3 4 ,,, 25 26 27} was chosen.

 

[jasen]

 

Spoiler

Here is my version :

I  will create 3 pairs of list

First pair : 
1A : 2,3,5,6,8,9,11,12,14,15,17,18,20,21,23,24,26,27
1B : 2,5,8,11,14,17,20,23,26

Then ask Mr. Smith : If I take a random list, Does the number in the list ?
If the answer is Yes then, write number 2 in first paper.
(because the number appears in both list)
If the answer is No then, write number 0 in first paper.
(because the number do not appears in both list)
If the answer is I do not know then, write number 1 in first paper.
(because the number only appears in one list)

then

2nd pair :
2A : 4,5,6,7,8,9,13,14,15,16,17,18,22,23,24,25,26,27
2B : 4,5,6,13,14,15,23,24,25

Then ask Mr. Smith : If I take a random list, Does the number in the list ?
If the answer is Yes then, write number 2 in the 2nd paper.
If the answer is No then, write number 0 in the 2nd paper.
If the answer is I do not know then, write number 1 in the 2nd paper.

then

3rd pair :
3A : 10 to 27
3B : 10 to 18 

Then ask Mr. Smith : If I take a random list, Does the number in the list ?
If the answer is Yes then, write number 2 in the 3rd paper.
If the answer is No then, write number 0 in the 3rd paper.
If the answer is I do not know then, write number 1 in the 3rd paper.

then

put the number in the order 3rd 2nd 1st as base 3 number, than convert it into base 10 number,do not forget to add 1. The result is Mr. Smith number.