Gaby and Jack's Number Guessing Game












6














Gaby, a good mathematician, thinks of a whole number between 1 and 100 inclusive. What is the least number of questions Jack needs to ask her, and what questions should those be, if he is to work out with absolute certainty Gaby's number. Gaby will answer Jack's questions truthfully, honestly, and to the best of her knowledge, and is only allowed to respond "Yes", "No", or "I don't know".



There is a difference with the apparent duplicate: Gaby has the option to honestly answer "I don't know". It makes a big difference! Puzzle inspired in another elsewhere in this site.










share|improve this question




















  • 2




    @GordonK: There is a difference: Gaby's option to truthfully answer "I don't know". Makes a big difference! Puzzle inspired in another elsewhere in this site.
    – Bernardo Recamán Santos
    Dec 6 '18 at 13:53










  • Ok, then I'm keen to see the intended answer and how a good mathematician's ignorance (and by implication the ignorance of mathematicians in general) can be used to faster reduce the search space!
    – Gordon K
    Dec 6 '18 at 13:57










  • I am assuming that what you are asking for is the least amount of questions in order to guarrantee a solution no matter what number Gaby is thinking of? Because if the question is what the least amount of questions that could lead to the correct answer, then clearly 1 is the solution. "Is your number 42?" is a solution if Gaby's number indeed is 42.
    – Phil
    Dec 7 '18 at 15:05










  • @Phil: Indeed, least number to guarantee Jack working out the number, the worst come to the worst. I have clarified the issue.
    – Bernardo Recamán Santos
    Dec 7 '18 at 15:24
















6














Gaby, a good mathematician, thinks of a whole number between 1 and 100 inclusive. What is the least number of questions Jack needs to ask her, and what questions should those be, if he is to work out with absolute certainty Gaby's number. Gaby will answer Jack's questions truthfully, honestly, and to the best of her knowledge, and is only allowed to respond "Yes", "No", or "I don't know".



There is a difference with the apparent duplicate: Gaby has the option to honestly answer "I don't know". It makes a big difference! Puzzle inspired in another elsewhere in this site.










share|improve this question




















  • 2




    @GordonK: There is a difference: Gaby's option to truthfully answer "I don't know". Makes a big difference! Puzzle inspired in another elsewhere in this site.
    – Bernardo Recamán Santos
    Dec 6 '18 at 13:53










  • Ok, then I'm keen to see the intended answer and how a good mathematician's ignorance (and by implication the ignorance of mathematicians in general) can be used to faster reduce the search space!
    – Gordon K
    Dec 6 '18 at 13:57










  • I am assuming that what you are asking for is the least amount of questions in order to guarrantee a solution no matter what number Gaby is thinking of? Because if the question is what the least amount of questions that could lead to the correct answer, then clearly 1 is the solution. "Is your number 42?" is a solution if Gaby's number indeed is 42.
    – Phil
    Dec 7 '18 at 15:05










  • @Phil: Indeed, least number to guarantee Jack working out the number, the worst come to the worst. I have clarified the issue.
    – Bernardo Recamán Santos
    Dec 7 '18 at 15:24














6












6








6







Gaby, a good mathematician, thinks of a whole number between 1 and 100 inclusive. What is the least number of questions Jack needs to ask her, and what questions should those be, if he is to work out with absolute certainty Gaby's number. Gaby will answer Jack's questions truthfully, honestly, and to the best of her knowledge, and is only allowed to respond "Yes", "No", or "I don't know".



There is a difference with the apparent duplicate: Gaby has the option to honestly answer "I don't know". It makes a big difference! Puzzle inspired in another elsewhere in this site.










share|improve this question















Gaby, a good mathematician, thinks of a whole number between 1 and 100 inclusive. What is the least number of questions Jack needs to ask her, and what questions should those be, if he is to work out with absolute certainty Gaby's number. Gaby will answer Jack's questions truthfully, honestly, and to the best of her knowledge, and is only allowed to respond "Yes", "No", or "I don't know".



There is a difference with the apparent duplicate: Gaby has the option to honestly answer "I don't know". It makes a big difference! Puzzle inspired in another elsewhere in this site.







mathematics arithmetic






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 7 '18 at 15:19

























asked Dec 6 '18 at 13:41









Bernardo Recamán Santos

2,3281141




2,3281141








  • 2




    @GordonK: There is a difference: Gaby's option to truthfully answer "I don't know". Makes a big difference! Puzzle inspired in another elsewhere in this site.
    – Bernardo Recamán Santos
    Dec 6 '18 at 13:53










  • Ok, then I'm keen to see the intended answer and how a good mathematician's ignorance (and by implication the ignorance of mathematicians in general) can be used to faster reduce the search space!
    – Gordon K
    Dec 6 '18 at 13:57










  • I am assuming that what you are asking for is the least amount of questions in order to guarrantee a solution no matter what number Gaby is thinking of? Because if the question is what the least amount of questions that could lead to the correct answer, then clearly 1 is the solution. "Is your number 42?" is a solution if Gaby's number indeed is 42.
    – Phil
    Dec 7 '18 at 15:05










  • @Phil: Indeed, least number to guarantee Jack working out the number, the worst come to the worst. I have clarified the issue.
    – Bernardo Recamán Santos
    Dec 7 '18 at 15:24














  • 2




    @GordonK: There is a difference: Gaby's option to truthfully answer "I don't know". Makes a big difference! Puzzle inspired in another elsewhere in this site.
    – Bernardo Recamán Santos
    Dec 6 '18 at 13:53










  • Ok, then I'm keen to see the intended answer and how a good mathematician's ignorance (and by implication the ignorance of mathematicians in general) can be used to faster reduce the search space!
    – Gordon K
    Dec 6 '18 at 13:57










  • I am assuming that what you are asking for is the least amount of questions in order to guarrantee a solution no matter what number Gaby is thinking of? Because if the question is what the least amount of questions that could lead to the correct answer, then clearly 1 is the solution. "Is your number 42?" is a solution if Gaby's number indeed is 42.
    – Phil
    Dec 7 '18 at 15:05










  • @Phil: Indeed, least number to guarantee Jack working out the number, the worst come to the worst. I have clarified the issue.
    – Bernardo Recamán Santos
    Dec 7 '18 at 15:24








2




2




@GordonK: There is a difference: Gaby's option to truthfully answer "I don't know". Makes a big difference! Puzzle inspired in another elsewhere in this site.
– Bernardo Recamán Santos
Dec 6 '18 at 13:53




@GordonK: There is a difference: Gaby's option to truthfully answer "I don't know". Makes a big difference! Puzzle inspired in another elsewhere in this site.
– Bernardo Recamán Santos
Dec 6 '18 at 13:53












Ok, then I'm keen to see the intended answer and how a good mathematician's ignorance (and by implication the ignorance of mathematicians in general) can be used to faster reduce the search space!
– Gordon K
Dec 6 '18 at 13:57




Ok, then I'm keen to see the intended answer and how a good mathematician's ignorance (and by implication the ignorance of mathematicians in general) can be used to faster reduce the search space!
– Gordon K
Dec 6 '18 at 13:57












I am assuming that what you are asking for is the least amount of questions in order to guarrantee a solution no matter what number Gaby is thinking of? Because if the question is what the least amount of questions that could lead to the correct answer, then clearly 1 is the solution. "Is your number 42?" is a solution if Gaby's number indeed is 42.
– Phil
Dec 7 '18 at 15:05




I am assuming that what you are asking for is the least amount of questions in order to guarrantee a solution no matter what number Gaby is thinking of? Because if the question is what the least amount of questions that could lead to the correct answer, then clearly 1 is the solution. "Is your number 42?" is a solution if Gaby's number indeed is 42.
– Phil
Dec 7 '18 at 15:05












@Phil: Indeed, least number to guarantee Jack working out the number, the worst come to the worst. I have clarified the issue.
– Bernardo Recamán Santos
Dec 7 '18 at 15:24




@Phil: Indeed, least number to guarantee Jack working out the number, the worst come to the worst. I have clarified the issue.
– Bernardo Recamán Santos
Dec 7 '18 at 15:24










3 Answers
3






active

oldest

votes


















15














If




the numbers are between $1$ to $3$




then




use this answer.




If




the numbers are between $1$ to $9$




then




ask "I'm thinking a number between $4$ to $7$, is your number strictly less than my number?"




because




if she says "Yes" that means her number is between $1$ to $3$,

else if she says "I don't know" that means her number is between $4$ to $6$,

else if she says "No" that means her number is between $7$ to $9$,




then




use previous strategy to guess the number between the group of $3$ numbers.




Hence,




We may use this iteratively to split the numbers into $3$ same-size groups, by asking similar questions. We need around $log_3(100) = 5$ questions in total.







share|improve this answer

















  • 1




    A somewhat more intuitive way to think about this is to ask: if $f_1(x)=begin{cases}textrm{Yes} & x leq 33 \ text{No} & x geq 67 \ text{I don't know} & text{otherwise}end{cases}$, what is $f_1(text{your number})$? etc.
    – ace
    Dec 7 '18 at 14:11






  • 1




    You seem to be an ace in stackexchange formatting...how did you even fit a multi-part function in the comments!
    – George Menoutis
    Dec 7 '18 at 14:18










  • @GeorgeMenoutis It's just standard LaTeX, $begin{cases} value1 & condition1 \ value2 & condition2 end{cases}$
    – ace
    Dec 7 '18 at 14:32



















2














This is a partial answer.



The trick is to




Devise a way to map the three options yes/no/I don't know in a way to split a number range to 3 parts. Eg: Find a question about the target number such that the answer would be "no" For 1-33, "yes" for 34-66, and "I don't know" for 67-100




After we establish this, it's just a matter of




Splitting the possible number set to one-third each time




which will take a maximum of




ceil(log3(100))=ceil(4.19...)=5 questions







share|improve this answer





















  • Indeed, and to work out a number between 1 and 1000 would require only 7 such questions (as opposed to 11 "yes or no" questions)! What then are the questions that allow this?
    – Bernardo Recamán Santos
    Dec 7 '18 at 12:52










  • Athin's answer covers this, I think, by using a sum between target number and another which is unknow but within known range.
    – George Menoutis
    Dec 7 '18 at 13:04



















1














Other answers, especially @athin's, argue convincingly that the target number can be identified with as few as




5




questions, but it remains to show that that is the minimum.



It is. We can see that by observing




in order to distinguish each of 100 possibilities from the others via N questions, we need at least 100 distinct patterns of N answers.




With three possible answers to each of N questions,




there are 3N distinct patterns of answers. 34 = 81 is not enough for our purposes, but 35 = 273 yields nearly three times as many answer patterns as we need.




The minimum number of questions needed is therefore certainly




5







share|improve this answer





















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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    15














    If




    the numbers are between $1$ to $3$




    then




    use this answer.




    If




    the numbers are between $1$ to $9$




    then




    ask "I'm thinking a number between $4$ to $7$, is your number strictly less than my number?"




    because




    if she says "Yes" that means her number is between $1$ to $3$,

    else if she says "I don't know" that means her number is between $4$ to $6$,

    else if she says "No" that means her number is between $7$ to $9$,




    then




    use previous strategy to guess the number between the group of $3$ numbers.




    Hence,




    We may use this iteratively to split the numbers into $3$ same-size groups, by asking similar questions. We need around $log_3(100) = 5$ questions in total.







    share|improve this answer

















    • 1




      A somewhat more intuitive way to think about this is to ask: if $f_1(x)=begin{cases}textrm{Yes} & x leq 33 \ text{No} & x geq 67 \ text{I don't know} & text{otherwise}end{cases}$, what is $f_1(text{your number})$? etc.
      – ace
      Dec 7 '18 at 14:11






    • 1




      You seem to be an ace in stackexchange formatting...how did you even fit a multi-part function in the comments!
      – George Menoutis
      Dec 7 '18 at 14:18










    • @GeorgeMenoutis It's just standard LaTeX, $begin{cases} value1 & condition1 \ value2 & condition2 end{cases}$
      – ace
      Dec 7 '18 at 14:32
















    15














    If




    the numbers are between $1$ to $3$




    then




    use this answer.




    If




    the numbers are between $1$ to $9$




    then




    ask "I'm thinking a number between $4$ to $7$, is your number strictly less than my number?"




    because




    if she says "Yes" that means her number is between $1$ to $3$,

    else if she says "I don't know" that means her number is between $4$ to $6$,

    else if she says "No" that means her number is between $7$ to $9$,




    then




    use previous strategy to guess the number between the group of $3$ numbers.




    Hence,




    We may use this iteratively to split the numbers into $3$ same-size groups, by asking similar questions. We need around $log_3(100) = 5$ questions in total.







    share|improve this answer

















    • 1




      A somewhat more intuitive way to think about this is to ask: if $f_1(x)=begin{cases}textrm{Yes} & x leq 33 \ text{No} & x geq 67 \ text{I don't know} & text{otherwise}end{cases}$, what is $f_1(text{your number})$? etc.
      – ace
      Dec 7 '18 at 14:11






    • 1




      You seem to be an ace in stackexchange formatting...how did you even fit a multi-part function in the comments!
      – George Menoutis
      Dec 7 '18 at 14:18










    • @GeorgeMenoutis It's just standard LaTeX, $begin{cases} value1 & condition1 \ value2 & condition2 end{cases}$
      – ace
      Dec 7 '18 at 14:32














    15












    15








    15






    If




    the numbers are between $1$ to $3$




    then




    use this answer.




    If




    the numbers are between $1$ to $9$




    then




    ask "I'm thinking a number between $4$ to $7$, is your number strictly less than my number?"




    because




    if she says "Yes" that means her number is between $1$ to $3$,

    else if she says "I don't know" that means her number is between $4$ to $6$,

    else if she says "No" that means her number is between $7$ to $9$,




    then




    use previous strategy to guess the number between the group of $3$ numbers.




    Hence,




    We may use this iteratively to split the numbers into $3$ same-size groups, by asking similar questions. We need around $log_3(100) = 5$ questions in total.







    share|improve this answer












    If




    the numbers are between $1$ to $3$




    then




    use this answer.




    If




    the numbers are between $1$ to $9$




    then




    ask "I'm thinking a number between $4$ to $7$, is your number strictly less than my number?"




    because




    if she says "Yes" that means her number is between $1$ to $3$,

    else if she says "I don't know" that means her number is between $4$ to $6$,

    else if she says "No" that means her number is between $7$ to $9$,




    then




    use previous strategy to guess the number between the group of $3$ numbers.




    Hence,




    We may use this iteratively to split the numbers into $3$ same-size groups, by asking similar questions. We need around $log_3(100) = 5$ questions in total.








    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Dec 6 '18 at 14:28









    athin

    7,13912368




    7,13912368








    • 1




      A somewhat more intuitive way to think about this is to ask: if $f_1(x)=begin{cases}textrm{Yes} & x leq 33 \ text{No} & x geq 67 \ text{I don't know} & text{otherwise}end{cases}$, what is $f_1(text{your number})$? etc.
      – ace
      Dec 7 '18 at 14:11






    • 1




      You seem to be an ace in stackexchange formatting...how did you even fit a multi-part function in the comments!
      – George Menoutis
      Dec 7 '18 at 14:18










    • @GeorgeMenoutis It's just standard LaTeX, $begin{cases} value1 & condition1 \ value2 & condition2 end{cases}$
      – ace
      Dec 7 '18 at 14:32














    • 1




      A somewhat more intuitive way to think about this is to ask: if $f_1(x)=begin{cases}textrm{Yes} & x leq 33 \ text{No} & x geq 67 \ text{I don't know} & text{otherwise}end{cases}$, what is $f_1(text{your number})$? etc.
      – ace
      Dec 7 '18 at 14:11






    • 1




      You seem to be an ace in stackexchange formatting...how did you even fit a multi-part function in the comments!
      – George Menoutis
      Dec 7 '18 at 14:18










    • @GeorgeMenoutis It's just standard LaTeX, $begin{cases} value1 & condition1 \ value2 & condition2 end{cases}$
      – ace
      Dec 7 '18 at 14:32








    1




    1




    A somewhat more intuitive way to think about this is to ask: if $f_1(x)=begin{cases}textrm{Yes} & x leq 33 \ text{No} & x geq 67 \ text{I don't know} & text{otherwise}end{cases}$, what is $f_1(text{your number})$? etc.
    – ace
    Dec 7 '18 at 14:11




    A somewhat more intuitive way to think about this is to ask: if $f_1(x)=begin{cases}textrm{Yes} & x leq 33 \ text{No} & x geq 67 \ text{I don't know} & text{otherwise}end{cases}$, what is $f_1(text{your number})$? etc.
    – ace
    Dec 7 '18 at 14:11




    1




    1




    You seem to be an ace in stackexchange formatting...how did you even fit a multi-part function in the comments!
    – George Menoutis
    Dec 7 '18 at 14:18




    You seem to be an ace in stackexchange formatting...how did you even fit a multi-part function in the comments!
    – George Menoutis
    Dec 7 '18 at 14:18












    @GeorgeMenoutis It's just standard LaTeX, $begin{cases} value1 & condition1 \ value2 & condition2 end{cases}$
    – ace
    Dec 7 '18 at 14:32




    @GeorgeMenoutis It's just standard LaTeX, $begin{cases} value1 & condition1 \ value2 & condition2 end{cases}$
    – ace
    Dec 7 '18 at 14:32











    2














    This is a partial answer.



    The trick is to




    Devise a way to map the three options yes/no/I don't know in a way to split a number range to 3 parts. Eg: Find a question about the target number such that the answer would be "no" For 1-33, "yes" for 34-66, and "I don't know" for 67-100




    After we establish this, it's just a matter of




    Splitting the possible number set to one-third each time




    which will take a maximum of




    ceil(log3(100))=ceil(4.19...)=5 questions







    share|improve this answer





















    • Indeed, and to work out a number between 1 and 1000 would require only 7 such questions (as opposed to 11 "yes or no" questions)! What then are the questions that allow this?
      – Bernardo Recamán Santos
      Dec 7 '18 at 12:52










    • Athin's answer covers this, I think, by using a sum between target number and another which is unknow but within known range.
      – George Menoutis
      Dec 7 '18 at 13:04
















    2














    This is a partial answer.



    The trick is to




    Devise a way to map the three options yes/no/I don't know in a way to split a number range to 3 parts. Eg: Find a question about the target number such that the answer would be "no" For 1-33, "yes" for 34-66, and "I don't know" for 67-100




    After we establish this, it's just a matter of




    Splitting the possible number set to one-third each time




    which will take a maximum of




    ceil(log3(100))=ceil(4.19...)=5 questions







    share|improve this answer





















    • Indeed, and to work out a number between 1 and 1000 would require only 7 such questions (as opposed to 11 "yes or no" questions)! What then are the questions that allow this?
      – Bernardo Recamán Santos
      Dec 7 '18 at 12:52










    • Athin's answer covers this, I think, by using a sum between target number and another which is unknow but within known range.
      – George Menoutis
      Dec 7 '18 at 13:04














    2












    2








    2






    This is a partial answer.



    The trick is to




    Devise a way to map the three options yes/no/I don't know in a way to split a number range to 3 parts. Eg: Find a question about the target number such that the answer would be "no" For 1-33, "yes" for 34-66, and "I don't know" for 67-100




    After we establish this, it's just a matter of




    Splitting the possible number set to one-third each time




    which will take a maximum of




    ceil(log3(100))=ceil(4.19...)=5 questions







    share|improve this answer












    This is a partial answer.



    The trick is to




    Devise a way to map the three options yes/no/I don't know in a way to split a number range to 3 parts. Eg: Find a question about the target number such that the answer would be "no" For 1-33, "yes" for 34-66, and "I don't know" for 67-100




    After we establish this, it's just a matter of




    Splitting the possible number set to one-third each time




    which will take a maximum of




    ceil(log3(100))=ceil(4.19...)=5 questions








    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Dec 6 '18 at 14:10









    George Menoutis

    920211




    920211












    • Indeed, and to work out a number between 1 and 1000 would require only 7 such questions (as opposed to 11 "yes or no" questions)! What then are the questions that allow this?
      – Bernardo Recamán Santos
      Dec 7 '18 at 12:52










    • Athin's answer covers this, I think, by using a sum between target number and another which is unknow but within known range.
      – George Menoutis
      Dec 7 '18 at 13:04


















    • Indeed, and to work out a number between 1 and 1000 would require only 7 such questions (as opposed to 11 "yes or no" questions)! What then are the questions that allow this?
      – Bernardo Recamán Santos
      Dec 7 '18 at 12:52










    • Athin's answer covers this, I think, by using a sum between target number and another which is unknow but within known range.
      – George Menoutis
      Dec 7 '18 at 13:04
















    Indeed, and to work out a number between 1 and 1000 would require only 7 such questions (as opposed to 11 "yes or no" questions)! What then are the questions that allow this?
    – Bernardo Recamán Santos
    Dec 7 '18 at 12:52




    Indeed, and to work out a number between 1 and 1000 would require only 7 such questions (as opposed to 11 "yes or no" questions)! What then are the questions that allow this?
    – Bernardo Recamán Santos
    Dec 7 '18 at 12:52












    Athin's answer covers this, I think, by using a sum between target number and another which is unknow but within known range.
    – George Menoutis
    Dec 7 '18 at 13:04




    Athin's answer covers this, I think, by using a sum between target number and another which is unknow but within known range.
    – George Menoutis
    Dec 7 '18 at 13:04











    1














    Other answers, especially @athin's, argue convincingly that the target number can be identified with as few as




    5




    questions, but it remains to show that that is the minimum.



    It is. We can see that by observing




    in order to distinguish each of 100 possibilities from the others via N questions, we need at least 100 distinct patterns of N answers.




    With three possible answers to each of N questions,




    there are 3N distinct patterns of answers. 34 = 81 is not enough for our purposes, but 35 = 273 yields nearly three times as many answer patterns as we need.




    The minimum number of questions needed is therefore certainly




    5







    share|improve this answer


























      1














      Other answers, especially @athin's, argue convincingly that the target number can be identified with as few as




      5




      questions, but it remains to show that that is the minimum.



      It is. We can see that by observing




      in order to distinguish each of 100 possibilities from the others via N questions, we need at least 100 distinct patterns of N answers.




      With three possible answers to each of N questions,




      there are 3N distinct patterns of answers. 34 = 81 is not enough for our purposes, but 35 = 273 yields nearly three times as many answer patterns as we need.




      The minimum number of questions needed is therefore certainly




      5







      share|improve this answer
























        1












        1








        1






        Other answers, especially @athin's, argue convincingly that the target number can be identified with as few as




        5




        questions, but it remains to show that that is the minimum.



        It is. We can see that by observing




        in order to distinguish each of 100 possibilities from the others via N questions, we need at least 100 distinct patterns of N answers.




        With three possible answers to each of N questions,




        there are 3N distinct patterns of answers. 34 = 81 is not enough for our purposes, but 35 = 273 yields nearly three times as many answer patterns as we need.




        The minimum number of questions needed is therefore certainly




        5







        share|improve this answer












        Other answers, especially @athin's, argue convincingly that the target number can be identified with as few as




        5




        questions, but it remains to show that that is the minimum.



        It is. We can see that by observing




        in order to distinguish each of 100 possibilities from the others via N questions, we need at least 100 distinct patterns of N answers.




        With three possible answers to each of N questions,




        there are 3N distinct patterns of answers. 34 = 81 is not enough for our purposes, but 35 = 273 yields nearly three times as many answer patterns as we need.




        The minimum number of questions needed is therefore certainly




        5








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Dec 6 '18 at 21:04









        John Bollinger

        1112




        1112






























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