Choosing a number between $1$ and $100$, and randomly guessing it. What is the expected value of the number...












8














I was watching Steve Balmer’s interview and he was talking about questions they’d ask from candidates. This is a question he gave, he says-



I choose a number between $1$ to $100$, the other person has to guess the number. If he gets it right the first time he gets $5$ bucks, if he misses the first time steve tells you whether the number is higher or lower( he does this every time you miss), if he gets it right the second time he gets $4$ bucks, third time $3$, fourth $2$ and so on and if he gets it right in the seventh guess the person has to give a buck to Steve and so on, the value goes decreasing. I am trying to calculate the expected value of this game, how can I solve this, I can’t seem to come up with a way.



P.S. I have edited the question with a slight variation, in the previous version steve doesn’t tell you anything after you have guessed the wrong number.










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  • Can you see why the probability you guess right during each time between $1$ and $100$ is the same?
    – Jorge Fernández
    Nov 30 '18 at 18:51






  • 6




    Are you sure you haven't left something out, eg Steve has to tell you whether the number is larger or smaller than your guess? 2^6=64, 2^7=128, and I don't think it's a coincidence that after 7 guesses you pay out.
    – Hong Ooi
    Nov 30 '18 at 20:50










  • @HongOoi is correct. Here's the actual interview (the problem statement starts at 00:35). Although this question has already been answered, so it's probably too late to edit it now.
    – NotThatGuy
    Nov 30 '18 at 21:13


















8














I was watching Steve Balmer’s interview and he was talking about questions they’d ask from candidates. This is a question he gave, he says-



I choose a number between $1$ to $100$, the other person has to guess the number. If he gets it right the first time he gets $5$ bucks, if he misses the first time steve tells you whether the number is higher or lower( he does this every time you miss), if he gets it right the second time he gets $4$ bucks, third time $3$, fourth $2$ and so on and if he gets it right in the seventh guess the person has to give a buck to Steve and so on, the value goes decreasing. I am trying to calculate the expected value of this game, how can I solve this, I can’t seem to come up with a way.



P.S. I have edited the question with a slight variation, in the previous version steve doesn’t tell you anything after you have guessed the wrong number.










share|cite|improve this question
























  • Can you see why the probability you guess right during each time between $1$ and $100$ is the same?
    – Jorge Fernández
    Nov 30 '18 at 18:51






  • 6




    Are you sure you haven't left something out, eg Steve has to tell you whether the number is larger or smaller than your guess? 2^6=64, 2^7=128, and I don't think it's a coincidence that after 7 guesses you pay out.
    – Hong Ooi
    Nov 30 '18 at 20:50










  • @HongOoi is correct. Here's the actual interview (the problem statement starts at 00:35). Although this question has already been answered, so it's probably too late to edit it now.
    – NotThatGuy
    Nov 30 '18 at 21:13
















8












8








8


3





I was watching Steve Balmer’s interview and he was talking about questions they’d ask from candidates. This is a question he gave, he says-



I choose a number between $1$ to $100$, the other person has to guess the number. If he gets it right the first time he gets $5$ bucks, if he misses the first time steve tells you whether the number is higher or lower( he does this every time you miss), if he gets it right the second time he gets $4$ bucks, third time $3$, fourth $2$ and so on and if he gets it right in the seventh guess the person has to give a buck to Steve and so on, the value goes decreasing. I am trying to calculate the expected value of this game, how can I solve this, I can’t seem to come up with a way.



P.S. I have edited the question with a slight variation, in the previous version steve doesn’t tell you anything after you have guessed the wrong number.










share|cite|improve this question















I was watching Steve Balmer’s interview and he was talking about questions they’d ask from candidates. This is a question he gave, he says-



I choose a number between $1$ to $100$, the other person has to guess the number. If he gets it right the first time he gets $5$ bucks, if he misses the first time steve tells you whether the number is higher or lower( he does this every time you miss), if he gets it right the second time he gets $4$ bucks, third time $3$, fourth $2$ and so on and if he gets it right in the seventh guess the person has to give a buck to Steve and so on, the value goes decreasing. I am trying to calculate the expected value of this game, how can I solve this, I can’t seem to come up with a way.



P.S. I have edited the question with a slight variation, in the previous version steve doesn’t tell you anything after you have guessed the wrong number.







probability probability-distributions random-variables






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edited Nov 30 '18 at 22:07







user601297

















asked Nov 30 '18 at 18:46









user601297user601297

1226




1226












  • Can you see why the probability you guess right during each time between $1$ and $100$ is the same?
    – Jorge Fernández
    Nov 30 '18 at 18:51






  • 6




    Are you sure you haven't left something out, eg Steve has to tell you whether the number is larger or smaller than your guess? 2^6=64, 2^7=128, and I don't think it's a coincidence that after 7 guesses you pay out.
    – Hong Ooi
    Nov 30 '18 at 20:50










  • @HongOoi is correct. Here's the actual interview (the problem statement starts at 00:35). Although this question has already been answered, so it's probably too late to edit it now.
    – NotThatGuy
    Nov 30 '18 at 21:13




















  • Can you see why the probability you guess right during each time between $1$ and $100$ is the same?
    – Jorge Fernández
    Nov 30 '18 at 18:51






  • 6




    Are you sure you haven't left something out, eg Steve has to tell you whether the number is larger or smaller than your guess? 2^6=64, 2^7=128, and I don't think it's a coincidence that after 7 guesses you pay out.
    – Hong Ooi
    Nov 30 '18 at 20:50










  • @HongOoi is correct. Here's the actual interview (the problem statement starts at 00:35). Although this question has already been answered, so it's probably too late to edit it now.
    – NotThatGuy
    Nov 30 '18 at 21:13


















Can you see why the probability you guess right during each time between $1$ and $100$ is the same?
– Jorge Fernández
Nov 30 '18 at 18:51




Can you see why the probability you guess right during each time between $1$ and $100$ is the same?
– Jorge Fernández
Nov 30 '18 at 18:51




6




6




Are you sure you haven't left something out, eg Steve has to tell you whether the number is larger or smaller than your guess? 2^6=64, 2^7=128, and I don't think it's a coincidence that after 7 guesses you pay out.
– Hong Ooi
Nov 30 '18 at 20:50




Are you sure you haven't left something out, eg Steve has to tell you whether the number is larger or smaller than your guess? 2^6=64, 2^7=128, and I don't think it's a coincidence that after 7 guesses you pay out.
– Hong Ooi
Nov 30 '18 at 20:50












@HongOoi is correct. Here's the actual interview (the problem statement starts at 00:35). Although this question has already been answered, so it's probably too late to edit it now.
– NotThatGuy
Nov 30 '18 at 21:13






@HongOoi is correct. Here's the actual interview (the problem statement starts at 00:35). Although this question has already been answered, so it's probably too late to edit it now.
– NotThatGuy
Nov 30 '18 at 21:13












4 Answers
4






active

oldest

votes


















10














The answer posted by Jorge is right. Just to add some clarifications.



In the first try you have $frac 1 {100}$ chance of guessing it right. On the second guess, your chance increases to $frac 1 {99}$ as you know the answer isn't your guess and you aren't going to make the same guess. However, the probability that you are going to make the second guess (i.e. you guess the first one wrong) is $frac {99} {100}$ so the probability is again, $frac 1 {99}$ * $frac {99} {100}$
= $frac 1 {100}$. With same logic, your probability of guessing it right on the nth try is always $frac 1 {100}$



The rest of the calculation checks out.



$$sumlimits_{i=1}^{100} frac{6-i}{100} = 6 - frac{100times 101}{2times 100}=6-50.5=-45.5$$






share|cite|improve this answer























  • Thank you for clarifying that. I thought it was 1/100 also every time, but I caught myself using the wrong reasoning.
    – SpiralRain
    Nov 30 '18 at 19:05












  • I’m having a hard time understanding this, i get how you come to $frac{1}{100}$ for the second try but can you give the calculation for getting it right on the third try, shouldn’t it be $frac{1}{98}*frac{99}{100}$
    – user601297
    Nov 30 '18 at 22:32










  • For 3rd try, you'll have (probability of getting to 3rd pick) * (probability of getting it right). Probability of getting to 3rd pick is $frac {99} {100}$ * $frac {98} {99}$ because you have to first get 1st one wrong , then the second one wrong, and 3rd one right. So you'll get $frac {99} {100} * frac {98} {99} * frac {1} {98} = frac {1} {100}$
    – Ofya
    Nov 30 '18 at 23:01












  • Thanks i got it.
    – user601297
    Dec 1 '18 at 20:07



















3














The probability you get it right at the i'th time is $frac{1}{100}$.



Thus the expected value of the game is $$sumlimits_{i=1}^{100} frac{6-i}{100} = 6 - frac{100times 101}{2times 100}=6-50.5=-44.5$$






share|cite|improve this answer























  • assuming someone won't ignorantly re-guess numbers they've already guessed
    – phdmba7of12
    Nov 30 '18 at 18:56










  • and that they stop guessing once they've hit the correct number ... lol
    – phdmba7of12
    Nov 30 '18 at 18:56






  • 1




    Well I thought that was a reasonable assumption
    – Jorge Fernández
    Nov 30 '18 at 18:57






  • 1




    That's the assumption for a candidate that Steve Balmer would hire, ;P
    – I like Serena
    Nov 30 '18 at 19:01










  • Thanks a lot, I got it.
    – user601297
    Nov 30 '18 at 22:11



















2














If Steve tells you whether the number is higher or lower then, you will at most lose a dollar.
Going by the binary approach, you start with 50, worst situation: more numbers on the side which Steve's number is, up you go 75(as 50 on higher side and 49 on the lower side), then 88, then 94, then 97, 99 and finally 98 or 100. Even if you start with 50 and the number is lower than 50, you still pay a dollar at most.



I have arbitrarily chosen the higher end in each case to have equal to or more numbers than the lower end.



Taking this to be the most conservative approach and that you only follow this, if you want to find the average amount gained or lost, find the value for each number (formulate, no need to solve for each) and take average.






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    1














    Let $X_t:=1{text{guess at stage $t$ is correct}}$ and let $T:=min{t:X_t=1}$. Then for $1le tle 100$,
    begin{align}
    mathsf{P}(T=t)&=mathsf{P}(X_t=1,X_{t-1}=0,ldots,X_1=0) \
    &=mathsf{P}(X_t=1|X_{t-1}=0,ldots,X_1=0)timesmathsf{P}(X_{t-1}=0,ldots,X_1=0) \
    &=frac{1}{100-(t-1)}timesbinom{99}{t-1}binom{100}{t-1}^{-1}=frac{1}{100}.
    end{align}

    Thus, the stopping time $T$ is uniformly distributed on ${1,ldots,100}$ and its mean is $50.5$. Since at each step you loose $1$$, the mean loss is $6-50.5=-44.5$.






    share|cite|improve this answer























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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      10














      The answer posted by Jorge is right. Just to add some clarifications.



      In the first try you have $frac 1 {100}$ chance of guessing it right. On the second guess, your chance increases to $frac 1 {99}$ as you know the answer isn't your guess and you aren't going to make the same guess. However, the probability that you are going to make the second guess (i.e. you guess the first one wrong) is $frac {99} {100}$ so the probability is again, $frac 1 {99}$ * $frac {99} {100}$
      = $frac 1 {100}$. With same logic, your probability of guessing it right on the nth try is always $frac 1 {100}$



      The rest of the calculation checks out.



      $$sumlimits_{i=1}^{100} frac{6-i}{100} = 6 - frac{100times 101}{2times 100}=6-50.5=-45.5$$






      share|cite|improve this answer























      • Thank you for clarifying that. I thought it was 1/100 also every time, but I caught myself using the wrong reasoning.
        – SpiralRain
        Nov 30 '18 at 19:05












      • I’m having a hard time understanding this, i get how you come to $frac{1}{100}$ for the second try but can you give the calculation for getting it right on the third try, shouldn’t it be $frac{1}{98}*frac{99}{100}$
        – user601297
        Nov 30 '18 at 22:32










      • For 3rd try, you'll have (probability of getting to 3rd pick) * (probability of getting it right). Probability of getting to 3rd pick is $frac {99} {100}$ * $frac {98} {99}$ because you have to first get 1st one wrong , then the second one wrong, and 3rd one right. So you'll get $frac {99} {100} * frac {98} {99} * frac {1} {98} = frac {1} {100}$
        – Ofya
        Nov 30 '18 at 23:01












      • Thanks i got it.
        – user601297
        Dec 1 '18 at 20:07
















      10














      The answer posted by Jorge is right. Just to add some clarifications.



      In the first try you have $frac 1 {100}$ chance of guessing it right. On the second guess, your chance increases to $frac 1 {99}$ as you know the answer isn't your guess and you aren't going to make the same guess. However, the probability that you are going to make the second guess (i.e. you guess the first one wrong) is $frac {99} {100}$ so the probability is again, $frac 1 {99}$ * $frac {99} {100}$
      = $frac 1 {100}$. With same logic, your probability of guessing it right on the nth try is always $frac 1 {100}$



      The rest of the calculation checks out.



      $$sumlimits_{i=1}^{100} frac{6-i}{100} = 6 - frac{100times 101}{2times 100}=6-50.5=-45.5$$






      share|cite|improve this answer























      • Thank you for clarifying that. I thought it was 1/100 also every time, but I caught myself using the wrong reasoning.
        – SpiralRain
        Nov 30 '18 at 19:05












      • I’m having a hard time understanding this, i get how you come to $frac{1}{100}$ for the second try but can you give the calculation for getting it right on the third try, shouldn’t it be $frac{1}{98}*frac{99}{100}$
        – user601297
        Nov 30 '18 at 22:32










      • For 3rd try, you'll have (probability of getting to 3rd pick) * (probability of getting it right). Probability of getting to 3rd pick is $frac {99} {100}$ * $frac {98} {99}$ because you have to first get 1st one wrong , then the second one wrong, and 3rd one right. So you'll get $frac {99} {100} * frac {98} {99} * frac {1} {98} = frac {1} {100}$
        – Ofya
        Nov 30 '18 at 23:01












      • Thanks i got it.
        – user601297
        Dec 1 '18 at 20:07














      10












      10








      10






      The answer posted by Jorge is right. Just to add some clarifications.



      In the first try you have $frac 1 {100}$ chance of guessing it right. On the second guess, your chance increases to $frac 1 {99}$ as you know the answer isn't your guess and you aren't going to make the same guess. However, the probability that you are going to make the second guess (i.e. you guess the first one wrong) is $frac {99} {100}$ so the probability is again, $frac 1 {99}$ * $frac {99} {100}$
      = $frac 1 {100}$. With same logic, your probability of guessing it right on the nth try is always $frac 1 {100}$



      The rest of the calculation checks out.



      $$sumlimits_{i=1}^{100} frac{6-i}{100} = 6 - frac{100times 101}{2times 100}=6-50.5=-45.5$$






      share|cite|improve this answer














      The answer posted by Jorge is right. Just to add some clarifications.



      In the first try you have $frac 1 {100}$ chance of guessing it right. On the second guess, your chance increases to $frac 1 {99}$ as you know the answer isn't your guess and you aren't going to make the same guess. However, the probability that you are going to make the second guess (i.e. you guess the first one wrong) is $frac {99} {100}$ so the probability is again, $frac 1 {99}$ * $frac {99} {100}$
      = $frac 1 {100}$. With same logic, your probability of guessing it right on the nth try is always $frac 1 {100}$



      The rest of the calculation checks out.



      $$sumlimits_{i=1}^{100} frac{6-i}{100} = 6 - frac{100times 101}{2times 100}=6-50.5=-45.5$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 30 '18 at 19:54









      Ruslan

      3,70921533




      3,70921533










      answered Nov 30 '18 at 19:00









      OfyaOfya

      5048




      5048












      • Thank you for clarifying that. I thought it was 1/100 also every time, but I caught myself using the wrong reasoning.
        – SpiralRain
        Nov 30 '18 at 19:05












      • I’m having a hard time understanding this, i get how you come to $frac{1}{100}$ for the second try but can you give the calculation for getting it right on the third try, shouldn’t it be $frac{1}{98}*frac{99}{100}$
        – user601297
        Nov 30 '18 at 22:32










      • For 3rd try, you'll have (probability of getting to 3rd pick) * (probability of getting it right). Probability of getting to 3rd pick is $frac {99} {100}$ * $frac {98} {99}$ because you have to first get 1st one wrong , then the second one wrong, and 3rd one right. So you'll get $frac {99} {100} * frac {98} {99} * frac {1} {98} = frac {1} {100}$
        – Ofya
        Nov 30 '18 at 23:01












      • Thanks i got it.
        – user601297
        Dec 1 '18 at 20:07


















      • Thank you for clarifying that. I thought it was 1/100 also every time, but I caught myself using the wrong reasoning.
        – SpiralRain
        Nov 30 '18 at 19:05












      • I’m having a hard time understanding this, i get how you come to $frac{1}{100}$ for the second try but can you give the calculation for getting it right on the third try, shouldn’t it be $frac{1}{98}*frac{99}{100}$
        – user601297
        Nov 30 '18 at 22:32










      • For 3rd try, you'll have (probability of getting to 3rd pick) * (probability of getting it right). Probability of getting to 3rd pick is $frac {99} {100}$ * $frac {98} {99}$ because you have to first get 1st one wrong , then the second one wrong, and 3rd one right. So you'll get $frac {99} {100} * frac {98} {99} * frac {1} {98} = frac {1} {100}$
        – Ofya
        Nov 30 '18 at 23:01












      • Thanks i got it.
        – user601297
        Dec 1 '18 at 20:07
















      Thank you for clarifying that. I thought it was 1/100 also every time, but I caught myself using the wrong reasoning.
      – SpiralRain
      Nov 30 '18 at 19:05






      Thank you for clarifying that. I thought it was 1/100 also every time, but I caught myself using the wrong reasoning.
      – SpiralRain
      Nov 30 '18 at 19:05














      I’m having a hard time understanding this, i get how you come to $frac{1}{100}$ for the second try but can you give the calculation for getting it right on the third try, shouldn’t it be $frac{1}{98}*frac{99}{100}$
      – user601297
      Nov 30 '18 at 22:32




      I’m having a hard time understanding this, i get how you come to $frac{1}{100}$ for the second try but can you give the calculation for getting it right on the third try, shouldn’t it be $frac{1}{98}*frac{99}{100}$
      – user601297
      Nov 30 '18 at 22:32












      For 3rd try, you'll have (probability of getting to 3rd pick) * (probability of getting it right). Probability of getting to 3rd pick is $frac {99} {100}$ * $frac {98} {99}$ because you have to first get 1st one wrong , then the second one wrong, and 3rd one right. So you'll get $frac {99} {100} * frac {98} {99} * frac {1} {98} = frac {1} {100}$
      – Ofya
      Nov 30 '18 at 23:01






      For 3rd try, you'll have (probability of getting to 3rd pick) * (probability of getting it right). Probability of getting to 3rd pick is $frac {99} {100}$ * $frac {98} {99}$ because you have to first get 1st one wrong , then the second one wrong, and 3rd one right. So you'll get $frac {99} {100} * frac {98} {99} * frac {1} {98} = frac {1} {100}$
      – Ofya
      Nov 30 '18 at 23:01














      Thanks i got it.
      – user601297
      Dec 1 '18 at 20:07




      Thanks i got it.
      – user601297
      Dec 1 '18 at 20:07











      3














      The probability you get it right at the i'th time is $frac{1}{100}$.



      Thus the expected value of the game is $$sumlimits_{i=1}^{100} frac{6-i}{100} = 6 - frac{100times 101}{2times 100}=6-50.5=-44.5$$






      share|cite|improve this answer























      • assuming someone won't ignorantly re-guess numbers they've already guessed
        – phdmba7of12
        Nov 30 '18 at 18:56










      • and that they stop guessing once they've hit the correct number ... lol
        – phdmba7of12
        Nov 30 '18 at 18:56






      • 1




        Well I thought that was a reasonable assumption
        – Jorge Fernández
        Nov 30 '18 at 18:57






      • 1




        That's the assumption for a candidate that Steve Balmer would hire, ;P
        – I like Serena
        Nov 30 '18 at 19:01










      • Thanks a lot, I got it.
        – user601297
        Nov 30 '18 at 22:11
















      3














      The probability you get it right at the i'th time is $frac{1}{100}$.



      Thus the expected value of the game is $$sumlimits_{i=1}^{100} frac{6-i}{100} = 6 - frac{100times 101}{2times 100}=6-50.5=-44.5$$






      share|cite|improve this answer























      • assuming someone won't ignorantly re-guess numbers they've already guessed
        – phdmba7of12
        Nov 30 '18 at 18:56










      • and that they stop guessing once they've hit the correct number ... lol
        – phdmba7of12
        Nov 30 '18 at 18:56






      • 1




        Well I thought that was a reasonable assumption
        – Jorge Fernández
        Nov 30 '18 at 18:57






      • 1




        That's the assumption for a candidate that Steve Balmer would hire, ;P
        – I like Serena
        Nov 30 '18 at 19:01










      • Thanks a lot, I got it.
        – user601297
        Nov 30 '18 at 22:11














      3












      3








      3






      The probability you get it right at the i'th time is $frac{1}{100}$.



      Thus the expected value of the game is $$sumlimits_{i=1}^{100} frac{6-i}{100} = 6 - frac{100times 101}{2times 100}=6-50.5=-44.5$$






      share|cite|improve this answer














      The probability you get it right at the i'th time is $frac{1}{100}$.



      Thus the expected value of the game is $$sumlimits_{i=1}^{100} frac{6-i}{100} = 6 - frac{100times 101}{2times 100}=6-50.5=-44.5$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 30 '18 at 19:54









      Ruslan

      3,70921533




      3,70921533










      answered Nov 30 '18 at 18:54









      Jorge FernándezJorge Fernández

      75.1k1190191




      75.1k1190191












      • assuming someone won't ignorantly re-guess numbers they've already guessed
        – phdmba7of12
        Nov 30 '18 at 18:56










      • and that they stop guessing once they've hit the correct number ... lol
        – phdmba7of12
        Nov 30 '18 at 18:56






      • 1




        Well I thought that was a reasonable assumption
        – Jorge Fernández
        Nov 30 '18 at 18:57






      • 1




        That's the assumption for a candidate that Steve Balmer would hire, ;P
        – I like Serena
        Nov 30 '18 at 19:01










      • Thanks a lot, I got it.
        – user601297
        Nov 30 '18 at 22:11


















      • assuming someone won't ignorantly re-guess numbers they've already guessed
        – phdmba7of12
        Nov 30 '18 at 18:56










      • and that they stop guessing once they've hit the correct number ... lol
        – phdmba7of12
        Nov 30 '18 at 18:56






      • 1




        Well I thought that was a reasonable assumption
        – Jorge Fernández
        Nov 30 '18 at 18:57






      • 1




        That's the assumption for a candidate that Steve Balmer would hire, ;P
        – I like Serena
        Nov 30 '18 at 19:01










      • Thanks a lot, I got it.
        – user601297
        Nov 30 '18 at 22:11
















      assuming someone won't ignorantly re-guess numbers they've already guessed
      – phdmba7of12
      Nov 30 '18 at 18:56




      assuming someone won't ignorantly re-guess numbers they've already guessed
      – phdmba7of12
      Nov 30 '18 at 18:56












      and that they stop guessing once they've hit the correct number ... lol
      – phdmba7of12
      Nov 30 '18 at 18:56




      and that they stop guessing once they've hit the correct number ... lol
      – phdmba7of12
      Nov 30 '18 at 18:56




      1




      1




      Well I thought that was a reasonable assumption
      – Jorge Fernández
      Nov 30 '18 at 18:57




      Well I thought that was a reasonable assumption
      – Jorge Fernández
      Nov 30 '18 at 18:57




      1




      1




      That's the assumption for a candidate that Steve Balmer would hire, ;P
      – I like Serena
      Nov 30 '18 at 19:01




      That's the assumption for a candidate that Steve Balmer would hire, ;P
      – I like Serena
      Nov 30 '18 at 19:01












      Thanks a lot, I got it.
      – user601297
      Nov 30 '18 at 22:11




      Thanks a lot, I got it.
      – user601297
      Nov 30 '18 at 22:11











      2














      If Steve tells you whether the number is higher or lower then, you will at most lose a dollar.
      Going by the binary approach, you start with 50, worst situation: more numbers on the side which Steve's number is, up you go 75(as 50 on higher side and 49 on the lower side), then 88, then 94, then 97, 99 and finally 98 or 100. Even if you start with 50 and the number is lower than 50, you still pay a dollar at most.



      I have arbitrarily chosen the higher end in each case to have equal to or more numbers than the lower end.



      Taking this to be the most conservative approach and that you only follow this, if you want to find the average amount gained or lost, find the value for each number (formulate, no need to solve for each) and take average.






      share|cite|improve this answer


























        2














        If Steve tells you whether the number is higher or lower then, you will at most lose a dollar.
        Going by the binary approach, you start with 50, worst situation: more numbers on the side which Steve's number is, up you go 75(as 50 on higher side and 49 on the lower side), then 88, then 94, then 97, 99 and finally 98 or 100. Even if you start with 50 and the number is lower than 50, you still pay a dollar at most.



        I have arbitrarily chosen the higher end in each case to have equal to or more numbers than the lower end.



        Taking this to be the most conservative approach and that you only follow this, if you want to find the average amount gained or lost, find the value for each number (formulate, no need to solve for each) and take average.






        share|cite|improve this answer
























          2












          2








          2






          If Steve tells you whether the number is higher or lower then, you will at most lose a dollar.
          Going by the binary approach, you start with 50, worst situation: more numbers on the side which Steve's number is, up you go 75(as 50 on higher side and 49 on the lower side), then 88, then 94, then 97, 99 and finally 98 or 100. Even if you start with 50 and the number is lower than 50, you still pay a dollar at most.



          I have arbitrarily chosen the higher end in each case to have equal to or more numbers than the lower end.



          Taking this to be the most conservative approach and that you only follow this, if you want to find the average amount gained or lost, find the value for each number (formulate, no need to solve for each) and take average.






          share|cite|improve this answer












          If Steve tells you whether the number is higher or lower then, you will at most lose a dollar.
          Going by the binary approach, you start with 50, worst situation: more numbers on the side which Steve's number is, up you go 75(as 50 on higher side and 49 on the lower side), then 88, then 94, then 97, 99 and finally 98 or 100. Even if you start with 50 and the number is lower than 50, you still pay a dollar at most.



          I have arbitrarily chosen the higher end in each case to have equal to or more numbers than the lower end.



          Taking this to be the most conservative approach and that you only follow this, if you want to find the average amount gained or lost, find the value for each number (formulate, no need to solve for each) and take average.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 22:46









          Delta UnicronDelta Unicron

          313




          313























              1














              Let $X_t:=1{text{guess at stage $t$ is correct}}$ and let $T:=min{t:X_t=1}$. Then for $1le tle 100$,
              begin{align}
              mathsf{P}(T=t)&=mathsf{P}(X_t=1,X_{t-1}=0,ldots,X_1=0) \
              &=mathsf{P}(X_t=1|X_{t-1}=0,ldots,X_1=0)timesmathsf{P}(X_{t-1}=0,ldots,X_1=0) \
              &=frac{1}{100-(t-1)}timesbinom{99}{t-1}binom{100}{t-1}^{-1}=frac{1}{100}.
              end{align}

              Thus, the stopping time $T$ is uniformly distributed on ${1,ldots,100}$ and its mean is $50.5$. Since at each step you loose $1$$, the mean loss is $6-50.5=-44.5$.






              share|cite|improve this answer




























                1














                Let $X_t:=1{text{guess at stage $t$ is correct}}$ and let $T:=min{t:X_t=1}$. Then for $1le tle 100$,
                begin{align}
                mathsf{P}(T=t)&=mathsf{P}(X_t=1,X_{t-1}=0,ldots,X_1=0) \
                &=mathsf{P}(X_t=1|X_{t-1}=0,ldots,X_1=0)timesmathsf{P}(X_{t-1}=0,ldots,X_1=0) \
                &=frac{1}{100-(t-1)}timesbinom{99}{t-1}binom{100}{t-1}^{-1}=frac{1}{100}.
                end{align}

                Thus, the stopping time $T$ is uniformly distributed on ${1,ldots,100}$ and its mean is $50.5$. Since at each step you loose $1$$, the mean loss is $6-50.5=-44.5$.






                share|cite|improve this answer


























                  1












                  1








                  1






                  Let $X_t:=1{text{guess at stage $t$ is correct}}$ and let $T:=min{t:X_t=1}$. Then for $1le tle 100$,
                  begin{align}
                  mathsf{P}(T=t)&=mathsf{P}(X_t=1,X_{t-1}=0,ldots,X_1=0) \
                  &=mathsf{P}(X_t=1|X_{t-1}=0,ldots,X_1=0)timesmathsf{P}(X_{t-1}=0,ldots,X_1=0) \
                  &=frac{1}{100-(t-1)}timesbinom{99}{t-1}binom{100}{t-1}^{-1}=frac{1}{100}.
                  end{align}

                  Thus, the stopping time $T$ is uniformly distributed on ${1,ldots,100}$ and its mean is $50.5$. Since at each step you loose $1$$, the mean loss is $6-50.5=-44.5$.






                  share|cite|improve this answer














                  Let $X_t:=1{text{guess at stage $t$ is correct}}$ and let $T:=min{t:X_t=1}$. Then for $1le tle 100$,
                  begin{align}
                  mathsf{P}(T=t)&=mathsf{P}(X_t=1,X_{t-1}=0,ldots,X_1=0) \
                  &=mathsf{P}(X_t=1|X_{t-1}=0,ldots,X_1=0)timesmathsf{P}(X_{t-1}=0,ldots,X_1=0) \
                  &=frac{1}{100-(t-1)}timesbinom{99}{t-1}binom{100}{t-1}^{-1}=frac{1}{100}.
                  end{align}

                  Thus, the stopping time $T$ is uniformly distributed on ${1,ldots,100}$ and its mean is $50.5$. Since at each step you loose $1$$, the mean loss is $6-50.5=-44.5$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 '18 at 22:28

























                  answered Nov 30 '18 at 20:23









                  d.k.o.d.k.o.

                  8,622528




                  8,622528






























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