Probaility of bruteforcing a four digit lock with and without memory












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I have a problem with a lock. The lock has a four digit code. The numbers span from 0 to 9 so the number of all possible combinations is $10^4$ I have 7200 guesses. And for now I can remember all of the codes I've inputted so far. What is my chance of finding the correct code?



My (probably incorrect) thought was that the chance of getting the correct one goes up after each wrong guess and since I have 7200 of these. I could simply add up an arithmetic sequence like this $frac{7200}{2}cdot(frac{1}{10^4}+frac{1}{(10^4)-7199})$ and get the probability. It equals 0.01285% in this case which just seems wrong.



And also I have the same problem but this time I don't remember which codes I've inputted and need to find out my probability of opening the lock this time.






probability







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    1












    $begingroup$


    I have a problem with a lock. The lock has a four digit code. The numbers span from 0 to 9 so the number of all possible combinations is $10^4$ I have 7200 guesses. And for now I can remember all of the codes I've inputted so far. What is my chance of finding the correct code?



    My (probably incorrect) thought was that the chance of getting the correct one goes up after each wrong guess and since I have 7200 of these. I could simply add up an arithmetic sequence like this $frac{7200}{2}cdot(frac{1}{10^4}+frac{1}{(10^4)-7199})$ and get the probability. It equals 0.01285% in this case which just seems wrong.



    And also I have the same problem but this time I don't remember which codes I've inputted and need to find out my probability of opening the lock this time.






    probability







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I have a problem with a lock. The lock has a four digit code. The numbers span from 0 to 9 so the number of all possible combinations is $10^4$ I have 7200 guesses. And for now I can remember all of the codes I've inputted so far. What is my chance of finding the correct code?



      My (probably incorrect) thought was that the chance of getting the correct one goes up after each wrong guess and since I have 7200 of these. I could simply add up an arithmetic sequence like this $frac{7200}{2}cdot(frac{1}{10^4}+frac{1}{(10^4)-7199})$ and get the probability. It equals 0.01285% in this case which just seems wrong.



      And also I have the same problem but this time I don't remember which codes I've inputted and need to find out my probability of opening the lock this time.






      probability







      share|cite|improve this question









      $endgroup$




      I have a problem with a lock. The lock has a four digit code. The numbers span from 0 to 9 so the number of all possible combinations is $10^4$ I have 7200 guesses. And for now I can remember all of the codes I've inputted so far. What is my chance of finding the correct code?



      My (probably incorrect) thought was that the chance of getting the correct one goes up after each wrong guess and since I have 7200 of these. I could simply add up an arithmetic sequence like this $frac{7200}{2}cdot(frac{1}{10^4}+frac{1}{(10^4)-7199})$ and get the probability. It equals 0.01285% in this case which just seems wrong.



      And also I have the same problem but this time I don't remember which codes I've inputted and need to find out my probability of opening the lock this time.






      probability




      probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 12 at 10:13









      LipovlanLipovlan

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          $begingroup$

          For the first problem:



          You are inputting $7200$ different codes. So the probability of one of them being the right code is exactly $$7200/10000=72%$$



          For the second problem:



          At each attempt, the probability of failure is $9999/10000$. So for $7200$ attempts, the probability of failing each time is $(9999/10000)^{7200}$. So the probability of success is



          $$1-(9999/10000)^{7200}approx 51.33%$$






          share|cite|improve this answer











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            $begingroup$

            For the first problem:



            You are inputting $7200$ different codes. So the probability of one of them being the right code is exactly $$7200/10000=72%$$



            For the second problem:



            At each attempt, the probability of failure is $9999/10000$. So for $7200$ attempts, the probability of failing each time is $(9999/10000)^{7200}$. So the probability of success is



            $$1-(9999/10000)^{7200}approx 51.33%$$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              For the first problem:



              You are inputting $7200$ different codes. So the probability of one of them being the right code is exactly $$7200/10000=72%$$



              For the second problem:



              At each attempt, the probability of failure is $9999/10000$. So for $7200$ attempts, the probability of failing each time is $(9999/10000)^{7200}$. So the probability of success is



              $$1-(9999/10000)^{7200}approx 51.33%$$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                For the first problem:



                You are inputting $7200$ different codes. So the probability of one of them being the right code is exactly $$7200/10000=72%$$



                For the second problem:



                At each attempt, the probability of failure is $9999/10000$. So for $7200$ attempts, the probability of failing each time is $(9999/10000)^{7200}$. So the probability of success is



                $$1-(9999/10000)^{7200}approx 51.33%$$






                share|cite|improve this answer











                $endgroup$



                For the first problem:



                You are inputting $7200$ different codes. So the probability of one of them being the right code is exactly $$7200/10000=72%$$



                For the second problem:



                At each attempt, the probability of failure is $9999/10000$. So for $7200$ attempts, the probability of failing each time is $(9999/10000)^{7200}$. So the probability of success is



                $$1-(9999/10000)^{7200}approx 51.33%$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 12 at 10:32

























                answered Jan 12 at 10:24









                TonyKTonyK

                42.6k355134




                42.6k355134






























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