A game of probability

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












An unbiased die having the numbers 1,2,3,4,5,6 is rolled 4 times. What is the probability that the minimum face value is 2?



According to my reasoning the answer should be $frac5^46^4$ because we have the options 2,3,4,5 and 6 . Which can be chosen for four trials in $5^4$ ways .



However the correct answer seems to be $frac5^4-4^46^4$ and I can’t reason it out. Please help me where I’m missing out



Thank you !










share|cite|improve this question





















  • $(5/6)^4$ is the probability that the minimum face value is $ge 2$.
    – ncmathsadist
    51 mins ago











  • @ncmathsadist so do you think that my answer is correct ? Please help
    – Aditi
    50 mins ago














up vote
1
down vote

favorite












An unbiased die having the numbers 1,2,3,4,5,6 is rolled 4 times. What is the probability that the minimum face value is 2?



According to my reasoning the answer should be $frac5^46^4$ because we have the options 2,3,4,5 and 6 . Which can be chosen for four trials in $5^4$ ways .



However the correct answer seems to be $frac5^4-4^46^4$ and I can’t reason it out. Please help me where I’m missing out



Thank you !










share|cite|improve this question





















  • $(5/6)^4$ is the probability that the minimum face value is $ge 2$.
    – ncmathsadist
    51 mins ago











  • @ncmathsadist so do you think that my answer is correct ? Please help
    – Aditi
    50 mins ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











An unbiased die having the numbers 1,2,3,4,5,6 is rolled 4 times. What is the probability that the minimum face value is 2?



According to my reasoning the answer should be $frac5^46^4$ because we have the options 2,3,4,5 and 6 . Which can be chosen for four trials in $5^4$ ways .



However the correct answer seems to be $frac5^4-4^46^4$ and I can’t reason it out. Please help me where I’m missing out



Thank you !










share|cite|improve this question













An unbiased die having the numbers 1,2,3,4,5,6 is rolled 4 times. What is the probability that the minimum face value is 2?



According to my reasoning the answer should be $frac5^46^4$ because we have the options 2,3,4,5 and 6 . Which can be chosen for four trials in $5^4$ ways .



However the correct answer seems to be $frac5^4-4^46^4$ and I can’t reason it out. Please help me where I’m missing out



Thank you !







probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 53 mins ago









Aditi

708314




708314











  • $(5/6)^4$ is the probability that the minimum face value is $ge 2$.
    – ncmathsadist
    51 mins ago











  • @ncmathsadist so do you think that my answer is correct ? Please help
    – Aditi
    50 mins ago
















  • $(5/6)^4$ is the probability that the minimum face value is $ge 2$.
    – ncmathsadist
    51 mins ago











  • @ncmathsadist so do you think that my answer is correct ? Please help
    – Aditi
    50 mins ago















$(5/6)^4$ is the probability that the minimum face value is $ge 2$.
– ncmathsadist
51 mins ago





$(5/6)^4$ is the probability that the minimum face value is $ge 2$.
– ncmathsadist
51 mins ago













@ncmathsadist so do you think that my answer is correct ? Please help
– Aditi
50 mins ago




@ncmathsadist so do you think that my answer is correct ? Please help
– Aditi
50 mins ago










3 Answers
3






active

oldest

votes

















up vote
3
down vote



accepted










There are $5^4$ roll sequences formed from the numbers 2, 3, 4, 5, 6 (and thus having minimum digit at least 2). Of these, $4^4$ roll sequences are formed from 3, 4, 5, 6, so cannot have minimum digit 2. The remaining $5^4-4^4$ sequences thus have minimum digit exactly 2, as required, and dividing by the $6^4$ rolls in total yields the correct probability.






share|cite|improve this answer




















  • Ohhh I didn’t know that it was strict to have 2 in the answer . They never mentioned that $2$ must be one of the outcomes though so I was a bit confused. How do you deduce that $2$ must be one of the outcomes ?
    – Aditi
    47 mins ago






  • 1




    @Aditi The minimum face value is stipulated to be 2. This means that at least one die actually shows 2, to realise the minimum.
    – Parcly Taxel
    46 mins ago











  • Ohh now I get it ! Thank you :)
    – Aditi
    45 mins ago

















up vote
3
down vote













Hint:



Your answer would be good if the question were "What is the probability that the minimum face is not $1$"?






share|cite|improve this answer




















  • Thank you , but how do we deduce that 2 must be one of the outcomes ?
    – Aditi
    46 mins ago










  • You can't. Every roll could be fives and sixes (for example). There it is the point.
    – ajotatxe
    45 mins ago











  • Thank you for the hint !
    – Aditi
    42 mins ago

















up vote
0
down vote













You can also calculate the numerator directly: 4*4^3 sequences have a single 2: 4C2*4^2 have two 2's; 4C3*4 sequences have 3 2's; 1 sequence of 2222 (all sequences containing no 1's). If you add these up, you get 369=5^4-4^4.






share|cite|improve this answer








New contributor




G. Thompson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















    Your Answer




    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2962362%2fa-game-of-probability%23new-answer', 'question_page');

    );

    Post as a guest






























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    There are $5^4$ roll sequences formed from the numbers 2, 3, 4, 5, 6 (and thus having minimum digit at least 2). Of these, $4^4$ roll sequences are formed from 3, 4, 5, 6, so cannot have minimum digit 2. The remaining $5^4-4^4$ sequences thus have minimum digit exactly 2, as required, and dividing by the $6^4$ rolls in total yields the correct probability.






    share|cite|improve this answer




















    • Ohhh I didn’t know that it was strict to have 2 in the answer . They never mentioned that $2$ must be one of the outcomes though so I was a bit confused. How do you deduce that $2$ must be one of the outcomes ?
      – Aditi
      47 mins ago






    • 1




      @Aditi The minimum face value is stipulated to be 2. This means that at least one die actually shows 2, to realise the minimum.
      – Parcly Taxel
      46 mins ago











    • Ohh now I get it ! Thank you :)
      – Aditi
      45 mins ago














    up vote
    3
    down vote



    accepted










    There are $5^4$ roll sequences formed from the numbers 2, 3, 4, 5, 6 (and thus having minimum digit at least 2). Of these, $4^4$ roll sequences are formed from 3, 4, 5, 6, so cannot have minimum digit 2. The remaining $5^4-4^4$ sequences thus have minimum digit exactly 2, as required, and dividing by the $6^4$ rolls in total yields the correct probability.






    share|cite|improve this answer




















    • Ohhh I didn’t know that it was strict to have 2 in the answer . They never mentioned that $2$ must be one of the outcomes though so I was a bit confused. How do you deduce that $2$ must be one of the outcomes ?
      – Aditi
      47 mins ago






    • 1




      @Aditi The minimum face value is stipulated to be 2. This means that at least one die actually shows 2, to realise the minimum.
      – Parcly Taxel
      46 mins ago











    • Ohh now I get it ! Thank you :)
      – Aditi
      45 mins ago












    up vote
    3
    down vote



    accepted







    up vote
    3
    down vote



    accepted






    There are $5^4$ roll sequences formed from the numbers 2, 3, 4, 5, 6 (and thus having minimum digit at least 2). Of these, $4^4$ roll sequences are formed from 3, 4, 5, 6, so cannot have minimum digit 2. The remaining $5^4-4^4$ sequences thus have minimum digit exactly 2, as required, and dividing by the $6^4$ rolls in total yields the correct probability.






    share|cite|improve this answer












    There are $5^4$ roll sequences formed from the numbers 2, 3, 4, 5, 6 (and thus having minimum digit at least 2). Of these, $4^4$ roll sequences are formed from 3, 4, 5, 6, so cannot have minimum digit 2. The remaining $5^4-4^4$ sequences thus have minimum digit exactly 2, as required, and dividing by the $6^4$ rolls in total yields the correct probability.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 49 mins ago









    Parcly Taxel

    36.6k137095




    36.6k137095











    • Ohhh I didn’t know that it was strict to have 2 in the answer . They never mentioned that $2$ must be one of the outcomes though so I was a bit confused. How do you deduce that $2$ must be one of the outcomes ?
      – Aditi
      47 mins ago






    • 1




      @Aditi The minimum face value is stipulated to be 2. This means that at least one die actually shows 2, to realise the minimum.
      – Parcly Taxel
      46 mins ago











    • Ohh now I get it ! Thank you :)
      – Aditi
      45 mins ago
















    • Ohhh I didn’t know that it was strict to have 2 in the answer . They never mentioned that $2$ must be one of the outcomes though so I was a bit confused. How do you deduce that $2$ must be one of the outcomes ?
      – Aditi
      47 mins ago






    • 1




      @Aditi The minimum face value is stipulated to be 2. This means that at least one die actually shows 2, to realise the minimum.
      – Parcly Taxel
      46 mins ago











    • Ohh now I get it ! Thank you :)
      – Aditi
      45 mins ago















    Ohhh I didn’t know that it was strict to have 2 in the answer . They never mentioned that $2$ must be one of the outcomes though so I was a bit confused. How do you deduce that $2$ must be one of the outcomes ?
    – Aditi
    47 mins ago




    Ohhh I didn’t know that it was strict to have 2 in the answer . They never mentioned that $2$ must be one of the outcomes though so I was a bit confused. How do you deduce that $2$ must be one of the outcomes ?
    – Aditi
    47 mins ago




    1




    1




    @Aditi The minimum face value is stipulated to be 2. This means that at least one die actually shows 2, to realise the minimum.
    – Parcly Taxel
    46 mins ago





    @Aditi The minimum face value is stipulated to be 2. This means that at least one die actually shows 2, to realise the minimum.
    – Parcly Taxel
    46 mins ago













    Ohh now I get it ! Thank you :)
    – Aditi
    45 mins ago




    Ohh now I get it ! Thank you :)
    – Aditi
    45 mins ago










    up vote
    3
    down vote













    Hint:



    Your answer would be good if the question were "What is the probability that the minimum face is not $1$"?






    share|cite|improve this answer




















    • Thank you , but how do we deduce that 2 must be one of the outcomes ?
      – Aditi
      46 mins ago










    • You can't. Every roll could be fives and sixes (for example). There it is the point.
      – ajotatxe
      45 mins ago











    • Thank you for the hint !
      – Aditi
      42 mins ago














    up vote
    3
    down vote













    Hint:



    Your answer would be good if the question were "What is the probability that the minimum face is not $1$"?






    share|cite|improve this answer




















    • Thank you , but how do we deduce that 2 must be one of the outcomes ?
      – Aditi
      46 mins ago










    • You can't. Every roll could be fives and sixes (for example). There it is the point.
      – ajotatxe
      45 mins ago











    • Thank you for the hint !
      – Aditi
      42 mins ago












    up vote
    3
    down vote










    up vote
    3
    down vote









    Hint:



    Your answer would be good if the question were "What is the probability that the minimum face is not $1$"?






    share|cite|improve this answer












    Hint:



    Your answer would be good if the question were "What is the probability that the minimum face is not $1$"?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 49 mins ago









    ajotatxe

    50.7k13185




    50.7k13185











    • Thank you , but how do we deduce that 2 must be one of the outcomes ?
      – Aditi
      46 mins ago










    • You can't. Every roll could be fives and sixes (for example). There it is the point.
      – ajotatxe
      45 mins ago











    • Thank you for the hint !
      – Aditi
      42 mins ago
















    • Thank you , but how do we deduce that 2 must be one of the outcomes ?
      – Aditi
      46 mins ago










    • You can't. Every roll could be fives and sixes (for example). There it is the point.
      – ajotatxe
      45 mins ago











    • Thank you for the hint !
      – Aditi
      42 mins ago















    Thank you , but how do we deduce that 2 must be one of the outcomes ?
    – Aditi
    46 mins ago




    Thank you , but how do we deduce that 2 must be one of the outcomes ?
    – Aditi
    46 mins ago












    You can't. Every roll could be fives and sixes (for example). There it is the point.
    – ajotatxe
    45 mins ago





    You can't. Every roll could be fives and sixes (for example). There it is the point.
    – ajotatxe
    45 mins ago













    Thank you for the hint !
    – Aditi
    42 mins ago




    Thank you for the hint !
    – Aditi
    42 mins ago










    up vote
    0
    down vote













    You can also calculate the numerator directly: 4*4^3 sequences have a single 2: 4C2*4^2 have two 2's; 4C3*4 sequences have 3 2's; 1 sequence of 2222 (all sequences containing no 1's). If you add these up, you get 369=5^4-4^4.






    share|cite|improve this answer








    New contributor




    G. Thompson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





















      up vote
      0
      down vote













      You can also calculate the numerator directly: 4*4^3 sequences have a single 2: 4C2*4^2 have two 2's; 4C3*4 sequences have 3 2's; 1 sequence of 2222 (all sequences containing no 1's). If you add these up, you get 369=5^4-4^4.






      share|cite|improve this answer








      New contributor




      G. Thompson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.



















        up vote
        0
        down vote










        up vote
        0
        down vote









        You can also calculate the numerator directly: 4*4^3 sequences have a single 2: 4C2*4^2 have two 2's; 4C3*4 sequences have 3 2's; 1 sequence of 2222 (all sequences containing no 1's). If you add these up, you get 369=5^4-4^4.






        share|cite|improve this answer








        New contributor




        G. Thompson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        You can also calculate the numerator directly: 4*4^3 sequences have a single 2: 4C2*4^2 have two 2's; 4C3*4 sequences have 3 2's; 1 sequence of 2222 (all sequences containing no 1's). If you add these up, you get 369=5^4-4^4.







        share|cite|improve this answer








        New contributor




        G. Thompson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|cite|improve this answer



        share|cite|improve this answer






        New contributor




        G. Thompson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered 29 mins ago









        G. Thompson

        11




        11




        New contributor




        G. Thompson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.





        New contributor





        G. Thompson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        G. Thompson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.



























             

            draft saved


            draft discarded















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2962362%2fa-game-of-probability%23new-answer', 'question_page');

            );

            Post as a guest













































































            Comments

            Popular posts from this blog

            Long meetings (6-7 hours a day): Being “babysat” by supervisor

            Is the Concept of Multiple Fantasy Races Scientifically Flawed? [closed]

            Confectionery