Calculating the probability of obtaining exactly four distinct values when a die is rolled six times












1












$begingroup$


Please could someone help me with determining the probability of getting $4$ distinct numbers (no order in the outcome e.g. 1,2,3,4 or 4,5,6,2 etc) or $5$ distinct from rolling a die $6$ times. So far I was able to calculate the probability of getting $6$ distinct numbers from $6$ rolls of a dice



$$frac{6!}{6^6}$$



But I am having issues determining if it's only $5$ or $4$ distinct numbers out of a $6$ rolls.



Additionally, is this type of probability binomial or hypergeometric?






probability combinatorics dice







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean "what is the probability of rolling at least one $4$ out of six tries with a fair dice?" If so, Hint; it is easier to compute the complement. That is, the probability of getting no $4's$.
    $endgroup$
    – lulu
    Jan 16 at 18:03










  • $begingroup$
    Are you asking for the probability of obtaining exactly four distinct numbers when six six-sided dice are rolled?
    $endgroup$
    – N. F. Taussig
    Jan 16 at 18:07










  • $begingroup$
    Yes. Sorry I edited. I was looking for exactly 4 distinct number when six sided dice is rolled 6 times. and also for 5 distinct numbers when a dice is rolled six times
    $endgroup$
    – nuune
    Jan 16 at 18:24










  • $begingroup$
    Thanks for the clarification. Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Jan 16 at 18:34
















1












$begingroup$


Please could someone help me with determining the probability of getting $4$ distinct numbers (no order in the outcome e.g. 1,2,3,4 or 4,5,6,2 etc) or $5$ distinct from rolling a die $6$ times. So far I was able to calculate the probability of getting $6$ distinct numbers from $6$ rolls of a dice



$$frac{6!}{6^6}$$



But I am having issues determining if it's only $5$ or $4$ distinct numbers out of a $6$ rolls.



Additionally, is this type of probability binomial or hypergeometric?






probability combinatorics dice







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean "what is the probability of rolling at least one $4$ out of six tries with a fair dice?" If so, Hint; it is easier to compute the complement. That is, the probability of getting no $4's$.
    $endgroup$
    – lulu
    Jan 16 at 18:03










  • $begingroup$
    Are you asking for the probability of obtaining exactly four distinct numbers when six six-sided dice are rolled?
    $endgroup$
    – N. F. Taussig
    Jan 16 at 18:07










  • $begingroup$
    Yes. Sorry I edited. I was looking for exactly 4 distinct number when six sided dice is rolled 6 times. and also for 5 distinct numbers when a dice is rolled six times
    $endgroup$
    – nuune
    Jan 16 at 18:24










  • $begingroup$
    Thanks for the clarification. Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Jan 16 at 18:34














1












1








1


1



$begingroup$


Please could someone help me with determining the probability of getting $4$ distinct numbers (no order in the outcome e.g. 1,2,3,4 or 4,5,6,2 etc) or $5$ distinct from rolling a die $6$ times. So far I was able to calculate the probability of getting $6$ distinct numbers from $6$ rolls of a dice



$$frac{6!}{6^6}$$



But I am having issues determining if it's only $5$ or $4$ distinct numbers out of a $6$ rolls.



Additionally, is this type of probability binomial or hypergeometric?






probability combinatorics dice







share|cite|improve this question











$endgroup$




Please could someone help me with determining the probability of getting $4$ distinct numbers (no order in the outcome e.g. 1,2,3,4 or 4,5,6,2 etc) or $5$ distinct from rolling a die $6$ times. So far I was able to calculate the probability of getting $6$ distinct numbers from $6$ rolls of a dice



$$frac{6!}{6^6}$$



But I am having issues determining if it's only $5$ or $4$ distinct numbers out of a $6$ rolls.



Additionally, is this type of probability binomial or hypergeometric?






probability combinatorics dice




probability combinatorics dice






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 18:32









N. F. Taussig

45.1k103358




45.1k103358










asked Jan 16 at 17:56









nuune nuune

73




73












  • $begingroup$
    Do you mean "what is the probability of rolling at least one $4$ out of six tries with a fair dice?" If so, Hint; it is easier to compute the complement. That is, the probability of getting no $4's$.
    $endgroup$
    – lulu
    Jan 16 at 18:03










  • $begingroup$
    Are you asking for the probability of obtaining exactly four distinct numbers when six six-sided dice are rolled?
    $endgroup$
    – N. F. Taussig
    Jan 16 at 18:07










  • $begingroup$
    Yes. Sorry I edited. I was looking for exactly 4 distinct number when six sided dice is rolled 6 times. and also for 5 distinct numbers when a dice is rolled six times
    $endgroup$
    – nuune
    Jan 16 at 18:24










  • $begingroup$
    Thanks for the clarification. Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Jan 16 at 18:34


















  • $begingroup$
    Do you mean "what is the probability of rolling at least one $4$ out of six tries with a fair dice?" If so, Hint; it is easier to compute the complement. That is, the probability of getting no $4's$.
    $endgroup$
    – lulu
    Jan 16 at 18:03










  • $begingroup$
    Are you asking for the probability of obtaining exactly four distinct numbers when six six-sided dice are rolled?
    $endgroup$
    – N. F. Taussig
    Jan 16 at 18:07










  • $begingroup$
    Yes. Sorry I edited. I was looking for exactly 4 distinct number when six sided dice is rolled 6 times. and also for 5 distinct numbers when a dice is rolled six times
    $endgroup$
    – nuune
    Jan 16 at 18:24










  • $begingroup$
    Thanks for the clarification. Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Jan 16 at 18:34
















$begingroup$
Do you mean "what is the probability of rolling at least one $4$ out of six tries with a fair dice?" If so, Hint; it is easier to compute the complement. That is, the probability of getting no $4's$.
$endgroup$
– lulu
Jan 16 at 18:03




$begingroup$
Do you mean "what is the probability of rolling at least one $4$ out of six tries with a fair dice?" If so, Hint; it is easier to compute the complement. That is, the probability of getting no $4's$.
$endgroup$
– lulu
Jan 16 at 18:03












$begingroup$
Are you asking for the probability of obtaining exactly four distinct numbers when six six-sided dice are rolled?
$endgroup$
– N. F. Taussig
Jan 16 at 18:07




$begingroup$
Are you asking for the probability of obtaining exactly four distinct numbers when six six-sided dice are rolled?
$endgroup$
– N. F. Taussig
Jan 16 at 18:07












$begingroup$
Yes. Sorry I edited. I was looking for exactly 4 distinct number when six sided dice is rolled 6 times. and also for 5 distinct numbers when a dice is rolled six times
$endgroup$
– nuune
Jan 16 at 18:24




$begingroup$
Yes. Sorry I edited. I was looking for exactly 4 distinct number when six sided dice is rolled 6 times. and also for 5 distinct numbers when a dice is rolled six times
$endgroup$
– nuune
Jan 16 at 18:24












$begingroup$
Thanks for the clarification. Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 16 at 18:34




$begingroup$
Thanks for the clarification. Welcome to MathSE. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 16 at 18:34










1 Answer
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1












$begingroup$

Let's see how many ways we can obtain $5$ distinct numbers from $6$ rolls. First choose which $5$ numbers will appear in $binom{6}{5}$ ways. Now observe that the only way to obtain $5$ distinct numbers in $6$ rolls is to have $4$ of the values appear exactly once, and one to appear exactly twice. There are $5$ choices for which value will appear twice and the value will appear in $binom{6}{2}$ locations. Then the remaining $4$ rolls can be ordered in $4!$ ways. This gives a total of
$$binom{6}{5}cdot 5cdotbinom{6}{2}cdot 4!$$
desirable rolls. There are $6^{6}$ possible rolls so the probability of obtaining a roll with $5$ distinct numbers is
$$frac{binom{6}{5}cdot 5cdotbinom{6}{2}cdot 4!}{6^{6}}$$



Now let's see how many ways we can obtain $4$ distinct numbers from $6$ rolls. First choose which $4$ numbers will appear in $binom{6}{4}$ ways. Now there are two cases to consider: one value appears $3$ times and each of the other values appears once, or two values appear twice and two values appear once. We'll handle these separately.




  • Case 1: There are $4$ choices for which value will appear $3$ times, and it will appear in $binom{6}{3}$ locations. We can then order the remaining rolls in $3!$ ways. So there are $4cdotbinom{6}{3}cdot 3!$ outcomes of this form.

  • Case 2: There are $binom{4}{2}$ choices for which values will appear twice, and we can place them in $binom{6}{2}cdotbinom{4}{2}$ ways. There are then $2$ possible orders for the remaining two values. So there are $binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2$ outcomes of this type.


Combining these cases, the probability of obtaining $4$ distinct numbers is
$$frac{binom{6}{4}left[4cdotbinom{6}{3}cdot 3! + binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2right]}{6^{6}}$$






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    $begingroup$

    Let's see how many ways we can obtain $5$ distinct numbers from $6$ rolls. First choose which $5$ numbers will appear in $binom{6}{5}$ ways. Now observe that the only way to obtain $5$ distinct numbers in $6$ rolls is to have $4$ of the values appear exactly once, and one to appear exactly twice. There are $5$ choices for which value will appear twice and the value will appear in $binom{6}{2}$ locations. Then the remaining $4$ rolls can be ordered in $4!$ ways. This gives a total of
    $$binom{6}{5}cdot 5cdotbinom{6}{2}cdot 4!$$
    desirable rolls. There are $6^{6}$ possible rolls so the probability of obtaining a roll with $5$ distinct numbers is
    $$frac{binom{6}{5}cdot 5cdotbinom{6}{2}cdot 4!}{6^{6}}$$



    Now let's see how many ways we can obtain $4$ distinct numbers from $6$ rolls. First choose which $4$ numbers will appear in $binom{6}{4}$ ways. Now there are two cases to consider: one value appears $3$ times and each of the other values appears once, or two values appear twice and two values appear once. We'll handle these separately.




    • Case 1: There are $4$ choices for which value will appear $3$ times, and it will appear in $binom{6}{3}$ locations. We can then order the remaining rolls in $3!$ ways. So there are $4cdotbinom{6}{3}cdot 3!$ outcomes of this form.

    • Case 2: There are $binom{4}{2}$ choices for which values will appear twice, and we can place them in $binom{6}{2}cdotbinom{4}{2}$ ways. There are then $2$ possible orders for the remaining two values. So there are $binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2$ outcomes of this type.


    Combining these cases, the probability of obtaining $4$ distinct numbers is
    $$frac{binom{6}{4}left[4cdotbinom{6}{3}cdot 3! + binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2right]}{6^{6}}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let's see how many ways we can obtain $5$ distinct numbers from $6$ rolls. First choose which $5$ numbers will appear in $binom{6}{5}$ ways. Now observe that the only way to obtain $5$ distinct numbers in $6$ rolls is to have $4$ of the values appear exactly once, and one to appear exactly twice. There are $5$ choices for which value will appear twice and the value will appear in $binom{6}{2}$ locations. Then the remaining $4$ rolls can be ordered in $4!$ ways. This gives a total of
      $$binom{6}{5}cdot 5cdotbinom{6}{2}cdot 4!$$
      desirable rolls. There are $6^{6}$ possible rolls so the probability of obtaining a roll with $5$ distinct numbers is
      $$frac{binom{6}{5}cdot 5cdotbinom{6}{2}cdot 4!}{6^{6}}$$



      Now let's see how many ways we can obtain $4$ distinct numbers from $6$ rolls. First choose which $4$ numbers will appear in $binom{6}{4}$ ways. Now there are two cases to consider: one value appears $3$ times and each of the other values appears once, or two values appear twice and two values appear once. We'll handle these separately.




      • Case 1: There are $4$ choices for which value will appear $3$ times, and it will appear in $binom{6}{3}$ locations. We can then order the remaining rolls in $3!$ ways. So there are $4cdotbinom{6}{3}cdot 3!$ outcomes of this form.

      • Case 2: There are $binom{4}{2}$ choices for which values will appear twice, and we can place them in $binom{6}{2}cdotbinom{4}{2}$ ways. There are then $2$ possible orders for the remaining two values. So there are $binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2$ outcomes of this type.


      Combining these cases, the probability of obtaining $4$ distinct numbers is
      $$frac{binom{6}{4}left[4cdotbinom{6}{3}cdot 3! + binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2right]}{6^{6}}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let's see how many ways we can obtain $5$ distinct numbers from $6$ rolls. First choose which $5$ numbers will appear in $binom{6}{5}$ ways. Now observe that the only way to obtain $5$ distinct numbers in $6$ rolls is to have $4$ of the values appear exactly once, and one to appear exactly twice. There are $5$ choices for which value will appear twice and the value will appear in $binom{6}{2}$ locations. Then the remaining $4$ rolls can be ordered in $4!$ ways. This gives a total of
        $$binom{6}{5}cdot 5cdotbinom{6}{2}cdot 4!$$
        desirable rolls. There are $6^{6}$ possible rolls so the probability of obtaining a roll with $5$ distinct numbers is
        $$frac{binom{6}{5}cdot 5cdotbinom{6}{2}cdot 4!}{6^{6}}$$



        Now let's see how many ways we can obtain $4$ distinct numbers from $6$ rolls. First choose which $4$ numbers will appear in $binom{6}{4}$ ways. Now there are two cases to consider: one value appears $3$ times and each of the other values appears once, or two values appear twice and two values appear once. We'll handle these separately.




        • Case 1: There are $4$ choices for which value will appear $3$ times, and it will appear in $binom{6}{3}$ locations. We can then order the remaining rolls in $3!$ ways. So there are $4cdotbinom{6}{3}cdot 3!$ outcomes of this form.

        • Case 2: There are $binom{4}{2}$ choices for which values will appear twice, and we can place them in $binom{6}{2}cdotbinom{4}{2}$ ways. There are then $2$ possible orders for the remaining two values. So there are $binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2$ outcomes of this type.


        Combining these cases, the probability of obtaining $4$ distinct numbers is
        $$frac{binom{6}{4}left[4cdotbinom{6}{3}cdot 3! + binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2right]}{6^{6}}$$






        share|cite|improve this answer









        $endgroup$



        Let's see how many ways we can obtain $5$ distinct numbers from $6$ rolls. First choose which $5$ numbers will appear in $binom{6}{5}$ ways. Now observe that the only way to obtain $5$ distinct numbers in $6$ rolls is to have $4$ of the values appear exactly once, and one to appear exactly twice. There are $5$ choices for which value will appear twice and the value will appear in $binom{6}{2}$ locations. Then the remaining $4$ rolls can be ordered in $4!$ ways. This gives a total of
        $$binom{6}{5}cdot 5cdotbinom{6}{2}cdot 4!$$
        desirable rolls. There are $6^{6}$ possible rolls so the probability of obtaining a roll with $5$ distinct numbers is
        $$frac{binom{6}{5}cdot 5cdotbinom{6}{2}cdot 4!}{6^{6}}$$



        Now let's see how many ways we can obtain $4$ distinct numbers from $6$ rolls. First choose which $4$ numbers will appear in $binom{6}{4}$ ways. Now there are two cases to consider: one value appears $3$ times and each of the other values appears once, or two values appear twice and two values appear once. We'll handle these separately.




        • Case 1: There are $4$ choices for which value will appear $3$ times, and it will appear in $binom{6}{3}$ locations. We can then order the remaining rolls in $3!$ ways. So there are $4cdotbinom{6}{3}cdot 3!$ outcomes of this form.

        • Case 2: There are $binom{4}{2}$ choices for which values will appear twice, and we can place them in $binom{6}{2}cdotbinom{4}{2}$ ways. There are then $2$ possible orders for the remaining two values. So there are $binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2$ outcomes of this type.


        Combining these cases, the probability of obtaining $4$ distinct numbers is
        $$frac{binom{6}{4}left[4cdotbinom{6}{3}cdot 3! + binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2right]}{6^{6}}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 18:08









        pwerthpwerth

        3,320417




        3,320417






























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