Calculating the probability of obtaining exactly four distinct values when a die is rolled six times
$begingroup$
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
edited Jan 16 at 18:32
N. F. Taussig
45.1k103358
asked Jan 16 at 17:56
nuune nuune
73
$endgroup$
add a comment |
$begingroup$
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
edited Jan 16 at 18:32
N. F. Taussig
45.1k103358
asked Jan 16 at 17:56
nuune nuune
73
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 16 at 18:32
N. F. Taussig
45.1k103358
asked Jan 16 at 17:56
nuune nuune
73
$endgroup$
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
edited Jan 16 at 18:32
N. F. Taussig
45.1k103358
asked Jan 16 at 17:56
nuune nuune
73
edited Jan 16 at 18:32
N. F. Taussig
45.1k103358
asked Jan 16 at 17:56
nuune nuune
73
edited Jan 16 at 18:32
N. F. Taussig
45.1k103358
edited Jan 16 at 18:32
N. F. Taussig
45.1k103358
edited Jan 16 at 18:32
N. F. Taussig
45.1k103358
45.1k103358
asked Jan 16 at 17:56
nuune nuune
73
asked Jan 16 at 17:56
nuune nuune
73
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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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
$$frac{binom{6}{4}left[4cdotbinom{6}{3}cdot 3! + binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2right]}{6^{6}}$$
answered Jan 16 at 18:08
pwerthpwerth
3,320417
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1 Answer
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1 Answer
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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
$$frac{binom{6}{4}left[4cdotbinom{6}{3}cdot 3! + binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2right]}{6^{6}}$$
answered Jan 16 at 18:08
pwerthpwerth
3,320417
$endgroup$
add a comment |
$begingroup$
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
$$frac{binom{6}{4}left[4cdotbinom{6}{3}cdot 3! + binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2right]}{6^{6}}$$
answered Jan 16 at 18:08
pwerthpwerth
3,320417
$endgroup$
add a comment |
$begingroup$
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
$$frac{binom{6}{4}left[4cdotbinom{6}{3}cdot 3! + binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2right]}{6^{6}}$$
answered Jan 16 at 18:08
pwerthpwerth
3,320417
$endgroup$
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
$$frac{binom{6}{4}left[4cdotbinom{6}{3}cdot 3! + binom{4}{2}cdotbinom{6}{2}cdotbinom{4}{2}cdot 2right]}{6^{6}}$$
answered Jan 16 at 18:08
pwerthpwerth
3,320417
answered Jan 16 at 18:08
pwerthpwerth
3,320417
answered Jan 16 at 18:08
pwerthpwerth
3,320417
answered Jan 16 at 18:08
pwerthpwerth
3,320417
3,320417
add a comment |
add a comment |
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$begingroup$
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