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By rolling a dice 7 times, what is the probability that we could get all the sides to come up at least once?

My attempt: Let the first role be any number.

2nd roll: 5/6 chance to get different number

3rd roll: 4/6 chance to get different number

4th roll: 3/6 chance to get different number

5th roll: 2/6 chance to get different number

6th roll: 1/6 chance to get different number

7th roll: Does not matter?

Ans: 5/6 * 4/6 * 3/6 * 2/6 * 1/6

Am i right? If wrong please explain with answer thanks!

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Since all sides come up at least once, only one can repeat. Thus the number of possibilities is $\frac{7!}{2!1!1!1!1!1!}\times 6 = 3\times 7!$. The probability is $\frac{3\times 7!}{6^7}$

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You haven't counted the cases when for example the 2nd and 3rd throw have the same face, but all faces appeared during the 7 throws.

To count all the faces consider the sets $\{1,2,3,4,5,6,6\},\{1,2,3,4,5,5,6\}$, $\{1,2,3,4,4,5,6\}$,$ \{1,2,3,3,4,5,6\}, \{1,2,2,3,4,5,6\}, \{1,1,2,3,4,5,6\}$ and using the multinomial formula find all different permutations of each of it. These are all "good" combinations and divide by the number of all possible combination to get the wanted probability.

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Number of valid results:

  • Choose the repeated result: $6$ possibilities.
  • Choose the order of the results (permutations with multisets): $\binom{7}{2,1,\ldots,1}=2520$ possibilities.

Number of possible results:

  • Seven throws of a die: $6^7$.

Result:

$$\frac{150120}{6^7}$$

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You are correct about the six rolls, however the "seventh" roll may also occur at one of the other rolls (besides the first, because that one is always unique).

You can add all probabilities of the different location of the not unique rolls. So for the second roll you have one of of six which is not unique, for the third roll you have two out of six, ect.

$$ P=\frac{6!}{6^6}\left(\frac16 +\frac26 +\frac36 +\frac46 +\frac56 +\frac66\right). $$

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