Dice !!

The probability that each of the first 6 rolls has a 1, 2, 3, 4, 5, or 6 is 6!/6^6.  The next three rolls can be anything, so the probability is 6!/6^6 = 5/324.

Thanks guest for the attempt.  

I have not come up with a likely answer but I am reasonably confident that yours is not correct.

Here is my attempt:

There are:[6+3-1]C3 =56 combinations of 3 remaining numbers with repeats as follows:
111, 222, 333, 444, 555, 666 = 6 - [3 of a kind.]

(112, 113, 114, 115, 116, 122, 133, 144, 155, 166, 223, 224, 225, 226, 233, 244, 255, 266, 334, 335, 336, 344, 355, 366, 445, 446, 455, 466, 556, 566)=30 - [2 of a kind ]

(123, 124, 125, 126, 134, 135, 136, 145, 146, 156, 234, 235, 236, 245, 246, 256, 345, 346, 356, 456) = 20 - [3 of different kind.]

123456111=9!/4! =15120 x 6 =90,720......The 6 is from  the above count.
123456112=9!/3!2!=30,240 x 30 =907,200......The 30 is from  the above count.
123456123=9!/2!2!2!=45,360 x 20 =907,200.......The 20 is also from the above{56 - 6 - 30 =20}.

Therefore, the probability is:
90,720 + 907,200 +907,200 =1,905,120 / 6^9 =18.90 % 

We got the permutations right!. It is the combinations of the remaining 3 numbers that we disagree. You have 6.5.4 =120, while I count only 20! Should yours be 1.5.4 ? There are only 20 combinations possible with 6C3, but with repeats (such as 222, 555, 112, 334.....etc.), there are  56 such combinations, while your total is 126, hence the large discrepancy. 

WOW!! I didn't know Wolfram did dice probabilities !! Believe or not their number agrees with mine !!

https://www.wolframalpha.com/input/?i=9+dice%2C+all+faces+show

 fair 6-sided dice are rolled. What is the probability of all 6 numbers showing (1, 2, 3, 4, 5, 6)? The remaining 3 numbers can take any value. Thank you.

Here is an adaption of my first answer.

I have corrected all silly errors and I have corrected one logic error.

Now it is correct

Number of permutations in total = 6^9 = 10, 077,696

3 doubles the others different =        \( 6C3 *\frac{9!}{2!2!2!}=45360=20*45360 = 907200\)

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The next one was the one that gave me the most trouble but it should not have

the 6C2 is because to of the numbers must be chosen as the duplicate ones.

Then i had to double it because one occured 3 times and the other only 2 and it could be either way around.

so

 1 double, 1 triple =       \(  6C2*2*\frac{9!}{3!2!}=30*30240=907200\)

-------------------------
1 quad = \(6*1*1*\frac{9!}{4!}=6*15120=90720 \)

907200+907200+90720 = 1905120 

1905120/10077696 = 0.1890432098765432

Approx   18.90%

Finally our answers agree and they also agree with Wolfram Alpha!

Micacle of miracles.

There are 56 combinations of the 3 remaining numbers as follows:

Ordinarily, the combinations should be 6C3 =20 DISTINCT combinations. But, with repeats permitted, the formula 6C3 becomes:[6 + 3 - 1] =8 C 3, which gives 56 combinations possible, which I have listed in my answer. 6 are 3 of a kind, 30 are 2 of a kind, and 20 are 3 of any kind for a total of:6 + 30 + 20 =56 distinct combinations.

Hi Melody: There is a formula to solve this kind of probability, even though I don't quite understand its logic that well; Here it is:

n =9; 6^n -  ((6 nCr 1 *5^n - (6 nCr 2*4^n) + (6 nCr 3*3^n) - (6 nCr 4*2^n) + (6 nCr 5*1^n)))

1,905,120 / 6^9 =18.90 %

Thanks for brainstorming this with me guest.

I really enjoy interacting with other people here!