9 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.

Guest Jan 11, 2020

#1**+1 **

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.

Guest Jan 11, 2020

#5**+1 **

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 % **

Guest Jan 12, 2020

#7**0 **

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.

Guest Jan 12, 2020

#9**+1 **

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

Guest Jan 12, 2020

#10**+1 **

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 = \( 6C3 *\frac{9!}{2!2!2!}=45360=20*45360 = 907200\)

-----------------------

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.**

Melody Jan 12, 2020

#12**+1 **

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.

Guest Jan 12, 2020

#13**+1 **

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 %

Guest Jan 12, 2020

#14**0 **

Thanks guest,

I almost never remember combination formulas.

They are great if you want to write programs to solve general situations but for me I am only usually interested in one off solutions so I want to understand how to get the answer purely from my own logic.

I will try and work out why this formula works. If I do then I will try to include the technique in my future one of problem solving.

Melody
Jan 12, 2020