If 7 dice are thrown simultaneously, then what does the probability that all six digit appears on the upper face equal to?

Question

I've approached the problem the following way :

Out of the 7 dice, I select any 6 which will have distinct numbers : 7C6.

In the 6 dice, there can be 6! ways in which distinct numbers appear.

And lastly, the last dice will have 6 possible ways in which it can show a number.

So the required answer should be : 7C6 * 6! * 6/(6^7) which on simplifying becomes : 70/(6^3 * 3).

However, the answer given is 35/(6^3 * 3).

Where exactly am I going wrong?

Answer 1

Here's a very simple, structured way of doing it. Consider a multinomial distribution with 6 outcomes. In $n=7$ trials, you want exactly two of one face and exactly one of all other faces. There are 6 equally likely situations, since it is equally likely that a face will be the one showing up twice.

For a single case, say, the probability that there are two 1's, and one of 2,3,4,5,6. The probability of this happening is $$ {7\choose 2,1,1,1,1,1} (1/6)^7. $$ Then accounting for all six equally likely situations, the final answer is $$ 6\cdot {7\choose 2,1,1,1,1,1} (1/6)^7=\frac{7!}{2!} (1/6)^6, $$ which simplifies to the correct answer.

Answer 2

You probably noticed that your answer differs from the correct answer by a factor 2, so apparently you count everything twice.

Suppose your dice are labeled A, B, C, D, E, F, G and you throw:

A:1

B:2

C: 3

D:4

E: 5

F: 6

G: 1

Then you count this throw twice: one time with ABCDEF as the 'special' dice showing 6 different figures and G as the redundant die, and once with BCDEFG as the special dice showing 6 different figures and A as the redundant die.

Answer 3

As noted in my comment above, you are double counting. Here's a different way of approaching the problem: let's compute the probability of throwing two 1's and one each of the other five numbers. So first choose the two dice that come up 1, in $\binom{7}{2}$ ways. The remaining five digits can be distributed in $5!$ ways, so the probability of throwing this combination is $$\frac{5!\binom{7}{2}}{6^7}.$$ Thus the answer you are looking for is six times this, or $$\frac{6!\binom{7}{2}}{6^7},$$ which comes out to $\frac{35}{6^3\cdot 3}$.

Answer 4

We need 2 alike and 5 distinct numbers on the die out of {$1,2,3,4,5,6$}, which can be selected in C(6,1).C(5,5) ways. Now 2 alike and 5 distinct numbers can be arranged in $\frac{7!}{2!}$ ways. Total number of points in the sample space as you said $6^7$

So, the required probability =$ \frac {C(6,1).C(5,5)\frac{7!}{2!}}{6^7}$ which is $\frac{35}{6^3.3}$

In your case when you are considering the arrangement you need to divide it with 2! as the permutation of 2 alike numbers will not result in new configuration.