Proof: Probability of a pair when rolling 7 dice is 1.

Question

I know that the probability should be over or exactly 1 since out 6 possible values the 7th dice will always be a duplicate.

My calculations are wrong though:

$\frac{{7\choose2} 5! 6*1}{6^7}$

Why? I'm guessing my error is in the denominator.

Edit: any one pair excluding triplets, full house etc

Edit 2: My way of thinking was: I need 6 distinct values on 6 of the dice and the 7th dice's value has to be equal to one of the other 6.

${7\choose2}$ = the different combination of dice pairs out of 7 dice

5! = the other combination of distinct values from the non pair dice

6 = how many numbers I get to choose my pair from

1 = ${6\choose6}$ choose value out of the distinct 6 values.

Answer 1

You seem to calculate the probability of exactly a pair (i.e., no triplet, no two pairs, etc.) and that is indeed smaller than $1$.

Answer 2

The probability of rolling two $1´s$ at first is $\frac{1}{6}\cdot\frac{1}{6}$. The probability of rolling $ 2,3,4,5,6$ next is $\frac{1}{6}\cdot\frac{1}{6} \cdot\frac{1}{6} \cdot\frac{1}{6} \cdot \frac{1}{6} $.

The first two rolls can be also a pair of $ 2,3,4,5$ or $6$. Therefore an extra factor of $6$ is needed. And the outcome of the $7$ rolls can be arranged in $\frac{7!}{2!}$ ways.

In total the probability to get $\texttt{exactly}$ one pair is $ \frac{7!\cdot 6}{2!\cdot 6^7}=\frac{35}{648}\approx 5.4\%$