If we consider $D_1,D_2,D_3$, and that two dices must show the same value, we have a total of $^3C_2$ ways this can happen.
That being said, the first die can be anything ($6/6$), then second die must be the same number $(1/6)$ chance, and the third die must be something else $(5/6)$. Therefore, the solution is:
$(^3C_2)(6/6)(1/6)(5/6)=5/12$
I would like to point out that there is already a question same as mine:
Rolling three dice…am I doing this correctly?
My question is that, we must roll two dices which show the same value. The question doesn't give us an exact value (like $1,2$). Therefore, wouldn't it make sense to perform $^6C_1$ to select any of the values beforehand? This changes our answer to become $6\cdot(^3C_2)(6/6)(1/6)(5/6)=5/2>0$
This clearly does not make sense as the probability is greater then $0$, but why else must we not multiply by 6.
On this website, it says the following
The probability of one dice being a particular number is 1/6. The probability of two dice being the same particular number is 1/6 x 1/6 = 1/36. This is not the same as saying that both dice are the same number. There are six different possible numbers, so that would be 6/36 or 1/6.
"This is not the same as saying that both dice are the same number. There are six different possible numbers, so that would be 6/36 or 1/6."
Following this information I multiply by $6$? What is my point of confusion here?
Best Answer
You have already multiplied by $6$ ways to select a number for the pair, and the $5$ ways to select a different number.$$\def\cbinom#1#2{{^{#1}\mathsf C_{#2}}} \cbinom 32\cdot\cbinom 61(\tfrac 16)^2\cdot\cbinom 51\tfrac 16 = \cbinom 32\tfrac{6\cdot 5}{6^3}$$