<s>[INST] How to understand the following result? I am beginning to study probability and I found the following problem:
What is the probability of to get a number twice after two throwings?
The answer was that since we have six distinct results in a fair die every possible result has $1/6$ of probability of happen and since that the result is:
$$ P(\text{rolling the same number twice in a row on a fair die} ) = (1/6)(1/6).$$
I understand till the part of the probabilities, I know that since every event is independent of the other we are applying the multiplication rule, However I would like to go more deep in the analysis and not to take this as a strict rule, I would like to understand why we had to multiply the numbers, I mean why not to sum them or divide?, I need to understand why the rule of multiplication works, I would like to receive a empirical or formal explanation of this result in order to understand more, thanks for the support.
I would like to state that my question is different since I know that if I have the scenario of only one die, thrown out in two different times my result is correct but I don't understand why we have to multiply the probabilities I know that we are using the multiplication rule but why it works? what is the reason to multiply the numbers I mean why not to divide it or to sum it, I would like to go more deep in the analysis. 
 [/INST] The probability of drawing a $1$ is $1/6$, because there is a single favorable outcome and six possible ones.
The probability of drawing two $1$s in a row is $1/36=1/6^2=(1/6)^2$ because there are now thirty-six possible outcomes. The product appears for combinatorial reasons. Specifically, the set of possible cases is the Cartesian product of the possible outcomes at the first drawing and those at the second drawing.
So just like $6^2=36$, the probabilites are multiplied, $1/6^2=(1/6)^2$. For three $1$s in a row, the probability would be $1/6^3=(1/6)^3$.

Now assume that you throw a die and a coin. The outcomes are from the sets $\{1,2,3,4,5,6\}$ and $\{H,T\}$. By the Cartesian product, there are $6\cdot2$ combinations and the probability of drawing a one and a head is $1/(6\cdot2)=(1/6)(1/2)$.

In the example with a die, probabilites aren't just multiplied, because there are several favorable cases ($6$ ways to draw the same value).
The probabilities for $1,2$ and $3$ identical drawings would be
$$\frac66=1,\\\frac6{6^2}=\frac16,\\\frac6{6^3}=\frac1{36}.$$

More "difficult": we throw the die twice and the coin three times and we look for equal numbers and sides.
As regards the dies, there are $6^2$ possible outcomes, and $2^3$ for the coin. Among these, $6$ combinations are favorable for the die and $2$ for the coin, forming a total of $6\cdot2$ distinct configurations (again by the Cartesian product).
Hence the probability
$$\frac{6\cdot2}{6^2\cdot2^3}=\left(\frac6{6^2}\right)\left(\frac2{2^3}\right).$$
$$\begin{matrix}
&& HHH & THH & HTH & TTH & HHT & THT & HTT & TTT \\
& 11 &\times & & & & & & &\times \\
& 12 & & & & & & & & \\
& 13 & & & & & & & & \\
& 14 & & & & & & & & \\
& 15 & & & & & & & & \\
& 16 & & & & & & & & \\
& 21 & & & & & & & & \\
& 22 &\times & & & & & & &\times \\
& 23 & & & & & & & & \\
& 24 & & & & & & & & \\
& 25 & & & & & & & & \\
& 26 & & & & & & & & \\
& 31 & & & & & & & & \\
& 32 & & & & & & & & \\
& 33 &\times & & & & & & &\times \\
& 34 & & & & & & & & \\
& 35 & & & & & & & & \\
& 36 & & & & & & & & \\
& 41 & & & & & & & & \\
& 42 & & & & & & & & \\
& 43 & & & & & & & & \\
& 44 &\times & & & & & & &\times \\
& 45 & & & & & & & & \\
& 46 & & & & & & & & \\
& 51 & & & & & & & & \\
& 52 & & & & & & & & \\
& 53 & & & & & & & & \\
& 54 & & & & & & & & \\
& 55 &\times & & & & & & &\times \\
& 56 & & & & & & & & \\
& 61 & & & & & & & & \\
& 62 & & & & & & & & \\
& 63 & & & & & & & & \\
& 64 & & & & & & & & \\
& 65 & & & & & & & & \\
& 66 &\times & & & & & & &\times \\
\end{matrix}$$
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