Probability question of multiple dice roll against a single dice roll I'm confused with the explanation of this problem. 
Problem:
which event has bigger probability?
Event 1:
Rolling 2 dice at a time for 5 times. Both dice need to have different value. Either 1 or 6. For example: If first die is 6, the second die has to be 1 (can't be another 6). if first die is 1, second die has to be 6. 
Event 2:
Rolling 1 dice 10 times. Order doesn't matter. 5 dice roll 1, the other 5 roll 6. 
The answer is Event 2 is more probable because it's less restrictive. I don't quite understand the intuition, so i try to work out the probability calculation. 
My calculation is as follow:
Event 1:
$(\frac{2}{6} * \frac{1}{6})^5 $ = $(\frac{2}{36})^5 =5.29*10^{-7} $.
The first die has 2 chance, hence $\frac{2}{6}$. The second die only has 1 chance. Hence $\frac{1}{6}$. 
Event 2:
${10 \choose 5} (\frac{1}{6})^5 * (\frac{1}{6})^5 = {10 \choose 5} (\frac{1}{6})^{10} =4.17*10^{-6}$ . 
Is this correct?
 A: Yes, your calculation is correct. I'd like to try to explain the "intuitive" answer a little more. 
The outcomes of both experiments can be thought of as a list of 10 numbers. In the first experiment, say the outcomes are:


*

*Die 1: 1, Die 2: 6

*Die 1: 1, Die 2: 6

*Die 1: 6, Die 2: 1

*Die 1: 1, Die 2: 6

*Die 1: 6, Die 2: 1


Then we could string the outcomes together into $(1,6,1,6,6,1,1,6,6,1)$ (the first two elements correspond to the outcomes of the first roll, the third and fourth correspond to the outcomes of the second roll, etc). Here I am assuming the dice are distinguishable, so there is a "die 1" and a "die 2". 
For the second experiment, we could label the dice die 1 through die 10, and just list their outcomes in order, e.g., $(1,1,1,6,6,6,1,6,1,6)$. 
Notice that the set of possible outcomes are the same in both experiments (any string $(x_1,\dots,x_{10})$ such that $x_i \in \{1,2,3,4,5,6\}$), and every possible outcome has the same probability, since they are all outcomes of 10 independent die rolls. The probability of an event when all outcomes are equally likely is the number of favorable outcomes over the number of total outcomes, so we just need to compare the number of favorable outcomes in event 1 with the number of favorable outcomes in event 2. 
To have a favorable outcome in event 1, we need $x_1$ and $x_2$ to be a $1$ and $6$ (in any order), $x_3$ and $x_4$ to be a $1$ and $6$ (in any order), etc. In particular this means there will be five $1$s and five $6$s. So if $(x_1,\dots,x_{10})$ is favorable for event 1, it is also favorable for event 2.
However, not every favorable outcome for event 2 is favorable for event 1. An example is the string I gave above: $(1,1,1,6,6,6,1,1,6,6)$. This is not favorable for event 1 because the first two elements don't have a $1$ and a $6$. It is favorable for event 2, though, because it has five $1$s and five $6$s. 
Therefore event 2 has strictly more favorable outcomes than event 1, so event 2 is more probable.
