Probability question from Brilliant Which is more likely?

*

*You roll two dice 5 times and, every time, one of the two comes up as 1 and the other as 6.

*You roll 10 dice all at once. 5 come up as 1s and the other 5 come up as 6s.

The answer is 2 and I'm confused by it.
Here's my thinking. When rolling two dice the probability of getting 1 and 6 or 6 and 1 is $2\cdot(1/6\cdot 1/6) = 2/36 = 1/18.$ This means that probability of getting this 5 times in a row is $(2/36)^5$.
In the second case there are $2^5$ dice combinations of 1s and 6s where we get 5 of each. The probability of every combination is $(1/6)^{10}$ and so the probability of the second case is $2^5\cdot (1/6)^{10}=(2/36)^5$ which is exactly the same as the first case.
What's wrong with my thinking?
 A: Here's some easy intuition as to why $2$ has to be more probable than $1$.  Suppose you roll the $10$ dice two at a time.  If the first pair comes up as a pair of $1$s, you're out of the running for scenario $1$, but you're still alive for scenario $2$.  But there is no sequence of rolls you can make that knocks you out of the running for scenario $2$ but leaves it possible to achieve scenario $1$.
A: Which has more ways of occurring?

*

*Obtaining five ‘1’s and five ‘6’s among ten concurrent die rolls in a row, with the ‘1’s spaced apart.

*Obtaining five ‘1’s and five ‘6’s among ten concurrent die rolls in a row, with no restriction on the numbers’ order of appearance.

(Whether the die rolls are successive or concurrent is immaterial to the answer.)
A: Look at the type of patterns of two pictures, and it will be quite clear:
Two dice rolled 5 times each
$\fbox1\fbox6\,\fbox6\fbox1\,\fbox1\fbox6\,\fbox6\,\fbox1\,\fbox6\fbox1$
Ten dice rolled together
$\fbox1\fbox1\fbox6\fbox6\fbox6\fbox1\fbox1\fbox6\fbox6\fbox1$
The second case can generate all of the types the first one can, and then more...
A: the second is
$$\binom{10}{5,5}\times \left(\frac{1}{6}\right)^5\times \left(\frac{1}{6}\right)^5=\frac{10!}{5!5!}\times \left(\frac{1}{6}\right)^5\times \left(\frac{1}{6}\right)^5$$
thus it is most likely being
$$\frac{10!}{5!5!}>2^5$$
A: The mistake is subtle: When rolling the dice 5 times in 1), you are required for each roll to include a 1 and a 6. However, if you roll ten dice, you allow for the two dice to be 6, or 1, or both. So the correct probability for the first five dice in 2) is in fact higher, than the first five dice in 1). So while your calculation in 1) is correct, yours in 2) is incorrect. The probability of 2) will range between $(\frac{2}{6})^9*(\frac{1}{6})^1$ at most, and $(\frac{2}{6})^5*(\frac{1}{6})^5$ at worst, depending on what you roll. Whatever the actual answer, 2) will always have a higher probability.
A: Alternative approach.
Label the dice $D_1, D_2, \cdots, D_{10}$.
Let $E_1$ denote the event represented by problem 1. 
Let $E_2$ denote the event represented by problem 2. 
Assume that for the purposes of the first problem, when the dice are rolled, $D_1,D_2$ are rolled together, $D_3,D_4,$ are rolled together, and so forth.
Assume that each pair of dice, re above paragraph are rolled.  $E_2$ will be deemed a success if exactly $5$ of the $10$ dice are a $1$.
Note that if $E_2$ does not occur, then $E_1$ can not occur.
Assume that $E_2$, succeeded.  Given that,  what is the probability that $E_1$ succeeded?
Without loss of generality (WLOG) $D_1$ was a $1$. Then the chance that $D_2$ was a $6$ is $(5/9)$.  Note that if $D_1$ had been a $6$, the chance that $D_2$ was a $1$ would similarly have been $(5/9)$.
Assuming that $D_1$ and $D_2$ have been navigated successfully, then WLOG, $D_3$ was a $1$ and the chance that $D_4$ was a $6$ is $(4/7)$.
Continuing to the end, you see that under the assumption that $E_2$ succeeded, the chance that $E_1$ then succeeded is
$$(5/9) \times (4/7) \times (3/5) \times (2/3).$$
