Could you please stop shuffling the deck and play already? Challenge:
Input: A list of distinct positive integers within the range \$[1, \text{list-size}]\$.
Output: An integer: the amount of times the list is riffle-shuffled. For a list, this means the list is split in two halves, and these halves are interleaved (i.e. riffle-shuffling the list [1,2,3,4,5,6,7,8,9,10] once would result in [1,6,2,7,3,8,4,9,5,10], so for this challenge the input [1,6,2,7,3,8,4,9,5,10] would result in 1).
Challenge rules:

You can assume the list will only contain positive integers in the range \$[1, \text{list-size}]\$ (or \$[0, \text{list-size}-1]\$ if you choose to have 0-indexed input-lists).
You can assume all input-lists will either be a valid riffle-shuffled list, or a sorted list which isn't shuffled (in which case the output is 0).
You can assume the input-list will contain at least three values.

Step-by-step example:
Input: [1,3,5,7,9,2,4,6,8]
Unshuffling it once becomes: [1,5,9,4,8,3,7,2,6], because every even 0-indexed item comes first [1, ,5, ,9, ,4, ,8], and then all odd 0-indexed items after that [ ,3, ,7, ,2, ,6, ].
The list isn't ordered yet, so we continue:
Unshuffling the list again becomes: [1,9,8,7,6,5,4,3,2]
Again becomes: [1,8,6,4,2,9,7,5,3]
Then: [1,6,2,7,3,8,4,9,5]
And finally: [1,2,3,4,5,6,7,8,9], which is an ordered list, so we're done unshuffling.
We unshuffled the original [1,3,5,7,9,2,4,6,8] five times to get to [1,2,3,4,5,6,7,8,9], so the output is 5 in this case.
General rules:

This is code-golf, so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
Default Loopholes are forbidden.
If possible, please add a link with a test for your code (i.e. TIO).
Also, adding an explanation for your answer is highly recommended.

Test cases:
Input                                                   Output

[1,2,3]                                                 0
[1,2,3,4,5]                                             0
[1,3,2]                                                 1
[1,6,2,7,3,8,4,9,5,10]                                  1
[1,3,5,7,2,4,6]                                         2
[1,8,6,4,2,9,7,5,3,10]                                  2
[1,9,8,7,6,5,4,3,2,10]                                  3
[1,5,9,4,8,3,7,2,6,10]                                  4
[1,3,5,7,9,2,4,6,8]                                     5
[1,6,11,5,10,4,9,3,8,2,7]                               6
[1,10,19,9,18,8,17,7,16,6,15,5,14,4,13,3,12,2,11,20]    10
[1,3,5,7,9,11,13,15,17,19,2,4,6,8,10,12,14,16,18,20]    17
[1,141,32,172,63,203,94,234,125,16,156,47,187,78,218,109,249,140,31,171,62,202,93,233,124,15,155,46,186,77,217,108,248,139,30,170,61,201,92,232,123,14,154,45,185,76,216,107,247,138,29,169,60,200,91,231,122,13,153,44,184,75,215,106,246,137,28,168,59,199,90,230,121,12,152,43,183,74,214,105,245,136,27,167,58,198,89,229,120,11,151,42,182,73,213,104,244,135,26,166,57,197,88,228,119,10,150,41,181,72,212,103,243,134,25,165,56,196,87,227,118,9,149,40,180,71,211,102,242,133,24,164,55,195,86,226,117,8,148,39,179,70,210,101,241,132,23,163,54,194,85,225,116,7,147,38,178,69,209,100,240,131,22,162,53,193,84,224,115,6,146,37,177,68,208,99,239,130,21,161,52,192,83,223,114,5,145,36,176,67,207,98,238,129,20,160,51,191,82,222,113,4,144,35,175,66,206,97,237,128,19,159,50,190,81,221,112,3,143,34,174,65,205,96,236,127,18,158,49,189,80,220,111,2,142,33,173,64,204,95,235,126,17,157,48,188,79,219,110,250]
                                                        45
JavaScript (ES6), 44 bytes
Shorter version suggested by @nwellnhof
Expects a deck with 1-indexed cards as input.

f=(a,x=1)=&gt;a[x]-2&amp;&amp;1+f(a,x*2%(a.length-1|1))

Try it online!
Given a deck \$[c_0,\ldots,c_{L-1}]\$ of length \$L\$, we define:
$$x_n=\begin{cases}
2^n\bmod L&amp;\text{if }L\text{ is odd}\\
2^n\bmod (L-1)&amp;\text{if }L\text{ is even}\\
\end{cases}$$
And we look for \$n\$ such that \$c_{x_n}=2\$.

JavaScript (ES6),  57 52  50 bytes
Expects a deck with 0-indexed cards as input.

f=(a,x=1,k=a.length-1|1)=&gt;a[1]-x%k&amp;&amp;1+f(a,x*-~k/2)

Try it online!
How?
Since JS is lacking native support for extracting array slices with a custom stepping, simulating the entire riffle-shuffle would probably be rather costly (but to be honest, I didn't even try). However, the solution can also be found by just looking at the 2nd card and the total number of cards in the deck.
Given a deck of length \$L\$, this code looks for \$n\$ such that:
$$c_2\equiv\left(\frac{k+1}{2}\right)^n\pmod k$$
where \$c_2\$ is the second card and \$k\$ is defined as:
$$k=\begin{cases}
L&amp;\text{if }L\text{ is odd}\\
L-1&amp;\text{if }L\text{ is even}\\
\end{cases}$$