JavaScript (ES6), 44 bytes

Shorter version suggested by @nwellnhof

Expects a 1-indexed deck as input.

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

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JavaScript (ES6),  57 52  50 bytes

Expects a 0-indexed deck as input.

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

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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:

$$c_2equivleft(frac{k+1}{2}right)^npmod k$$

where $c_2$ is the second card and $k$ is defined as:

$$k=begin{cases}
L&text{if }Ltext{ is odd}\
L-1&text{if }Ltext{ is even}\
end{cases}$$

Python 2, 39 bytes

f=lambda x:x[1]-2and-~f(x[::2]+x[1::2])

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-4 thanks to Jonathan Allan.

Returns False instead of 0.

APL (Dyalog Unicode), 35 26 23 bytes^SBCS

{∧/2≤/⍵:0⋄1+∇⍵[⍒2|⍳⍴⍵]}

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Thanks to Adám for the help and Erik the Outgolfer for -3.

The TIO link contains two test cases.

Old solution:
{i⊣{i+←1⋄⍵[⍒2|⍳⍴⍵]}⍣{∧/2≤/⍵}⍵⊣i←¯1}

Jelly, 8 bytes

ŒœẎ$ƬiṢ’

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How?

ŒœẎ$ƬiṢ’ - Link: list of integers A

    Ƭ    - collect up until results are no longer unique...

   $     -   last two links as a monad:

Œœ       -     odds & evens i.e. [a,b,c,d,...] -> [[a,c,...],[b,d,...]]

  Ẏ      -     tighten                         -> [a,c,...,b,d,...]

     Ṣ   - sort A

    i    - first (1-indexed) index of sorted A in collected shuffles

      ’  - decrement

05AB1E (legacy), 9 bytes

[DāQ#ι˜]N

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Explanation

[   #  ]     # loop until

  ā          # the 1-indexed enumeration of the current list

 D Q         # equals a copy of the current list

     ι˜      # while false, uninterleave the current list and flatten

        N    # push the iteration index N as output

Perl 6, 34 bytes

{(0...{.[2**$^i%($_-1+|1)]==2})-1}

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Similar to Arnauld's approach. The index of the second card after n shuffles is 2**n % k with k defined as in Arnauld's answer.

Java (JDK), 59 bytes

a->{int c=0;for(;a[(1<<c)%(a.length-1|1)]>2;)c++;return c;}

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Works reliably only for arrays with a size less than 31 or solutions with less than 31 iterations. For a more general solution, see the following solution with 63 bytes:

a->{int i=1,c=0;for(;a[i]>2;c++)i=i*2%(a.length-1|1);return c;}

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Explanation

In a riffle, the next position is the previous one times two modulo either length if it's odd or length - 1 if it's even.

So I'm iterating over all indices using this formula until I find the value 2 in the array.

Credits

• -8 bytes thanks to Kevin Cruijssen. (Previous algorithm, using array)


• -5 bytes thanks to Arnauld.

Perl 6, 34 bytes

-2 bytes thanks to nwellnhof

{($_,*.sort({$++%2})...*[1]==2)-1}

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Reverse riffle shuffles by sorting by the index modulo 2 until the list is sorted, then returns the length of the sequence minus one.

Explanation:

{                                }   # Anonymous code block

  $_,                                # Starting from the input list

     *.sort({$++%2})                 # Sort by the parity of each index

                    ...              # Until

                       *[1]==2       # The list is in ascending order

 (                            )-1    # And return the length of the list minus 1

APL(NARS), chars 49, bytes 98

{0{∧/¯1↓⍵≤1⌽⍵:⍺⋄(⍺+1)∇⍵[d],⍵[i∼d←↑¨i⊂⍨2∣i←⍳≢⍵]}⍵}

why use in the deepest loop, one algo that should be nlog(n), when we can use one linear n? just for few bytes more?
[⍵≡⍵[⍋⍵] O(nlog n) and the confront each element for see are in order using ∧/¯1↓⍵≤1⌽⍵ O(n)]test:

  f←{0{∧/¯1↓⍵≤1⌽⍵:⍺⋄(⍺+1)∇⍵[d],⍵[i∼d←↑¨i⊂⍨2∣i←⍳≢⍵]}⍵}

  f ,1

0

  f 1 2 3

0

  f 1,9,8,7,6,5,4,3,2,10

3

  f 1,3,5,7,9,11,13,15,17,19,2,4,6,8,10,12,14,16,18,20

17

Japt, 13 bytes

Taking my shiny, new, very-work-in-progress interpreter for a test drive.

eUñ)?T:ÒßUñÏu

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eUñ)?T:ÒßUñÏu     :Implicit input of integer array U

e                 :Test U for equality with

 Uñ               :  U sorted

   )              :End test

    ?             :If true

     T            :  Return 0

      :           :Else return

       Ò          :  Negation of bitwise NOT of

        ß         :  A recursive call to the programme with input

         Uñ       :    U sorted

           Ï      :    By 0-based indices

            u     :    Modulo 2

Wolfram Language (Mathematica), 62 bytes

c=0;While[Sort[a]!=a,a=a[[1;;-1;;2]]~Join~a[[2;;-1;;2]];c++];c

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Explanation

The input list is a . It is unriffled and compared with the sorted list until they match.

Pyth, 18 bytes

L?SIb0hys%L2>Bb1

y

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-2 thanks to @Erik the Outgolfer.

Red, 87 79 78 bytes

func[b][c: 0 while[b/2 > 2][c: c + 1 b: append extract b 2 extract next b 2]c]

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Python 3, 40 bytes

f=lambda x:x[1]-2and 1+f(x[::2]+x[1::2])  # 1-based

f=lambda x:x[1]-1and 1+f(x[::2]+x[1::2])  # 0-based

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I need to refresh the page more frequently: missed Erik the Outgolfer's edit doing a similar trick =)

J, 28 bytes

1#@}.(/:2|i.@#)^:(2<1{])^:a:

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C (GCC) 64 bytes

i,r;f(c,v)int*v;{for(i=1,r=0;v[i]>2;++r)i=i*2%(c-1|1);return r;}

This is a drastically shorter answer based on Arnauld's and Olivier Grégoire's answers. I'll leave my old solution below since it solves the slightly more general problem of decks with cards that are not contiguous.

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C (GCC) 162 bytes

a[999],b[999],i,r,o;f(c,v)int*v;{for(r=0;o=1;++r){for(i=c;i--;(i&1?b:a)[i/2]=v[i])o=(v[i]>v[i-1]|!i)&o;if(o)return r;for(i+=o=c+1;i--;)v[i]=i<o/2?a[i]:b[i-o/2];}}

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a[999],b[999],i,r,o; //pre-declare variables

f(c,v)int*v;{ //argument list

    for(r=0;o=1;++r){ //major loop, reset o (ordered) to true at beginning, increment number of shuffles at end

        for(i=c;i--;(i&1?b:a)[i/2]=v[i]) //loop through v, split into halves a/b as we go

            o=(v[i]>v[i-1]|!i)&o; //if out of order set o (ordered) to false

        if(o) //if ordered

            return r; //return number of shuffles

        //note that i==-1 at this point

        for(i+=o=c+1;i--;)//set i=c and o=c+1, loop through v

            v[i]=i<o/2?a[i]:b[i-o/2];//set first half of v to a, second half to b

    }

}

Perl 5 -pa, 77 bytes

map{push@{$_%2},$_}0..$#F;$_=0;++$_,@s=sort{$a-$b}@F=@F[@0,@1]while"@F"ne"@s"

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R, 70 bytes

x=scan();i=1;while(any((x=c(x[y<-seq(x)%%2>0],x[!y]))>sort(x)))i=i+1;i

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