Jelly, 17 14 bytes

ċⱮ6Ḣ©+$®aĖUṀ)M

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A monadic link that takes a list of the two lists as its argument and returns [1] for player 1 wins, [2] for player 2 wins and [1, 2] for a tie. TIO link tidies this up for display.

Thanks to @JonathanAllan for saving 3 bytes!

Explanation

 ) | For each input list (e.g. of one list 1,1,3,3,4)
ċⱮ6 | - Count each of 1..6 (e.g. 2,0,2,1,0,0)
 $ | - Following as a monad:
 Ḣ | - Head (number of 1s)
 ©︎ | - Copy to register
 + | - Add (e.g. 2,4,3,0,0)
 ®a | - Register logical and with this
 Ė | - Enumerate (e.g. [1,2],[2,4],[3,3],[4,0],[5,0])
 U | - Reverse each (e.g. [2,1],[4,2],[3,3],[0,4],[0,5])
 Ṁ | - Max (e.g. 4,2)
 M | Index/indices of maximal score

05AB1E, 16 15 bytes

-1 byte thanks to JonathanAllan

εWΘ*6L¢ć+°ƶà}ZQ

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Returns [1, 0] for P1 wins, [1, 1] for ties, [0, 1] for P2 wins.

Rather than using lexicographic order on a 2-tuple (dice count, dice value), this computes the score as 10**dice count * dice value. Hands without a 1 score 5.

ε } # map each hand to its computed score
 WΘ # minimum == 1 (1 if the hand contains a 1, 0 otherwise)
 * # multiply (sets hands without 1 to [0, 0, 0, 0, 0])
 6L # range [1..6]
 ¢ # count occurences of each in the hand
 ć # head extract (stack is now [2-count, ..., 6-count], 1-count)
 + # add the 1-count to all the other counts
 ° # 10**x
 ƶ # multiply each element by its 1-based index
 à # take the maximum

Z # maximum
 Q # equality test (vectorizes)

JavaScript (ES6),  97  90 bytes

Takes input as (a)(b). Returns +Infinity for P1, -Infinity for P2 or NaN for a tie.

a=>b=>((g=(x,m)=>x>6?m*/1/.test(a):g(-~x,a.map(k=>n+=k<2|k==x,n=x/6)|m>n?m:n))()-g(a=b))/0

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Commented

a => b => ( // a[] = dice of P1; b[] = dice of P2
 ( g = ( // g is a recursive function taking:
 x, // x = dice value to test; initially, it is either undefined
 // or set to a non-numeric value
 m // m = maximum score so far, initially undefined
 ) => //
 x > 6 ? // if x is greater than 6:
 m * /1/.test(a) // return m, or 0 if a[] does not contain any 1 
 : // else:
 g( // do a recursive call:
 -~x, // increment x (or set it to 1 if it's non-numeric)
 a.map(k => // for each dice value k in a[]:
 n += // add 1 to n if:
 k < 2 | // k is equal to 1
 k == x, // or k is equal to x
 n = x / 6 // start with n = x / 6
 ) | // end of map()
 m > n ? // if m is defined and greater than n:
 m // pass m unchanged
 : // else:
 n // update m to n
 ) // end of recursive call
 )() // first call to g, using a[]
 - g(a = b) // subtract the result of a 2nd call, using b[]
) / 0 // divide by 0 to force one of the 3 consistent output values

R, 115 102 bytes

-6 bytes thanks to Giuseppe.

-2 bytes thanks to Arnauld, following the output format in his JavaScript answer.

function(i,j)(f(i)-f(j))/0
f=function(x,s=tabulate(x,6),l=s[1]+s[-1]*!!s[1])max(l)*6+max.col(t(l),"l")

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Returns Inf for P1, NaN for a tie, -Inf for P2.

Uses the helper function f which computes a score for each hand. The score is defined as follows: let d be the digit which is repeated the most, and n the number of times it is repeated. Then the score is 6*n+d if there is at least one ace, and 0 if there are no aces. We then just need to find the player with the highest score.

Ungolfed:

f = function(x) 
 s = tabulate(x, 6) # number of occurrences of each integer
 l = s[1] + s[-1] * !!s[1] # add the number of wild aces to all values; set to 0 if s[1] == 0
 max(l) * 6 + # highest number of repetitions (apart from aces)
 max.col(t(l), "last") # most repeated integer (take largest in case of tie)

function(i, j)
 sign(f(i) - f(j))

Python 2, 85 81 bytes

lambda p:cmp(*[max((h.count(i)+h.count(i>1),i)*(1 in h)for i in[6]+h)for h in p])

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Returns 1 for P1, 0 for tie, and -1 for P2.

PowerShell, 112 126 bytes

Takes input as (a)(b). Returns -1 for P1 win, 1 for P2 or 0 for a tie.

$args|%$d=(7*(($o=($_-eq1).Count)+($m=$_-ne1
[math]::Sign($d)

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Unrolled:

$args|%select -Last 1 # element with the widest group and the maximum digit except 1
 $score=7*($ones+$max.Count)+($max.Name,6)[!$max]
 $diff=$score[!$ones]-$diff # score is 0 if the array does not contains 1

[math]::Sign($diff) # output the score difference

Perl 6, 60 bytes

&[cmp]o*.map:.1&&.1+max .kv.map:o&bag

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Returns More, Same, Less for P1 Wins, Tie, P2 Wins.

Explanation

 *.map: # Map input lists
 &bag # Convert to Bag
 o # Pass to block
 .1&& # Return 0 if no 1s
 .1+ # Number of 1s plus
 max # Maximum of
 .kv.map: # Key-value pairs mapped to
 2>$^k|| # If k==1, return 1 
 $^v+$k/6 # Otherwise v+k/6
&[cmp]o # Compare mapped values

Jelly, 27 bytes

’o5Rṗ5¤ṢŒr$€UẎṀ1e⁸¤×µ€_/Ṡo/

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1 for P1, -1 for P2, 0 for tie

Explanation

’o5Rṗ5¤ṢŒr$€UẎṀ1e⁸¤×µ€_/Ṡo/ Main link
 µ€ For each list
’ Decrement each value (so 1s become falsy)
 o Vectorized logical or (this replaces previous 1s (now 0s) with the test values)
 5Rṗ5¤ 1..5 cartesian-power 5 (1,1,1,1,1; 1,1,1,1,2; 1,1,1,1,3; ...)
 $€ For each test list
 ṢŒr Sort and run-length encode (gives [digit, #digit])
 U Reverse each list (gives [#digit, digit])
 Ẏ Tighten by one (gives a list containing each possible hand for each possible wildcard)
 Ṁ Take the maximum
 1e⁸¤× Multiply the list values by (whether or not the original contained a 1) - becomes [0, 0] if not
 _/Ṡ Take the sign of the difference between the #digits and the digits
 o/ If the number of digits differs, then 1/-1 is returned; otherwise, check the value of the digit (could still be 0)

Java 8, 244 240 bytes

a->b->int c[][]=new int[2][7],m[]=new int[2],s,p,i=5;for(;i-->0;c[1][b[i]]++)c[0][a[i]]++;for(i=7;i-->1;)for(p=2;p-->0;m[p]=s>m[p]?s:m[p])s=(int)Math.pow(9,c[p][i]+(i>1?c[p][1]:0))*(i<2?6:i)*(c[p][1]>0?1:0);return Long.compare(m[0],m[1]);

Sigh.. This byte-count.. >.> Ah well, Java I guess ¯_(ツ)_/¯

Returns 1 if P1 wins; -1 if P2 wins; 0 if it's a tie.

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Explanation:

a->b-> // Method with 2 integer-array parameters & integer return-type
 int c[][]=new int[2][7], // Create a count-array for each value of both players,
 // initially filled with 0s
 m[]=new int[2], // The maximum per player, initially 0
 s,p, // Temp-values for the score and player
 i=5;for(;i-->0; // Loop `i` in the range (5, 0]:
 c[1] // For player 2:
 [b[i] // Get the value of the `i`'th die,
 ]++) // and increase that score-count by 1
 c[0][a[i]]++; // Do the same for player 1
 for(i=7;i-->1;) // Then loop `i` in the range (7, 1]:
 for(p=2;p-->0 // Inner loop `p` over both players:
 ; // After every iteration:
 m[p]=s>m[p]?s:m[p]) // Update the maximum score if the current score is higher
 s= // Set the current score to:
 (int)Math.pow(9, // 9 to the power:
 c[p][i] // The amount of dice for value `i`
 +(i>1? // And if the current die `i` is not 1:
 c[p][1]:0) // Add the amount of dice for value 1
 *(i<2?6:i) // Multiply that by 6 if `i` is 1; or `i` otherwise
 *(c[p][1]>0?1: // And if no 1s were rolled:
 0); // Multiply it by 0 to make the score 0
 return Long.compare(m[0],m[1]);
 // Finally compare the maximum scores of the players,
 // resulting in -1 if a<b; 0 if a==b; 1 if a>b

Jelly, 21 bytes

^{crushed before I even posted it by Nick Kennedy :)}

’-oÆṃṀ$$Ạ?fÆṃ$L;$o5)M

A monadic Link accepting a list of players which outputs a list of (1-indexed) winners.

So P1 is [1], P2 is [2] and a tie is [1,2].

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Charcoal, 48 45 bytes

ＵＭθ⮌Ｅ⁷№ι∨λ¹ＵＭθ׬¬⊟ι⁺⊟ιιＵＭθ⟦⌈ι±⌕ι⌈ι⟧Ｉ⁻⌕θ⌈θ⌕θ⌊θ

Try it online! Link is to verbose version of code. Takes input as an array of arrays and outputs -1 if player 1 wins, 0 for a tie, and 1 if player 2 wins. Explanation:

ＵＭθ⮌Ｅ⁷№ι∨λ¹

Replace each hand with the count of how many times the values 6..1 appear in the hand. The list is reversed because a) it makes it easier to find the highest value with the highest count and b) it makes it easier to remove the count of 1s. The count of 1s is doubled because it needs to be removed twice, once to check that it is nonzero and once to add it to the other counts.

ＵＭθ׬¬⊟ι⁺⊟ιι

Add the count of 1s to the counts for 6..2, but set all of the counts to zero if the count of 1s was zero.

ＵＭθ⟦⌈ι±⌕ι⌈ι⟧

For each hand find the highest count and the highest value with that count. (Actually we find value minus 6 as that's golfier.)

Ｉ⁻⌕θ⌈θ⌕θ⌊θ

Determine which hand won by subtracting the positions of the winning and losing hands. (If the hands are tied then the first hand is both winning and losing so the result is 0 as desired.)

Wolfram Language (Mathematica), 78 75 bytes

Order@@(FreeQ[#,1]||Last@Sort[Reverse/@Tally@Flatten[#/. 1->Range@6]]&)/@#&

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Outputs -1 when player 1 wins, 1 when player 2 wins, and 0 for a tie.

 Helper function to score a list:
FreeQ[#,1] || If there are 0 1s, score is True
Last@Sort[ Otherwise, take the largest element of
 Reverse/@Tally@ the frequency, number pairs in the flat list
 Flatten[ #/. 1->Range@6] where each 1 is replaced by 1,2,3,4,5,6.
]& e.g. 1,3,3,5,5 -> 1,2,3,4,5,6,3,3,5,5 -> 4,5

Order @@ (...) /@ #& Apply this function to both lists,
 then find the ordering of the result.

Sledgehammer 0.4, 27 bytes

⢱⢙⢂⠠⡾⢃⠐⢈⠸⣞⠴⠻⠎⡥⡳⡐⢒⠘⢛⣩⡓⣮⡕⡠⣢⣡⠿

Decompresses into this Wolfram Language function:

Order @@ (FreeQ[#1, 1] || Last[Sort[Reverse[Tally[Flatten[#1 /. 1 -> Range[6]]], 2]]] & ) /@ #1 & 

which turns out to be exactly the same as my Mathematica answer.