Stacks: y1 b3 g5 w5
Alice: r1 g1 b1 w1
Bobby: y1 r5 y5 b5
Carol: g2 y3 g3 g4
David: g1 b1 w1 y3
Wins: 6
Deck: b4 kt y2 .. y4 kt r2 r3 r4 kt
Line: r1 dc dc .. b4 b5 y2 y3 y4 y5 r2 r3 r4 r5
Plan: A  B  I  G  F  U  E  L  V  E  S  S  E  L
Display: A big fuel vessel

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[A] Turn  1 ..... Alice plays [A] red 1 and draws [S] blue 4
[B] Turn  2 ..... Bobby discards a stale card and draws any card
[I] Turn  3 ..... Carol discards a stale card and draws [G] yellow 2
[G] Turn  4 ..... David stalls by spending a clue
[F] Turn  5 ..... Alice plays [S] blue 4 and draws [Q] yellow 4
[U] Turn  6 ..... Bobby plays [X] blue 5 and draws any card
[E] Turn  7 ..... Carol plays [G] yellow 2 and draws [F] red 2
[L] Turn  8 ..... David plays [L] yellow 3 and draws [K] red 3
[V] Turn  9 ..... Alice plays [Q] yellow 4 and draws [P] red 4
[E] Turn 10 ..... Bobby plays [V] yellow 5 and draws any card
[S] Turn 11 ..... Carol plays [F] red 2 on their last turn
[S] Turn 12 ..... David plays [K] red 3 on their last turn
[E] Turn 13 ..... Alice plays [P] red 4 on their last turn
[L] Turn 14 ..... Bobby plays [U] red 5 on their last turn

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PROOF OF UNIQUENESS

The starting pace is +2.
However, there is another limiting factor.

We claim that the number of turns that can occur is at most 17.
There are four turns in the final round, and up to nine plays/discards.
Also, one can get at most four clues before the final round:
two from 5's and two from discards.
Thus, the upper bound on the number of turns is 9+4+2+2 = 17.

This means Bobby has at most four turns.
As Bobby plainly cannot play on his first turn,
he MUST play a 5 on every other turn to finish in time.

We can then reason that the b5 must be played on turn 6,
because at most one card not already visible can be played
before the turn 6 (namely, the card Alice draws on turn 1),
and the b5 is the only 5 which is one-away from playable.

We claim that Alice cannot discard.
If she does, then between turns 6 through 10 must
involve the plays y2, y3, y4, y5
in order to make the y5 playable
(the r5 requires four cards, and so cannot be gotten).
Now on turns 7 through 11,
the team has four red cards needed but only three turns to play them.
This gives a proof that Alice cannot discard.

Thus, on her first turn, Alice plays r1.
This means Bobby must discard as he has no other option.

One of Carol and David must then discard and spend a clue respectively,
since neither of them have currently playable cards.
Looking ahead, this means the red 5 can't be played on turn 10
because Carol and David cannot both draw,
therefore the yellow 5 is played on turn 10,
and as before it must involve plays y2, y3, y4, y5 in that order.
So David must play y3 on turn 8,
and thus Carol should discard and try to draw y2 to play on turn 7.
For now, David spends a clue.

Now Alice plays b4 on turn 5, and Bobby plays b5.

The yellow cards promised now play: y2 from Carol, y3 from David,
y4 from Alice (which must be her newly drawn card),
and y5 from Bobby.
This starts the final round.

In order to finish, the team must now play all the red cards:
r2 from Carol, r3 from David, r4 from Alice, r5 from Bobby.

Examining the deck order, every card drawn was needed immediately
except for whichever three cards Bobby draws,
which are arbitrary trash cards and can come in any order.
So 3! = 6 of the deck orderings lead to wins.