How Many Turns A Game Of Gwent Lasts?


The number of turns in a game of Gwent is a vital topic (end of turn effects and engines…) and a good mind exercise at the same time. This article was inspired by one of Shinmiri streams, where the number of turns was demonstrated based on a game against AI.

The topic is rather simple overall, but deceiving – it is easy to make mistakes in calculations. In this tutorial we would make a systematic study, which hopefully would leave no space for ambiguity.


  • b/r notation

To start with, let’s define ‘b’ as player going first in a round and ‘r’ as player going second. While being an abbreviation for blue/red coin, this notation obeys independently for each round; i.e. player going first in any round is ‘b’ no matter the initial coinflip.

  • finishing player

Finishing player is the one who plays the last card given particular scenario of round development.

  • sequence notation

A turn sequence starts with b and ends when both players pass. The turn when a player passed is denoted with (pass), turns elapsed after pass are distinguished with ‘p’ suffix. For example:

b, r, b, r, b(pass), r, bp, r(pass)

  • lead cards
Lead cards is the number of cards played by more contributing player. For example, if (b) played 3 cards and (r) played 6 cards, then lead cards = 6.


We would start from some examples which may seem naive and unnecessary at the first sight. The simplest thing b could do is instant pass. Let’s see turns sequence in such scenario:

b(pass), r(pass)

There is 1b turn and 1r turn. Both sides played no cards, but counted one turn. The second most elementary possibility is r passing instantly after b plays a card:

b, r(pass), b(pass)

2b, 1r. As we could see the number of turns is not symmetric – ‘b’ could count one turn more than ‘r’. Could it work in reverse direction if r plays cards instead of b?

b(pass), r, bp, r, bp, … ,r(pass)

No, bp’s even out the number of turns for both players! Could not be different in a binary sequence starting from b. At this point we could already make crucial generalizations and observations.

Two Laws Of Turns

  1. If player going first (b) is finishing then they count one turn more than opponent. Otherwise the number of turns is equal.

  2. The number of turns in rounds with pass (R1 and R2 most often) is always equal to lead cards+1 for (b). For (r) look at Law 1.

Consequences (Chosen Case Studies)

Number of turns in a normal (3 rounds) game
  • Minimal

The minimal number of turns in a normal game must occur for a (r) player, who loses R1 after passing at 7 cards (4 turns). Then in R2 (b) player drypasses (=plays no cards), while (r) drops one card (2 turns). R3 would be a normal 10 cards contest (10 turns).

\[N_{turns}(min,normal) = 4+2+10 = 16\]

The minimal number of turns for a player in a normal gameplay scenario is equal to 16! (which is by accident also the number of cards played every model game)

  • Maximal

The maximal number of turns in a normal scenario is forced when Round 1 is deep enough (finishing player wins at 4 cards or less).

Otherwise, R2 must be played down at least to 7 cards in a way that maximal possible leading cards count is achieved. For example if R2 is b(9) and r(10), then r has to go down to (7), while b has complete freedom of when to pass (number of cards in hand in brackets).

\[N_{turns}(max,normal) = 7+2+10 = 19\]

  • Casual

In practice the ‘maximal’ scenario occurs more often than ‘minimal’. The common motif is (r) in R2 deciding if to play down to 7 or stay at 8, which is effectively decision between 18 and 19 turns for the player who was (b) in R1.

Assuming that 50% of R1’s is won by blue (which also means being finishing player in the normal scenario), blue coin plays 0.5 more turns than red coin on average!

Maximal number of turns (excluding card effects)

Only lead cards are relevant for the number of turns. Maximal number of turns is achieved when Player 1 and Player 2 invest maximal number of their cards in different rounds as lead cards.

For example in R1 (b) could go down to 1 card, while (r) passes on 10 (or 7; +10 turns). Then in R2 (b) insta passes, while (r) goes down to 1 card this time (+10 turns). In R3 there are 7 lead cards left (+7 turns).

\[N_{turns}(max) = 10+10+7 = 27\]

There is more than one way to achieve this number. Let me know if you could do better 😉

Hero Pass

Could the difference in the number of turns be greater than 1 between (b) and (r) after 2 rounds?

Let’s explore the Two Laws Of Turns again. It would be possible if (b) plays one more card than (r) in both R1 and R2. The R1 part is obvious and was discussed already. R2 though is tricky.

That’s because whenever (b) plays a card (r) would reply and become the finishing player; otherwise (r) gets 2:0ed.

The only exception very rarely occuring in real play is the Hero Pass. It happens when (r) recognises that 2:0 is no longer a threat with the cards possibly left in (b) hand. If brave enough, (r) would pass at b(1):r(3) cards count. This way (b) played one card more and became the finishing player. The turns count of (b) would be +2 with respect to (r) after R2.

Missing Card R3 Paradox

Not always everything proceeds according to the normal scenario. Sometimes you end up with card disadvantage in a long R3 due to lack of point reach in one of rounds. Would you rather see yourself in a 9 vs 10 R3 as going first or second?

The common answer to this question shall be: going second. If going first opponent may play two last cards with no response (double last say), which in many matchups is abusable.

How does the turn count look like though if we apply the Two Laws Of Turns? It intuitively looks even in both cases, but let’s see.

When playing as (b) according to 1st Law, the number of turns would be equal for both players and in a non-pass round equal to number of leading cards (10).

On the other hand, when opponent is (b), then they play one turn more according to 1st Law. Therefore (r) must be (9) turns.

In counter to intuition, ‘end of turn engines’ get not only order advantage when playing as (b), but also an extra turn with respect to (r)! This information could be used in practice when planning rounds for your engine deck.


Thanks for reading! Hope that various scraps of number of turns knowledge got systematized with this article and the Two Laws.

At the same time I still think I might have overlooked something very basic – please let me know if spotted any blunder.

Have fun counting turns and see you on Gwent ladder!

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