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What happens in an untimed match if neither player can deterministically win (neither can score a point or force the opponent to burn out)?

verified
FAQ #2573·asked by 111128236661964800·January 4, 2026·Rules v1.4·Asked before Vendetta
Answer

Ruling: There is currently no explicit rule covering this situation. The closest applicable rule is TR 505.6 regarding loops, which states that if no player chooses to break a loop and there were no maintaining players, the game ends in a draw. However, this rule may not perfectly cover situations where both players are maintaining the loop.

Sequence:

  • If both players establish a deterministic loop where neither can win, judges can invoke loop rules
  • Both players would be asked how many times they want to continue the loop
  • If neither chooses to break it, the game could be ruled a draw under TR 505.6 (by analogy)
  • In a best-of-3 match, draws don't count toward match completion, so a new game would start
  • Ultimate resolution falls to head judge/tournament organizer discretion per TR 204.4.f

Nuances:

  • The loop must be truly deterministic (same actions repeated with predictable outcomes)
  • Drawing cards typically makes loops non-deterministic unless the draw is completely predictable (e.g., single card recycled)
  • TR 505.11 gives judges final authority on what constitutes a loop when secret information is involved
  • In untimed matches, if the situation persists indefinitely, tournament organizers would need to intervene
  • Players cannot be forced to agree to a draw; both must consent
  • A feedback request has been submitted to Riot to clarify rules for this situation
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