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Ticket to Ride Tie Breakers
Posted: 03 January 2026, 17:09
by Knightime98
This is pretty straight forward and so is my question at the end.
Tie breaks (in order)
1. Highest points at end of game. (if tied go to 2)
2. Most completed destinations. (if tied still go to 3)
3. Longest continuous path (between the tied players). (if still tied here, no further explanation is given as to how to determine who wins).
In 99% + games these 3 tie breaks will have a clear winner.
The question here now remains what happens if you are still tied after the 3rd tie break. Below, are suggestions that I saw and I don't believe are official.
The below excert is from Spliced Online (which is all that I could find with further explanation of tie breakers).
While the official tie breaker criteria are the first three listed above, there are some additional options that can be used if necessary. These include:
" Random Draw: In some cases, a random draw may be used to break the tie. This can be done by drawing a card from a deck or rolling a die.
Additional Rounds: In some cases, the game may be extended by one or more additional rounds, allowing players to continue building routes and earning points.
Custom Rules: Some players may choose to create their own custom tie breaker rules, such as using a combination of the official criteria or introducing new elements to the game."
* I am looking to see what BGA TTR online official outcome rules are if tied after the 3rd tie break and the game is still tied?
From here how is the winner determined?
Thanks one and all for any clarity.
Re: Ticket to Ride Tie Breakers
Posted: 03 January 2026, 20:41
by Emerson_Matt
I don't know about TTR specifically, but the general policy is to call it a tie if you're still level after working through all the stated tie-breakers. I don't see why it would be any different here.
Re: Ticket to Ride Tie Breakers
Posted: 03 January 2026, 23:07
by ExaltedAngel
I don’t know what Spliced Online is or who suggested that, but playing additional rounds is the worst possible idea for a tiebreaker, because it would just reward the player who played worse. A tiebreaker that actually makes sense could be the lowest number of turns played, but that only applies in less than half of the cases.
Also, most good players would agree that the official #2 tiebreaker should be #3 and vice versa, but those are the official rules, so what can you do... I'm not a fan of house rules anyways.
To get back to the main point of the thread - on BGA, if two players are still tied after all three official tiebreakers, they are simply declared tied. In tournaments where one of the two has to be eliminated, if I recall correctly, the system takes the age of the account into account, but I can’t remember who that favors.
Re: Ticket to Ride Tie Breakers
Posted: 03 January 2026, 23:44
by Ceaseless
ExaltedAngel wrote: ↑03 January 2026, 23:07
I don’t know what Spliced Online is or who suggested that, but playing additional rounds is the worst possible idea for a tiebreaker, because it would just reward the player who played worse. A tiebreaker that actually makes sense could be the lowest number of turns played, but that only applies in less than half of the cases.
If we're choosing tiebreakers, I say eliminate the previous tiebreakers and just have the last player to take their first turn win if there's a tie. Simpler and reduces first turn advantage a bit more.
ExaltedAngel wrote: ↑03 January 2026, 23:07
if I recall correctly, the system takes the age of the account into account, but I can’t remember who that favors.
My memory says it's oldest account.
Re: Ticket to Ride Tie Breakers
Posted: 04 January 2026, 20:34
by Wistama
Ceaseless wrote: ↑03 January 2026, 23:44
If we're choosing tiebreakers, I say eliminate the previous tiebreakers and just have the last player to take their first turn win if there's a tie. Simpler and reduces first turn advantage a bit more.
That has merit. Another possibility: The tied player with most remaining (unplayed) wagons wins. This player got the same number of points but did it more efficiently, i.e., by using fewer wagons.
If players tied on remaining wagons, then the first-turn criterion would always produce a winner.
Re: Ticket to Ride Tie Breakers
Posted: 04 January 2026, 20:47
by Ceaseless
Wistama wrote: ↑04 January 2026, 20:34
That has merit. Another possibility: The tied player with most remaining (unplayed) wagons wins. This player got the same number of points but did it more efficiently, i.e., by using fewer wagons.
That likely says more about the point value from your tickets. More efficient placement tends to mean using wagons faster due to larger payments having greater points per wagon efficiency.
I'd rather just have the going second player win if they're tied. Going first would still obviously be better, but by a bit less.
Re: Ticket to Ride Tie Breakers
Posted: 11 January 2026, 16:34
by The Cult Of Skaro
Re: Ticket to Ride Tie Breakers
Posted: 12 January 2026, 23:37
by Cournot
IMHO there's should be NO tiebreakers in TTR as longuest is worth 10 pts, not 10.5 pts or whatever.
Tickets have a nominal value, so it shouldn't be used either.
Re: Ticket to Ride Tie Breakers
Posted: 13 January 2026, 00:55
by Ceaseless
Cournot wrote: ↑12 January 2026, 23:37
IMHO there's should be NO tiebreakers in TTR as longuest is worth 10 pts, not 10.5 pts or whatever.
Tickets have a nominal value, so it shouldn't be used either.
Not even an alternative tiebreaker like going last wins?
Re: Ticket to Ride Tie Breakers
Posted: 15 January 2026, 01:06
by Cournot
Ceaseless wrote: ↑13 January 2026, 00:55
Cournot wrote: ↑12 January 2026, 23:37
IMHO there's should be NO tiebreakers in TTR as longuest is worth 10 pts, not 10.5 pts or whatever.
Tickets have a nominal value, so it shouldn't be used either.
Not even an alternative tiebreaker like going last wins?
No, a draw is a draw.
But, there should be mandatory alternated starts, meaning in 2P you HAVE to play 2 games back to back to mitigate the start factor; 2-0 is a 1.0 result /// 1.5-0.5 is 0.75 /// 1-1 is 0.5 /// etc. when you calculate the ELO changes, regardless of the sum of the games’ score.