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Dead of Winter: A Crossroads Game» Forums » General

Subject: Formula for hidden objectives stack during setup rss

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(ɹnʎʞ)
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Hey everyone,

the rulebooks recommends: number of players x2 = amount of hidden objective cards + 1 traitor card.

There is also a special traitor variant, which just adds 1 regular hidden objective card per player +1 traitor objective.

Both variants sound a bit too extreme on both ends to me.
My first game will be with 5 players, resulting in a stack of 10 regular and 1 traitor objective.

As I said, I have no personal experience with this game so far, but this sounds quite diluted to me. Somehow. Maybe.

Am I wrong ? What is your formula during setup ?
 
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Ossian Grr aka "Josh"
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The Secret Objectives are all selfish in some way or another. The Betrayer's real only difference is that they want to be selfish *and* reduce Morale to 0, instead of being selfish *and* completing the Main Objective.
Even without a Betrayer, there are people who are going to need to "play suboptimally" anyway.
And (with the basic formula) there's just under a 50% chance of there being a Betrayer in the game at all, so you have to be on your toes whether people are playing suboptimally because of their regular Objective or because they're the Betrayer.

The chance of Exiling a non-Betrayer is a risk that makes the game interesting.

So, no I don't think this is a problem at all.

In practice, less than half of my games have had a Betrayer and the game was still full of doubt and finger-pointing.
 
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Nicholas Hjelmberg
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None of the variants is extreme if you calculate the probabilities.

With 10 regular objectives and 1 traitor objective, the 1st player gets a regular objective in 10 of 11 cases, the 2nd player in 9 of 10 cases and so on. (10/11)*(9/10)*(8/9)*(7/8)*(6/7) = 55% so in 55% of the games, there will be no betrayer.

Edit: Forgot to multiply with 1/2

With 5 regular objectives and 1 traitor objective, the calculation is (5/6)*(4/5)*(3/4)*(2/3)*(1/2) = 17% so in 17% of the games, there will be no betrayer.

Have I calculated correctly?
 
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Peter Mulholland
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jozxyqk wrote:
In practice, less than half of my games have had a Betrayer and the game was still full of doubt and finger-pointing.


There is nothing better than a game full of finger pointing, doubt, exiling, arguing and lack of trust which (win or lose) finishes with the discovery that there was no traitor all along.

Dont. Trust. Anyone.
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Antonia
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If my math did not leave me, the chances of no one being a betrayer are
3 players - 57.1%
4 players - 55.6%
5 players - 54.5%

As already said, roughly half of the games go without an betrayer and are still full of finger pointing. The game creates a tension being cautious and finger pointing at someone, not sure if he is an betrayer or has a pretty selfish objective and I liked that a lot.
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David Williams
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To clear up the probabilities - it's easier to just look at the final distribution of cards.

For the Betrayer variant, there are 6 cards one of which is a Betrayer. 1 card is left over, and it could equally be any 1 of the 6. Therefore there is a 1/6 probability this is the betrayer card. So the odds of having no betrayer is 1/6 = 17%, and the probability of having a betrayer is the remaining 5/6 = 83%.

For non-betrayer, there are 6/11 cards not used, and 5/11 used. There's an equal probability for the betrayer to be in any one of those 11 locations. Therefore there will be no betrayer in 6/11 games = 55%, and a betrayer in the remaining 45%.


If neither of these probabilities is to your taste, simply pick a medium ground. The probabilities of having a betrayer are:

5p, 11 cards: 5/11 = 45.5%
5p, 10 cards: 5/10 = 50%
5p, 9 cards: 5/9 = 55.5%
5p, 8 cards: 5/8 = 62.5%
5p, 7 cards: 5/7 = 71.4%
5p, 6 cards: 5/6 = 83.3%

And of course 5p 5 cards is 100%.

Take your pick!! I don't see that using any of these different probabilities will significantly affect the game. The greater the probability of a traitor, the more people will be looking out for them and the greater the paranoia. Which could make things easier or harder for the betrayer, it all depends on the player dynamics.
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David Williams
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nhjelmberg wrote:
None of the variants is extreme if you calculate the probabilities.

With 10 regular objectives and 1 traitor objective, the 1st player gets a regular objective in 10 of 11 cases, the 2nd player in 9 of 10 cases and so on. (10/11)*(9/10)*(8/9)*(7/8)*(6/7) = 55% so in 55% of the games, there will be no betrayer.

With 5 regular objectives and 1 traitor objective, the calculation is (5/6)*(4/5)*(3/4)*(2/3) = 33% so in 33% of the games, there will be no betrayer.

Have I calculated correctly?


Not quite - the first calculation is correct. The 2nd one should give a result of 1/6, which it would if you remembered to include the 5th probability of 1/2.

Your method is sound though, and necessary for more complex problems, but in this case there are simply 6 cards and 1 not used, which could equally be any 1 of them. Thus a 1/6 chance the discarded card was the betrayal card.
 
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Nicholas Hjelmberg
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Thanks, I stand corrected and of course it's simpler to just count the one dicarded card.
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David Williams
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nhjelmberg wrote:
Thanks, I stand corrected and of course it's simpler to just count the one dicarded card.


It is when there's only 1 card - if there were 2 betrayal cards (eek!) then I think it's easier to use the method you used. Otherwise you have to count all the possible combinations and how many of those end up with both betrayer cards being discarded. Useful for most deck calculations, but in this case it's simple enough not to bother.
 
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Ryan M
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Kyur wrote:
Hey everyone,

the rulebooks recommends: number of players x2 = amount of hidden objective cards + 1 traitor card.

There is also a special traitor variant, which just adds 1 regular hidden objective card per player +1 traitor objective.

Both variants sound a bit too extreme on both ends to me.
My first game will be with 5 players, resulting in a stack of 10 regular and 1 traitor objective.

As I said, I have no personal experience with this game so far, but this sounds quite diluted to me. Somehow. Maybe.

Am I wrong ? What is your formula during setup ?


As others have stated, it isn't as extreme as it sounds. It really depends how you feel about having a betrayor in the game. If you follow the standard game rules, you will be playing with no betrayor more than half the time. If you play the betrayor variant, you will almost always have a betrayor. Almost.

But the fun of the game comes from not knowing whether there really is a betrayor or not. Standard objectives may be hording food or letting players die. Most standard objectives are designed to ensure at some point you will be making a move that the group doesn't like (or may be harmful to the group) but you need it if you want to win.

The second thing that makes the fun is how these dynamics play out with or without a betrayor. How do you convince someone else to pick up the slack while you knowingly lie about the food you are hording? How do you convince the group to starve for a round or two, or fail a crises or two, so that you can better your own position at the end of the game? How far are you willing to risk blowing the whole game for everyone to pursue your own selfish goals?

Of course, if you still feel the odds are too extreme either way you can modify the objective deck to your liking. Maybe instead of 2 for every player, 1.5?
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(ɹnʎʞ)
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Thanks for the feedback.

Considering the calculations and personal experience you all provided, I guess following the rules will be less extreme than I first thought it would be.

The chances of a present traitor being just a bit under 50% sounds actually really good and reasonable to me. thumbsup
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Cameron McKenzie
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The game is always a little disappointing without a traitor. It's just a cake walk. Same issue as Shadows Over Camelot.

Sure, you may not be sure until the end, but typically if all players are loyal and good at the game, things will go so smoothly that you aren't even really worried about it anyway by that time.

And if it is a close game, and you end up winning and saying "We got lucky there wasn't a betrayer," that feels cheap to me. If you aren't that good, the entire game comes down to this coin flip? Meh.
 
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Ossian Grr aka "Josh"
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MasterDinadan wrote:
The game is always a little disappointing without a traitor. It's just a cake walk. Same issue as Shadows Over Camelot.

Sure, you may not be sure until the end, but typically if all players are loyal and good at the game, things will go so smoothly that you aren't even really worried about it anyway by that time.

And if it is a close game, and you end up winning and saying "We got lucky there wasn't a betrayer," that feels cheap to me. If you aren't that good, the entire game comes down to this coin flip? Meh.


Are you sure you're talking about Dead of Winter and not BSG here?

In DoW, even without a traitor, you are working on your own personal objective. Even if the Main Objective seems too easy -- and *especially* if the Main Objective gets finished too quickly -- you have to worry about not meeting your own goal.
There's no "we win" in DoW. There's 5 "I win/I didn't win" individual results.
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Nicholas Hjelmberg
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jozxyqk wrote:

Are you sure you're talking about Dead of Winter and not BSG here?

In DoW, even without a traitor, you are working on your own personal objective. Even if the Main Objective seems too easy -- and *especially* if the Main Objective gets finished too quickly -- you have to worry about not meeting your own goal.
There's no "we win" in DoW. There's 5 "I win/I didn't win" individual results.


Four games so far: 2 losses and 2 betrayer victories where the colony fell without the betrayer's "help".

Surely we can play better but with each new survivor needing 1/2 food (requires 1/2 search action) and attracting 1/2 zombie (requires another 1/2 action to kill) AND having to manage wounds, crises and crossroad cards with only 1 action die, I think Dead of Winter will remain difficult enough for many more sessions.
 
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Chris Witt
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If you're new to the game, definitely try the "normal" version first (# of players x 2 + 1 betrayer card). Heck, some of my friends wanted to try it first with no betrayer at all.

After several games, we've settled on somewhere between the two. For some reason with the standard, we ended up with no betrayer close to 90% of the time, which was definitely not the expected distribution. Now for 5 players (our usual), we do 7 normal and 1 betrayer cards.
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Ryan Bohm
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My group has taken this approach and we are quite fond if it. We use one non-betrayer objective for each player in the game, plus one additional non-betrayer objective as well as the betrayer objective.

So for example, if playing a 5 player game, there will be a total of 7 secret objective cards to choose from, with one being the betrayer card. We like the idea of it being more probable that a betrayer will be present, yet slightly increasing the odds that there is no betrayer.
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Dreadknot Knotdread
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I agree with the 7-9 cards per 5 players route. It's better to have a higher chance of a betrayer.
 
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Nicholas Hjelmberg
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All of you who prefer a high betrayer probability, how does it change your gameplay? To me, 50% sounds like a good balance as it is the suspicion of a betrayer that is the main source of tension.

Do you have a lower acceptance of "non-cooperative" actions? Do you always exile a player just to be sure? Or do you simply want as many games as possible to end with a betrayal, whether successful or not? I'm curious to learn from your experience.
 
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