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Subject: 2 Faced D6 Question rss

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Scott Allen
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I was thinking about this idea for gaining resources in a game. I have since moved on to something else, but this is still interesting to me.

Imagine rolling 2D6 to gain 1 of 4 resources in a game. Let's call the resources A, B, C, D to keep it simple.

Each face of each die has 2 resources on it, in the following distribution:
4 A
3 B
3 C
2 D

So, each die would have the following 6 faces: AB AC AD AB BC CD

Roll 2 dice, then get (or in rare instances choose) the resource that shows up twice. For example, my roll might be: AC AB, so I get 1 "A" resource.

You roll AC AC, you can choose to get an A or a C.

I roll AD BC, I get nothing.

This seems like it may be a useful way to gather resources.


When I tried to figure out the odds of getting each resource, I get stuck.

The odds of getting an A resource (rolling 2 As) is 4/6 * 4/6 right? So, 16/36?

The odds of getting a B would be 3/6 * 3/6 = 9/36

The odds of getting a C would be the same, 9/36

The odds of getting a D would be 2/6 * 2/6 = 4/36

All seems logical, then I add them all up and get 38/36, which I guess is OK, since occasionally a player will get a choice of 2 resources to pick from.

Then, I think there are 2 ways to get 0 resources, so 1/36 + 1/36 = 2/36


I guess I'm just looking for someone to double check my math. The 38/36 threw me a bit. I probably need to do some "1 minus the odds of NOT getting something" thing somewhere, or no?

Any other thoughts on this approach? Is it used in any games already?

Thanks.
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Cheb
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You are pretty close with your maths

Here are the chances of each result in 36:

NONE 6
A 10
B 4
C 6
D 2
A OR B 4
A OR C 1
A OR D 1
B OR C 1
C OR D 1

So you are right that the chances of being able to get an A is 16/36.

You were slightly off on your calculations of getting no resources - it's actually 6/36 (1/6) [the combos are CD-AB BC-AD CD-AB AD-BC AB-CD AB-CD]

Also when you start combining probabilities you need to take into account that some rolls could get you either - e.g. the probability of being able to get an A is 16/36, the probability of being able to get a B is 9/36, but the probability of being able to get an A or a B is 21/36.

Hope that helps
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Graham Muller
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Thought would take a quick attempt at it
https://www.dropbox.com/s/k1595anllg6jnqr/DiceTest.xlsx?dl=0

It is an excel file that breaks it down, you can change the dice faces to test alternatives
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James Bowie
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My thought is that I'd much rather see three resources for this system and not four. So you'd have three combinations, A+B, A+C and B+C, and each die would have the same face twice. The reason to do it this way is that there's no longer a possibility of failure, and each resource has the exact same distribution method instead of being inherently imbalanced toward one or more symbols. The more resources you want, you can adjust the quantity of symbols on the dice faces or even the number of dice being rolled to make up for it, or maybe have one die face designated as a wild spot or some other effect, all to ensure there's no chance of failure.

Really though, it depends on the kind of game you'd be going for. If some resources are meant to be scarcer than others, the original method does make sense, but you can also make resources scarcer via other means, such as increasing the amount required to do things or making extra ways in the game to acquire the other, more common resources and thus making one symbol in higher demand than the others on the dice.

Looking at your example again, you also doubled up on a face rather than granting free distribution. A+B is there twice while B+D is missing, and after changing this you'd actually make each symbol appear an equal number of times (though there's still a chance of whiffing).

If you want to keep the uneven distribution and yet at the same time avoid the whiff chance, you could actually use four resources and two dice, but three symbols on some faces.

Example:

A+B+C
A+B+D
A+C+D
A+B
B+C
B+D


Using this formula, there are no whiffs, yet each result has a different set of odds of being possible.

Only A: 2/36
Only B: 8/36
Only C: 2/36
Only D: 2/36

Can be A: 16/36
Can be B: 25/36
Can be C: 9/36
Can be D: 9/36

If you assume that B is the least useful resource, then A, then C, with D being most useful, and that players always take the most useful result, this becomes as follows:

A: 10/36
B: 8/36
C: 9/36
D: 9/36

These numbers are important because they show that

1. The player will never have to choose between C or D, making the two of them equally precious.

2. The most common resource the player will have access too with slightly over 2/3rds of all rolls, yet if the player shafts it always in favor of C or D, the most common becomes the least common. In other words, even though the rarer resources are more precious individually, if the game requires some amount of all four to achieve victory, the player would frequently be forced to choose between the two.

While this sounds good, it probably isn't due to speed and game flow. Imagine if in Catan, instead of numbers on tiles they were on the nodes and rolling the number your city was on let you pick from any of the 2-3 adjacent hexes to collect from. That extra layer of decision making would slow the game to a crawl, even if it might lead to interesting choices or an increased skill ceiling.

Sadly, I'd wager this is going to be true for all versions of this system you try to come up with. if the roll happens frequently over the course of the game, it would probably get bogged down, hard.

The better alternative would probably be to have different TYPES of dice, based on the decisions the player has already made throughout the game. So the choice is made early on and persists later and throughout, instead of happening every turn. This is the equivalent of upgrading a city in Catan and instead of a second resource being granted from it, you pick a specialization for the city when you upgrade it and when one of its plots are rolled, you roll a die associated with that city type which may grant an additional bonus.
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