How to calculate dice chance?

Lurk · May 2010

I have a boring job with a lot of time spent doing nothing. To keep me sane, I have been designing a board game and recording all my thoughts in a small notebook. I am getting close to producing a prototype but I need to refine one major mechanic before I can confidently test.

Combat.

The system I have in mind is fairly simple. Each unit has a power value from 1 to 6. When two units enter combat, the owners of those units roll 3 6-sided dice (or d6s). A roll result of 5 or 6 is considered a hit.

If neither players roll any hits, the unit with the higher power value wins.
If one player rolls a hit and the other player does not, the player who rolled the hit wins even if his unit had a lower power value.
If both players roll a hit, each hit rolled is added to their units power value. So if player A rolled 2 hits with his unit of 2 power, that units power would be 4 for determining the outcome of the battle. The unit with the higher power value after hit bonuses are added wins.

Each unit has their own set of abilities and there are cards to influence the outcome of the battle. Common effects include increasing or decreasing the number of dice rolled by 1, adding 1to dice results and re-rolling results.

Is there any convenient way of calculating dice chances or would I be better served by writing a program to do this for me?

enlightenedbum · May 2010

Better off writing a program. It's not awful to do by hand with three dice, but it's far far easier to just write the program.

matt has a problem · May 2010

You'd have a 1 in 216 chance of any given roll, as there are 216 possible combinations using 3 dice with 6 sides.

admanb · May 2010

matthasaproblem wrote: »

You'd have a 1 in 216 chance of any given roll, as there are 216 possible combinations using 3 dice with 6 sides.

But "a given roll" is a specific permutation. So 2,1,1 is distinct from 1,2,1.

Two dice are easy, three dice are a bitch. Write a program.

SkyCaptain · May 2010

1 Hit: 6 out of 216 (5xx, x5x, xx5, 6xx, x6x, xx6)
2 Hits: 12 out of 216 (55x, 56x, 66x, 65x, 5x5, 5x6, 6x6, 6x5, x55, x56, x66, x65)
3 Hits: 8 out of 216 (555, 556, 566, 666, 665, 655, 656, 565)

I think that covers all the permutations.

khain · May 2010

SkyCaptain wrote: »

1 Hit: 6 out of 216 (5xx, x5x, xx5, 6xx, x6x, xx6)
2 Hits: 12 out of 216 (55x, 56x, 66x, 65x, 5x5, 5x6, 6x6, 6x5, x55, x56, x66, x65)
3 Hits: 8 out of 216 (555, 556, 566, 666, 665, 655, 656, 565)

I think that covers all the permutations.

This isn't correct. The chance to roll only a single hit is (5xx, x5x, xx5, 6xx, x6x, xx6), but that isn't 6 out of 216 as x can be any number 1-4 which changes the chance to 96 out of 216 I believe. The second example is also wrong for the same reason and I believe it changes to 48 out of 216. If my numbers are correct this leads to 156 out of 216 that you roll 1 or more 5 or 6 with 3d6.

Sir Headless VII · May 2010

Something like This Website?

Tag · May 2010

Since 5 or 6 count as a hit, each die has a 1/3 chance to hit and a 2/3 chance to miss.

3 misses:
(2/3)(2/3)(2/3) = .296296
2 misses:
(2/3)(2/3)(1/3) = .148148
+
(2/3)(1/3)(2/3) = .148148
+
(1/3)(2/3)(2/3) = .148148
1 miss
(2/3)(1/3)(1/3) = .074074
+
(1/3)(2/3)(1/3) = .074074
+
(1/3)(1/3)(2/3) = .074074
0 misses:
(1/3)(1/3)(1/3) = .037037

If I did that right, that should account for every probability and add up to 1 (it does). Therefore the odds that you get
0 Hits: 29.63%
1 Hit: 44.44%
2 Hits: 22.22%
3 Hits: 3.70%

But... that's a lot of work to do for each mod. I would recommend taking Headless' advice and learn to use AnyDice so that you can quickly make adjustments.

DirtyDirtyVagrant · May 2010

I dunno. Your most probable outcome is something like 10.5, and the closer you get to either end of the range (3, 18), the smaller the odds of rolling that particular result. I want to say that the odd of 3 and 18 are purely based on the probability of the die rolls, whereas middle values such as 10 and 11 have multiple rolls that will result in their outcome, so the odds of them appearing is effectively much greater.

I don't know if this is helping or not. You could try writing a program, but if you search around you might just be able to find this information on the internet.

A single d6 has an average roll of 3.5 (not 3), because 0 is not a possible outcome.

Mojo the Avenger · May 2010

What you're looking for is called the Binomial Theorem. It's used to calculate probabilities associated with flipping coins, rolling dice. Anything where you want to know the probability of an outcome and the pool of possibilities doesn't change. Some brute force application of the theorem is probably less efficient than programming it, but it's probably worth a look.

GothicLargo · May 2010

From left field:

Why not design your game around using d20's instead?

Nostregar · May 2010

GothicLargo wrote: »

From left field:

Why not design your game around using d20's instead?

This is what I'm thinking too.

If you want to stick with the "single, double, triple hit" thing you could do rolls of 1-5 as no hits, 6-10 as 1 hit, 11-15 as 2 hits, and 16-20 as 3 hits.

The odds aren't really the same, but it is a lot simpler. You could shift the numbers however you want, too - maybe 3 hits is only a range of 18-20 to make it more difficult to get. Maybe the range of numbers for hits starts at 10 instead of 0 (for example, 1-10 is no hit, 11-14 is 1 hit, 15-18 is 2 hits, and 19 and 20 are 3 hits) to make them harder to get in general, meaning you only have a 50% chance of hitting at all.

There are a lot of different ways to arrange the numbers, but I do think that going to larger dice is a good idea.

Edit: You might also check out the way that Mechaton does battle, because I kind of like that system. Basically, each unit has a certain number of "attachments" of various types and in an attack they get to roll 1d6 per combat attachment and add the rolls up. This number is then compared to the defenses of the attacked unit.

So yeah, not sure what kind of game you're doing, but that's one thing that has worked out pretty well.

If you use the d20 idea, maybe make different weapons have different chances to hit (so, different arrangements of rolls for the different numbers of hits)?

admanb · May 2010

I disagree. Large dice create a significantly different feel to the game. They are simpler, but also less predictable. Less predictability in a strategy game means less ability to make tactical plans and more reliance on luck. The more dice you're rolling and the smaller they are the more they tend towards the mean, and therefore the better the players ability to plan around them and the more effect player skill has on the game.

Obviously randomness should be a factor, or you wouldn't be using dice, but using a D20 over 3D6 significantly changes the tone of the game.

dispatch.o · May 2010

admanb wrote: »

I disagree. Large dice create a significantly different feel to the game. They are simpler, but also less predictable. Less predictability in a strategy game means less ability to make tactical plans and more reliance on luck. The more dice you're rolling and the smaller they are the more they tend towards the mean, and therefore the better the players ability to plan around them and the more effect player skill has on the game.

Obviously randomness should be a factor, or you wouldn't be using dice, but using a D20 over 3D6 significantly changes the tone of the game.

It also removes the option to add in abilities or modifiers that allow you to reroll a single or pair of dice for specific units.

Tag · May 2010

It's not really any less tactical, it's just different ranges and distribution. A 1-20 die will have an equal chance to roll any of those numbers, a 3d6 setup will have a bell curve with a very low chance to roll 3 or 18 and a very high chance to roll around 10.

Thus, if you want equal probabilities or an easier to adjust scale, a larger die works better. If you want a bell curve setup (with, say, critical fails and success at low probability numbers), 3d6 is better. For the system described I would lean toward the d20 too. It would easily and intuitively allow one to set up, say, 30% miss, 40% one hit, 20% 2 hit, 10% three hit (or any other combination) and it makes things like "add 1 to the roll" easier to balance and plan for.

Of course, the average person / house is more likely to have a few D6s just laying around than D20s or D100s.

MrMonroe · May 2010

you just said "calculating dice chances" without indicating really what you want to calculate. How likely a hit is? How many hits you are likely to get in a certain scenario? The statistical effect of a particular ability?

any of these questions is going to be solved by a single expression on a calculator, it would be silly to try and program something to brute force every single possibility

Sir Headless VII · May 2010

As a more concrete example of the webstite I posted earlier (and because I wanted to mess with it more)
This equation will calculate who wins a given encounter between a player with power 1 and a player with power 2 in this situation:

Lurk wrote:

If both players roll a hit, each hit rolled is added to their units power value. So if player A rolled 2 hits with his unit of 2 power, that units power would be 4 for determining the outcome of the battle. The unit with the higher power value after hit bonuses are added wins.

If there is a negative result the player with power 2 won and a positive the player with power 1 won. You can replace the 1 and 2 in the equation with anything you want obviously. The equation itself counts the number of results greater than or equal to 5 and then adds the power value. The results of the two players are then subtracted (so you can see which is higher).

The number of hits seems to be the only other equation you would need for that small ruleset.

The website is pretty powerful and set up to do this type of thing.

EggyToast · May 2010

Since your job is so boring, perhaps a more interesting approach to the game problem would be a way to resolve combat without resorting to dice? Removing the random aspect of combat increases the overall strategy of combat significantly, because the tactics employed must match what's on the table -- just look at M:TG for an example of how a lack of dice influences combat. When you *know* what will happen, it forces the combat to be balanced on the part of the designer -- hence, more work for you.

Plus I think diceless combat is more interesting overall. The boardgame Risk is dreadful, and D&D players strive to modify their equipment and skills in a way that minimizes the randomness of the dice overall. That's not to say that you eliminate all forms of chance -- just that you, as the designer, can create specific ways to influence combat and the chance involved, rather than letting dice do it for you.

Lurk · May 2010

I am designing this game as an exercise and trying to keep it as realistic as possible, which means that I want to keep costs as low as possible. It is highly unlikely that I will ever sell this game (unless it turns to be magically fun), but keeping the theoretical costs means that it will be easier to prototype as well.

That said, d20s are more expensive to produce/acquire than d6s, so I am going to stick with them. Often the alternative to dice is cards and specifically producing a deck to determine combat outcome is much more expensive than having a handful of d6s.

Also, I have changed some of the mechanics to be more simple/intuitive, the possible results are now:
A) If neither players roll any hits, the unit with the higher power value wins.

If only one player rolls a hit, the player who rolled the hit wins even if his unit had a lower power value.
C) If both players roll a hit, the unit with the higher power value wins.

A fight between a unit of power 3 and power 4 has the same possible probability higher power unit winning as a fight between a unit of power 3 and power 100. You have combat advantage or you don't.

Rend · May 2010

Lurk wrote: »

I am designing this game as an exercise and trying to keep it as realistic as possible, which means that I want to keep costs as low as possible. It is highly unlikely that I will ever sell this game (unless it turns to be magically fun), but keeping the theoretical costs means that it will be easier to prototype as well.

That said, d20s are more expensive to produce/acquire than d6s, so I am going to stick with them. Often the alternative to dice is cards and specifically producing a deck to determine combat outcome is much more expensive than having a handful of d6s.

Also, I have changed some of the mechanics to be more simple/intuitive, the possible results are now:
A) If neither players roll any hits, the unit with the higher power value wins.
If only one player rolls a hit, the player who rolled the hit wins even if his unit had a lower power value.
C) If both players roll a hit, the unit with the higher power value wins.

A fight between a unit of power 3 and power 4 has the same possible probability higher power unit winning as a fight between a unit of power 3 and power 100. You have combat advantage or you don't.

In that case your randomness will be very chaotic. When the power of a unit is only useful for breaking ties, your random is much much different.

Lurk · May 2010

Given that a lower power unit can beat a higher power unit only when the higher power unit rolls 0 hits (and it rolls at least 1), I think it won't be extremely chaotic. Units of a same power employ a tie-breaking mechanism that I am fiddling with still.

Having hits treated as increases in power value adds complication with very little benefit. A unit of power value 1 needs to roll 3 hits to beat a unit of power value 2 that rolled just 1 hit.

piL · May 2010

First, I will say that there are plenty of opportunities for me to err in the following description. I write, however, with an authoritative voice, because adding "I think" or "if I calculated it correctly" after every calculation would just make it harder to read.

This rule set could be simplified. This isn't to make it easier to understand, it just reduces the number of calculations. If both players score hits, then you simply compare power (P) + hits (H). When nobody hits, it still uses this comparison, but in this case, H = 0.

This gives us two categories: One where P + H is compared, and one where one player scores hits and the other player does not.

Lets talk about this second category first.

So you start thinking about 216. This is useful, but it's not the whole picture; the odds are actually a subset of 6^3 * 6^3, which equals 6^6 = 46,656, since the outcome is dependant on /both/ players rolls.

So what are the odds of the second category even occuring? They are the sum of the odds of each player rolling a hit without the opponent rolling a hit.

The odds of a player not rolling a hit is 4/6 * 4/6 * 4/6 = (4/6)^3
The odds of the other player rolling a hit is the odds of not rolling the above, or ( 1 - (4/6)^3 ).

The odds of player 1 winning via category 2 is (4/6)^3 * ( 1 - (4/6)^3 ) = 64/216 - (64/216)^2. 20.85%. The odds of player 2 winning via category 2 is the same, so 20.85%. So now we've accounted for 41.70% of the outcomes!

That leaves 58.30% of the outcomes for category 1, which is dependent on respective power. Here, I will refer to P as Power of player 1 minus the power of player 2. So if player 1 has a power of 5 and player 2 has a power of 4, then P = 1. If they have the same, P = 0. If player 1 has a power of 3 and player 2 a power of 6, then P = -3.

If P > 2, then player 1 wins 100% of the time, except for category 2 wins by player 2. The following are the possibilities:
P1 Win: 148/216 = 79.15%
P2 Win: (4/6)^3 * ( 1 - (4/6)^3 ) = 64/212 - (64/212)^2 = 20.85%

If P = 2, then two possible outcomes exist: P1 win. P1 will win when there is no tie. Tie occurs only when P2 scores 3 hits and P scores only 1.

P2 scores three hits with a probability of (2/6)^3, or 8/216.
P1 scores one hit and only one hit when (2/6)(4/6)(4/6) + (4/6)(2/6)(4/6) + (4/6)(4/6)(2/6) = 3((4*4*2)/(6^3)) = 96/216

The odds of this tie event are the multiplication of the two. (8/216) * (96/212) = 768/46656 = 1.65%

So, if P = 2, the following are the odds of possible outcomes:
Tie: 768/46656 = 1.65%
P2 Win: (4/6)^3 * ( 1 - (4/6)^3 ) = 64/212 - (64/212)^2 = 20.85%
P1 Win: 77.50% (everything else)

If P = 1, then P2 wins with 3 hits vs 1 which (fortunately) we've already calculated to be 768/46656 = 1.65%.
The other way for P2 to win is by category 2, making the odds of a P2 victory 20.85% + 1.65% = 22.50%

Tie will occur when P2 scores 3 hits and P1 scores 2 or when P2 scores 2 hits and P1 scores 1.

3v2: (2/6)^2 * ( (3(4*2*2)/212 ) [calculated similarly to above]) = (8/216)*(48/216) = 384/46656 = 0.82%

2v1: ( (3(4*2*2)/216 ) * ( (3(4*4*2)/216 ) = (48/216) * (96/216) = 4608 / 46656 (I hope you expected this to be more likely then 3v2, due to having many more possibilities) = 9.88%

So the odds of Tie are 9.88% + 0.82% = 10.7%

Whatever is left is P1's odds of winning.

Tie: 10.70%
P2 win: 22.50%
P1 win: 66.80%

And now, P = 0. We can think about this differently. P1's chance of winning is equal to P2's chance of winning. The outcomes are: Tie or someone wins. Because the two are equal, then the odds of either winning are someone wins divided by 2. That means all we have to figure is the odds of Tie.

Tie occures when both players score the same number of hits, or score no hits.

3v3: (2/6)^3 * (2/6)^3 = 8/216 * 8/216 = 64/46656 = 0.14%
2v2: ( (3(4*2*2)/216 ) * ( (3(4*2*2)/216 ) = 48/216 * 48/216 = 2304/46656 = 4.94%
1v1: ( (3(4*4*2)/216 ) * ( (3(4*4*2)/216 ) = 96/216 * 96/216 = 9216/46656 = 19.75%
0v0: ( (4*4*4)/216 ) * ( (4*4*4)/216 ) = 64/216 * 64/216 = 4096/46656 = 8.78%

P1's chance of victory = P2's chance of victory = 1 - (chance of Tie).

Tie: 33.61%
P1 win: 33.20%
P2 win: 33.20%

To figure out values for which P is negative, just flip the values. So, to summarize:

If P > 2
P1 Win: 79.15%
P2 Win: 20.85%

If P = 2
Tie: 1.65%
P2 Win: 20.85%
P1 Win: 77.50%

If P = 1
Tie: 10.70%
P2 win: 22.50%
P1 win: 66.80%

If P = 0
Tie: 33.61%
P1 win: 33.20%
P2 win: 33.20%

If P = -1
Tie: 10.70%
P2 win: 66.80%
P1 win: 22.50%

If P = -2
Tie: 1.65%
P2 Win: 77.50%
P1 Win: 20.85%

If P > -2
P1 Win: 20.85%
P2 Win: 79.15%

So yeah, that was kind of complex, but I have a lot of fun playing with probabilities. As for the effect of variations due to special abilities, I'm not going to actually do that myself, but if you can get a decent grasp of what I've illustrated in the above, you could probably slip changes into there and see what happens, and you can definitely see what happens when you do simple changes to a unit's power.

Fake edit: Also go to hell for changing the rules :P

piL · May 2010

New rules. Three possibilities

P > 0
P = 0
P < 0

Odds sort of match the P = 0 from before, so I'll just take that...

Tie: 33.61%
P1 win: 33.20%
P2 win: 33.20%

If P > 0
P1 win: 66.81%
P2 win: 33.20%

If P = 0
Tie: 33.61%
P1 win: 33.20%
P2 win: 33.20%

If P < 0
P1 win: 33.20%
P2 win: 66.80%

Edit: This is wrong.

Tag · May 2010

Well you could use common playing cards like War, Ace-King is 1-13. Mix suit in and you can get 1-52, which isn't bad single "roll" granularity, albeit with a lot of shuffling. However, with shuffling limits, you can add card counting to tilt battle probabilities in your favor as part of the skill.

But with the current system as you've just described, the die are relatively unimportant and will probably frustrate more than help. If only a single hit on each side matters and there is only a base 30% that either side will miss, the odds that both players will miss are only ~9%. That means the straight powerlevel only *won't* win less than 9% of the time, which may be the point (sort of a hail mary for weak units) but seems like it will just be eating up time the other 91 out of 100 battles or frustrating someone who gets really unlucky. It also potentially unbalances the rest of your game toward swarm rather than quality.

I do think the idea is cool though, I love mechanics thinking.

Edit: Misread:
Odds that only a single side will miss are 22%. Odds that it is the inferior side are 11%. So it's slightly more important but not really :-/

piL · May 2010

Lurk wrote: »

Given that a lower power unit can beat a higher power unit only when the higher power unit rolls 0 hits (and it rolls at least 1), I think it won't be extremely chaotic. Units of a same power employ a tie-breaking mechanism that I am fiddling with still.

Having hits treated as increases in power value adds complication with very little benefit. A unit of power value 1 needs to roll 3 hits to beat a unit of power value 2 that rolled just 1 hit.

The benefit, as it were, is the addition of different levels of advantage based on different levels of power; P1 wins in a 3-1 much more often than in a 3-2.

Lurk · May 2010

Jesus Christ piL, you're amazing! I am going to carefully study the detailed explanation you gave

I find it strange that the win percentage of the P2 would go up under the new rule set when it gives less a chance for the weaker power value unit to win.

The unfairness of dice is a pretty large issue, though I feel that randomness is required to give weaker units a chance. Each player controls a faction that has 2 possible units that it could make, a basic unit and a special unit. As an example, here are two factions of the ten I have so far(stock refers to the maximum amount of a unit you can have):

Cabal
Cultist [power 1, cost 2, initiative 2, stock 10]: When cultist is killed, take 1 gold from the coffers.
Shrine [power 4, cost 4, initiative 0, stock 1]: Structure (cannot move, can only defend)

Legion
Soldier [power 3, cost 3, initiative 2, stock 5]: If soldier wins a combat, replace it with a veteran if able.
Veteran [power 4, cost n/a, initiate 3, stock 5]

Each faction has a few cards that focus on their special units, Shrines (sacrificing cultists for effects) and Veterans (being awesome), in addition to cards that focus around each faction's theme. Cabal are based around area control while Legion is focused around combat (and needs to win combat to increase army size).

Under a system where power is the only that is compared, the Cabal would always lose to the Legion in terms of combat (outside Shrine vs Soldier). That is the main reason why I think dice are important, to give the little guy a chance to win the fight. Yes, the Legion has a much more expensive units, but since they combat themed faction, they will have card effects that will allow them to not lose their unit investments.

This is an explanation for people who were wondering about context.

Sir Headless VII · May 2010

Pil something seems off with your numbers for the newer rules. This is what I got using nuymbers calculated here

A) If neither players roll any hits, the unit with the higher power value wins.

chance of neither getting any hits: 8.78%

If only one player rolls a hit, the player who rolled the hit wins even if his unit had a lower power value.
it should be

P1		P2		# rolls		%

1 hit 		0 hits		6,144		13.17
2 hits		0 hits		3,072		6.58		
3 hits		0 hits 		512  		1.10
0 hit 		1 hits		6,144		13.17
0 hits		2 hits		3,072		6.58		
0 hits		3 hits 		512  		1.10

				total:		41.7%

C) If both players roll a hit, the unit with the higher power value wins.

P1		P2		# rolls		%

1 hit		1 hits		9,216		19.75	
1 hit		2 hits		4,608		9.88	
1 hit		3 hits 		768	           1.65	
2 hit		1 hits		4,608		9.88		
2 hit		2 hits		2,304		4.94	
2 hit		3 hits 		384		   0.82
3 hit		1 hits		768          	1.65	
3 hit		2 hits		384   		0.82
3 hit		3 hits 		64                    0.14

				         Total		49.53%

If powers are not equal:

Chance of unit with higher power winning: 79.11 % (conditions A and C and half of

Chance of unit with lower power winning: 20.85 % (half of

if powers are eqaul:
From what I understand of the rules a tie will occur when both units have the same power and condition B is not met.
That will be the same as letting power decide and should occur 58.3 % of the time.
Chance of a random player winning: 41.7 %

soxbox · May 2010

Lurk wrote: »

As an example, here are two factions of the ten I have so far(stock refers to the maximum amount of a unit you can have):

If you're going to have players select units with different abilities, you're really going to need to playtest each faction against every other faction - you ideally want to have either every faction being mathematically equivalent to every other or have some form of earth/wind/fire/water circle of power so that each one is more powerful than the next but having a specific weakness.

With 10 factions, you're going to have ummm 45 (n*(n-1)/2 ?) combinations to playtest each faction against each other faction, and you'll need to playtest that a few times each... which sounds very time consuming.

Can you program? If not, learn to. Being able to program and run simulations to answer these sorts of questions is a really really really useful skill to have in game design. It will also help train your brain to feel out your game design a bit.

piL · May 2010

Sir Headless VII wrote: »

Pil something seems off with your numbers for the newer rules.

You're right, I rush jobbed the newer rules without giving it the proper thought before going to sleep. It bugged me in bed last night :P I measured the chance of ties, but that doesn't matter; all that matters is the chance of both players rolling one or more hits. I just did the following math to check what you said, and you're right. Then I just finished it up to sort of give Lurk an idea of how you came up with the numbers.

P > 0
P1 wins when there are no hits, when there are both hits, and when he gets any hits and his opponent receives none. The other results are P2 wins.

No hits:
(4/6)^3 * (4/6)^3 = 8.78% (chance of each player getting no hits)

Both hits: (1 - (4/6) ^3) * (1 - (4/6)^3) = 49.20% (chance of each player not getting no hits)

Only P1 hits: (1 - (4/6) ^3) * (4/6) ^ 3 = 20.85% (chance of player 1 not getting no hits, but player 2 gets no hits)

P1 wins 8.78% + 49.20% + 20.85% = 78.83%
P2 wins 11.17%

P = 0
Tie when there are no hits, or both hits. P1 wins when he receives a hit, but opponent does not. P2 wins when receives a hit, but opponent does not.

Only P1 hits: (1 - (4/6) ^3) * (4/6) ^ 3 = 20.85% (chance of player 1 not getting no hits, but player 2 gets no hits)
Only P2 hits: (1 - (4/6) ^3) * (4/6) ^ 3 = 20.85% (chance of player 2 not getting no hits, but player 1 gets no hits)

Chance of tie: Remainder

P1 wins 20.85%
P2 wins 20.85%
Tie: 58.30%

soxbox wrote:

Can you program? If not, learn to. Being able to program and run simulations to answer these sorts of questions is a really really really useful skill to have in game design. It will also help train your brain to feel out your game design a bit.

I feel that the skills I used for this are influenced by learning to program. I don't program, and it would take me forever to make anything useful these days, because I don't remember the language parts, but the skill of going through possibilities stepwise and trying to consider all inputs was a brain-muscle I certainly flexed while learning to program many years ago. Makes Game Theory make a whole lot more sense as well, which is pretty useful for game design imo.

Nostregar · May 2010

Lurk wrote: »

That said, d20s are more expensive to produce/acquire than d6s, so I am going to stick with them. Often the alternative to dice is cards and specifically producing a deck to determine combat outcome is much more expensive than having a handful of d6s.

It's your game so do whatever you want, obviously, but I just want to let you know that you can pick up a d20 at a local game/hobby shop for like $0.50.

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