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D10 math (game design)

15 replies [Last post]
Sat, 06/11/2005 - 11:22
Anonymous

HI This is my first post here.

I'm trying to make my own game and I really suck at math.
Can anyone solve this riddle?

Ken and Dan want to win, to win, they must throw a variable on xd10.
Before the game starts, Ken and Dan have a number of $ that they can spend. With these $ they can buy d10's and the variable (chance) that they must roll to win.

EX
Ken paid for 1d10 with a variable of 5 (50% chance of victory).
To win, Ken must throw 1d10 and get that result or lower.
Let's say that +10%=1$

Players can also invest in penalties for all their adversaries.
They can invest their $ in dice, bonusses(their own percent) and penalties.

Question 1:
How much would -10% cost?

Question 2:
When -10% and +10% point values were determined.

How much would it cost the players to "buy" 1d10?
What is the price of 1d10?
(The price for getting a chance at winning, they still have to pay for the % on those dice seperatly.)

Let's say, that Dan wants two shots. (Two dice rolls.)
Does this cost him $? (He is already paying for the percent of each roll.) In the above example: if 10%=1, two rolls (each 50% would cost him a total of 5+5points=10points, wich would allow Ken to buy 100% and win the game.

AN ANSWER FROM BBG, but it looks incorrect...

"For the first question, I'd say +10% and -10% should be equal in cost, purely mathematically speaking. Perhaps thematically you might want to adjust it to something like: pay $x+1 to lower the opponents odds by x*10%. In plain English, you pay $1 just to be able to mess with him, and then every additional $1 lowers his odds by 10%.
Second question:
Buying 2 rolls (50% and 50%) doesn't add up to 100%. You multiply the chance of failure, so 0.5 (or 50%) times 0.5 = 0.25 (or 25%) chance of failure. Two rolls = 75% chance of success. So an extra d10 would be worth 2.5; you'd have to choose if you wanted to value it at $2 or $3.

So the 3rd d10 would be .5 * .5 * .5 = .125 chance of failure, equals 87.5% chance of success. So the 2nd and 3rd d10's together are worth 37.5%, or ~$4.
Thus, if you valued a d10 at $2, the first one purchased would be worth a little more than you paid, the second a little less, and each successive die worth less and less.
If you value it at $3, and only allow the player to purchase one extra die, then it is a risky move, since the odds gain is a little less than the expense. This might be a good thing, depending on the intent of your design.
Even if you had infinite d10's, the chance of success never reaches 100%. It gets close, but there's always the chance that you'll roll all >5's.

MY PROBLEM WITH THIS
(Yes, I was too stupid to understand his answer, so please keep yours clean and easy..)

10percent chance on 1d10= 2$(d10)+1$(10%)=3$
100percent chance on 1d10= 2$(d10)+10$(100%)=12$

System correct? Do all of these players have the same chance at winning?
100percent chance on 1d10 (player A) = 12£
20percent chance on 1d10 + all other players -80% (player B) = 12£
10percent chance on 2d10 (player C) + 60percent chance on 1d10 = 12£

(Player C can't win, player B destroyed his chances with his -80%)
With the penalty of 1point for screwing over other players (would the "-1point for screwing other players" be multiplied with each -10% or a base value 1+ "number of -10% penalties"?)

Thanx in advance and for reading my post.
If you know a link where I can get this info - please post it.

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Sat, 06/11/2005 - 15:54
#1
Hedge-o-Matic
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D10 math (game design)

This sort of calculation is exactly the major flaw I first pointed out when introduced to the Vampire RPG when it came out long ago. My question at the time was whethr it had ever been playtested, since it's not obvious which is more difficult: 4+ on three dice, or 5+ on two dice.

what makes this sort of culculation cumbersome and inaccurate in an RPG, however, might be the core of a good game of a different type, such as your idea. Basically, you're dealing with two variables: D (the number of dice rolled) and C (the % chance of success on that particular die. The value if any given mix of non-additive dice (each considered individually) is merely (100/(d x 10)) x (total v).

Since this is the heart of the game, I'd say that this value should be a percentage of some sliding scale, to make it impossible to buy 100%. Maybe the cost would be the percentage of the player's current money, so the more money you've got, the more each percentage costs.

One problem, game-wise is that if the players can buy higher chances of success with their winnings, you've got a built in runaway leader problem. Early wins make subsequent wins more likely, which make even larger payoffs likely, and so on. I'm not sure if even my "sliding scale" idea would eliminate this, though it would help.

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Sat, 06/11/2005 - 16:47
#2
Anonymous
?

The value if any given mix of non-additive dice (each considered individually) is merely (100/(d x 10)) x (total v).

what is "v" ?

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Sat, 06/11/2005 - 16:59
#3
hpox
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D10 math (game design)

total v is total value.

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Sat, 06/11/2005 - 17:21
#4
FastLearner
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D10 math (game design)

Here was my post on the BGG in response to you... I'd rather continue the conversation here, as it's easier and more appropriate:
I suggest making each die modifier more expensive (using triangular numbers, but those words aren't important).

+ or - 1 costs 1

+ or - 2 costs 3

+ or - 3 costs 6

+ or - 4 costs 10

+ or - 5 costs 15

This allows minor modifications to be affordable, while greater ones are quite expensive -- but do-able, if it's important enough to you in the game at that moment.

As noted, the additional die adds 25% if the dice aren't modified, but the math gets trickier if one die is modified and another is not.

If, for example, 1 player is going to roll and he pays for a +1, and the other player pays for a -4, the net chance of success is now 20%. If he adds another die that has no modifiers, his net chance of success becomes 60%, so the die tripled his chance of success (increase of 200%).

However, if 1 player pays for a +1 and the other for, say, a -1, and the player buys another die, his odds improve to 75% (increase of 50%). So buying a die when your odds are bad is a really great move, which means an extra die should be expensive. Except when the odds are close to even, at which point it shouldn't be as expensive.

HOWEVER

If the modifiers always affect ALL dice, no matter how many are added, then in the above scenario with the +1 and the -4, the additional die changes the odds to 36%, an increase of about 80%.

With the +1 and -1 example, it's as above, a 50% increase. Still the extra die is more valuable in relative terms if your odds are bad, though not as effective in absolute terms (16% increase vs. 25% increase).

I think you need to get a real handle on the math for this game, as such things are pretty tricky, and odds are that without doing so you'll end up with something where a reasonably math-skilled player will be able to kick the ass of a player with poor math skills, being able to see that, for example, an extra die is an awesome value if x and y are true, but a terrible value if z is true, things that the math-challenged player can't see at all. Which, after one playing, won't be fun for either of them.

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Sat, 06/11/2005 - 17:22
#5
FastLearner
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D10 math (game design)

And in response to your followup question there:

Quote:

Your triangular system, is it still viable if players must pay(gamble) in advance (not knowing what the other players will "field" as d10s?

Yes, it works just as well when you don't know what the other guy is spending, you bet.

In regards to this:

Quote:

Start game; player choose a basic percentage (for all dice).
They get one "d10" for free (the first one).

Stat of 5: first die costs 5. (?yes)
Extra die costs 2,5 and keeps lowering in value (so a basic value of 2)

Startstat of 4: would cost 4.
Extra dice cost 2,4?

Startstat of 6: cost 6
Extra dice cost 4 * 4 = 16 (100-16=84. already payed 60) 2,4.

etc?

Unfortunately I don't quite understand what you're asking.

-- Matthew

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Sat, 06/11/2005 - 17:37
#6
Anonymous
thanx and more...

Well, I'm asking if this is correct.

If players get their starting 'unit' OR d10 for free (but have to pay for the X% variable that comes with that free d10), will the extra dice cost 2,5 whatever the base-% value?

strength = % value

as in: "unit strength 5, 1d10" = 5$
"unit strength 2, 1d10" = 2$
and any extra dice for both of these would cost 2,5$?
or would the 2 be = 1,6$

if i want the correct value for enemy-penalties? i dont use the triangular concept. BUT then, (if im correct?..) all other values (like the extra dice) that those enemies bought would also decrease in value.

(If the strength five player gets -3, his dice would suddenly be overcosted, as they initially cost him 2,5 and are only worth 1,6$).
?
please confirm that i was correct about the 1,6$ value and the penalty stuff, it would mean that i have understood the basic formula for determining the secondary dice costs.

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Sat, 06/11/2005 - 17:56
#7
sedjtroll
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D10 math (game design)

Frankly many of the descripions lost me. And I'm pretty good at math and stuff.

The guy who you quoted from BGG seemed to not understand something though... he seemed to think that once you bought a 50% chance, ALL your dice had a 50% chance, so adding dice worked th way he said.

I believe what you actually said was that EACH 10% for EACH die costs $1, so (ignoring the cost of the die itself), buying a 50% chance on 2 different dice costs $5+$5=$10, and buying 100% on a single die costs $10 as well.

I think that would make for a lousy game. Like Matthew said, I would recommend increasing osts for additional %. Either triangular numbers as Matthew listed (1, 3, 6, 10, etc) or just linear (1, 2, 3, 4, etc) depending on how much money people are likely to have.

The important math here in making the decision (as a player) is what the BGG guy said... 2 dice with 50% chance each costs a certain amount, and has a final chance of success of 75% (half the time the first die doean't win, and of those times, half the time the second die doesn't win either - so 25% of the time you lose, which means 75% of the time you win).

So understanding the 1 - (.5 * .5) = .75 thing will help you decide what you should do... buy another die with lower chances or increase your chances on the die you already have.

I'm not clear on how your penalty is supposed to work. If you spend whatever plus $1 per 10% penalty, does that apply to EVERYONE, or one player? Or just a specific die?

- Seth

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Sat, 06/11/2005 - 18:04
#8
Anonymous
D10 math (game design)

im not saying this to be rude but.....
this game is even if you balance it "perfect" gonna get very boaring very fast by people sitting down counting on the best options and calculating how to conter each other the best way ..

Maybee if its a pure battle solider game it might work ..

Im just saying what first pops to my mind

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Sat, 06/11/2005 - 18:08
#9
Anonymous
more and more

it goes for one player

Setup:
All players get same $ amount.
Player gets 1d10 for free and must start by choosing a value (% chance of winning), that chosen value will be used for all of his dice.
The starting value is zero, so they must pay for a value (and chance of victory)

Options:
Value for all d10s -------> 10%=1$
Additional dice= (failure * failure.......formula explained earlier)
Penalty on all enemy dice ---------> -10%=1$ (or triangular if more correct)

?
It must be completely balanced. Equal between both players.

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Sat, 06/11/2005 - 18:12
#10
Anonymous
game quote

eweric wrote:
im not saying this to be rude but.....
this game is even if you balance it "perfect" gonna get very boaring very fast by people sitting down counting on the best options and calculating how to conter each other the best way ..

Maybee if its a pure battle solider game it might work ..

Im just saying what first pops to my mind

This is not the complete game. It is one mechanic that I could not solve,
because I suck at math. The other game-elements have already been solved.
It is VERY IMPORTANT however that this part of the game is completely balanced (even if it is boring) for the other game-elements to stay balanced as well.

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Sun, 06/12/2005 - 01:25
#11
Hedge-o-Matic
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D10 math (game design)

Oops! My bad:

Basically, you're dealing with two variables: D (the number of dice rolled) and C (the % chance of success on that particular die. The value if any given mix of non-additive dice (each considered individually) is merely (100/(d x 10)) x (total v)

V should indeed be C. Sorry 'bout that.

But I'd be careful to spell all this out during the game, even as far as to have each possability given on anlittle table, considering the fact that everyone's going to do the calculations in their heads anyway. Like you pointed out in reply to eweric, this is just a mechanic and not the soul of the game. so, sice these costs are going to be fixed, I'd advise spelling them out so they don't bog down the game.

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Sun, 06/12/2005 - 06:55
#12
Anonymous
Again

....chance of success on that particular die. The value if any given mix of non-additive dice (each considered individually) is merely (100/(d x 10)) x (total v)

So if I buy 1d10 value 5 = 50$
Every extra d10 will be 25$ --- perfect thanx.

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Thu, 07/21/2005 - 17:21
#13
Anonymous
dice

It sounds too much complex for my brain, sorry.
Since the probability of taking a value such as 3 or more on different dices is:

3+ for 3 side dice:
{3} = 1/3 = aprox. 33%

3+ for 6 side dice:
{3, 4, 5, 6} => 4/6 = aprox. 67%

3+ for 10 side dice:
{3, 4, 5, 6, 7, 8, 9, 0} = 8/10 = aprox. 80%

The number of sides really matters, because is easy taking 3+ for 10 side dice than 3 side dice.

Just a suggestion: offer diferent dices for each situation. Let´s suppose all players start game with 3 side dice and when they pay for bonus, may use 6, 10 and so on. It would be more practical than dealing with probabilities explicitelly.
Remember that probabilities are important for game designer, not fun for players.

Good luck,
R. Biondi

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Mon, 08/08/2005 - 16:49
#14
Anonymous
NEED HELP - FRUSTRATED LIKE HELL

This is an old post of mine. I never totally grasped this mathematical 'solution' for my problem=

The game is fought out (yes, this was an earlier try, for the miniature wargame) - this game is fought between multiple units.

One unit consists of models (each model represents 1d10) and their base-combat trait. If they throw the trait (in my example above: it was 5), they destroy any enemy unit within range.

But I had to give the units costs. This was the solution someone gave me:

"Basically, you're dealing with two variables: D (the number of dice rolled) and C (the % chance of success on that particular die. The value if any given mix of non-additive dice (each considered individually) is merely (100/(d x 10)) x (total c)"

But:

I don't get it:

(100/(2x10)) x (50) = 250. (Or should I have used 5 and not 50?)

Anyway = that was a unit of Combat 5 with 2 models (2d10)

(100/(1x10)) x (50) = 50. (What?)

This unit, wich rolls only 1d10 and has 50% of success costs 1/5 of a unit that has 75% of success.

Is my math wrong or the formula?

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Mon, 08/08/2005 - 19:41
#15
Anonymous
2 dice

1. Let's consider 2D6 because I'm using this combination in my game called "Dogma":

d1 {1, 2, 3, 4, 5, 6}
d2 {1, 2, 3, 4, 5, 6}

2. Let's take the combinations of them {d1, d2}:

{1, 1}, {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6} ... {6, 6} = 36 combinations

3. Sum the values of each {d1, d2}: 1+1 = 2, 1+2=3, 1+3=4 and so on... 6+6=12

4. You will have:

2: appears 1 time
3: appears 2 times
4: appears 3 times
5: appears 4 times
6: appears 5 times
7: appears 6 times
8: appears 5 times
9: appears 4 times
10: appears 3 times
11: appears 2 times
12: appear 1 time

These frequencies mean that is very uncommon taking 2 and 12 in two 6-side dice. 3 a bit more common, 4 a bit more, 5 a bit more... 7 is the most common of them.

The probabilities:

2: 1 time / 36 possible combinations = 0,02 = 2%
3: 2 times / 36 = 0,05 = 5%
...
7: 6 times / 36 = 0,16 = 16~17%

If you have the following condition:

The attacker is successfull takes <= 6 in 2D6. So, you have to sum all probabilities from 2 until 6: 2% + 5% +...13%
And you will get 41% of probability.

It's not difficult follow the same steps for solving your problem. If you use D10, you will have 2 dice:

d1 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
d2 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Combine these two dice and you will get 100 combinations. It's a boring work, but you will have to do it once. Math equations are great, but don't confuse yourself with them. I hope it helps.

Good Luck,
Rogerio Biondi

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