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Subject: Dice probability question rss

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Carlo
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Quick stat question cause I'm not good with that kind of thing so I'd appreciate some help.

The general idea is modifying D6s around a target number, in an attack vs defense scenario. Attacker gets a high number of D6s, say 6D6. Defender rolls a smaller number, say 3D6, and the defender uses those dice to determine the target number (X+). IF they choose a lower target number it reduces the number of dice the attacker gets.

For example if the Defender has 3D6 and rolls 6, 4, 2, they could make the target 6+, but the attacker would get their full 6D6. If they chose 4+ as the target number the attacker would reduce their number of dice by 6-TN, or in this case 6-4=2, so they'd be rolling 4D6 at 4+ instead of 6D6 at 6. Similarly if the defender chose 2 it'd be 2D6 at 2+.

What has better odds? What about if the defender only gets 2 dice and the attacker gets more, say 7? Is there a clear "winning choice" of reducing TN but also reducing dice, or keeping a high TN and high number if dice.
 
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Andrew Walters
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Not sure how the target number is matched. If I have to roll a target number of 6 or more on 6d6 I will probably exceed that number. If you need to *match* the target number it does not matter what the target number is: ie I am exactly as likely to roll 3s on six dice as sixes on those same dice.

So are you trying to roll multiples of the target number? If so, the odds are the same. I am as likely to roll two, three or four times two on two dice as four on four dice. There are some stochastic differences that will make the game feel different, but I'm not qualified to talk about those without putting a lot of time into it.

Have you read the rules to Frag by SJ Games? Those should be available somewhere. They use a lot of dice to roll multiples of what's rolled on some other dice. They might be interesting to you.
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Carlo
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Sorry I should have explained better, the target number would be PER DIE, not total. So 6D6 needing 6s. Then you count the number of dice greater than or equal to the target number as successes, which end up being hits on the defender.

Basically how do the odds compare between 6D6 looking for 6s, or 4D6 looking for 4+, or 2D6 looking for 2+.
Say you need 2 successes to kill an enemy, would the 6D6 looking for 6s be better than 2D6 looking for 2+?

Think of the 6D6/4D6/2D6 being the number of attacks by the attacker, and the target number being the style of defense the defender has used. The defender can go for a style that reduces incoming attacks (lower attack dice like 2D6) but those attacks have a higher chance of hitting per die (2+), or let all attacks (ie: full 6D6 attack dice) through but each individual attack is harder (6+).
 
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Ben Crane
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As Andrew says, we need more information.

What do you mean by target number? Does the attacker add his dice together, or does he compare each against the target number to see how many "successes" he has rolled?

If the attacker is adding his dice together, he will always win in your current scenario (rolling nd6 will always result in a sum of at least n).

If he is going for successes, then how many successes does he need? If it's any given number n, then a target number less than n will always result in a failure (can't roll 3 dice at 2+ if you're only rolling 2 dice!). If it is always 1, then here are the odds of him getting at least 1 success with each target number, rolling that many dice:

1 - 1/1 = 100%
2 - 35/36 ~= 97%
3 - 26/27 ~= 96%
4 - 15/16 ~= 94%
5 - 211/243 ~=87%
6 - 31031/46656 ~=67%


EDIT: Ninja'd by your answer. Redoing numbers
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Jeremy Lennert
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Does the attacker add dice together, or consider each one separately? (I'm assuming separately because otherwise your numbers are silly.) If separately, do they get extra credit for multiple hits, or does it only matter whether they get at least one?

The average number of hits the attacker will get is (number of dice)*(7-TN)/6; if the attacker always gets a number of dice equal to the TN, that becomes TN*(7-TN)/6 giving an average of 1 hit for TN=1 or 6, 1.67 hits for TN=2 or 5, and 2 hits for TN=3 or 4, so the defender wants to choose a high or low number and avoid a middle number.

If getting 2 hits isn't exactly twice as good as getting 1 hit, then things get more complicated, but basically the defender will tend to pick high numbers if the second hit is less valuable than the first and low numbers if the second hit is more valuable than the first.

My instinct is that this system isn't that great; its overcomplicated for the amount of fidelity it offers, difficult to tweak for balance, and counter-intuitive but easily-solvable. I suspect you'd be better off looking at other options.
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Craig C
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In general terms, the chance you have of rolling a certain number or higher is [(7-number)/6], using 7 instead of 6 to include that specific number, as well as every number higher than it.

If all you want is the chance to roll higher than a specific number, use [(6-number)/6] instead of 7.

So, rolling a 4+ is a (7-4)/6= 3/6 = 50% chance. If you only need to roll that number one time to be successful, the chance increases by [its original value] - [half of the current chance you'll fail] for each die you add.

To roll a 4+ with one d6 is a 50% chance (original value), with 2d6 is 75% chance [50% + ([100%-50%]/2)], with 3d6 it's 87.5% [75% + ([100%-75%]/2)], and so on.

So, you can compare the multiple d6 chance with the single d6 chance [(7-number)/6] and see which is the better move.
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Ben Crane
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TN 1:
1 hit, 100% chance

TN2:
2 hits, 69% chance
1 hit, 28% chance
0 hits, 3% chance

TN3:
3 hits, 29% chance
2 hits, 44% chance
1 hit, 22% chance
0 hits, 4$ chance

TN4:
4 hits, 6% chance
3 hits, 25% chance
2 hits, 38% chance
1 hit, 25% chance
0 hits, 6% chance

EDIT: someone above me has shown that the average hits start going down from here. I don't care to do those numbers.
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Craig C
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Dang. Jeremy beat my by one minute.

Similar to what he said at the end of his post, what's the thematic reason for attacking in this manner? Will it "make sense" to the attacker that these are his two options?

If there's a game reason for doing it this way, keep it, but if it's just there because it sounds cool or is an intriguing math problem (i.e. a Eurogame reason), you might look at a different combat mechanic.
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Kevin Jeong
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Let's look at the math (feel free to correct me if (more like when) I make a mathematical error).

Let's look at the probability of either side winning.

Assumptions: 4+ means anything 4 and above. Thus a defender result of 1 is never chosen or considered since it is impossible for the defender to win with a choice of 1.
_________________________________________________________________________
Defender Chooses 6
Let's see if a defender chooses a high number such as 6:

Number of Attacker Dice: 6-(6-6) = 6
Minimum win threshold: 6+

Probability:
The probability that you lose is easier to calculate.
(5/6)^6 = 0.3348
Thus the probability that you win is 0.6652
About a 66% chance that you win if the defender chooses 6

_________________________________________________________________________
Defender Chooses 2
On the other end of the spectrum is the defender choosing 1:

Number of Attacker Dice: 6-(6-2)= 2
Minimum win threshold: 2+

Probability:
This one is even easier
(1/6)^2 for defender to win.
About a 97% chance that you win if the defender chooses 2

_________________________________________________________________________
The Formula

Defender's Choice: x;x>=2 AND x less than 6 (BGG hates that symbol apparently)
Formula: ((x-1)/6)^x


Defender has no benefit from choosing small values.

EDIT: new information makes this not so useful
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Russ Williams
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To solve such questions, you need to know only very basic stuff:

The probability of rolling at least N on a D6 is N/6 (for N from 1 to 6).

The probability of something NOT happening = 1 - the probability of it happening.

Throwing multiple dice, their probabilities of rolling at least N are independent of each other.

The probability of independent events X and Y and Z all happening is the product of their probabilities P(X) * P(Y) * P(Z).

So e.g. the probability of NOT rolling a 6 on 6 dice is the probability that all of them fail to roll 6:
5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 = roughly 0.33
and so the probability of rolling a 6 on 6 dice is roughly 1 - 0.33 = 0.67.

Similarly you can work out the probability of rolling a 4 on 4 dice, etc.
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Carlo
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Thanks for all the responses so far!

Yeah just kind of brainstorming around mechanical ideas, and I wanted to see if the math meant it was a horrible idea. I don't have a theme or game in mind for this at all, just a basic attacker and defender.

I see the process like attacker gets XD6 based on weapon, I was just using 6D6 as an example. Range would probably be 2-8 dice.

Defender gets less dice, probably 1-3.

Defender rolls 3D6 to see what his defensive options are. Results are 2, 3, 6. He could choose the 6 as the target number, which means the attacker gets to roll their full dice. If the attacker has a bunch of dice this might be a bad call (say 8D6), whereas it'd be perfect for a smaller number of attack dice.

If the defender chooses a lower number, the attack dice are reduced by 6-number, so if he chose 3 the attacker would lose 3 dice, if he chose 1 the attacker would lose 4 dice, etc.

Then the attacker rolls their modified number of dice and counts each result that is greater than or equal to the target number as a success. 1 success wounds, 2 successes kill, 3+ successes have no extra effect.

The idea was to introduce some interesting choices based on what the defender rolls, and also to have a reason to choose something other than 6 as the target number. As well the defender might just roll garbage, like 1, 1, 3, and not even have the option to set the target number that high.

EDIT: Thanks Russ and Kevin, I found your examples really clear. It looks like 2+ on 2D6 is always a better choice than 6s on 6D6. I was hoping it'd be a bit closer than 97% vs 66%, haha.
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Jeremy Lennert
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So if the attacker is using a 5-dice weapon, and the defender rolls a 1, then the attacker loses 5 dice, leaving him with 0 dice for an automatic miss?

If that's how it works, then your idea that high numbers are bad against large numbers of dice is almost backwards: defender wants high numbers to reduce the value of each die when there are a lot of them, or low numbers to reduce the attacker from few to zero dice.
 
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Carlo
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Yeah I'm not sure how minimum dice would be handled, I'd probably cap at 1D6 minimum to prevent auto misses, similar to how the defender minimum is 2+ (as they can't choose 1+).

That also provides the rare possibility of the defender rolling ALL 1s on their dice to represent some kind of automatic critical hit by the attacker.
 
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Kevin Jeong
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Here's the math behind the new information.
________________________________________________________________________
Formula

The defender chooses a die from the 3 they have rolled and we can call that TN.

Number of dice rolled: 6-(6-TN) = TN
Average number of hits: y
Probability of each hit: P

TN number of dice rolled with each of them having a P probability to hit.

y = TN*P

P is calculated identically to my previous post except without the exponent since we are looking at one die at a time.

P = (6-(TN-1))/6
= (7-TN)/6

Combine the two equations to find:

y = TN*(7-TN)/6
= (7TN-TN^2)/6

Now let's plot that to see if there's a deviation in average number of hits.



Choosing 6 is the best choice when you get it. Followed by 5 and 2 being closely tied for second. 3 and 4 are equally the worst choice.

EDIT: But if you look at the number of average hits themselves, There is only a range of 1. Which means that there really isn't much difference based on what choice you make. It's pretty balanced but also means that no matter what you pick, there isn't really going to be a big impact on the number of hits you expect to receive. Obviously if you have 3 health left and you roll a 2, you might want to pick it to prevent the risk of dying but besides that, 6 is slightly better but not by much.
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Steve Zagieboylo
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It does allow for one possible strategic choice -- If I know I can take one hit, but not two, and I'm pretty sure I can kill you on my next turn, then I want to give you only 1 die, even if that guarantees that you hit with it.

What happens in your situation if the attacker has fewer than 6 dice? If he has only 5 dice, then I can say that target number is 1, so he gets zero dice to roll to hit it, according to your formula. Can this happen?

I'm inclined to think that having the defender roll to see what his choices are is just going to be time-consuming and frustrating. If your mechanic doesn't work if he just chooses a number from 1 to 6, then rolling to reduce his choices is not going to fix it.
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Carlo
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@Kevin: Thanks again for the detailed stat work, graphs and all the math done for me were exactly what I was hoping for when I made the post laugh

@Steve: I think I'd have a minimum of 1D6 for the attacker, so they can't be reduced to 0 or less. That's a good point you have about the "choosing 1-6 vs rolling". The reason I considered the defender rolling dice would be to have to strategically decide what to do with the numbers they have available, based on what the attacker has (which would hopefully vary between attackers). Plus that element "I really need a 6 here!" as the defender is always fun for people who like dice.
 
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Carlo
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The last dice mechanic that's been rattling around in my head is generally for swordfighting / swashbuckling, perhaps in a pirate or naval game. Don't need probability stuff as much as general feedback on it.

Again might be a bit wonky, but I like the dice mechanics to provide something beyond just 2D6 > 8

Each weapon has a Risky/Safe rating, which is a number of dice (terms would be changed for theme). Such as 6D6 Risky vs 4D6 Safe.

To begin with the attacker secretly chooses a 1-6 target number on 1D6, which represents the fencing style they're using, like Lunge or Thrust. They need to roll EXACTLY this number on each die for it to count as a success.
If they want to attack Risky, they get more dice to do so, but have to tell the defender "Attack Low" (meaning their target number was 1-3) or "Attack High" meaning 4-6.
This represents giving away some muscle movement or tell that the defender can see. If the attacker plays it Safe they don't have to declare the range, but roll less dice.

So say the attacker decides on 5, and goes Risky, they say "Attack High" and roll 6D6. Then the defender has to guess the attack style, and respond with their own (Riposte, Parry, etc. still represented as a 1-6 range), so he guesses 4. That number is different than the attacker target number, so he failed to defend.

Now the attacker looks at his 6D6 rolls for EXACT matches to his target number of 5. Same idea of 1 success = wound, 2 success = kill.

IF the defender guess had been the same as the attacker target number the attack is completely blocked and nullified.

The idea is to introduce an element of bluffing, kind of like Liar's Dice. Like the defender might notice the attacker tends to use 2 during Attack Low, or 6 during Attack High, and may try to guess that. Knowing that he's been providing a pattern, the attacker instead mixes it up and chooses 5. So you might get into fun situations of second guessing each other. I also think the reveal of the attackers number could be a fun and tense moment.

As mentioned I could see the target numbers being mapped to "Feint", "Lunge", "Parry", "Riposte", etc. to give a bit more flavor. Almost like the duels in the video game Nidhogg if anyone has experience with that.

Summarized:
1. Attacker choose technique (1-6)
2. Attacker choose style (Risky/Safe)
3. If Risky say Attack Low (technique 1-3) or Attack High (technique 4-6)
4. Attacker rolls dice
5. Defender chooses technique (1-6)
6. If techniques don't match, count successes (dice rolls = attacker technique)
7. Wound on 1 success, kill on 2 success
8. If techniques match no damage is done
 
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