For Beginners Edit

How many dice to roll ? Edit

It all depends on your attack power and the monster's health total. There are three situations :

  • You completely overpower the monster - Attack until you have enough damage to overkill it with 50+ damage overkill, and earn the maximum of 50 coins for overkill.
  • The monster is a good match - Attack until it's near death, and overkill it with +50 damage on a second attack.
  • The monster has a lot of HP, but you're not scared for your life - Stick with a certain target damage. For example, attack when you reach 20, or 50, or 100, or 200 damage.
  • You're not sure you'll win this one - Same as the above, but don't try to overkill it. 50 coins isn't worth losing your progress.
  • For regular dice, 5-6 attacks will deal the most average damage (purple for 1 die, green for two dice, the graph was made by simulating each point 10000000 times)

Earning coins and DNA as a low level Edit

Have one pet on which you will equip items to lower it's attack as much as possible. You want to have -85% or -70%, so that a die roll can only raise your attack by 1 or 2. It also helps if your monster has a >1 die. Then raise your attack on your main character as much as possible. Overkill monsters as much as possible (overkill coin bonus caps at 50 or so), and capture those you can with your pet, who will be able to capture them fairly often without renting the monster catcher. You'll earn 50+ coins per monster killed or captured without the space to store it, and stock up on DNA to use on leveling later on when you have monsters worth spending DNA on.

x2 and x3 die vs >1 die Edit

While a x3 die may seem impressive, the >1 die is actually better for characters that have a x1 core die. Rolling a x1 and a x3 will give you at most 4x6 damage, with two chances of rolling a 1. Rolling a x1 and a >1, twice, then the >1 and attacking will grant you at most 5x6 damage, with two chances of rolling a 1.

Optimizing your dice setup Edit

To optimize your dice setup, consider the damage potential versus the chances of failure. For instance, using a single [1x] die would be a setup with a factor 1. Adding a [x1-2] die would up the damage potential by 50%, or a factor of 1.5.

To calculate the "damage factor" of your dice setup, add each die times their quality, and divide by the number of "risk" dice (dice that can make an attack fail)

Another example, using a [1x] and a [3x] die would have 4x more damage potential than a single [1x] die, but twice the chance of failing. Which gives a factor of 2.0 : 4x / 2 = 2.0

This is useful in determining if adding a die will increase the damage potential more than another. For instance, you have [1x] [<1] [1-3x]. You wonder if replacing the >1 by a 3x  will increase your damage output.

[1x] [<1] [1-3x] = (1+1)x2 /1 chance of failing. Initial setup has a factor of 4.

[1x] [3x] [1-3x] = (1+3)x2 /2 chances of failing. Setup still has a factor of 4.

Take an early game setup - [1x] and [>1]. Two rolls, one fail chance. (2/1)=2 Your setup yields twice as much as a [1x] die, so factor 2.0. Although you have to consider that your last roll will always be the >1 dice, so if you roll your two dice, then one, you had a 3.0 factor. Rolling 5 dice will be a 2.5, 7 dice a 2.33, and so on, closing in on 2.0

Should you use [x1] [>1] or [x1] [x1] [>1] ? As expressed earlier, two identical dice are worth 40% more - so we have (2x1.4+1)/2 = 1.9. If we consider the next die that is free, (2x1.4+1+1)/2 = 2.4. Which means [x1] [>1] is the better setup, as the factor is higher.

Then you have multiplication dice. It's fairly simple - [1-2x] will give you 1.5x on average. Just multiply your factor by the average of your multiplication dice, you get your dice setup value.

Catching high level monsters Edit

  • Sort your pets from the strongest to the weakest. Have the last one weak enough so that it cannot roll more than your catcher upgrade value +1. So if your catcher is at +5, a monster at 100% attack will do.
  • It helps if said monster has a >1 die. Side jaw, with its >3 die, is a pretty good monster catcher.
  • Invest dice shards into capture dice, but spend carefully. Invest only in your character's capture dice, not on your pets'. If you must, invest only in one monster, preferable one that you will keep around and use only for catching monsters. Again, you could pick Side Jaw for such use.
  • If you are after a particular monster, and it doesn't show up, you can press pause and "restart". This will cancel and kills or loot you made, but you will retain the catcher that you rented. You can repeat this quickly until your monster shows up on the first stage. You can also do that if a fight seems hopeless and you invested a bunch of coins in xp upgrades or such, or if you do not want to lose a rest square.
  • Another way to capture monsters with 100% accuracy is to have a monster catcher of level 5 or above, and set your hero with attack >x, since they have core 1x attack. Weaken the monster you want to capture to almost near death, 2 health guarentees it if you have a >1 (Adventurer example), and then attack. You can risk it a bit so you have to roll more attack to capture said monster, but there is a chance it may fail. Monsters with health dice in the wild can be tougher to deal with though.
  • Leave a secondary monster on the field with very low health, that you will finish off with the unit that goes right before your Finisher/Capture-Unit, so that you don't have to risk over-rolling while getting ready.

Math Edit

The following includes a lot of math on the game, and helps understand the reasoning behind certain strategies.

Keeping it Simple Edit

Early in the game you not only want to maximize attack, you also want to get some coin. So, instead of worrying about whether to roll one more time or not after each and every roll, start with the strategy of rolling to the yellow die (5th one) and stopping. Or rather stopping before reaching the sixth die (see the >1 example). This doubles the coin from rolls and from the overkill from monsters.  It also makes the math much easier. If you want complex math please read "Rolling Math" below.

On a single non-failing roll of 2-6, the average value will be 4.  If you just roll 5 (1x) die, you may get as high as 30 or as low as 10, but on average you will get 20. Simple enough. Now since you're going to stop rolling at the 5th, it actually makes sense to use as few die that count towards that as possible.

Just FYI, mathematically failure is supposed to happen ~60% of the time. The game has some manipulation of how often you fail though (evidenced through rewards from Kingdom of Courage), and in my personal experience, seems closer to 40-50%. Don't get discouraged. Stopping at 4 rolls still is a 53% failure rate and misses out on 20% of the damage and that extra coin.

Which die to choose for your second slot Edit

Consider the following secondary dice.

1x and 2x - Do three rolls on the 2x and two on the 1x -> (3 x 8) + (2 x 4) = 32

1x and >1 - Five rolls on the 1x and six on the >1 as you can sneak an extra in -> (5 x 4) + (6 x 4) = 44

1x and 3x = Three for the 3x and two on the 1x -> (3 x 12) + (2 x 4) = 44

  • Because of the 3x's chance of failure, it has the same value as the >1 for 5 rolls. However this will not be the case if you always stop at 4 rolls or if you always do 6.

1x and 1-2x - Five rolls on the 1x and the value of the 1-2 is 1.5 -> (5 x 4)x1.5 = 30

1x and 1-3x - Five rolls on the 1x and the value of the 1-3 is 2 -> (5 x 4)x2 = 40

  • Poison is a bit of a special case.  If you kill a monster in a single turn, it's nothing special. In 2 turns it gets 50% more damage, and in 3 turns it does 1.75. If it is the only damage done and given enough time it approaches 2x.  Here I am just going to do the math for both 2 turns (6 damage) and double (8 damage), which are the likely scenarios when fighting ordinary monsters and bosses.

1x and 1xP - Three rolls on poison and two on 1x -> (3 x 6) + (2 x 4) = 26, (3 x 8) + (2 x 4) = 32

1x and 2xP - Three rolls on 2xpoison and two on 1x -> (3 x 12) + (2 x 4) = 44, (3 x 16) + (2 x 4) = 56

NOTE: You need poison multipliers to increase poison dice. Attack multipliers will not work.

There are many more possibilities and other starting dice as well which might make different secondary dice better (such as a pair of 2x), but this is just a start for beginners.

Which die to choose for your third (and fourth) slot Edit

The answer here is much simpler. Choose a multiplier, and choose the best one you can craft. I'll just do a couple examples.

1x, 2x, 3x - Two 3x, two 2x, and one 1x - (2 x 12) + (2 x 8) + (1 x 4) = 44

1x, 2x, 1-2x - Three rolls on the 2x and two on the 1x then multiply by 1.5 -> ((3 x 8) + (2 x 4))x1.5 = 48

Notice how even 2 'weak' dice are great when multiplied.

1x, >1, 1-3x - Five rolls on the 1x and six on the >1 then multiply by 2 -> ((5 x 4) + (6 x 4))x2 = 88

For multipliers, take whatever two dice total above and multiply it by 1.5 (1-2x), 2 (1-3x), or 2.5 (1-4x) for average damage.

Monster's core dice Edit

All of the above really just applies to the player characters and the most basic monsters.  HOWEVER, it also makes it very clear that the core dice of a monster is one of the most important aspects when choosing which to keep and which to let go.  With a 2x, 3x, or even 4x die being the only one generating the 5 marked dice adding in >X, and especially multipliers add much greater power. Even duplicating a 2x for a pair is a huge improvement with the bonus damage for rolling a double.

Get one as soon as you can.

Rolling Math Edit

1 Atk DieEdit

Even the simplest case will turn out to be somewhat non trivial. Start with some assumptions - a) the dice are equally likely to roll 1-6 b) it is desirable to maximize the average damage at the cost of all other variables.

Define a class of strategies (not necessarily ideal) by 2 variables: N and Mindmg. The plan is to keep rolling until you reach either N rolls, or the damage equals or exceeds Mindmg... or you fail. Setting either variable too low causes you to do a lot of weak attacks, while setting them too high result in a high chance of failure.

One case to consider is to optimize for N, and disregard Mindmg (same as setting Mindmg \geq 6N). This is a simple matter of maximizing 4N \left( \frac{5}{6} \right)^N. For integer values, this function reaches the same max value at N=5 and N=6: \frac{15625}{1944} \sim 8.038.

Unfortunately, it is less obvious how to optimize for Mindmg, disregarding N. Computer simulation allows for the use of brute force. The result is to set N=7-10, and Mindmg=20-21, with an average damage of  \sim 8.14 . Additional testing is underway to narrow these windows.

A quick way to estimate an optimal value for Mindmg is to roll if rolling will increase the expected damage. So we want (\text{current dmg} + \Delta)(\text{chance of success}) > \text{current dmg} , where \Delta is the expected increase in damage. In principle, both \Delta and chance of success depends on the strategy being used, and thus this equation is recursive. To keep things simple, consider the case where we consider rolling one extra die. In that case, \Delta=4 and chance of success = 5/6. Thus, (\text{current dmg} + 4)\frac{5}{6} > \text{current dmg}. This reduces to \text{current dmg} < 20 , in agreement with the above brute force approach.

Despite all this analysis, this doesn't exhaust the set of all possible strategies, so the optimal strategy may involve something even more complicated... and this is just for 1 attack die!


While these results are useful when fighting a monster with billions of hp, and you want to kill it as quickly as possible, it is less useful for monsters with 50 or so hp, due to randomness. The above analysis neglects to mention that the "optimal" parameters for that class of strategies involve over 60% failure rate - ie 0 damage. There is thus good reason to consider trading average damage for higher certainty.

Randomness is especially an issue, considering that averaging over 100 attack cycles for N=4-10, Mindmg=15-25 results in no clear pattern or maximum. Even averaging over 10000 attack cycles, the pattern is not obvious; it takes ~1 million attack cycles for a pattern to emerge, and >10 million to locate a global maximum.

2 Identical Attack DiceEdit

The first thing to note is that the expectation value of the 2nd die is much higher than the first. With a little analysis, it can be shown that the 2nd one should always be rolled if the 1st one is rolled.

Instead of bogging down this page with math, only the results will be displayed.
Always roll 6: avg ~ 9.65, 66.5% failure rate
Stop rolling at 20 or higher: avg ~ 9.87, 60.5% failure rate
Stop rolling at 21 or higher: avg ~ 9.91, 62.4% failure rate
Stop rolling at 22 or higher: avg ~ 9.93, 63.3% failure rate


A pair should be rolled if

\frac{5}{6}\left( dmg + \text{avg dmg of a pair of dice} \right) &> dmg \\
\frac{240}{11} \sim 21.8 &> dmg

If the first die is known to be d, the 2nd should be rolled if

\frac{5}{6}\left( dmg + 4 + \frac{2}{5}d \right) &> dmg \\
20 + 2d = (21 + d) + (d-1)&> dmg

Assuming the first condition, the 2nd condition will always be true.

1 Attack Die with Multiplier DiceEdit

This situation is more complicated than it sounds. Basically you want to set the target damage based on what the current and average multipliers are. This 2nd part is quite challenging.

1 Attack die with 1-2x multiplier
Always roll 5: avg ~12.06, 59.8% failure rate
Stop rolling at 31 or higher: avg ~12.34, 62.7% failure rate
Stop rolling at 20*current multiplier or higher: avg ~12.45, 56% failure rate
If multiplier is 1, stop rolling at 25 or higher; else stop rolling at 40 or higher: avg ~12.65, 61.8% failure rate

Dice Configuration Edit

Since the multiplier dice stack multiplicatively, you want to take advantage of the power of exponential growth. More multipliers is better.

Using the above data as reference, dmg bonuses granted by the following dice are
2nd die: ~ 22%
1-2x: ~55%

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