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This question is in reference to a game called "Tiny Dice Dungeon." In this game, damage is calculated based on rolling dice. I want to know how I can achieve the highest average damage.

This is the setting:

There are following dice types:

  1. "Attack dice": have the values 1-6; rolling a one on the dice causes the whole attack to fail.
  2. "Greater 1 attack dice": have the values 2-6.
  3. "Damage multiplier dice": Multiply the sum of all attack dice the value of the multiplier. Rolling a one with this die results in multiplying the sum by one but does not end the round.

The types have different sub-types:

  1. Attack dice are split into normal, double, triple and quadruple attack dice, which multiply their base value accordingly (1x,2x,3x,4x).
  2. Multiplier dice exist with ranges of 1-2, 1-3, and 1-4.

Additionally if I take two attack dice of the same kind (e.g. two double attack dice) and they roll the same value, they will not be summed; instead the multiplier becomes 10.

Following rules also apply:

  1. Every die set can have a maximum of 4 dice.
  2. There can only be two attack dice of the same kind, e.g. only two triple attack dice.
  3. There has to be at least one attack die, otherwise you obviously would never fail a turn.
  4. Before you can roll a die again, you have to have rolled all the other dice.

The damage is calculated the following way:

The values shown on the attack dice are first multiplied by their own multiplier, then they are summed together, and then the multiplier value of the multiplier dice is applied.

Example:

  • 1x attack dice rolling 5
  • 1x attack dice rolling 5
  • 2x multiplier rolling 2
  • 4x multiplier rolling 4

The calculation becomes

  1. 5x1 = 5
  2. 5x1 = 5
    • These are the same die type showing the same value, so the shwon value gets multiplied by 10 instead of summed so we have
  3. 5*10 = 50
  4. 50*2 = 100
  5. 100*4 = 400

If you could follow me so far, now my specific questions:

  1. Which dice should I choose for the highest average damage when I roll every attack dice 4 times?
  2. How high would the probability be to NOT fail to get to the fourth roll in the setup?
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  • $\begingroup$ @AidanF.Pierce I would love to, maybe any idea how? $\endgroup$ – print x div 0 Jun 6 '14 at 13:19
  • $\begingroup$ Oh, never mind. $\endgroup$ – Aidan F. Pierce Jun 6 '14 at 13:22
  • $\begingroup$ Some clarifying questions: If I have two attack dice and one rolls 1, does the attack fail? Is a greater 1 attack die 5-sided, or is there a repeated value? Is there any reason to prefer a double attack die over a triple or quadruple (or a triple over a quadruple)? Same two questions for damage multiplier dice. Are you asking about a set of four dice rolled four times each? $\endgroup$ – Teepeemm Jun 8 '14 at 20:09
  • $\begingroup$ Yes, rolling a 1 on any attack die ends the whole turn. A greater 1 die has 5 sides. A double attack die does multiply its value by 2 and so on, leading to higher values. Same goes for multiplier, they have 2,3 or 4 sides, according to their highest mulitply value. Yes, a number from 1-4, rolled 4 times each $\endgroup$ – print x div 0 Jun 8 '14 at 21:33
  • $\begingroup$ But is there a reason to choose a double attack die when I could instead choose a triple attack die? Is there a reason to choose a 2x multiplier when I could instead choose a 3x multiplier? I don't know what you mean by "a number from 1-4". You can have a maximum of four dice. Is your question about rolling all four dice four times each (for a total of 16 die rolls)? $\endgroup$ – Teepeemm Jun 9 '14 at 12:50
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tl;dr: The best expectation of 96.5 is with one attack die, two 1-4 multiplier dice, and one 1-3 multiplier die. This would have a probability of .48 of the attack die not rolling a 1.

When it comes down to it, any attack die has a 5/6 chance of success. Rolling it four times means that you would only have a $(5/6)^4=625/1296\approx .48$ chance of success. Consequently, two attack dice would mean only a .23 chance of success, three dice would mean a .11 chance of success, and four dice would mean a .054 chance of success. Since we don’t have many options, we need to make sure we maximize this chance, and consequently try to minimize the attack dice.

As a first attempt, let’s try one attack die (with value $D_1$), two 1-4 multiplier dice (with values $D_2$ and $D_3$), and one 1-3 multiplier die (with value $D_4$). The value $V$ of a roll is the product of the four dice (or 0, if $D_1=1$). Since the dice are all independent, the expectation (aka average) of the value is the product of the expected values: $\mathbb{E}(V)=\mathbb{E}(D_1D_2D_3D_4)=\mathbb{E}D_1\mathbb{E}D_2\mathbb{E}D_3\mathbb{E}D_4=\frac{10}3\cdot\frac52\cdot\frac52\cdot 2=\frac{125}3$.

The complication, of course, is that we need the attack die to avoid 1 for four rolls (three rolls if you just need “to get to the fourth roll”). We can calculate this with a conditional probability: $\mathbb{E}(\sum_{i=1}^4 V_i)=\mathbb{E}(\sum_{i=1}^4 V_i\big|\prod_{i=1}^4 V_i>0)\mathbb{P}(\prod_{i=1}^4 V_i>0)=4\mathbb{E}(V\big|V>0)\mathbb{P}(V>0)^4=4\mathbb{E}(V)\mathbb{P}(V>0)^3=4\cdot\frac{125}3\cdot(\frac56)^3=\frac{15625}{162}\approx 96.5.$

A second attempt would be to use two attack dice (with values $D_1$ and $D_2$) and two 1-4 multiplier dice (with values $D_3$ and $D_4$). Now, $\mathbb{E}(V)=\mathbb{E}(v(D_1,D_2)D_3D_4)=\mathbb{E}(v(D_1,D_2))\mathbb{E}D_3\mathbb{E}D_4$. The expectation of the attack dice is easiest to find with a table:

0  0  0  0  0  0
0 20  5  6  7  8
0  5 30  7  8  9
0  6  7 40  9 10
0  7  8  9 50 11
0  8  9 10 11 60

This sums to 360, so the $\mathbb{E}(v(D_1,D_2))=10$, $\mathbb{E}(V)=10\cdot \frac52\cdot\frac52=\frac{125}2$, and $\mathbb{E}(\sum_{i=1}^4 V_i)=4\mathbb{E}(V)\mathbb{P}(V>0)^3=4\cdot\frac{125}2\cdot(\frac56)^6=\frac{1953125}{23328}\approx 83.7$. By sacrificing a multiplier die, we increased the expectation of a single roll of four dice. But the chance of failure increase faster, and we had a net loss.

It’s not clear if you’re allowed to use a quadruple attack die as the required attack die (http://www.tinydicedungeon.com/Guide/GuideIndex.html#typesofdice indicates that a 1 on a quadruple attack die still results in no attack). If so, then obviously you want to do that.

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  • $\begingroup$ "...two 1-4 multiplier dice, and one 1-3 multiplier die." Why not three 1-4 multipliers? $\endgroup$ – Austin Mohr Jun 10 '14 at 22:50
  • $\begingroup$ I was assuming "Following rules" #2 (at most two attack dice of the same kind) applied to multipliers as well, but now I see that it doesn't necessarily. If OP indicates it would be allowed, then that would obviously be the way to go, and I'll adjust things accordingly. But I think I'll wait to hear from him about this (and if the only attack die can be a quadruple). $\endgroup$ – Teepeemm Jun 10 '14 at 22:56
  • $\begingroup$ Thank you for that answer, it does answer my points perfectly. Yes, the quadruple attack die is a valid die, since it can cause the turn to fail. Any the "rule of two die" would apply to the multipliers as well. $\endgroup$ – print x div 0 Jun 11 '14 at 6:14

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