Combinatorics Analysis of Game “Olaf Hits The Dragon With His Sword”

combinatoricsdiceprobability

I recently encountered the short tabletop roleplaying game Olaf Hits The Dragon With His Sword and I've been trying to do a combinatorics analysis of its dice mechanics. The full rules are available behind the link I've provided but I'll summarize them here.

  • There are six colors of dice and two players.
  • Any number of sides can be used for the dice as long as they are all the same.
  • The players make a series of choices which add dice to a pool and then all of the dice in the pool are rolled to determine the outcome.
  • There are seven possible outcomes – each color can win or the pool can be emptied.

The game proceeds in three phases:

  1. In the first phase each player picks a color and adds a die of that color to the pool. The players can pick the same color.
  2. In the second phase each player picks three different colors and adds a die of each chosen color to the pool. These colors can include a repeat of the color chosen in the first round.
  3. The eight dice in the pool are rolled. The color with the single highest die wins. If multiple colors are tied then discard all dice showing the highest result and repeat until there is a winner or the pool is empty.

Given these rules I'd like to answer a couple of questions:

  1. How many pools are possible?
  2. For a given pool and choice of number of sides what is the probability of each outcome?
  3. For each of the seven outcomes what's the pool that maximizes the probability it occurs?

I'm not sure how to approach the first two questions. For the third question it seems clear that the pool which maximizes the probability that a given color wins is the pool with 4 dice of that color and 4 dice which each have a different color. However I'm not sure what pool will maximize the probability that the pool is emptied.

I'd appreciate any help with solving this.

Best Answer

For the first, you can have a maximum of four of one color, which only occurs if they both pick that color in the first round and then both pick it as one of the three in the second round. If there are four of one color, there cannot be three of any other color, but they could be split $(2,2), (2,1,1), (1,1,1,1)$. If there are at most three of one color, any distribution is possible. To get the number of possibilities for a given color distribution, take the number of permutations of six things taking the number of colors and divide by the factorials of the numbers of any duplicates. For a distribution of $3,2,1,1,1$ we have $\frac {6\cdot 5 \cdot 4 \cdot 3 \cdot 2}{3!}=120$ possibilities. We just have to run down the list. $$\begin {array} {r|r} \text{colors}&\text{number}\\ \hline 4,2,1,1&180\\ 4,1,1,1,1&30\\ 3,3,2&60\\ 3,3,1,1&90\\ 3,2,2,1&180\\ 3,2,1,1,1&120\\ 3,1,1,1,1,1&6\\ 2,2,2,2&15\\ 2,2,2,1,1&60\\ 2,2,1,1,1,1&15\\ \hline \text{Total}& 756\end {array}$$

To get the probability of each outcome, I would just write a program to model it. I am sure that $4,1,1,1,1$ maximizes the winning chance for a color. It is better than $4,2,1,1$ because presumably if the $2$ dice tie for the highest number they win, while if they are different colors they will be thrown out and the $4$ might win.

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