Details

Example Game

The following is an example game play with three players. Read the rules first.

Preparing the game:

• Put 3x25 = 75 stones in the central stash.
• All players agree to play the default of 10 games.
• All players 6 stones from the central stash.
• Player 1 rolls the highest total number with 3 dice (1+4+5=10) and may start the game. Player 1 receives an extra stone.

First game:

• Player 1 starts the game and bets on rolling a Wunsch. Player 1 then proceeds to roll 1,2,5 which is das Unvermeidliche and therefore loses 2 stones. Player 1 started with 7 stones and now has 5. Now it is the next players turn.
• The 2nd player predicts an Einhorn and rolls a 1,3,6 and therefore gains 5 stones as well as the unicorn figure.
• The 3rd player choses to not predict anything and rolls 2,2,2 which is Dreifaltigkeit. In this case the player gains one stone.
• It is player 1 turn again and predicts das Unvermeidliche, then rolls 3,1,4 which is das Unvermeidliche. Player 2 takes one stone from the central stash.
• Player 2 now has the Einhorn and predicts das Unvermeidliche as well and rolls 4,5,6 which is das Unvermeidliche. Player 2 takes the 2 stones from player 1 instead of the central stash.

The game goes on like this until, in our example, player 2 loses the first game. At this point player 1 has 8 stones and player 3 has 4 stones.

The Einhorn is now returned (player 3 had it last) and placed on the dice tray. The Einhorn is now open for auction. Since this is the first game and player 2 lost the game there are no stones to bet with for this player. Player 1 bets 3 stones and player 3 bets 1 stone. Both players put the stones they bet in the central stash and player 1 receives the unicorn figure. Player 1 now has 8-3=5 stones in their own stone stash and player 3 has 4-1 stones in their stash.

All players now receive 6 new stones. Player 1 decides to add 2 stones from the own stash to the current play stash. Player 1 now has 4 stones remaining in their own stash and 6+2=8 stones to play with in this game as well as the unicorn figure. Both other players do not add stones and start with 6 stones.

Player 1 with the unicorn figure starts the new game and the game continues.

At the end of game 10, player 1 has 20 stones, player 2 has 15 stones and player 3 has 30 stones. Player 3 therefore wins the game.

Statistics

Roll outcome

With 3x D6 we get a total of 216 (6x6x6 or 6^3) possible outcomes.

Let us now look at unique rolls. To get all possible combinations we can use this formula for combination with repetition:

To learn more read about Binomial coefficient and combinatorics.

With 3x D6 we get n = 6, k = 3 resulting in 56 combinations.

You can copy(6+3-1)!/(3!*(6-1)!) to WolframAlpha to calculate it or use this link.

These 56 combinations distribute as follows:

However, this is not the probability of each combination since the distinct combinations are not evenly distributed over the 216 possible outcomes.

I do not know how to calculate this, so I created a python script and analyzed of all possible outcomes. To view the full list, expand "Full list of possible dice combinations". The outcome is as follows:

The result shows that Das Unvermeidliche has less distinct variations than Wunsch but more possible combinations.

All possible roll attempts

This python script calculates all possible dice results of the game Einhorn, counts the combinations and prints the probability of each of the four roll outcomes.

from itertools import product # pip install itertools
from collections import OrderedDict
import math

# Dice settings
diceFaces = 6
diceSmallestNumber = 1
diceHighestNumber = 6
diceAmount = 3

# Counter
Dreifaltigkeit = 0
Wunsch = 0
Unvermeidlich = 0
Einhorn = 0

# Create a list of all possible roll attempts
rolls = list(product(range(diceSmallestNumber,diceHighestNumber+1), repeat=diceAmount))

# Iterate roll attempts and check result
for roll in rolls:
    # Prepare list
    removedDoubleRoll = tuple(OrderedDict.fromkeys(roll).keys())
    sortedRoll = list(removedDoubleRoll)
    sortedRoll.sort() 

    # Count and print result
    if (len(sortedRoll)==2):
        print(roll, " Wunsch")
        Wunsch += 1
    elif (len(sortedRoll)==1):
        print(roll, " Dreifaltigkeit")
        Dreifaltigkeit += 1
    else:
        if ((sortedRoll[1]-sortedRoll[0]==1) or (sortedRoll[2]-sortedRoll[1]==1)):
            print(sortedRoll, " Das Unvermeidliche")
            Unvermeidlich += 1
        else:
            print(sortedRoll, " Einhorn")
            Einhorn += 1

# Print stats
print("\nTotal amount of possible rolls: ", len(rolls))
print("Total amount of distint rolls: ", int((math.factorial(diceFaces+diceAmount-1))/(math.factorial(diceAmount)*(math.factorial(diceFaces-1)))))
print("\nDreifaltigkeit: ", Dreifaltigkeit, "({:.1f}".format(Dreifaltigkeit / len(rolls) * 100), "%)")
print("Wunsch: ", Wunsch, "({:.1f}".format(Wunsch / len(rolls) * 100), "%)")
print("Das Unvermeidliche: ", Unvermeidlich, "({:.1f}".format(Unvermeidlich / len(rolls) * 100), "%)")
print("Einhorn: ", Einhorn, "({:.1f}".format(Einhorn / len(rolls) * 100), "%)")

Full list of possible dice combinations

This is a full list of all possible roll combinations in this game and the result based on the described rules above.