This is part 2 of a series on building an AI for the dice game “10.000”. To read part 1 go here.
The mission
My goal is that at the end of this post I’ll have a set of formal rules that can be coded into the AI. I want to know what the odds are for scoring when rolling any number of dice and how many points I can expect to score. I also want to make a simple decision tree for when to roll and when not to roll.
Calculating the odds
Since I’m no genius when it comes to statistics, I read a few articles on the dice probabilities at different rolls. For those of you that are interested you can find some mathematics explained here, here and here. These articles give us a basis for determining what the chances are of not being able to score points on a given roll.
In the following I’ll be listing tables showing the number of combinations that score points and the probability of these when rolling X dice (the columns “# successes” and “Success %” respectively). Conversely I’ve also listed the chance of not rolling points (the “Failure %” column). In the first table I’ll assume no 3-of-a-kind has been rolled or what I’d like to call basic probability.
The total number of combinations is calculated as 6^x, where x is the number of dice rolled.
Basic probability
| Dice |
# Successes |
Success % |
Failure % |
| 6 |
45.576 |
97,69 % |
2,31 % |
| 5 |
7.176 |
92,28 % |
7,72 % |
| 4 |
1.092 |
84,26 % |
15,74 % |
| 3 |
156 |
72,22 % |
27,78 % |
| 2 |
20 |
55,56 % |
44,44 % |
| 1 |
2 |
33,33 % |
66,67 % |
When rolling 1, 2 or 3 dice it also becomes relevant if there has been rolled 3-of-a-kind previously, since it increases your odds of rolling dice that can score points. In the second table I’ll assume that there has been rolled 3-of-a-kind of 2’s, 3’s, 4’s or 6’s, since these affect the odds.
3-of-a-kind already rolled
| Dice |
# Successes |
Success % |
Failure % |
| 3 |
192 |
88,89 % |
11,11 % |
| 2 |
27 |
75,00 % |
25,00 % |
| 1 |
3 |
50,00 % |
50,00 % |
Diving deeper into the combinations
These odds can be divided further into the different combinations of dice. I’ll list a separate table for every number of dice rolled, sorted by their success probability in ascending order. The column “Combinations” is the name of the combination you scored. “# Successes” and “Success %” express the same as before.
The row “Two 3-of-a-kind” have been included since we use all the dice and unlock all 6 on successive rolls. The row “1 or 5” is for the cases when none of the other combinations occurs.
Rolling 6 dice
| Combinations |
# Successes |
Success % |
| 6-of-a-kind |
6 |
0,01 % |
| 5-of-a-kind |
180 |
0,39 % |
| Two 3-of-a-kind |
300 |
 0,64 % |
| Straight |
720 |
 1,54 % |
| Three pairs |
1.800 |
 3,86 % |
| 4-of-a-kind |
2.250 |
 4,82 % |
| 3-of-a-kind |
14.400 |
 30,86 % |
| 1 or 5 |
25.920 |
 55,56 % |
If you sum up “# Successes” you’ll luckily end up with 45.576, which is the same number as listed in the basic probability table above. In the next two tables the process is repeated for four or five dice.
Rolling 5 dice
| Combinations |
# Successes |
Success % |
| 5-of-a-kind |
6 |
0,08 % |
| 4-of-a-kind |
150 |
1,93 % |
| 3-of-a-kind |
1.500 |
19,29 % |
| 1 or 5 |
5.520 |
70,99 % |
Rolling 4 dice
| Combinations |
# Successes |
Success % |
| 4-of-a-kind |
6 |
0,46 % |
| 3-of-a-kind |
120 |
9,26 % |
| 1 or 5 |
966 |
74,54 % |
The next three tables are for rolling one, two or three dice without having rolled a 3-of-a-kind previously.
Rolling 3 dice
| Combinations |
# Successes |
Success % |
| 3-of-a-kind |
6 |
2,78 % |
| 1 or 5 |
150 |
69,44 % |
Rolling 2 dice
| Combinations |
# Successes |
Success % |
| 1 or 5 |
20 |
55,56 % |
Rolling 1 dice
| Combinations |
# Successes |
Success % |
| 1 or 5 |
2 |
33,33 % |
You could also be rolling one, two or three dice when you’ve already rolled 3-of-a-kind. The following tables shows, where I’ve added the row “Add to 3-of-a-kind” for this. Remember the odds only change when the 3-of-a-kind is 2’s, 3’s 4’s or 6’s.
Rolling 3 dice with 3-of-a-kind already rolled
| Combinations |
# Successes |
Success % |
| 3-of-a-kind |
5 |
2,31 % |
| Add to 3-of-a-kind |
91 |
42,13 % |
| 1 or 5 |
96 |
44,44 % |
Rolling 2 dice with 3-of-a-kind already rolled
| Combinations |
# Successes |
Success % |
| Add to 3-of-a-kind |
11 |
30,56 % |
| 1 or 5 |
16 |
44,44 % |
Rolling 1 dice with 3-of-a-kind already rolled
| Combinations |
# Successes |
Success % |
| Add to 3-of-a-kind |
1 |
16,67 % |
| 1 or 5 |
2 |
33,33 % |
This covers the odds for every combination of dice.
Weighting the odds
We now have some probabilities down that we can use to guide our decisions (and the AI’s) on whether or not we want to roll the dice at any given point. However if you just decided to roll based on the probability of scoring alone, you would not be playing optimally i.e. scoring the most points.
Enter Expected Value and Opportunity Cost.
So what is this?
Expected Value (or EV) is the value we statistically can anticipate getting over the long run from our roll. In this case the score we can add on average. The Opportunity Cost is the value we give up by rolling; being the points we could have banked instead of rolling.
How do we use that when deciding whether or not to roll?
What we basically want to be doing is weighting out the two, by saying if my opportunity cost greater than my EV, I shouldn’t roll. But why is this correct? Let’s take an example. You have one dice left and have already scored a 1, a 5 and 3-of-a-kind of 2’s. Your chance of rolling and scoring is 50%. You could either be rolling a 1 scoring 100, a 5 scoring 50 or a 2 scoring 200 more. This would give you an average score of 116,67, but only on 50% of your rolls, giving you an EV of 58,33. Your opportunity cost is your current unbanked points of 350 times the chance you’ll miss out on points on the next roll giving you 116,67. Since the opportunity cost is greater than the EV you should be banking instead of rolling here.
To put it simply:
If EV > Opportunity cost, then roll.
Otherwise bank the points.
This is not the complete answer though. It would be if we always banked after this roll, but we could choose to roll again and again. Therefore we need to factor in the EV of successive rolls, at least to some extent. But how do we do this, since you can score and therefore remove a different number of dice after each roll? I’m certain that smarter and more mathematically experienced people would be able to solve this puzzle, but I accepted that the heuristic I’m building will not guarantee that the optimal solution would be found every time, which I why I’ll believe I could employ a short cut here. Calculate the chance of having scored all six dice at the end of the roll when rolling any number of remaining dice. Then weight this by multiplying by the EV of rolling all six dice. This post is however getting quite lengthy, so I’ll save this part for now.
The Expected Value of a roll
Working on the combinations I listed earlier I will now map out the EV of every combination on every number of dice rolled. By doing this we get the last set of numbers we need to make an equation for when to roll and when to not. The assumption is that you’ll score the maximum number of dice.
Rolling 6 dice
| Combinations |
EV |
| 6-of-a-kind |
2.000 |
| 5-of-a-kind |
1.525 |
| Two 3-of-a-kind |
1.000 |
| Straight |
1.000 |
| Three pairs |
750 |
| 4-of-a-kind |
1.050 |
| 3-of-a-kind |
575 |
| 1 or 5 |
156 |
Rolling 5 dice
| Combinations |
EV |
| 5-of-a-kind |
1.500 |
| 4-of-a-kind |
1.025 |
| 3-of-a-kind |
550 |
| 1 or 5 |
139 |
Rolling 4 dice
| Combinations |
EV |
| 4-of-a-kind |
1.000 |
| 3-of-a-kind |
525 |
| 1 or 5 |
121 |
Rolling 3 dice
| Combinations |
EV |
| 3-of-a-kind |
500 |
| 1 or 5 |
105 |
Rolling 2 dice
| Combinations |
EV |
| 1 or 5 |
90 |
Rolling 1 dice
| Combinations |
EV |
| 1 or 5 |
75 |
Rolling 3 dice with 3-of-a-kind already rolled
| Combinations |
EV |
| 3-of-a-kind |
500 |
| Add to 3-of-a-kind |
3-of-a-kind value * 108/91 |
| 1 or 5 |
105 |
The fraction in the “Add to 3-of-a-kind” is to account for doubles and triples. I counted these by hand.
Rolling 2 dice with 3-of-a-kind already rolled
| Combinations |
EV |
| Add to 3-of-a-kind |
3-of-a-kind value * 12/11 |
| 1 or 5 |
94 |
Rolling 1 dice with 3-of-a-kind already rolled
| Combinations |
EV |
| Add to 3-of-a-kind |
3-of-a-kind value |
| 1 or 5 |
75 |
That should be enough numbers to feed into the equation.
What if any kind of special rules take effect when playing the last round?
During the last round we could use the equation, but it is not enough to bank any number of points. It only matters if your total score exceeds that of all of your opponents. A simple solution is to keep rolling until you have more points (both banked and unbanked) than your opponent.
Put simply:
If banked + unbanked points <= opponents score, then roll.
Otherwise bank
That should be enough to implement a set of rules for the AI.
