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

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

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

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

Rolling 4 dice

The next three tables are for rolling one, two or three dice without having rolled a 3-of-a-kind previously.

Rolling 3 dice

Rolling 2 dice

Rolling 1 dice

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

Rolling 2 dice with 3-of-a-kind already rolled

Rolling 1 dice with 3-of-a-kind already rolled

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

Rolling 5 dice

Rolling 4 dice

Rolling 3 dice

Rolling 2 dice

Rolling 1 dice

Rolling 3 dice with 3-of-a-kind already rolled

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

Rolling 1 dice with 3-of-a-kind already rolled

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.

Hi again,

After being in hiatus for quite some time I’ve gotten back to programming a bit. I guess there is an ebb and flow with everything.

Back in November I was looking into how I could be Building an AI with ASyncTask, but I never got to feed the thing some rules with regards to how to play the game. This I aim to remedy now. Since this could get pretty lengthy I’m planning a short series as I go though the process of building an AI for the dice game “10.000”.

Choosing a method for problem solving

Building an AI can be done in different ways. I haven’t explored the topic that much, but I could see going about problem solving in different ways; neural networks or some other kinds of fuzzy logic, search algorithms or a basic heuristic. I do however have some programming experience with heuristics and I believe that, since we’re talking about a limited problem space like this a heuristic can do the job just fine.

One of the issues at hand is whether or not we want to get the optimal solution i.e. winning as fast as possible. I can see an argument for building an AI that can do this, but I don’t think it would be very fun to play against. The probability built into the game somewhat remedies this, but the AI should strike a balance between being challenging to play against and still being beatable. Another plus for using a heuristic is that they can be made to give us an answer in short time consistently, thus giving the user a better experience.

A basic set of rules

When implementing the heuristic I need to come up with some rules that it will consist of. In principle I want the AI mimics the process of a human player asking himself a set of questions at every decision point during his turn and answering them based on a predefined logic (as humans we have the advantage of being able to adjust our evaluation during a game, whereas my AI will have a fixed set of rules).

Questions you could be asking yourself?

• What is the probability of being able to make points with X dices?
• What is the reward for rolling the dices?
• What is the risk for rolling the dices?
• Is it the last round and do I have the most points?

Joining these together my heuristic becomes:

Rule A) If my chance of success times the reward for rolling the dices is bigger than the chance of failure times the risk, then roll the dice. Otherwise bank the points (if able)

Rule B) If it is the last round and I haven’t got the most points, then keep rolling.

What remains to be done?

For now I’ve established a basic set of rules. To determine whether or not the AI should keep rolling I need to feed some numbers into rule A. These numbers I’ll try and calculate in my next post using statistical analysis.

That’s it for now 🙂

EDIT: To read part 2 go here.