*In which I do a little bit of maths to demonstrate the surprising, and to the game designer surprisingly useful, nature of probabilities. And pay attention! I shall be asking questions later!*

One measure of the nature of any game is to ask: How random is it?

A game such as Chess, for example, is entirely devoid of chance. It is a game where the choices of the players, and their choices alone, determine the outcome. In this sense it can be called wholly **deterministic**.

A game such as Snakes and Ladders, on the other hand, is a game where the actions taken by the players are determined completely at random; the outcome is entirely dependent on the roll of a dice and the players make no choices during the game whatsoever. I’ve recently learnt that a good word to describe this state of affairs is **stochastic**.

Most games are, of course, a mix of these two, being neither wholly deterministic nor wholly stochastic. Most games strike a balance between how the players’ choices and the players’ chances combine to determine the eventual winner.

### Choice vs. chance

All genuinely pure strategy games — such as Chess, Go, Reversi and Draughts to name just a few — are wholly deterministic. There are fewer obvious examples of games that are entirely stochastic although one I recall fondly from my own childhood is the curiously named **Beat Your Neighbour Out of Doors** played with a standard deck of 52 cards (Wikipedia lists the game’s other incarnations).

It’s pertinent to note that the random outcome of Snakes and Ladders is manifest during the game by successive dice rolls, whereas in Beat Your Neighbour the eventual outcome is ‘built in’ at the very beginning by the random order of the deck. When playing Beat Your Neighbour the players simply go through an entirely predetermined set of operations to discover who has won (cards are never shuffled during the game). The sequence of these operations is unknowable in advance by the players, but their order and eventual outcome is fixed by the state of the deck at the very beginning.

Each of these two games, then, can be interpreted as a random sequence of events that, taken to their conclusion, eventually reveals a winner. The players take part by initiating these events and responding to their outcome, but at no point in either game does any player make a choice.

There is, however, an important difference between the nature of the randomness in Snakes & Ladders and Beat Your Neighbour. Each dice throw is a separate, unconnected event; each turn of a card from a single 52-card deck is not. In the latter the events, though randomly **ordered**, have a known **distribution**: we know that there are exactly four Aces, four Kings, four Queens, four Jacks and 36 other cards (in Beat Your Neighbour, only the Aces and the face cards are important).

Each time you roll a dice you have exactly the same chance of rolling any particular result. And if you were to roll a dice 52 times, you would be no more equipped to predict the result of the 52nd roll than you were to predict the first.

In contrast, when you shuffle a 52-card deck and turn cards up one by one, the situation is very different. The result of turning up the first card is indeed genuinely random, since the card is just as likely to be, say, an Ace as any other. But when eventually you turn up the 52nd card, provided you have been keeping count, you would know absolutely what card it was before it was revealed.

### Probable cause

It is this distinction, between the mathematical nature of the randomness of a single dice and that of a deck of cards, which can be a great help to the game designer. In both cases the outcome of an individual event (the roll of a dice, the turn of a card) may be considered unpredictable, but when we consider the **distribution** of the results of **multiple** outcomes the situations could not be more different.

To make the comparison even simpler, and the distinction even more distinct, let us imagine, alongside our standard 6-sided dice, a reduced deck of just six cards, comprising the Ace of spades, plus the 2, 3, 4, 5 and 6. We shuffle the cards and then play a very simple game. In each turn I will turn a card up from the deck and you will roll the dice. I won’t reshuffle the deck after each turn, so the game will only last six turns until my deck is exhausted.

- What chance do I have of turning up an Ace before the end of the game?
- What chance do you have of rolling a ‘1’ before the end of the game?

I’m not pretending these are difficult questions, and indeed the answer to the first question should be obvious: I can’t fail. By the end of the game I will have turned up all six cards, one of which must be the Ace. My chance of turning up the Ace is, mathematically speaking, 100%. It’s a sure thing.

Another way of stating the same result is to say that my chance of *not* turning up the Ace is 0%; that result can *never* happen. Always, the chance of a certain thing happening and the opposite chance of the very same thing *not* happening must add up to make a full 100%.

The answer to the second question is less obvious. You will roll the dice six times, and every time you roll you have a 1-in-6 chance of rolling a ‘1’, but you cannot simply add up these individual probabilities and say that after 6 rolls you have a 100% chance of having rolled a ‘1’. It’s not a certainty.

After all, you might have rolled a ‘6’ every time. Or you might have rolled a ‘1’ every time. Indeed, you might have rolled any possible sequence of the numbers from 1 to 6 in any order (each specific sequence is just as likely as any other). A good few of those sequences do not contain any ‘1’s at all, so your chances of rolling a ‘1’ cannot be a full 100%. But if not, what are they?

We know the chances of rolling a ‘1’ with a single dice roll is 1-in-6, or 1/6th, or 16.7%, so we can know immediately that our chances of rolling at least one ‘1’ with multiple dice rolls has to be higher than that. We also know that whatever the answer it can’t be 100%. Therefore the answer we are looking for has to be between these two limits. The exact probability is very close to one of the following simple fractions. Take your pick!

A:around^{1}⁄_{3}B:around^{1}⁄_{2}C:around^{2}⁄_{3}D:around^{5}⁄_{6}

There are several ways to calculate the answer. It is relatively simple to work out, for example, that there are exactly 46,656 different sequences of six dice rolls: each successive roll has six possible results, so each additional roll multiplies the number of possible sequences by six. If we multiply six by itself six times we get this number: 6 × 6 × 6 × 6 × 6 × 6 = 46,656.

One way of calculating the chances of rolling at least one ‘1’ would be to count up all the possible sequences that contain **one or more ‘1’s** (how many contain precisely one ‘1’? how many contain precisely two ‘1’s? and so on) and then work out what proportion of the 46,656 possible combinations this total represents. That proportion, whatever it turned out to be, would be your chances of rolling at least one ‘1’. Sounds like a lot of counting up, doesn’t it?

Thankfully there is a simpler approach, which is to ask the opposite question: What are the chances of not rolling any ‘1’s at all? The chance of a single dice roll not being a ‘1’ is 5-in-6, or 5/6ths, or 83.3%. Indeed, the individual chance of each successive dice roll not being a ‘1’ is exactly the same, so to work out the overall chances of not rolling a ‘1’ six times in a row we can simply multiply these six similar probabilities together.

^{5}⁄_{6}×^{5}⁄_{6}×^{5}⁄_{6}×^{5}⁄_{6}×^{5}⁄_{6}×^{5}⁄_{6}=33.49%

Since the chances of *not* rolling a ‘1’ with six rolls is just slightly more than 1/3rd (1/3rd as a percentage is 33.33%), we know that the chances of rolling at least one ‘1’ must be the opposite of this, which is 66.51%, or just a little less than **2/3rds**. Answer C was therefore the correct one. How did you do?

### The weight of numbers

There are many ways in which the game designer can incorporate randomness into games, and it is up to the designer to decide whether and how to limit its scope or control its effects. When designing any game with a chance-driven mechanism it is important to consider just how likely or unlikely specific outcomes are. There is always a risk that the randomness can take over, swamping the influence of other more deliberate aspects of the game, and rendering the players’ strategic and tactical choices immaterial. Some games, of course, both rely on and revel in such chaos, but for others it can be slow poison.

One method of control, as already illustrated, is to create a known distribution of possible results, and then allow the order in which the results are revealed to be randomly determined. In our example of the six-card deck, the distribution of those results is **flat**: each of the possible results (the numbers 1 to 6) is represented the same number of times (that is, once) within the deck.

A distribution within which the results are not represented in equal number is called a **weighted** distribution; indeed, any distribution that isn’t perfectly flat is, by definition, a weighted one. The probabilities involved, and hence the level of control the designer can exercise over the likelihood of specific outcomes, can now be very different indeed.

To put flesh on the bones of this notion, let me explain a very real game design issue that recently came up as part of my reworking an old tile-laying prototype. In its existing incarnation the game continued until a very clear and very predictable point was reached: specifically until the tiles ran out. In this sense the end-game generated neither tension nor surprise, and so as part of a more wide-ranging reengineering of the game I began to think about ways of injecting a little unpredictability into the timing of the end-game.

My idea was to add into the tile mix four new tiles that had an additional feature on them and to additionally structure the game as a series of rounds. At the beginning of each round a fixed number of the randomly selected tiles would be turned face-up, and the game would finish at the end of the round in which the last of the four special tiles appeared.

Now, this *seemed* a perfectly viable solution which held the prospect of manufacturing likely game lengths that felt ‘about right’ — but the very real risk was that my intuition was simply unreliable. The possibility existed that this change would create games that were, too often, far too short; or would conversely, because of overwhelming probabilities, make little difference to the length of a majority of games and hence not generate the tension I was looking for. In this case it wasn’t enough simply to guess, and even repeated play-testing could easily be misleading; I had to *know*.

### Good idea? You do the math

To make things simple let’s use a regular pack of cards to model my proposed tile-game. Let’s imagine shuffling a standard 52-card pack and then playing a game in which, in each successive round, we turn up 4 cards from the deck. What happens to those cards during the game is immaterial; what matters is that the game ends in the round in which the fourth Ace is revealed.

An important thing to appreciate is that the deck can now be seen as representing a simple but significantly weighted distribution. We are only concerned with whether each card is an Ace or a ‘Not-Ace’ (that is, any other card). Within the deck there are 4 Aces and 48 ‘Not-Aces’, so the likely result of an individual card turn is heavily weighted towards the result being a ‘Not-Ace’.

Since there are a total of 52 cards, and exactly 4 cards are turned up in each round, any game will last somewhere *between and including* 1 to 13 rounds. But what are the odds?

What, for example, are the odds of the game ending in the first round? In other words: what are the chances that the four Aces are the first four cards drawn from a shuffled deck? Most people would intuitively put this possibility in the ‘very unlikely’ category but just how unlikely is it?

Fortunately the maths in this first example is straightforward. We know that we have 4 chances in 52 of the first card being an Ace (there are only 4 Aces in a regular deck of 52 cards). Assuming the first card is an Ace, we know we would then have 3 chances in 51 of the second card being an Ace (now only 3 Aces in a deck of 51 cards remain). In the same way, assuming each time that we do indeed draw an Ace, there are 2 chances in 50 of the third card being one, and just 1 chance in 49 of the fourth. Since these four probabilities represent the successive and cumulative chances of drawing four aces from a freshly shuffled deck we can calculate the overall likelihood by simply multiplying them together:

^{4}⁄_{52}×^{3}⁄_{51}×^{2}⁄_{50}×^{1}⁄_{49}=^{1}⁄_{270,725}

So that’s 1 in 270,725, which is, I think you’ll agree, pretty unlikely. It is, for example, very roughly equivalent to the chance of flipping a coin 18 times in a row and it coming up ‘heads’ every time.

This means that we can, for all practical purposes, discount the possibility of the game ending in the first round. We know it’s feasible, but the odds are vanishingly small. As I said, you might have guessed that the game ending in the first round would be highly unlikely, but here are a few far-less obvious questions to test your powers of probabilistic intuition (or, if you want to give them a spin, your maths skills). In each I have indicated four possible percentage ranges within which the answer may lie.

**What proportion of games are likely to end in the 13th round?**

In other words: what are the chances that the game will last the maximum number of rounds and that therefore at least one Ace will be amongst the final four cards drawn from the deck?

A:less than5%B:5%to15%C:15%to25%D:more than25%

**What proportion of games are likely to end after 6 rounds or less?**

In other words: what is the cumulative probability that any game will end in one of the first 6 rounds?

A:less than5%B:5%to15%C:15%to25%D:more than25%

**What proportion of games are likely to end after 10 rounds or more?**

In other words: what is the cumulative probability that any game will end in round 10, 11, 12 or 13?

A:less than25%B:25%to50%C:50%to75%D:more than75%

Before I tell you the answers I urge you to think about what you might naturally expect the probabilities in such a relatively simple scenario to be, and to make your own choices. Almost everyone has some experience of playing games with a regular deck of cards, and all of those games depend to some degree on the innate probabilities of a freshly shuffled deck. Hence we all have a ‘feel’ for those probabilities, even if we have never taken any time to think about them.

Regardless of how you answer the questions — be it guesswork, preternatural intuition or more formal mathematical analysis — I’m sure you’ll agree that the probabilities involved are anything but simple or obvious; which is, of course, precisely why they are so useful to the game designer.

The amount of control the designer can exercise here is immense. And this example is just one possible application (4 cards drawn in each round, for a total possible game length of 13 rounds) of just one possible distribution (a mix of 4 Aces and 48 ‘Not-Aces’). Change any of those numbers, even a little, and you will shift the probabilities, possibly dramatically.

### So what *are* the odds?

Here are the answers. If you find them surprising then I can only agree with you. Certainly, they were not the numbers I was expecting when I set about calculating them, so if you got any of these right: well done!

**What proportion of games are likely to end in the 13th round?**

**D: more than 25%**(the exact proportion is 28.13%)**What proportion of games are likely to end after 6 rounds or less?**

**A: less than 5%**(the exact proportion is 3.93%)**What proportion of games are likely to end after 10 rounds or more?**

**D: more than 75%**(the exact proportion is 78.24%)

Do these probabilities represent a viable way to reengineer the timing of my tile game? I would say yes, even though the possibility of a frustratingly short game must of course remain.

After all, almost 4 out of 5 of all games would last 10 rounds or more, with more than 1 in 4 of all games lasting the maximum 13 rounds. Conversely, in only around 1 in 20 games would less than half the tiles come into play (which happens if the game ends in round 6 or earlier). Additionally, my calculations tell me that only about 1 game in 150 would last 4 rounds or less, and that fewer than 1 in 550 would last 3 rounds or less.

### Possible vs. probable

It is certainly true that in the case of my tile prototype, games lasting only 3 or 4 rounds would indeed be annoyingly short; far too short for the game to be satisfying. As a designer I have to weigh that risk against the reward of being able to define a game length that is — at least most of the time! — both long enough to be interesting, and unpredictable enough to be surprising.

To properly balance the risk and reward of leaving any part of a game’s engine to chance the designer must first establish the exact nature of its consequences, and must understand the possible, the probable — and the difference between the two!

‘

Chance, too, which seems to rush along with slack reins, is bridled and governed by law.’

Developing a heightened, more informed and more reliable sense of intuition is something all game designers will find invaluable, but this can only come with experience. Most of us, lacking as we do the prescience and insight of true genius, will have to settle for something a little more mundane: doing the math.

*This article is part of a series examining various aspects of board game design. The story so far can be found at the following locations:*

Philip said...

January 15, 2010 11:23 pm

Very interesting post. So how does the math work for the last examples (for people who struggle with, like, math)?

Brett said...

January 17, 2010 12:53 am

@Philip: I'll try and put together a post soon going through the (relatively) straightforward maths. As they used to say in maths class: Always show your working!

It's still a matter of multiplying simple fractions together, but you have to throw in a multiple based on the number of combinations of possible ways in which cards are drawn to make everything work out.

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