Is This Game Fair?

Six games on paper

One real game is pure luck: you win Rs 40 (the Rs 50 prize minus your Rs 10) or you lose Rs 10. Ten games are still mostly luck.

So imagine a perfectly average run instead, a run on paper. A die has six faces, each as likely as the others, so in six throws imagine each face comes up exactly once:

Throw	1	2	3	4	5	6	Six games
You pay	Rs 10	Rs 10	Rs 10	Rs 10	Rs 10	Rs 10	Rs 60
You win	0	0	0	0	0	Rs 50	Rs 50

Six games cost Rs 60 and win Rs 50. The man with the die keeps Rs 10 in every six games, which is Rs 10 ÷ 6 ≈ Rs 1.67 a game.

A real run of six games won’t have exactly one six: it might have none, or three. But over 600 games the sixes come to about 100, and over 6,000 closer still in proportion. The run on paper is what real runs get closer and closer to.

Your line goes downhill

Play a few players to 1,000 games in the widget, and look at their lines. In the first hundred games they wander: one player goes up, another down. But by a thousand games every line has gone downhill at about the same slope, and nearly every player has ended a long way below zero: nine in ten of them between about Rs 700 and Rs 2,650 down, near the Rs 1,667 of the run on paper. That slope is the Rs 1.67 a game from the run on paper, shown by the dashed line.

The sign was true: the prize really is five times the price. What it doesn’t say is that you win it only one time in six. Five times the price, but only one chance in six: on average you get back 5 ÷ 6 of what you pay.

What would make it fair?

The stall keeps Rs 10 in every six games. To make it fair, so that on average nobody gains, the six games must pay back exactly what they cost. There are two ways.

• Raise the prize. Six games cost Rs 60, so one prize must be Rs 60: six times the price, because a six comes once in six throws.
• Lower the price. One prize of Rs 50 in six games is Rs 50 ÷ 6 ≈ Rs 8.33 a game. (Not a price anyone can charge in coins, but it is the fair one.)

Now design your own stall. Give any faces you like a prize.

The average of a game

What you found has a name. The average result of one game, over a very long run of games, is the game’s expected value. For the stall it is −Rs 1.67: on average, every game costs the player Rs 1.67.

You can find it without playing at all. Grade 8 finds a mean by adding the values and dividing by how many there are. The run on paper does exactly that: it adds up the results of six games, one for each face, and divides by six.

$$\frac{40 + (-10) + (-10) + (-10) + (-10) + (-10)}{6} = \frac{40 - 50}{6} = \frac{-10}{6} \approx -1.67$$

There is another way to write the same sum. You win Rs 40 with a chance of 1 in 6, and lose Rs 10 with a chance of 5 in 6. Multiply each result by its chance, and add:

$$40 \times \frac{1}{6} + (-10) \times \frac{5}{6} = \frac{40}{6} - \frac{50}{6} \approx -1.67$$

Both say the same thing: each result, weighted by how often it happens. The second way is quicker when the results don’t all have the same chance. With three dice below, each row of the table is one result times the number of ways it happens.

A game is fair when its expected value is zero: over a long run, neither side wins. A game with a negative expected value for the player has a house edge, the share of each rupee bet that the stall (or casino, or bank) keeps on average. For our stall it is Rs 1.67 of every Rs 10 played: about 17%.

What “expected” means, and what it doesn’t

The word is a trap. Nobody expects to lose Rs 1.67 in one game: one game can only win Rs 40 or lose Rs 10. The expected value is what the games average out to in a long run.

And “long” matters. After 10 games, a player’s average per game can be −Rs 10 (no sixes), +Rs 5 (three sixes) or, once in a while, even more. After 1,000 games, nine players in ten have an average between about −Rs 2.65 and −Rs 0.70, close to the −Rs 1.67. The more games, the closer the average comes to the expected value. This is called the law of large numbers.

Why? Each player’s money is the run on paper plus some luck: a few sixes more than average, or a few fewer. After 10 games, two extra sixes change the average per game by Rs 10. After 1,000 games, a few extra sixes are shared out over a thousand games and hardly change it. The luck doesn’t disappear (your total can still be hundreds of rupees from −Rs 1,667), but it grows much more slowly than the number of games, so the average per game, the slope of the line, settles down.

A game between two friends has no stall and no fee, but it can still be unfair.

Gita and Bibek throw two dice and multiply the numbers. If the product is even, Gita gets a point; if it is odd, Bibek does. “Even or odd, half and half,” says Bibek. Is it fair?

Each die can show 6 numbers, so there are 6 × 6 = 36 equally likely ways for two dice to fall (Figure 14.1). 27 products are even and only 9 odd: for an odd product both numbers must be odd (1, 3 or 5 on each die: 3 × 3 = 9 ways). So Gita wins 3 games in every 4.

Can it be made fair without changing the dice? Yes, by changing the points. Bibek wins only 9 times in 36 to Gita’s 27, a third as often. Give him 3 points for each win and Gita 1, and in a run of 36 games on paper each scores 27 points. The game is fair again: not because they win equally often, but because they win equal amounts on average.

Now a harder stall, an old game that sailors played. You pay Rs 10 and choose a number, say six. Then three dice are thrown.

• If your number shows on one die, you get your Rs 10 back and Rs 10 more.
• If it shows on two dice, you get your Rs 10 back and Rs 20 more.
• If it shows on all three, you get your Rs 10 back and Rs 30 more.
• If it shows on none, you lose your Rs 10.

It sounds fair, and here is the argument that it is: “Each die has a 1 in 6 chance of showing a six. There are three dice, so 3 × 1 in 6 = 1 in 2. Half the time I win Rs 10, and half the time I lose Rs 10. Fair!”

The run on paper for three dice

Three dice can fall in 6 × 6 × 6 = 216 ways, all equally likely, so the run on paper is 216 games. In how many of them does your six show on no die, one, two or three?

• No six: each die shows one of the other five numbers, 5 × 5 × 5 = 125 ways.
• Three sixes: only 1 way.
• Exactly two sixes: choose which die isn’t a six (3 ways), and it shows one of 5 other numbers: 3 × 5 = 15 ways.
• Exactly one six: choose which die is the six (3 ways), and the other two show any of 5 numbers each: 3 × 5 × 5 = 75 ways.

Check: 125 + 75 + 15 + 1 = 216. Now the money in those 216 games:

Sixes showing	Ways in 216	Your result	Total
none	125	−Rs 10	−Rs 1,250
one	75	+Rs 10	+Rs 750
two	15	+Rs 20	+Rs 300
three	1	+Rs 30	+Rs 30
	216		−Rs 170

In 216 games the player loses Rs 170: about Rs 0.79 a game, or 7.9% of every rupee bet.

So where did “half the time” go wrong? 3 × 1 in 6 counts sixes, not games. The 216 games have 75 + 2 × 15 + 3 × 1 = 108 sixes, exactly half a six a game, and since the prizes pay Rs 10 for each six, that part is right: Rs 1,080 in prizes. But a game with two or three sixes uses up several at once, so the sixes are crowded into fewer games: only 91 games have a six, and 125 have none, not 108. You lose your Rs 10 in each of those 125 games, Rs 1,250. Prizes of Rs 1,080 against losses of Rs 1,250: −Rs 170.

So far the rule has been simple: a game with a negative expected value loses you money in the long run, so don’t play it. But here is one that sensible people play anyway.

Suppose (the numbers are made up, to keep them simple) a farmer has a buffalo worth Rs 150,000, and in any one year there is a 2 in 100 chance that it dies of an illness or an accident. An insurance company offers a deal: pay Rs 4,000 a year, and if the buffalo dies, the company pays the farmer Rs 150,000.

Do the run on paper: 100 farmers, each paying Rs 4,000, is Rs 400,000 paid in. About 2 buffaloes die, and the company pays out 2 × Rs 150,000 = Rs 300,000. On average the company keeps Rs 1,000 a farmer. For the farmer the expected value of buying the insurance is −Rs 1,000 a year: an unfair game, and the farmer is the one losing.

So is buying it a mistake? No. The expected value is an average over a long run, and the farmer has one buffalo. Without insurance, most years cost nothing, but one bad year costs Rs 150,000, which could mean selling land or borrowing at a high rate. With insurance, every year costs Rs 4,000, and no year is a disaster. The farmer pays Rs 1,000 a year on average to swap a small chance of ruin for a small, certain cost. The company can take the risk because it has thousands of farmers: for it, the law of large numbers works.

That is where the rule breaks. The expected value tells you how a game averages out over many plays. It doesn’t tell you how bad one play can be. When one bad result would be a disaster, the safer choice can be worth more than the better average.

• Which Total Comes Up Most with Two Dice? The 36 ways two dice can fall, and why 7 beats 12.
• How Many Ways Can You Walk to School? Counting paths, and a board where balls bounce left or right at random.
• How Do You Cut a Cake Fairly for Three? Another meaning of fair: sharing so that nobody feels cheated.
• What Is a Game Worth on Average? (Grades 9–12) Expected value with algebra, longer games and the St Petersburg paradox.
• Should You Switch Doors? (Grades 9–12) A game-show puzzle where the run on paper surprises almost everyone.
• Why Do Snakes and Ladders Games Take So Long? (Computing, Going Further) An average found by having a program play ten thousand games.
• Can You Beat a Computer at Rock-Paper-Scissors? (Computing, Going Further) A fair game, until one player stops being random.