The Mathematics of Block Dominoes: Counting Tiles and Odds

Everything a Block Dominoes player can work out at the table follows from how the 28-tile set is built. This article goes through that arithmetic: tiles and pips per number, what a seven-tile deal leaves unseen, the odds that a particular opponent holds a number you care about, and why a blocked hand tends to go to the player who got rid of heavy tiles. Every calculation is written out so you can check it.

What is in the set

A double-six set has one tile for every unordered pair of numbers from 0 to 6, including pairs of the same number. There are C(7,2) = 21 pairs of two different numbers, the mixed tiles, plus 7 doubles, for 21 + 7 = 28 tiles. The figure lays them out by their lower number.

Take any one number, say six. It sits on 6-0, 6-1, 6-2, 6-3, 6-4 and 6-5, which is six mixed tiles, and on 6-6, its double. Seven tiles carry a six. If you count pip-ends rather than tiles, the answer is eight, because 6-6 shows a six at both ends. The distinction matters when you count at the table: once five tiles carrying a six have been played, two are still out, whether or not 6-6 was among the five. The same is true for every number, blank included.

The right-hand column is how the 168 total arises: each number appears on eight pip-ends, so it contributes eight times its value, and 8 × (0 + 1 + 2 + 3 + 4 + 5 + 6) = 8 × 21 = 168. The seven doubles account for 2 × (0 + 1 + 2 + 3 + 4 + 5 + 6) = 42 of those pips and the 21 mixed tiles for the other 126. Averaged over the set, a tile weighs 168 / 28 = 6 pips.

The deal, and what you cannot see

Under the Block Dominoes rules used here, every player is dealt seven tiles whatever the number of players. With four players that is 4 × 7 = 28, the whole set. With three it is 21 tiles dealt and 7 set aside; with two, 14 dealt and 14 set aside. Set-aside tiles stay out of play for the whole hand; nobody ever picks them up, because there is no drawing in Block Dominoes.

From your seat the count of unseen tiles is the same in all three cases: 28 − 7 = 21. What differs is where they are. With four players all 21 are in opponents' hands, so every unseen tile will eventually be played or counted against someone. With three players, 14 are in hands and 7 are out of the game. With two, only 7 of the 21 are in your opponent's hand and 14 are out of the game. The odds below are worked for one particular seven-tile opponent hand, so they apply in every seat count; the difference is what it means when the tile you are worried about is not in that hand.

Two averages are worth memorising. First, the expected pip count of a seven-tile hand is 7 × 168 / 28 = 7 × 6 = 42, so a hand weighing 50 is heavy and one weighing 30 is light. Second, each tile you are dealt has a 7 / 28 = 1 / 4 chance of carrying any particular number, so a seven-tile hand carries a given number on 7 / 4 = 1.75 tiles on average. Holding four tiles of one number, as in the first example, is well above the average; holding none of some number is common, as the next calculation shows.

That last figure comes from the same method used throughout this article. The chance that your own seven tiles include no six is the chance that all seven come from the 21 tiles without a six: C(21,7) / C(28,7) = 116,280 / 1,184,040 = 323 / 3,289, about 9.8%. Since there are seven numbers and each has the same 9.8% chance of being absent, a dealt hand is missing 7 × 0.098, roughly 0.7, of its seven numbers on average: not quite one number per hand, but close.

Worked example: will the next player have a six?

Four players. You are dealt the hand below and you hold 6-6, the highest double in the set, so you must open with it. After the opening tile both ends of the line show a six, and the player on your left must play a six or pass.

Seven tiles carry a six and you hold four of them, so 7 − 4 = 3 sixes are somewhere among the 21 tiles you cannot see. The player on your left holds 7 of those 21. The chance that their hand contains none of the three sixes is the number of seven-tile hands that avoid them, divided by the number of seven-tile hands altogether:

C(18,7) / C(21,7)

C(21,7) is the number of ways to choose 7 tiles from 21. Written out: (21 × 20 × 19 × 18 × 17 × 16 × 15) / (7 × 6 × 5 × 4 × 3 × 2 × 1) = 586,051,200 / 5,040 = 116,280. C(18,7) is the number of ways to choose 7 from the 18 unseen tiles that have no six: (18 × 17 × 16 × 15 × 14 × 13 × 12) / 5,040 = 160,392,960 / 5,040 = 31,824. So the chance the left-hand player holds no six is 31,824 / 116,280 = 26 / 95, which is 0.2737, or a little over 27%. The chance they hold at least one six, and so can answer your 6-6, is 1 − 26 / 95 = 69 / 95, about 73%.

There is a quicker route to the same fraction that avoids the large numbers. Deal the opponent's seven tiles one at a time. The first is not a six with probability 18 / 21; given that, the second is not a six with probability 17 / 20; and so on down to 12 / 15 for the seventh. Multiply the seven fractions, cancel, and you get (14 × 13 × 12) / (21 × 20 × 19) = 2,184 / 7,980 = 26 / 95 again.

The same method gives the odds for any number of unseen tiles of a number, still at the deal, with 21 unseen tiles and a seven-tile opponent hand. Replace the 18 by 21 minus the number still out.

The figures are for one named opponent, not for all of them together: in the four-player example, the three unseen sixes are certainly in somebody's hand, so at least one of your three opponents can answer 6-6, and 27% is only the chance that the particular player to your left is the one who cannot. In a two-player game the same 26 / 95 applies to your single opponent, but if they hold no six, the three sixes are among the 14 tiles set aside and nobody will play them for the rest of the hand; both ends of the line are then closed to everyone but you.

Later in the hand, change the numbers

Mid-hand the method is unchanged but the inputs are not. The number of unseen tiles is 28 minus the tiles on the line minus the tiles in your hand, and the opponent's hand is however many tiles they have left. If 12 tiles are unseen, the opponent holds 4 of them, and 2 unseen tiles carry a five, then the chance the opponent has no five is C(10,4) / C(12,4) = 210 / 495 = 14 / 33, about 42%. As the sequential product: (10 / 12) × (9 / 11) × (8 / 10) × (7 / 9) = 14 / 33. The arithmetic is quick enough to do at the table once you are used to it, and the tactics article covers what to do with the answer before you leave a number on an open end.

Weight, and why blocked hands go to the light hand

A tile's weight is its pip total, from 0 for 0-0 to 12 for 6-6. The three heaviest tiles, 6-6, 6-5 and 6-4, weigh 12 + 11 + 10 = 33 between them; the three lightest, 0-0, 1-0 and 2-0, weigh 0 + 1 + 2 = 3. When a hand is blocked, each player adds up what they still hold, the lowest total wins the hand, and the winner scores the pips left in every other hand. Three tiles can therefore be worth anything from 3 to 33 at the count, and the difference is decided entirely by which tiles you chose to play earlier.

Here is a two-player hand that has just blocked. Both players were dealt seven tiles of almost equal weight, both have played four and hold three, and both ends of the line show a blank that neither can match.

Check the totals. Player A was dealt 4-4, 6-5, 5-3, 3-2, 2-1, 3-1 and 2-2: 8 + 11 + 8 + 5 + 3 + 4 + 4 = 43 pips. Player B was dealt 4-6, 1-4, 0-1, 2-0, 6-3, 6-2 and 5-4: 10 + 5 + 1 + 2 + 9 + 8 + 9 = 44. The deals were a single pip apart and both close to the 42 average. The 4-4 opened because it was the highest double in either hand; 5-5 and 6-6 were among the 14 tiles set aside. Four plays each later, A has shed 32 and holds 11; B has shed 18 and holds 26. The 15-pip gap in what remains is the 14-pip gap in what was shed plus the one pip B was dealt heavier.

B's three light plays were all choices, not forced moves. When the ends showed 4 and 5, B held both 1-4 and 5-4 and played 1-4, keeping 9 pips instead of 5. When the ends showed 1 and 3, B held 0-1 and 6-3 and played 0-1, keeping 9 instead of 1. When the ends showed 0 and 2, B held 2-0 and 6-2 and played 2-0, keeping 8 instead of 2. Each time the heavier tile was legal, and each time B kept it. The tiles B chose not to play are exactly the tiles B is now counting. A, by contrast, put down 6-5 the moment a six was open and chose 5-3 over 2-1 and 3-1 when a five and a one were both available.

This is the whole argument for shedding weight. A block does not care how many tiles you hold, only what they add up to, and a heavy tile stranded at the block costs you more than a light one: the same 6-5 that was worth playing at move three is worth 11 points to your opponent if it is still in your hand at the end. Whether to go out or to force a block when you are down to your last two or three tiles is a separate decision, worked through in the endgame article.

The numbers to carry to the table

Seven tiles carry each number. From any seat there are 21 tiles you cannot see at the deal; with four players they are all in hands, with two players only seven of them are. A random seven-tile hand weighs 42 pips and carries 1.75 tiles of each number. The chance that a named opponent holds none of k unseen tiles of a number is C(21 − k, 7) / C(21, 7) at the deal, and C(U − k, h) / C(U, h) later, where U is the number of unseen tiles and h the opponent's hand size. At a block the count is decided by what you kept, so the tile to play, when you have a choice, is usually the heavier one. None of this needs a calculator; it needs the discipline to count what has been played.