Counting 7-bag: how far apart can the same piece be?
A column that puts numbers on the limits of luck in 7-bag: repeat odds, the gap until a piece returns, the longest S and Z run and the first 10 pieces.
7-bag is a very simple mechanism: put the 7 kinds of piece in a bag and draw them one at a time. Yet once you start counting, the feelings you get while playing, such as “the I never comes” or “I keep getting S and Z”, turn out to have clear upper limits and probabilities. This column lays out the 7-bag numbers that the lessons did not cover, staying within what you can check on a calculator.
A bag can come out in 5040 orders
7 pieces can be ordered in 7×6×5×4×3×2×1 = 5040 ways, and every order is equally likely. The chance that the first piece is an I is 1 in 7; the chance that it is an S, Z or O is 3 in 7 (about 43%). This is why so many openers branch with rules like “if the first piece is S, Z or O, build a different shape”: it happens too often to ignore.
The same piece twice in a row: 1 in 49
A bag holds only one of each piece, so the same piece can come twice in a row only at a bag boundary. The chance that the last piece of one bag and the first piece of the next are the same is 1 in 7.
A boundary comes only once every 7 pieces, so at any random moment the chance that the next piece is the same again is 1/7 × 1/7 = 1/49, about 2%. With a fully random generator it would happen 1 time in 7 (about 14%), so in 7-bag a repeat is quite a rare event. And the same piece never comes three times in a row.
How far apart can the same piece be?
Count the pieces from one appearance of a piece to the next appearance of the same piece. If it comes straight away, that is “1 piece later”; the farthest is “13 pieces later”. Its position in the earlier bag and its position in the next bag each have 7 equally likely values, so counting all 49 cases gives the table below.
| How many pieces later | Probability |
|---|---|
| 1 piece later (in a row) | 1/49 (about 2%) |
| 4 pieces later | 4/49 (about 8%) |
| 7 pieces later | 7/49 (about 14%) |
| 10 pieces later | 4/49 (about 8%) |
| 13 pieces later (the longest) | 1/49 (about 2%) |
The probabilities form a hill with its peak at 7 pieces later. Each step of one piece away from the peak takes off 1/49, and the average is exactly 7 pieces later. Behind the obvious fact that “an I comes once every 7 pieces on average”, the real gap varies from 1 piece to 13.
When you open a well and wait for an I to score a Quad, in the worst case 12 pieces come in between. Can you stack those 12 pieces, that is 48 cells, into the 9 columns outside the well without leaving a hole? Whether to wait for the I comes down to whether your stack is low enough to absorb this worst case.
Compared with fully random
What would happen if every piece were chosen independently, each with a 1 in 7 chance?
| Event | Fully random | 7-bag |
|---|---|---|
| The next 7 pieces include every kind | About 0.6% | 100% from the start of a bag |
| No I for 13 pieces in a row | About 13% | 0% |
| The same piece twice in a row | About 14% | About 2% |
| The same piece three times in a row | About 2% | 0% |
With a fully random generator, roughly 1 time in 7 you would wait 13 pieces and still not get an I. Opening a well and waiting would not work as a tactic. By putting an upper limit on the swings of luck, 7-bag makes this a game in which planning is worth the effort.
S and Z in a row: 4 at most
The awkward thing in flat stacking is S and Z pieces coming one after another. In 7-bag this has an upper limit too. Each bag holds one S and one Z, so the longest run is 4 pieces: the last 2 pieces of one bag are S and Z, and the first 2 pieces of the next bag are S and Z as well.
The chance that the last 2 pieces of a bag are S and Z is 1 in 21, and the chance that the first 2 pieces of the next bag are S and Z is also 1 in 21, so the chance of this worst order at any one boundary is 1 in 441. It hardly ever happens, but knowing the worst that can happen gives you a rule of thumb: “keep room to place 2 S or Z pieces and you will almost never be in trouble”.
The first 10 pieces: about 1.06 million orders
An opening perfect clear is decided by the first 10 pieces. The 7 pieces of the first bag come in 5040 orders and the first 3 pieces of the second bag in 7×6×5 = 210, which makes 5040×210 = 1,058,400 orders in total.
That looks like a lot, but for a computer it is a number that can be checked in full. When the lessons on this site give a figure such as “about 85% when every order is checked”, it is the result of counting them all in this way. 7-bag is “easy to read” not only for people but for computation as well.
Summary
- The same piece comes twice in a row only at a bag boundary, with a probability of about 2%. It never comes three times in a row.
- The gap between two appearances of the same piece is 1 to 13 pieces, 7 on average. If you wait for an I, in the worst case 12 pieces come in between.
- A run of S and Z pieces is 4 long at most.
- The first 10 pieces come in about 1.06 million orders, and all of them can be counted.