Requirements

How Many Unique Codes Do We Need?#

From estimation: 1M pastes/day × 3,650 days = 3.65B pastes over 10 years. Add safety margin → call it 5 billion unique codes needed.

Why Base62?#

Short codes appear in URLs — they must be URL-safe and human-readable. Base62 uses:

a-z  → 26 chars
A-Z  → 26 chars
0-9  → 10 chars
Total: 62 chars

No special characters, no encoding needed in URLs, easy to type and share. Base64 adds + and / which are not URL-safe without percent-encoding. Base62 is the clean choice.

How Many Characters Do We Need?#

Same logic as any number system. In base 10 with N digits you can represent 10^N combinations. In Base62 with N characters you can represent 62^N combinations.

1 char → 62^1 = 62
2 chars → 62^2 = 3,844
3 chars → 62^3 = 238,328
4 chars → 62^4 = 14,776,336        (~14 million)
5 chars → 62^5 = 916,132,832       (~900 million)
6 chars → 62^6 = 56,800,235,648    (~56 billion)

We need 5 billion unique codes.

5 chars → max 900 million  → NOT enough (5B > 900M)
6 chars → max 56 billion   → ENOUGH     (5B < 56B) ✓

6 Base62 characters gives 56 billion combinations — 11× headroom over our 10-year requirement.

The Binary Intuition#

To understand why 6 chars is enough, think in binary first.

In binary, each position has 2 choices (0 or 1). N bits = 2^N combinations.

32 bits → 2^32 = 4.3 billion  → not enough for 5B
33 bits → 2^33 = 8.6 billion  → enough

In Base62, each position has 62 choices instead of 2. So each Base62 character carries more information than a single bit — roughly log2(62) ≈ 5.95 bits per character.

6 Base62 chars ≈ 6 × 5.95 = 35.7 bits of information
2^35.7 ≈ 56 billion combinations ✓

This is why 6 chars in Base62 comfortably covers what would take 36 bits in binary.

Final Answer#

Code length:    6 characters
Alphabet:       Base62 (a-z, A-Z, 0-9)
Combinations:   56 billion
Requirement:    5 billion
Headroom:       11×