This is a table of the number of guesses required to generate a key with a certain prefix length (code here).

So to have a 90% chance of guessing a 7-bit prefix means generating 294 keypairs. This corresponds to a network size of about 6K vaults.

```
| probability of generating a valid key |
| 0.1 0.5 0.9 0.99 0.9999 |
Prefix Length | | Sections Vaults
3 | 1 5 17 34 69 | 8 400
4 | 2 11 36 71 143 | 16 800
5 | 3 22 73 145 290 | 32 2K
6 | 7 44 146 292 585 | 64 3K
7 | 13 88 294 587 1K | 128 6K
8 | 27 177 588 1K 2K | 256 13K
9 | 54 355 1K 2K 5K | 512 26K
10 | 108 709 2K 5K 9K | 1K 51K
11 | 216 1K 5K 9K 19K | 2K 102K
12 | 432 3K 9K 19K 38K | 4K 205K
13 | 863 6K 19K 38K 75K | 8K 410K
14 | 2K 11K 38K 75K 151K | 16K 819K
15 | 3K 23K 75K 151K 302K | 33K 2M
16 | 7K 45K 151K 302K 604K | 66K 3M
17 | 14K 91K 302K 604K 1M | 131K 7M
18 | 28K 182K 604K 1M 2M | 262K 13M
19 | 55K 363K 1M 2M 5M | 524K 26M
20 | 110K 727K 2M 5M 10M | 1M 52M
21 | 221K 1M 5M 10M 19M | 2M 105M
22 | 442K 3M 10M 19M 39M | 4M 210M
23 | 884K 6M 19M 39M 77M | 8M 419M
24 | 2M 12M 39M 77M 155M | 17M 839M
25 | 4M 23M 77M 155M 309M | 34M 2B
26 | 7M 47M 155M 309M 618M | 67M 3B
27 | 14M 93M 309M 618M 1B | 134M 7B
28 | 28M 186M 618M 1B 2B | 268M 13B
29 | 57M 372M 1B 2B 5B | 537M 27B
30 | 113M 744M 2B 5B 10B | 1B 54B
31 | 226M 1B 5B 10B 20B | 2B 107B
32 | 453M 3B 10B 20B 40B | 4B 215B
33 | 905M 6B 20B 40B 79B | 9B 429B
34 | 2B 12B 40B 79B 158B | 17B 859B
35 | 4B 24B 79B 158B 316B | 34B 1718B
36 | 7B 48B 158B 316B 633B | 69B 3436B
37 | 14B 95B 316B 633B 1266B | 137B 6872B
38 | 29B 191B 633B 1266B 2532B | 275B 13744B
39 | 58B 381B 1266B 2532B 5063B | 550B 27488B
40 | 116B 762B 2532B 5063B 10127B | 1100B 54976B
```

My logic for this table:

Define L as the number of bits in the prefix

Chance of a single key generation being a match

1/2^L

Chance of it not being a match

1-1/2^L

Chance of nth try not being a match

(1-1/2^L)^n

Chance of nth try being a match

p = 1-(1-1/2^L)^n

rearrange to find n for chance p of a match

1-p = (1-1/2^L)^n

log(1-p) = n * log(1-1/2^L)

n = log(1-p) / log(1-1/2^L)