I heard someone say that Bitcoin is dangerous because you could easily make a typo when entering an address, sending money to the wrong person, and have no recourse. There are dangers associated with Bitcoin, such as losing a private key, but address typos are not a major concern.
Checksums
There are several kinds of Bitcoin addresses. Each is at least 20 bytes (160 bits) long, with at least 4 bytes (32 bits) of checksum. The chances of a typo resulting in a valid checksum are about 1 in 232.
Used addresses
Let’s ignore the checksum for this section.
Because addresses are formed by cryptographic hash functions, we can assume the values are essentially randomly distributed in the space of possible addresses. The addresses are deterministic, but for modeling purposes, random is as random does.
This means a typo of an actual address is no more or less likely to be another actual address than an address typed at random. This is unlike, say, English words: a mistyped English word is more likely to be another English word than random keystrokes would be.
There have been on the order of a billion Bitcoin addresses used, in a space of 2160 possibilities. (Actually more since some addresses have more than 160 bits.) There’s about a 1 in 1039 chance that a random 160-bit sequence corresponds to an address somewhere on the Bitcoin blockchain.
Addresses close in edit distance
Someone with the Caesarean handle Veni Vidi Vici on X asked
What about the odds that out of those 1B addresses, two of them are one character swap away from each other?
That’s an interesting question. Let’s assume the addresses are Base58-encoded strings of length 26. Addresses could be longer, but assuming the minimum length increases the probability of addresses being close.
How many addresses are within one or two character swaps of another? I addressed a similar question here a couple weeks ago. If all the characters were unique, the number of strings within k swaps of each other would be
|S1(26, 26 − k)|
where S1 denotes Stirling numbers of the first kind. For k = 1 this would be 325 and for k = 2 this would be 50,050. This assumes all the characters are unique; I haven’t thought through the case where characters are repeated.
For round numbers, let’s say there are a billion addresses, and for each address there are a million other addresses that are close in some sense, plausible typos of the address. That would be 1012 addresses and typos, spread out in a space of ≈1045 (i.e. 5826) possible addresses.
Now there’s an implicit Birthday Problem here. No particular address is likely to collide with another, even when you allow typos, but what about the likelihood that some address collides?
Say we partition our space of 1045 addresses into N = 1029 addresses with a million possible typos for each address. Then as a rule of thumb, you’d need around √N random draws before you have a 50-50 chance of seeing a collision. Since 109 is a lot less than 1014.5, it’s unlikely that any two addresses collide, even when you consider each address along with a million associated typos.
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