user196: Is it known the minimum amount of characters you need to mistype to get another valid address?
Addresses contain checksums, and also must follow some specific rules, so I guess this could be calculated: I wonder if someone did.

user13751: Here's an example of a pair of valid addresses that differ by only one character.
1xxxxxxxxxxxxxxxxxxxy1wqmDWjatp7t
1xxxxxxxxxxxxxxxxxxxy1wqmDWgatp7t

another example pair
1xxxxxxxxxxxxxxxxxxxxyE4zW1N3nMx2
1xxxxxxxxxxxxxxxxxxxxyE4zW1U3nMx2

For both pairs, the 32-bit checksums are NOT the same, but they satisfy the condition that
(checksum1 % 656356768) == ((checksum2 + 356826688) % 656356768)


user10411: Check sum is a result of 2 hash-functions: SHA256 and RIPEMD160. The length of checksum is 4 bytes. Hash functions are indistinguishable from random oracle, so it is possible, if you make 1 mistake in non-checksum character, you get the same checksum. But probability of this is 1/2^32.
Added later: not exactly correct due to base58 encoding. Correction in the another answer.

196: <p>Is it known the minimum amount of characters you need to mistype to get another valid address?</p>

<p>Addresses contain checksums, and also must follow some specific rules, so I guess this could be calculated: I wonder if someone did.</p>
 13751: <p>Here's an example of a pair of valid addresses that differ by only one character.</p>

<pre><code>1xxxxxxxxxxxxxxxxxxxy1wqmDWjatp7t
1xxxxxxxxxxxxxxxxxxxy1wqmDWgatp7t
</code></pre>

<p>another example pair</p>

<pre><code>1xxxxxxxxxxxxxxxxxxxxyE4zW1N3nMx2
1xxxxxxxxxxxxxxxxxxxxyE4zW1U3nMx2
</code></pre>

<p>For both pairs, the 32-bit checksums are NOT the same, but they satisfy the condition that</p>

<pre><code>(checksum1 % 656356768) == ((checksum2 + 356826688) % 656356768)
</code></pre>
 10411: <p>Check sum is a result of 2 hash-functions: SHA256 and RIPEMD160. The length of checksum is 4 bytes. Hash functions are indistinguishable from random oracle, so it is possible, if you make 1 mistake in non-checksum character, you get the same checksum. But probability of this is 1/2^32.</p>

<p>Added later: not exactly correct due to base58 encoding. Correction in the another answer.</p>
