<p>Yes, exactly 1/2^32.</p>
<p>There isn't much more to say about it. Hashes of distinct values are uniformly randomly distributed. The chance for two independently uniformly randomly distributed hash to match in 32 bits is 1/2^32.</p>

CC BY-SA 4.0
2020-07-03T03:38:01.153Z
2020-07-03T03:38:01.153Z
Pieter Wuille
<p>checksum = 1st4byte(sha256(sha256(PubKeyHash)))</p>
<p>checksum1 = 1st4byte(sha256(sha256(PubKeyHash1)))</p>
<p>What is the P(checksum == checksum1)? where PubKeyHash != PubKeyHash1</p>
<p>Is it 1/(2^^32)?</p>

<checksum>
What is the probability of different public key hash that produce the same checksum value
Cisco Mmu
# What is the probability of different public key hash that produce the same checksum value

(Asked by: username_0 on 2020-07-03)

checksum = 1st4byte(sha256(sha256(PubKeyHash)))

checksum1 = 1st4byte(sha256(sha256(PubKeyHash1)))

What is the P(checksum == checksum1)? where PubKeyHash != PubKeyHash1

Is it 1/(2^^32)?

**Tags:** checksum


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## 1 answers

### Answer by username_1 on 2020-07-03. Score: 8

Yes, exactly 1/2^32.

There isn't much more to say about it. Hashes of distinct values are uniformly randomly distributed. The chance for two independently uniformly randomly distributed hash to match in 32 bits is 1/2^32.


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