 Another example of the number of attempts is in an encryption system a brute force attack on a key chosen by a user Let's say a user has chosen a random key a 128 bit key and An attacker is trying to guess that key So what the attacker has to do? They need to guess that key So how many possible keys do they need to guess from with 128 bits? There are two to the power of 128 possible keys So in the best case the attacker Chooses a key and they're very very lucky and they get the key that the user chose That would be the best case and it takes one attempt The worst case is that they try all keys and it's only into the last key that to the power of 128 key That they get the one that the user chose The average case the average number of attempts is The number of possible keys divided by two two to the power of 128 divided by two which is Simply two to the power of 127 so on average that attacker if the user chose a random key and The attacker knows nothing else about how they chose that key. It would take two to the power of 127 attempts to get the key if we modify the case and the user instead chose a 129 bit key That is they increase the key length by just one bit so in this case the average number of attempts becomes or that that the key space is two to the power of 129 That's the total number of possible keys and the average number of attempts will be two to the power of 129 divided by two or two to the power of 128 This illustrates by increasing the key length by just one bit Then we double the amount of attempts that the attacker must take to find the key