Finding Hash with 11 leading zeroes

Every attempted a hash has an independent probability of 1 in 16¹¹ of having 11 leading zeroes.

That means a probability of not finding such a hash after n attempts is equal to the probability that n attempts are all unsuccesful: (1 - 16^-11)ⁿ

That means the probability that you are successful increases with n, but never reaches 1:

• 1% chance after 176,807,378,144 attempts (~16^9.34).
• 5% chance after 902,361,177,698 attempts (~16^9.93).
• 20% chance after 3,925,582,869,332 attempts (~16^10.46).
• 50% chance after 12,193,974,156,573 attempts (~16^10.87).
• 80% chance after 28,313,531,182,477 attempts (~16^11.17).
• 95% chance after 52,701,479,495,622 attempts (~16^11.40).
• 99% chance after 81,015,010,678,098 attempts (~16^11.55).
• 99.9% chance after 121,522,516,017,148 attempts (~16^11.70).
• ...

No,the "nonce" could be alternate.Not limited in 1 to N. But the expect time of finding a target hash is 16^11,that meaning if u try 16^11 different "nonce" u are likely get a target.

there must be a nonce somewhere between 1 and 16^11 where the hash has 11 leading zeroes [in fixed-length hexadecimal encoding]?

1. You cannot assume this.

2. Bitcoin doesn't encode in hexadecimal and then test for number of leading zeroes. The test it uses is literally if (hash <= hashTarget)

As you probably know, for current hashTarget values, miners expect to exhaust all nonce values and vary other block values such as the so-called "extranonce" etc - and retry many many times.

See

you need difficulty * 2⁴⁸ / 65535 attempts.

Also note that's on average. You might never find one. The probability of finding a sufficiently small hash after n attempts is never 100% for any finite n.