0.00027 BCH

Anyone good at statistics/Math? What are the probabilities for a 4 letter word to appear somewhere random on a Bitcoin Cash address?

For example the word cake: A bitcoincash address consists of 42 characters: qpv44zw0....cake.....2elahdw5 Not sure if there is some kind of restriction on how many numbers can appear on a bitcoin cash address but for the shake of argument they are also in play completely random. A-Z , 1-9. TIL: Bitcoin addresses do not contain "0", "O", "l", or "I" to prevent mistyping addresses. I assume this probably will make it harder or easier :P Let's say that there aren't restrictions.

its a good question... but if you want a word, make a vanity address.

OCLVANITYGEN.

https://www.google.com/url?sa=t&source=web&rct=j&url=https://github.com/samr7/vanitygen/blob/master/oclvanitygen.c&ved=2ahUKEwi3oO3P5LrvAhU6QkEAHQFOCHwQFjAAegQIBBAC&usg=AOvVaw0lPgkA9NHF1N0wYJKyAnqB

OCLVANITYGEN.

https://www.google.com/url?sa=t&source=web&rct=j&url=https://github.com/samr7/vanitygen/blob/master/oclvanitygen.c&ved=2ahUKEwi3oO3P5LrvAhU6QkEAHQFOCHwQFjAAegQIBBAC&usg=AOvVaw0lPgkA9NHF1N0wYJKyAnqB

mind you, it can take a long time to generate an address with a word of 6 letters or more..

I am aware of vanity addresses. Can be handy. Not sure in terms of safety in comparison to some completely random.

the probability is 1 out of 77,408 as for if the four digit letter if combines or has numbers in between the probability becomes 1 out of 12,384,700

I try to answer your question is a parametric way, so you can use it more generally. Suppose there are "n" characters (you said 42 characters but I think it is actually about 34) each can have "c" options (including A-Z an 0-9 excluding the I or l or O characters, let's say 54 characters). You want a specific word with "k" characters (for example in case of the cake, "k" is equal to 4) happen somewhere in the string of the "n" characters.

To calculate the probability we have to calculate the number of preferred combinations and the number of all combinations and divide them together.

The number of all combinations is n^c

The number of preferred combinations is (assuming the chances of the same word occurring more than once is negligible): (n-k+1)*(n-k)^c.

So for n = 34, c = 54, k = 4, the answer will be:

all = 5.0102744987272601939484648898079e+82

preferred = 1.802641847094241850402095239e+81

probability is about 3% (to be more exact we should subtract the probability of the cases where the word appears twice (or even more) but the probability for that is very low).

If the word gets bigger, the changes go down very quickly.

For example for n = 34, c = 54, and k = 6 (a 6 letter word) the answer will be:

all = 5.0102744987272601939484648898079e+82

preferred = 4.0637940524343363736473214818622e+79

probability is about 0.08%

Hope that everything is clear.

To calculate the probability we have to calculate the number of preferred combinations and the number of all combinations and divide them together.

The number of all combinations is n^c

The number of preferred combinations is (assuming the chances of the same word occurring more than once is negligible): (n-k+1)*(n-k)^c.

So for n = 34, c = 54, k = 4, the answer will be:

all = 5.0102744987272601939484648898079e+82

preferred = 1.802641847094241850402095239e+81

probability is about 3% (to be more exact we should subtract the probability of the cases where the word appears twice (or even more) but the probability for that is very low).

If the word gets bigger, the changes go down very quickly.

For example for n = 34, c = 54, and k = 6 (a 6 letter word) the answer will be:

all = 5.0102744987272601939484648898079e+82

preferred = 4.0637940524343363736473214818622e+79

probability is about 0.08%

Hope that everything is clear.

Hi. I really liked your answer. I want to verify something and will mark yours as correct ASAP so you can earn the reward.

Hi. After checking my answer (due to Aldo Colombo comment) I realized I have made a mistake. The correct formula is c^n for total number of combinations and (n-k+1)*c^(n-k) for desired. So, for n = 34, c = 54, k = 4, the answer is:

total = 7.9687065478951823295709092093068e+58

desired = 2.9051896516352550449708691262119e+53

probability = 3.6457480698704089447370451282457e-6

Which is much less than 3% that I originally said.

total = 7.9687065478951823295709092093068e+58

desired = 2.9051896516352550449708691262119e+53

probability = 3.6457480698704089447370451282457e-6

Which is much less than 3% that I originally said.

total = 7.9687065478951823295709092093068e+58

desired = 2.9051896516352550449708691262119e+53

probability = 3.6457480698704089447370451282457e-6

Which is much less than 3% that I originally said.

Following your reasoning: shouldn't it be (c-k+1)*n^(c-k)? 3% is way too high probability for a given word to appear in a random address.

I made a mistake in my original answer. Thanks for pointing it out. I corrected it now. The chances for having a 4 character random word is much less than 3% that I had said. I should have suspected my answer was wrong!

Following your reasoning, shouldn't it be: (c-k+1)*n^(c-k)? 3% is way too high probability for a given word to appear in a random address.

The answer is 2284879/208827064576 if no letters are repeated. This is roughly 5×10−12" role="presentation" style="box-sizing: inherit; margin: 0px; padding: 0px; border: 0px; font-style: normal; font-variant: inherit; font-weight: normal; font-stretch: inherit; line-height: normal; font-family: inherit; font-size: 15px; vertical-align: baseline; display: inline; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">5×10−125×10−12 less than

Hi, was it a math equation? It doesn't show up properly...

Good quertions

Let's put it clear:

a) Suppose there are "n" characters (a bch address has 42 characters as the op said, to the best of my knowledge)

b) Each character can have "c" options (a-z lowercase and 1-9, excluding i , l , o -> it's 32 different characters, or 'options')

c) You want a given word with "k" characters (for example in case of the cake, "k" is equal to 4) happen somewhere in the string of the "n" characters.

d) To calculate the probability we have to calculate the number of preferred combinations and the number of all combinations and divide them together.

e) The number of all combinations is c^n (not n^c).

f) The number of preferred combinations is (assuming the chances of the same word occurring more than once is negligible): (n-k+1)*c^(n-k) (possible positions of 'cake', multiplied by all possible combinations of n-k characters).

g) So for n = 42, c = 32, k = 4, the answer will be:

all = 1.6455e+63

preferred = 6.1202e+58

Probability is about 3.72e-5 (to be more exact we should subtract the probability of the cases where the word appears twice or more, but the probability for that is way lower).

That is, for a given 4 letter word (not *any* 4 letter word) to appear in a bch address.

a) Suppose there are "n" characters (a bch address has 42 characters as the op said, to the best of my knowledge)

b) Each character can have "c" options (a-z lowercase and 1-9, excluding i , l , o -> it's 32 different characters, or 'options')

c) You want a given word with "k" characters (for example in case of the cake, "k" is equal to 4) happen somewhere in the string of the "n" characters.

d) To calculate the probability we have to calculate the number of preferred combinations and the number of all combinations and divide them together.

e) The number of all combinations is c^n (not n^c).

f) The number of preferred combinations is (assuming the chances of the same word occurring more than once is negligible): (n-k+1)*c^(n-k) (possible positions of 'cake', multiplied by all possible combinations of n-k characters).

g) So for n = 42, c = 32, k = 4, the answer will be:

all = 1.6455e+63

preferred = 6.1202e+58

Probability is about 3.72e-5 (to be more exact we should subtract the probability of the cases where the word appears twice or more, but the probability for that is way lower).

That is, for a given 4 letter word (not *any* 4 letter word) to appear in a bch address.

Am very interested in this

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answer is1

First generate a Bitcoin private key. ... seed Usually this private cryptographic key is a long string of numbers and letters. ... from seed words and provide an extended public key to derive bitcoin addresses

Try this site:

https://www.statista.com/statistics/807070/bitcoin-cash-price/

https://www.statista.com/statistics/807070/bitcoin-cash-price/

Suppose there are "n" characters (you said 42 characters but I think it is actually about 34) each can have "c" options (including A-Z an 0-9 excluding the I or l or O characters, let's say 54 characters).

Probability is about 3.72e-5 (to be more exact we should subtract the probability of the cases where the word appears twice or more, but the probability for that is way lower). That is, for a given 4 letter word (not *any* 4 letter word) to appear in a bch address.

Probability is about 3.72e-5 (to be more exact we should subtract the probability of the cases where the word appears twice or more, but the probability for that is way lower)

H0.000334=0.0334%

bitcoin addresses do not contain 0, 1 or O,I. I think this probably will make it easier or harder .completely random A to Z , 1 to 9

A-z,1-9

possibly is extremely low, I'm not sure myself its. Although if i had to choose i think it would be 3 to4 percent.

good question

There are 2 types of addresses.

P2SH - This address starts with number “3”, and we use “Pay to Script Hash” addresses when we choose to spend the bitcoin using a script hash from the other person.

P2PKH - This is the common address that we use daily for transactions and starts with “1”. I do believe that you want just the mathematical answer to this kind of addres.

Given the details that each P2PKH has “1” at the beginning and 26–35 characters.

There are addresses that have 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 characters.

Each character can be a letter or number,

Each letter can be uppercase or lowercase

So, we got 10 number (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) and 52 uppercase characters mixed with lowercase characters

For a 26 characters length address, you can have 62 ^ 26 possibilities.

For 27, there are another 62 ^ 27.

In the end, you would get around 62^(26+27+28+29+30+31+32+33+34+35) unique addresses.

P2SH - This address starts with number “3”, and we use “Pay to Script Hash” addresses when we choose to spend the bitcoin using a script hash from the other person.

P2PKH - This is the common address that we use daily for transactions and starts with “1”. I do believe that you want just the mathematical answer to this kind of addres.

Given the details that each P2PKH has “1” at the beginning and 26–35 characters.

There are addresses that have 26, 27, 28, 29, 30, 31, 32, 33, 34, 35 characters.

Each character can be a letter or number,

Each letter can be uppercase or lowercase

So, we got 10 number (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) and 52 uppercase characters mixed with lowercase characters

For a 26 characters length address, you can have 62 ^ 26 possibilities.

For 27, there are another 62 ^ 27.

In the end, you would get around 62^(26+27+28+29+30+31+32+33+34+35) unique addresses.

Ye_il I m 4'7'8 the nomber

the probability is 1 out of 77,408

0.009615

I just tried to generate 10M addresses with vanitygen, 332 of them contain "cake", so probability is 0.00003.

Not probable since the codes do not take a definite format

Please, remember that you are addressing a multilingual system. Therefore this must be thought of in a multilingual way. Such as a major group of alphabets; in cyrillic, in mandarin, in occidental romance (anglo-Hispanic western). You must address this conundrum in the language of mathematics, greek.

everything is possible, as it was randomly generated it is just a coincidence

0.0034%

A wallet consists of two very long sets of random numbers and letters. One of them is your wallet's public address which other