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EnglishAnyone 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.

tonyuk932 4 months ago
    Tags:
  • Randomness
  • Statistics
  • Probability
question_image


Ailuc 4 months ago
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
Ailuc 4 months ago
mind you, it can take a long time to generate an address with a word of 6 letters or more..
tonyuk932 4 months ago
I am aware of vanity addresses. Can be handy. Not sure in terms of safety in comparison to some completely random.
lama 4 months ago
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
Al 4 months ago
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.
tonyuk932 4 months ago
Hi. I really liked your answer. I want to verify something and will mark yours as correct ASAP so you can earn the reward.
Al 4 months ago
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.
Al 4 months ago
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.
Aldo Colombo 4 months ago
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.
Al 4 months ago
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!
Aldo Colombo 4 months ago
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.
SENA SEHO SUNDY 4 months ago
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
tonyuk932 4 months ago
Hi, was it a math equation? It doesn't show up properly...
ASMA WIJAYA 4 months ago
Good quertions
Aldo Colombo 4 months ago Correct
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.
Ogunbiyi Yusuf 4 months ago
Am very interested in this
Ibisiki Godspower 4 months ago
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
Aniekan 4 months ago
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Hebron 4 months ago
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
Hd2020 4 months ago
Try this site:

https://www.statista.com/statistics/807070/bitcoin-cash-price/
Emmanuel emmasing 4 months ago
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).
Farooq Ghaffar 4 months ago
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.
eze daniel 4 months ago
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)
Magart 4 months ago
H0.000334=0.0334%
trillions28 4 months ago
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
kali 4 months ago
possibly is extremely low, I'm not sure myself its. Although if i had to choose i think it would be 3 to4 percent.
alwiashari 4 months ago
good question
Pranav Nirwan 4 months ago
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.
bert2145 4 months ago
Ye_il I m 4'7'8 the nomber
Sina kanafkar 4 months ago
the probability is 1 out of 77,408
kimo2 4 months ago
I just tried to generate 10M addresses with vanitygen, 332 of them contain "cake", so probability is 0.00003.
Chrisphine Obado 4 months ago
Not probable since the codes do not take a definite format
Miss Mike 4 months ago
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.
Daniel Hernan Alves Galli 3 months ago
everything is possible, as it was randomly generated it is just a coincidence
Abatchari musa 3 months ago
A wallet consists of two very long sets of random numbers and letters. One of them is your wallet's public address which other