Digilocks are ornamental, on most of them you can simply shine a light at the right angle and see which buttons were pressed due to the marks left by fingers. Add to this the fact that the sequence is irrelevant and they're pointless.
What is the combination probability?
- Thread starter Lysander
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Associate
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hah , i always wondered how safe those safe's you get at a tourist hotel are.
Op wants to rob an off license.
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lol
Man of Honour
We have digilocks or the mechanical digilocks on every door in our Trust.
The digilocks only need 4 numbers and the mechanical digilocks require you to press C first and 4 numbers and a letter.
I had a debate on Wednesday with a colleague, I told another colleague the door code was CZ1234 but my colleague argued it was C4231Z.
I turned to the woman and said "use CZ1234" but the other colleague argued so I said go and try it.
Also like Diddums said, I was trying to get into Cardiology but forgot the number, I shined my phone light at the keypad and you could see the 4 most used ones.
The digilocks only need 4 numbers and the mechanical digilocks require you to press C first and 4 numbers and a letter.
I had a debate on Wednesday with a colleague, I told another colleague the door code was CZ1234 but my colleague argued it was C4231Z.
I turned to the woman and said "use CZ1234" but the other colleague argued so I said go and try it.
Also like Diddums said, I was trying to get into Cardiology but forgot the number, I shined my phone light at the keypad and you could see the 4 most used ones.
Soldato
2580X has got to be the most common with these anyway
Caporegime
The mechanical ones don’t “need” C pressing, only if someone has pressed an incorrect button beforehand.
Of course you have no way of knowing if someone has already mashed the keypad or not, so pressing C every time will certainly cut out the faffing those times the code doesn’t work, but that doesn’t mean C is needed.
Soldato
Surely the answer is ((4/10)^4)*(1/3) so about 0.9%.
Edit, actually do all the numbers have to be different?
Edit, actually do all the numbers have to be different?
Caporegime
Yes, you need 5 buttons total, all different to each other, but order is irrelevant.
Soldato
Ah, in which case it's
4/10 X 3/9 X 2/8 X 1/7 X 1/3
4/10 X 3/9 X 2/8 X 1/7 X 1/3
Man of Honour
Man of Honour
They'll protect your stuff from people who know nothing about the subject and haven't read the easily obtained manual for the safe and won't force the safe open mechanically. So they're probably generally somewhat safe.
Just for jollies, here's the lockpicking lawyer showing 3 ways to open a locked hotel safe. Two of which only require reading the easily obtained manual and 1 of which requires minimal skill in lockpicking.
There are worse "safes". He once opened a "safe" with a fork. Just a standard household eating fork. Someone jokingly mentioned a spoon because a fork was too easy, so he did another video showing that - opening the "safe" with a spoon. And opening another "safe" with a plastic child's toy figure. Or soft drink cans. There are a fair few "safes" that are so bad that a person with relevant skills doesn't even need any lockpicking tools.
Although it could be said that hotel "safe" in the video is worse because it can be opened by anyone who has read the easily obtained manual. Without any tools at all. Not even a spoon.
After watching a few of his videos it was surprising looking around how many every day lock I encounter could be opened with a simple shim and pretty much no skill :s
Reminds me of college and the doors with 4 pin combos and someone found if you put pressure on at a slight angle any combination would work with a little jiggling.
^^^ these guys have it
Though to make it easier, remember you can just use the nCr button on your calculator:
10C4 = 210
1/210 * 1/3 = 0.16% (approximately)
Soldato
Surely this is:
1/10 x 1/9 X 1/8 x 1/7 x 1/3
You start with a 1 in 10 chance of getting a correct number regardless of it needing to be the first in a set sequence and so on with a diminishing number of numbers to choose from.
So 10x9x8x7x3 = 15,120 : 1 chance of guessing correctly.
1/10 x 1/9 X 1/8 x 1/7 x 1/3
You start with a 1 in 10 chance of getting a correct number regardless of it needing to be the first in a set sequence and so on with a diminishing number of numbers to choose from.
So 10x9x8x7x3 = 15,120 : 1 chance of guessing correctly.
Caporegime
Which only happens if one has been clicked.
Take one apart, C is a reset, nothing more. It doesn’t need pressing after a successful unlock, just try entering the code without “C” straight after you’ve successfully unlocked the door.
the X is 'execute' pressed at the end of every attempt
the C is cancel and start again
the Y can be included in the 'code'
any code is meaningless, its just an arrangement of ones and zeros.
so the combination is
1234567890Y = each set to 0 or 1
00000000000
<snip> every combination of those inbetween
11111111111
= 11 bits, so 1024?
It's unlikely to are allowed nothing as the code
so 1023, much less if the code is fixed to 4 digits, that would be:
11 x 10 x 9 x 8 ? = 7920
1 in 11
2 in 10
3 in 9
4 in 8
ok thats wrong
lol stream of conscienesnss safe cracking
the chances of me guessing the first correct one is for a 4 length code
there is a row of 1s set, so eg the code was set like this:
1234567890Y
10010011000 = my 'code' is 1478 and its also 4187 etc
this is the failure of such a lock, I have a 4 in 11 chance of guessing the correctcode digit, every time I press a button. nuts.
That's not 40% as its 4 out of 11 so its slightly less than 40% less, erm how does maths work again. need coffee.
so its like 36% * length of code (4 digits vs 5 etc)
the C is cancel and start again
the Y can be included in the 'code'
any code is meaningless, its just an arrangement of ones and zeros.
so the combination is
1234567890Y = each set to 0 or 1
00000000000
<snip> every combination of those inbetween
11111111111
= 11 bits, so 1024?
It's unlikely to are allowed nothing as the code
so 1023, much less if the code is fixed to 4 digits, that would be:11 x 10 x 9 x 8 ? = 7920
1 in 11
2 in 10
3 in 9
4 in 8
ok thats wrong
lol stream of conscienesnss safe cracking
the chances of me guessing the first correct one is for a 4 length code
there is a row of 1s set, so eg the code was set like this:
1234567890Y
10010011000 = my 'code' is 1478 and its also 4187 etc
this is the failure of such a lock, I have a 4 in 11 chance of guessing the correct
That's not 40% as its 4 out of 11 so its slightly less than 40% less, erm how does maths work again. need coffee.
so its like 36% * length of code (4 digits vs 5 etc)
Man of Honour
I'll give it a go on Wednesday, there are 6 doors to go at on our floor at work
Soldato
If the order of numbers/letters does not matter then the chance of hitting the first one correct is 5/13 as it could be any 4 out of the 10 numbers OR any 1 out of the 3 letters. Need a pen and paper to work it out.