Jammed keysafe. Grrr.....

[mike]

e > positions, on or off. It's like one suit of cards and the only

The 'numbers' don't come into play in the sense that each button (unlike a dial) can't represent 0 to 9; it can only represent on or off, 0 or 1, face up or face down. And you can repeat them because, if you have ten buttons that are either on or off and four have been set to "on", then "on" is repe ated four times.

The button can have one of two states: selected or not selected, on or off. If you choose at random you'll have a one in two or 50:50 chance of getti ng the right one.

I'm not saying I'm right and anyone else is wrong. I made the original pos ting and I'm questioning whether I'm right.

As Dave TMH has already pointed out, Keysafe gave a different number of tot al possible combinations to the one that I suggested. And I suggested why they might be right.

It's possible to make a particular explanation seem plausible but then you have to prove that the other plausible explanations that give different res ults are wrong.

I agree with you that probabilties are not particularly intuative. The Mon ty Hall problem was mentioned on here not so long ago and that's obviously quite contentious.

[David]

es of getting the second right is also 1 in 2 and so on. This works out as 1/(2 to the power 10) with is 1/1024 which means there is only one combina tion that will work --- and that is to get all four right (but the order th at you push them in doesn't matter).

I think the problem with this analysis is that it would be correct if we di dn't know the number of digits in the "combination". In other words, if it could be anywhere between zero and ten digits long.

But we know that there are only four buttons that should be "on". The abov e analysis doesn't take that into account.

[docholliday93]

Which of course simplifies your choice for the first two digits... Looking at the maths it appears that when keysafe claim 1024 combinations t hey're using the entire range from no buttons pressed to all buttons presse d. If you include the three largest sets only (4, 5 and 6 buttons pressed) you'd get 772 combinations, but in fact it seems almost universal to use ju st 4.

[dennis]

They don't unless the person knows its a six digit code. If you don't know how long the code is you have 1024 possibles.

If they think its a four digit code their chances are zero if any other length is tried.

Anyway even at 252 variants its probably more than a typical Yale cylinder has and people don't worry about someone bringing a hundred keys and trying each one which is probably quicker than trying all the codes.

the main thing is to fit the keysafe where it can't be easily seen but the person using it can't hide. Don't fit them inside porches or rear gardens as nobody takes any notice of a person "working" there.

[Dave Liquorice]

Repeat a button how? Once it's pressed it's pressed (or turn a card, it's turned) and no longer available.

The odds for getting the combination by chance must surely be:

No buttons pressed you choose 1 from the 10 available, 4 of which could be right, odds of getting picking a correct one 1:2.5.

Next choose 1 from the 9 available, 3 of which could be right, odds of getting picking a correct one 1:3.

Next choose 1 from the 8 available, 2 of which could be right, odds of getting picking a correct one 1:4.

Next choose 1 from the 7 available, 1 of which could be right, odds of getting picking a correct one 1:7.

2.5 x 3 x 4 x 7 = 210

Same here, we need a real statistician/mathematician to give the definitive proof.

That's the other thing, maximum number of combinations given the rules of 10 buttons, each button can only be used once, a 4 digit code that can be entered in any order may not be the same as that that could be indicated by the odds of entering the correct combination by chance.

But proven, even if some very eminent mathematicians didn't believe the proof...

[docholliday93]

That agrees with the standard formula for picking 4 objects out of 10. The formula for picking n objects out of m is m!/(n!.(m-n)!) (n! is n factorial). Using that and adding up all the possible combinations then you do get 1024, as keysafe claim, though some of the combinations are probably not recommended! It builds up like this: no of ways of picking 0 buttons

[docholliday93]

Damn - posted prematurely - to continue... no of ways of picking 0 buttons 1 no of ways of picking 1 button 10 no of ways of picking 2 buttons 45 no of ways of picking 3 buttons 120 no of ways of picking 4 buttons 210 no of ways of picking 5 buttons 252 no of ways of picking 6 buttons will be the same as for 4, for fairly obvious reasons, and the same goes for 7 to 10 buttons. Add all this up and you get 1024. I wouldn't recommend choosing 0 or 10 buttons as your combination, though - that's taking security by obscurity a bit too far!