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A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings. |
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"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)"
1 in a million (10 to the power of 6).
"
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings. "
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random). |
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"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random)."
I'm sure it's more complicated than that  |
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"What's in the safe?
Mr ddc
£10000"
Buggar, not worth getting Jason Statham in then.
If you are all wearing stockings on your heads, I know how many you have to take out from the drawer before you get a matching pair.
That's what our maths was limited to
 |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that "
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one? |
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"
I'm sure it's more complicated than that "
There are six numbers; each can be in the range 0-9.
10 x 10 x 10 x 10 x 10 x 10 (or 10^6) = 1,000,000.
This is exactly how many combinations are possible. |
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"
I'm sure it's more complicated than that
There are six numbers; each can be in the range 0-9.
10 x 10 x 10 x 10 x 10 x 10 (or 10^6) = 1,000,000.
This is exactly how many combinations are possible."
That's Numberwang! |
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By (user no longer on site)
over a year ago
|
"So the odds of guessing the code is about 999999 to 1
1 000 000
You haven't counted 000000, I suspect.
There are 1 000 000 possibilities."
Oh yup I messed that up. For the way to see if it would be guessed.
I think you could just devideo it out but I'm not certain |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
I'm pretty sure Brummie is correct.
If 3000 unique combinations are tried the probability would be 3000/1000000.
If nobody knows what anyone guessed previously, the odds are 1 in 1000000.
But I am sleep deprived. |
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"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?"
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know) |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know) "
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses. |
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"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses."
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one. |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one. "
No, not if they don't know what's been guessed previously.
I know it seems counter-intuitive but that's how it is. |
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"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one. "
No because the wrong answers wont be deducted from the 1 to 1,000,000 odds. Basically every wrong answer resets the odds to 1,000,000 |
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By (user no longer on site)
over a year ago
|
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings. "
Semtex |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one.
No because the wrong answers wont be deducted from the 1 to 1,000,000 odds. Basically every wrong answer resets the odds to 1,000,000"
Good way of explaining it 
Yes, ^this. |
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By (user no longer on site)
over a year ago
|
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one.
No, not if they don't know what's been guessed previously.
I know it seems counter-intuitive but that's how it is."
wouldn't that only be everyones individual chance? because the chances of every individual entering the identical code is close to nill...It stands to reason more than one combination will be tried (though how many variations will be entered couldn't be known until after the fact) then it stands to reason the chances of thousands of attempts to crack it would be higher than if just one attempt was made..no?
I don't know..This is like a headache with pictures. why did I come in here? |
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"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one.
No, not if they don't know what's been guessed previously.
I know it seems counter-intuitive but that's how it is.
wouldn't that only be everyones individual chance? because the chances of every individual entering the identical code is close to nill...It stands to reason more than one combination will be tried (though how many variations will be entered couldn't be known until after the fact) then it stands to reason the chances of thousands of attempts to crack it would be higher than if just one attempt was made..no?
I don't know..This is like a headache with pictures. why did I come in here?"
This is exactly what I'm trying to say. (I think)
The no difference in chance thing makes absolutely no sense to me. |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one.
No, not if they don't know what's been guessed previously.
I know it seems counter-intuitive but that's how it is.
wouldn't that only be everyones individual chance? because the chances of every individual entering the identical code is close to nill...It stands to reason more than one combination will be tried (though how many variations will be entered couldn't be known until after the fact) then it stands to reason the chances of thousands of attempts to crack it would be higher than if just one attempt was made..no?
I don't know..This is like a headache with pictures. why did I come in here?"
It seems like that would be that case but no, it's not. |
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By (user no longer on site)
over a year ago
|
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one.
No, not if they don't know what's been guessed previously.
I know it seems counter-intuitive but that's how it is.
wouldn't that only be everyones individual chance? because the chances of every individual entering the identical code is close to nill...It stands to reason more than one combination will be tried (though how many variations will be entered couldn't be known until after the fact) then it stands to reason the chances of thousands of attempts to crack it would be higher than if just one attempt was made..no?
I don't know..This is like a headache with pictures. why did I come in here?
This is exactly what I'm trying to say. (I think)
The no difference in chance thing makes absolutely no sense to me. "
yeah..Each individuals chance of opening the safe would stay the same..I can see that. But the overall chance of the safe being opened would change..I think..But you couldn't measures the chance of it being opened until you knew how many variations of combination will be entered (I'd bet both balls it would be more than one)..so its not really work outable to me.
But then I'm better with hammers and spanners than I am calculators so  |
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"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one.
No, not if they don't know what's been guessed previously.
I know it seems counter-intuitive but that's how it is.
wouldn't that only be everyones individual chance? because the chances of every individual entering the identical code is close to nill...It stands to reason more than one combination will be tried (though how many variations will be entered couldn't be known until after the fact) then it stands to reason the chances of thousands of attempts to crack it would be higher than if just one attempt was made..no?
I don't know..This is like a headache with pictures. why did I come in here?
It seems like that would be that case but no, it's not."
Well that sucks... |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one.
No, not if they don't know what's been guessed previously.
I know it seems counter-intuitive but that's how it is.
wouldn't that only be everyones individual chance? because the chances of every individual entering the identical code is close to nill...It stands to reason more than one combination will be tried (though how many variations will be entered couldn't be known until after the fact) then it stands to reason the chances of thousands of attempts to crack it would be higher than if just one attempt was made..no?
I don't know..This is like a headache with pictures. why did I come in here?
This is exactly what I'm trying to say. (I think)
The no difference in chance thing makes absolutely no sense to me.
yeah..Each individuals chance of opening the safe would stay the same..I can see that. But the overall chance of the safe being opened would change..I think..But you couldn't measures the chance of it being opened until you knew how many variations of combination will be entered (I'd bet both balls it would be more than one)..so its not really work outable to me.
But then I'm better with hammers and spanners than I am calculators so "
No, trust me, unless each person knows the previous results and uses a unique code, the individual chance and overall chance are the same.
I'll try to find proof and a clearer explanation tomorrow. I can't be bothered now. |
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By (user no longer on site)
over a year ago
|
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one.
No, not if they don't know what's been guessed previously.
I know it seems counter-intuitive but that's how it is.
wouldn't that only be everyones individual chance? because the chances of every individual entering the identical code is close to nill...It stands to reason more than one combination will be tried (though how many variations will be entered couldn't be known until after the fact) then it stands to reason the chances of thousands of attempts to crack it would be higher than if just one attempt was made..no?
I don't know..This is like a headache with pictures. why did I come in here?
It seems like that would be that case but no, it's not."
nope. my head doesn't bend that way |
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"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one.
No, not if they don't know what's been guessed previously.
I know it seems counter-intuitive but that's how it is.
wouldn't that only be everyones individual chance? because the chances of every individual entering the identical code is close to nill...It stands to reason more than one combination will be tried (though how many variations will be entered couldn't be known until after the fact) then it stands to reason the chances of thousands of attempts to crack it would be higher than if just one attempt was made..no?
I don't know..This is like a headache with pictures. why did I come in here?
This is exactly what I'm trying to say. (I think)
The no difference in chance thing makes absolutely no sense to me.
yeah..Each individuals chance of opening the safe would stay the same..I can see that. But the overall chance of the safe being opened would change..I think..But you couldn't measures the chance of it being opened until you knew how many variations of combination will be entered (I'd bet both balls it would be more than one)..so its not really work outable to me.
But then I'm better with hammers and spanners than I am calculators so
No, trust me, unless each person knows the previous results and uses a unique code, the individual chance and overall chance are the same.
I'll try to find proof and a clearer explanation tomorrow. I can't be bothered now."
I don't like it. This is actually quite upsetting. I can't wrap my head around it and it's making my brain sad. |
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By (user no longer on site)
over a year ago
|
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one.
No, not if they don't know what's been guessed previously.
I know it seems counter-intuitive but that's how it is.
wouldn't that only be everyones individual chance? because the chances of every individual entering the identical code is close to nill...It stands to reason more than one combination will be tried (though how many variations will be entered couldn't be known until after the fact) then it stands to reason the chances of thousands of attempts to crack it would be higher than if just one attempt was made..no?
I don't know..This is like a headache with pictures. why did I come in here?
This is exactly what I'm trying to say. (I think)
The no difference in chance thing makes absolutely no sense to me.
yeah..Each individuals chance of opening the safe would stay the same..I can see that. But the overall chance of the safe being opened would change..I think..But you couldn't measures the chance of it being opened until you knew how many variations of combination will be entered (I'd bet both balls it would be more than one)..so its not really work outable to me.
But then I'm better with hammers and spanners than I am calculators so
No, trust me, unless each person knows the previous results and uses a unique code, the individual chance and overall chance are the same.
I'll try to find proof and a clearer explanation tomorrow. I can't be bothered now."
one of my teachers had a breakdown and had to leave lesson. I'm hard to teach..be aware |
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"Ok New question. Say every person knew the other numbers used (or they couldn't use ones already used or whatever), what would be the chances of somebody getting it right? "
Good question. It goes down by one for every wrong guess. If the previous 2,999 had all got it wrong it would leave 998,001 combinations. |
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Or, yet another way of looking at it is...
Your odds of winning the National Lottery are approximately 1 in 45,000,000 (my smartphone can't calculate the exact figure) if you buy one ticket.
If you buy two tickets your odds are 1 in 22,500,000 (unless you were dumb enough to use exactly the same numbers twice) of winning the jackpot.
But, individually, each ticket still has the same 1 in 45,000,000 chance of winning the jackpot. |
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"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that
The two parts of the question are the same, unless the first part should say that each person does know the combinations tried before and tries a unique one?
I don't know
Basically, there's a City Safe thing going on in Preston in December. There's a safe with a 6 digit code. There's a keypad 0-9 (like a pin on a card machine).
Its gonna be in town for 12 and a half hours and anyone can try once.
I worked out that if they processed 4 people a minute (welcome, enter code, thank and leave) then a maximum of 3000 people would be able to enter a combination of their choice.
So what are the odds of someone opening the safe on that day? And what are the odds of me guessing it right? (I'm not going, but I'm interested to know)
1 in 1 million. To both questions, as Brummie said.
Your chance is the same as anyone elses.
But isn't there more chance of somebody (with their 1 in a million chance) guessing it if there's more people? There's 3000 people to guess, instead of just one. So that's 3000 guesses, instead of one.
No, not if they don't know what's been guessed previously.
I know it seems counter-intuitive but that's how it is.
wouldn't that only be everyones individual chance? because the chances of every individual entering the identical code is close to nill...It stands to reason more than one combination will be tried (though how many variations will be entered couldn't be known until after the fact) then it stands to reason the chances of thousands of attempts to crack it would be higher than if just one attempt was made..no?
I don't know..This is like a headache with pictures. why did I come in here?
This is exactly what I'm trying to say. (I think)
The no difference in chance thing makes absolutely no sense to me.
yeah..Each individuals chance of opening the safe would stay the same..I can see that. But the overall chance of the safe being opened would change..I think..But you couldn't measures the chance of it being opened until you knew how many variations of combination will be entered (I'd bet both balls it would be more than one)..so its not really work outable to me.
But then I'm better with hammers and spanners than I am calculators so
No, trust me, unless each person knows the previous results and uses a unique code, the individual chance and overall chance are the same.
I'll try to find proof and a clearer explanation tomorrow. I can't be bothered now.
one of my teachers had a breakdown and had to leave lesson. I'm hard to teach..be aware "
I was a bolshy little shit in my R.E class because I felt the need to contradict everything the teacher was saying. He shut himself in the cupboard for last half an hour of the lesson. |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"Ok New question. Say every person knew the other numbers used (or they couldn't use ones already used or whatever), what would be the chances of somebody getting it right? "
3000 in 1 million (or 3 in 1000, or 1 in 333.3333333...) |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"Ok New question. Say every person knew the other numbers used (or they couldn't use ones already used or whatever), what would be the chances of somebody getting it right?
Good question. It goes down by one for every wrong guess. If the previous 2,999 had all got it wrong it would leave 998,001 combinations."
Overall though, if 3000 combinations were used, it's 3000 in 1 million.
Of course, in this case, as they would stop if the right combination was found, 3000 combinations may not be used. |
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By (user no longer on site)
over a year ago
|
"Or, yet another way of looking at it is...
Your odds of winning the National Lottery are approximately 1 in 45,000,000 (my smartphone can't calculate the exact figure) if you buy one ticket.
If you buy two tickets your odds are 1 in 22,500,000 (unless you were dumb enough to use exactly the same numbers twice) of winning the jackpot.
But, individually, each ticket still has the same 1 in 45,000,000 chance of winning the jackpot."
I get that..The individual persons chance is always the same.
I'm stuck on how so many tries effects the overall chance of it being opened.
or to continue your lottery analogy, does selling thousands of lottery tickets increase the chances of the lottery being won, doesn't matter who by. just if it gets won or not. If that makes sense? |
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"
Good question. It goes down by one for every wrong guess. If the previous 2,999 had all got it wrong it would leave 998,001 combinations.
Not quite...997,001 combinations  ."
Yeah lol. Its been a long night. |
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By (user no longer on site)
over a year ago
|
"Or, yet another way of looking at it is...
Your odds of winning the National Lottery are approximately 1 in 45,000,000 (my smartphone can't calculate the exact figure) if you buy one ticket.
If you buy two tickets your odds are 1 in 22,500,000 (unless you were dumb enough to use exactly the same numbers twice) of winning the jackpot.
But, individually, each ticket still has the same 1 in 45,000,000 chance of winning the jackpot.
I get that..The individual persons chance is always the same.
I'm stuck on how so many tries effects the overall chance of it being opened.
or to continue your lottery analogy, does selling thousands of lottery tickets increase the chances of the lottery being won, doesn't matter who by. just if it gets won or not. If that makes sense? "
if your chances as an individual are 14 million to one of winning it. That doesn't mean the chances of the lottery being won at all are at the same odds. even though millions of people all have a go not knowing everyone else's number combinations it doesn't keep the odds of it being won down to 14 million to one. It's won almost weekly because...The amount of attempts is so high no? even though nobody knows everyone else's guess. I think... |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
Ok, copied off another site I'm not allowed to link to:
"if you toss a coin nineteen times and it comes up heads each time, then it is not more likely that the next toss will be a tail. The odds stay the same, at 50%. The tosses are called 'independent events' which means that the coin can't remember what has happened to it. While twenty heads in a row is unlikely, once you have nineteen heads in a row, the unlikely event has already happened. The potential twentieth head has the same probability as the first head. Another way of looking at it is that any sequence of twenty tosses is unlikely as twenty heads in a row, even if it looks random. But you have to write down the sequence before you start tossing to see if you get it!"
Does that help? |
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By (user no longer on site)
over a year ago
|
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
1 in a million (10 to the power of 6).
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings.
Exactly the same if they don't know which combinations have already been tried (and without discounting likely combinations such as '000000', '123456' etc. if the combination was generated at random).
I'm sure it's more complicated than that "
No, it is the right answer, assuming it's a digital combination, if it's analogue I. e. a dial then there are more combinations as you have left and right turns to factor in. which at 3:30 am I think would be 10 to the power of 6 squared.
|
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By (user no longer on site)
over a year ago
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"Ok, copied off another site I'm not allowed to link to:
if you toss a coin nineteen times and it comes up heads each time, then it is not more likely that the next toss will be a tail. The odds stay the same, at 50%. The tosses are called 'independent events' which means that the coin can't remember what has happened to it. While twenty heads in a row is unlikely, once you have nineteen heads in a row, the unlikely event has already happened. The potential twentieth head has the same probability as the first head. Another way of looking at it is that any sequence of twenty tosses is unlikely as twenty heads in a row, even if it looks random. But you have to write down the sequence before you start tossing to see if you get it!
Does that help?"
it might when I look with fresh eyes tomorrow..right now I'm tired stubborn and not functioning properly |
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"
I get that..The individual persons chance is always the same.
I'm stuck on how so many tries effects the overall chance of it being opened.
"
It doesn't...
"
or to continue your lottery analogy, does selling thousands of lottery tickets increase the chances of the lottery being won, doesn't matter who by. just if it gets won or not. If that makes sense? "
It depends how many different combinations are purchased...it is no different really. A lot of people buy more than one ticket so already have two chances, despite how poor their chances may be. And the jackpot can rollover, and people tend to buy more tickets during a rollover...
In the game/con you described, everyone has one chance without knowing which combinations had already been attempted... |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"Ok, copied off another site I'm not allowed to link to:
if you toss a coin nineteen times and it comes up heads each time, then it is not more likely that the next toss will be a tail. The odds stay the same, at 50%. The tosses are called 'independent events' which means that the coin can't remember what has happened to it. While twenty heads in a row is unlikely, once you have nineteen heads in a row, the unlikely event has already happened. The potential twentieth head has the same probability as the first head. Another way of looking at it is that any sequence of twenty tosses is unlikely as twenty heads in a row, even if it looks random. But you have to write down the sequence before you start tossing to see if you get it!
Does that help?
it might when I look with fresh eyes tomorrow..right now I'm tired stubborn and not functioning properly "
Like I said, I'll try find a better explanation tomorrow.
My example above isn't of this exact scenario, but it is designed to illustrate why things that seem like they should be likely or unlikely, statistical, may not be. |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"
I get that..The individual persons chance is always the same.
I'm stuck on how so many tries effects the overall chance of it being opened.
It doesn't...
or to continue your lottery analogy, does selling thousands of lottery tickets increase the chances of the lottery being won, doesn't matter who by. just if it gets won or not. If that makes sense?
It depends how many different combinations are purchased...it is no different really. A lot of people buy more than one ticket so already have two chances, despite how poor their chances may be. And the jackpot can rollover, and people tend to buy more tickets during a rollover...
In the game/con you described, everyone has one chance without knowing which combinations had already been attempted..."
And the chance resets every time because nobody knows what anyone else chose. |
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"
Not quite...997,001 combinations .
Yeah lol. Its been a long night."
I'm like Rachel Riley.
But minus the great body and looks.
But plus a cock.
Well, if I met her, I'd love to get two from the top and one in the...
Ahem. |
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"
And the chance resets every time because nobody knows what anyone else chose."
Exactly, everyone has the same chance.
In terms of the lottery; if 10,000,000 tickets are bought and 8,000,000 contain unique number sequences then the chances of SOMEONE winning the lottery are 45,000,000/8,000,000 or 1 in 5.63.
Those odds might sound good, but the odds of any individual ticket winning remain the same. |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"
And the chance resets every time because nobody knows what anyone else chose.
Exactly, everyone has the same chance.
In terms of the lottery; if 10,000,000 tickets are bought and 8,000,000 contain unique number sequences then the chances of SOMEONE winning the lottery are 45,000,000/8,000,000 or 1 in 5.63.
Those odds might sound good, but the odds of any individual ticket winning remain the same."
Nobody is debating the individual chance though.
What is confusing people is why the number of people trying doesn't change the overall odds.
It seems like if 2 people try there should be a better chance of the safe being opened.
And if 10 people try, there should be a better chance still someone will get the right combination.
What we're saying is unless people all try unique combinations, the _overall_ chance is 1 in 1 million regardless of how many people try.
I'm struggling to find a way to explain it clearly. |
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"
Nobody is debating the individual chance though.
What is confusing people is why the number of people trying doesn't change the overall odds.
It seems like if 2 people try there should be a better chance of the safe being opened.
And if 10 people try, there should be a better chance still someone will get the right combination.
What we're saying is unless people all try unique combinations, the _overall_ chance is 1 in 1 million regardless of how many people try.
I'm struggling to find a way to explain it clearly."
I think it comes back to the lottery analogy. Millions of people buy lottery tickets, some buy more than one ticket, some buy a few, some buy their own ticket and are also part of a syndicate. Although the odds are 45,000,000 to one against for any individual ticket, the sheer number of tickets bought means the odds of someone winning is (relatively) high.
In the case of the safe there are one million combinations but only three thousand participants each capable of entering a choice only once. |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
Ok...
The probability of something happening is 1-(the probability of it not happening)
sooo
Each guess has a probability of 1/1000000 of being right.
First guess has a probability of 999999/1000000 of being wrong.
After 2 attempts
----------------
chance of failure:
(999999/1000000)^2
chance of success:
1 - ((999999/1000000)^2)
After 3 attempts:
chance of failure:
(999999/1000000)^3
1 - ((999999/1000000)^3)
and so on.
After n attempts,
chance of failure:
((999999/1000000)^n)
chance of success:
1 - ((999999/1000000)^n)
But I'm really going to have to come back to this after sleep. |
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"Ok, copied off another site I'm not allowed to link to:
if you toss a coin nineteen times and it comes up heads each time, then it is not more likely that the next toss will be a tail. The odds stay the same, at 50%. The tosses are called 'independent events' which means that the coin can't remember what has happened to it. While twenty heads in a row is unlikely, once you have nineteen heads in a row, the unlikely event has already happened. The potential twentieth head has the same probability as the first head. Another way of looking at it is that any sequence of twenty tosses is unlikely as twenty heads in a row, even if it looks random. But you have to write down the sequence before you start tossing to see if you get it!
Does that help?"
20 heads in a row is unlikely.
But tossing a coin 20 times, there's a high probability you'll toss a tail at some point, just as there's a high possibility you'll toss a head. The coin doesn't remember what it already did. As an independent event, yes there is a 50% chance of tossing a tail. But in 20 tosses you're more likely to toss a tail than not. So what are the chances of tossing a tail in those 20 tosses? More than 50% because there are more opportunities to toss a tail than if you were to only toss it once. |
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"
And the chance resets every time because nobody knows what anyone else chose.
Exactly, everyone has the same chance.
In terms of the lottery; if 10,000,000 tickets are bought and 8,000,000 contain unique number sequences then the chances of SOMEONE winning the lottery are 45,000,000/8,000,000 or 1 in 5.63.
Those odds might sound good, but the odds of any individual ticket winning remain the same."
Yes that's what I mean. What is the chance of SOMEONE opening the safe? |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"Ok, copied off another site I'm not allowed to link to:
if you toss a coin nineteen times and it comes up heads each time, then it is not more likely that the next toss will be a tail. The odds stay the same, at 50%. The tosses are called 'independent events' which means that the coin can't remember what has happened to it. While twenty heads in a row is unlikely, once you have nineteen heads in a row, the unlikely event has already happened. The potential twentieth head has the same probability as the first head. Another way of looking at it is that any sequence of twenty tosses is unlikely as twenty heads in a row, even if it looks random. But you have to write down the sequence before you start tossing to see if you get it!
Does that help?
20 heads in a row is unlikely.
But tossing a coin 20 times, there's a high probability you'll toss a tail at some point, just as there's a high possibility you'll toss a head. The coin doesn't remember what it already did. As an independent event, yes there is a 50% chance of tossing a tail. But in 20 tosses you're more likely to toss a tail than not. So what are the chances of tossing a tail in those 20 tosses? More than 50% because there are more opportunities to toss a tail than if you were to only toss it once. "
Now I've redefined the question to myself I've changed my answer. See above.
The overall chance does change.
I was answering based on individual chance, not overall chance. Even after reading you meant overall chance, it hadn't quite made it through to my brain.
I think my last answer is right. It has a name. It might be binomial something or other. I'll look again after sleep. |
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"Ok...
The probability of something happening is 1-(the probability of it not happening)
sooo
Each guess has a probability of 1/1000000 of being right.
First guess has a probability of 999999/1000000 of being wrong.
After 2 attempts
----------------
chance of failure:
(999999/1000000)^2
chance of success:
1 - ((999999/1000000)^2)
After 3 attempts:
chance of failure:
(999999/1000000)^3
1 - ((999999/1000000)^3)
and so on.
After n attempts,
chance of failure:
((999999/1000000)^n)
chance of success:
1 - ((999999/1000000)^n)
But I'm really going to have to come back to this after sleep."
After 3000 attempts
chance of success:
1-((999999/1000000)^3000)= 0.0029955059921347
Chance of failure:
((999999/1000000)^3000)= 0.9970044940078653 |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"Ok...
The probability of something happening is 1-(the probability of it not happening)
sooo
Each guess has a probability of 1/1000000 of being right.
First guess has a probability of 999999/1000000 of being wrong.
After 2 attempts
----------------
chance of failure:
(999999/1000000)^2
chance of success:
1 - ((999999/1000000)^2)
After 3 attempts:
chance of failure:
(999999/1000000)^3
1 - ((999999/1000000)^3)
and so on.
After n attempts,
chance of failure:
((999999/1000000)^n)
chance of success:
1 - ((999999/1000000)^n)
But I'm really going to have to come back to this after sleep.
After 3000 attempts
chance of success:
1-((999999/1000000)^3000)= 0.0029955059921347
Chance of failure:
((999999/1000000)^3000)= 0.9970044940078653"
I think so at the moment.
I'll look again tomorrow. |
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"Ok...
The probability of something happening is 1-(the probability of it not happening)
sooo
Each guess has a probability of 1/1000000 of being right.
First guess has a probability of 999999/1000000 of being wrong.
After 2 attempts
----------------
chance of failure:
(999999/1000000)^2
chance of success:
1 - ((999999/1000000)^2)
After 3 attempts:
chance of failure:
(999999/1000000)^3
1 - ((999999/1000000)^3)
and so on.
After n attempts,
chance of failure:
((999999/1000000)^n)
chance of success:
1 - ((999999/1000000)^n)
But I'm really going to have to come back to this after sleep.
After 3000 attempts
chance of success:
1-((999999/1000000)^3000)= 0.0029955059921347
Chance of failure:
((999999/1000000)^3000)= 0.9970044940078653
"
So there's a 99.7% chance of any of the 3000 people guessing incorrectly (the over all chances)
And a 0.3% chance of any of them guessing correctly
And a 0.001 chance of an individual guessing correctly
Is that right? |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"Ok...
The probability of something happening is 1-(the probability of it not happening)
sooo
Each guess has a probability of 1/1000000 of being right.
First guess has a probability of 999999/1000000 of being wrong.
After 2 attempts
----------------
chance of failure:
(999999/1000000)^2
chance of success:
1 - ((999999/1000000)^2)
After 3 attempts:
chance of failure:
(999999/1000000)^3
1 - ((999999/1000000)^3)
and so on.
After n attempts,
chance of failure:
((999999/1000000)^n)
chance of success:
1 - ((999999/1000000)^n)
But I'm really going to have to come back to this after sleep.
After 3000 attempts
chance of success:
1-((999999/1000000)^3000)= 0.0029955059921347
Chance of failure:
((999999/1000000)^3000)= 0.9970044940078653
So there's a 99.7% chance of any of the 300 people guessing correctly (the over all chances)
And a 0.3% chance of any of them guessing.
And a 0.001 chance of an individual guessing it?
Is that right? "
99.7% chance nobody will guess it.
0.3% chance someone will guess it. |
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"Ok...
The probability of something happening is 1-(the probability of it not happening)
sooo
Each guess has a probability of 1/1000000 of being right.
First guess has a probability of 999999/1000000 of being wrong.
After 2 attempts
----------------
chance of failure:
(999999/1000000)^2
chance of success:
1 - ((999999/1000000)^2)
After 3 attempts:
chance of failure:
(999999/1000000)^3
1 - ((999999/1000000)^3)
and so on.
After n attempts,
chance of failure:
((999999/1000000)^n)
chance of success:
1 - ((999999/1000000)^n)
But I'm really going to have to come back to this after sleep.
After 3000 attempts
chance of success:
1-((999999/1000000)^3000)= 0.0029955059921347
Chance of failure:
((999999/1000000)^3000)= 0.9970044940078653
So there's a 99.7% chance of any of the 300 people guessing correctly (the over all chances)
And a 0.3% chance of any of them guessing.
And a 0.001 chance of an individual guessing it?
Is that right?
99.7% chance nobody will guess it.
0.3% chance someone will guess it."
I realised my mistake and changed it. I'm also tired |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"Ok...
The probability of something happening is 1-(the probability of it not happening)
sooo
Each guess has a probability of 1/1000000 of being right.
First guess has a probability of 999999/1000000 of being wrong.
After 2 attempts
----------------
chance of failure:
(999999/1000000)^2
chance of success:
1 - ((999999/1000000)^2)
After 3 attempts:
chance of failure:
(999999/1000000)^3
1 - ((999999/1000000)^3)
and so on.
After n attempts,
chance of failure:
((999999/1000000)^n)
chance of success:
1 - ((999999/1000000)^n)
But I'm really going to have to come back to this after sleep.
After 3000 attempts
chance of success:
1-((999999/1000000)^3000)= 0.0029955059921347
Chance of failure:
((999999/1000000)^3000)= 0.9970044940078653
So there's a 99.7% chance of any of the 300 people guessing correctly (the over all chances)
And a 0.3% chance of any of them guessing.
And a 0.001 chance of an individual guessing it?
Is that right?
99.7% chance nobody will guess it.
0.3% chance someone will guess it.
I realised my mistake and changed it. I'm also tired "
That's if it is 3000 people.
The overall probability changes with the number of attempts made, obviously. |
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"Ok...
The probability of something happening is 1-(the probability of it not happening)
sooo
Each guess has a probability of 1/1000000 of being right.
First guess has a probability of 999999/1000000 of being wrong.
After 2 attempts
----------------
chance of failure:
(999999/1000000)^2
chance of success:
1 - ((999999/1000000)^2)
After 3 attempts:
chance of failure:
(999999/1000000)^3
1 - ((999999/1000000)^3)
and so on.
After n attempts,
chance of failure:
((999999/1000000)^n)
chance of success:
1 - ((999999/1000000)^n)
But I'm really going to have to come back to this after sleep.
After 3000 attempts
chance of success:
1-((999999/1000000)^3000)= 0.0029955059921347
Chance of failure:
((999999/1000000)^3000)= 0.9970044940078653
So there's a 99.7% chance of any of the 300 people guessing correctly (the over all chances)
And a 0.3% chance of any of them guessing.
And a 0.001 chance of an individual guessing it?
Is that right?
99.7% chance nobody will guess it.
0.3% chance someone will guess it.
I realised my mistake and changed it. I'm also tired
That's if it is 3000 people.
The overall probability changes with the number of attempts made, obviously."
Yes because of a million people tried, the possibility would be 1.
If 500000 people tried, the possibility would be 0.5. Etc |
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|
By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"Ok...
The probability of something happening is 1-(the probability of it not happening)
sooo
Each guess has a probability of 1/1000000 of being right.
First guess has a probability of 999999/1000000 of being wrong.
After 2 attempts
----------------
chance of failure:
(999999/1000000)^2
chance of success:
1 - ((999999/1000000)^2)
After 3 attempts:
chance of failure:
(999999/1000000)^3
1 - ((999999/1000000)^3)
and so on.
After n attempts,
chance of failure:
((999999/1000000)^n)
chance of success:
1 - ((999999/1000000)^n)
But I'm really going to have to come back to this after sleep.
After 3000 attempts
chance of success:
1-((999999/1000000)^3000)= 0.0029955059921347
Chance of failure:
((999999/1000000)^3000)= 0.9970044940078653
So there's a 99.7% chance of any of the 300 people guessing correctly (the over all chances)
And a 0.3% chance of any of them guessing.
And a 0.001 chance of an individual guessing it?
Is that right?
99.7% chance nobody will guess it.
0.3% chance someone will guess it.
I realised my mistake and changed it. I'm also tired
That's if it is 3000 people.
The overall probability changes with the number of attempts made, obviously.
Yes because of a million people tried, the possibility would be 1.
If 500000 people tried, the possibility would be 0.5. Etc"
No, only if each used a unique combination would that be the case. |
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"Ok...
The probability of something happening is 1-(the probability of it not happening)
sooo
Each guess has a probability of 1/1000000 of being right.
First guess has a probability of 999999/1000000 of being wrong.
After 2 attempts
----------------
chance of failure:
(999999/1000000)^2
chance of success:
1 - ((999999/1000000)^2)
After 3 attempts:
chance of failure:
(999999/1000000)^3
1 - ((999999/1000000)^3)
and so on.
After n attempts,
chance of failure:
((999999/1000000)^n)
chance of success:
1 - ((999999/1000000)^n)
But I'm really going to have to come back to this after sleep.
After 3000 attempts
chance of success:
1-((999999/1000000)^3000)= 0.0029955059921347
Chance of failure:
((999999/1000000)^3000)= 0.9970044940078653
So there's a 99.7% chance of any of the 300 people guessing correctly (the over all chances)
And a 0.3% chance of any of them guessing.
And a 0.001 chance of an individual guessing it?
Is that right?
99.7% chance nobody will guess it.
0.3% chance someone will guess it.
I realised my mistake and changed it. I'm also tired
That's if it is 3000 people.
The overall probability changes with the number of attempts made, obviously.
Yes because of a million people tried, the possibility would be 1.
If 500000 people tried, the possibility would be 0.5. Etc
No, only if each used a unique combination would that be the case."
Oh yeah. Oops.
Your calculation that led me to 99.7% chance of failure. Is that a unique number each or not? |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"Ok...
The probability of something happening is 1-(the probability of it not happening)
sooo
Each guess has a probability of 1/1000000 of being right.
First guess has a probability of 999999/1000000 of being wrong.
After 2 attempts
----------------
chance of failure:
(999999/1000000)^2
chance of success:
1 - ((999999/1000000)^2)
After 3 attempts:
chance of failure:
(999999/1000000)^3
1 - ((999999/1000000)^3)
and so on.
After n attempts,
chance of failure:
((999999/1000000)^n)
chance of success:
1 - ((999999/1000000)^n)
But I'm really going to have to come back to this after sleep.
After 3000 attempts
chance of success:
1-((999999/1000000)^3000)= 0.0029955059921347
Chance of failure:
((999999/1000000)^3000)= 0.9970044940078653
So there's a 99.7% chance of any of the 300 people guessing correctly (the over all chances)
And a 0.3% chance of any of them guessing.
And a 0.001 chance of an individual guessing it?
Is that right?
99.7% chance nobody will guess it.
0.3% chance someone will guess it.
I realised my mistake and changed it. I'm also tired
That's if it is 3000 people.
The overall probability changes with the number of attempts made, obviously.
Yes because of a million people tried, the possibility would be 1.
If 500000 people tried, the possibility would be 0.5. Etc
No, only if each used a unique combination would that be the case.
Oh yeah. Oops.
Your calculation that led me to 99.7% chance of failure. Is that a unique number each or not? "
No, that's with them not knowing what anyone else chose and being able to duplicate combinations previously tried. |
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"Your calculation that led me to 99.7% chance of failure. Is that a unique number each or not?
No, that's with them not knowing what anyone else chose and being able to duplicate combinations previously tried."
So if they did know the others' choices, would the probability be 3%? |
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By (user no longer on site)
over a year ago
|
"I'd just blow the bloody door off 
Don't ask me to calculate the probability you'd blow the money into confetti too!  "
you seen robocop? Clarence boddika's lads burn the money on one heist |
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"Your calculation that led me to 99.7% chance of failure. Is that a unique number each or not?
No, that's with them not knowing what anyone else chose and being able to duplicate combinations previously tried.
So if they did know the others' choices, would the probability be 3%? "
Sorry, 3.333r% |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"Your calculation that led me to 99.7% chance of failure. Is that a unique number each or not?
No, that's with them not knowing what anyone else chose and being able to duplicate combinations previously tried.
So if they did know the others' choices, would the probability be 3%?
Sorry, 3.333r% "
Honestly... no idea right now!
Not sure I can spell my own name! |
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"Your calculation that led me to 99.7% chance of failure. Is that a unique number each or not?
No, that's with them not knowing what anyone else chose and being able to duplicate combinations previously tried.
So if they did know the others' choices, would the probability be 3%?
Sorry, 3.333r%
Honestly... no idea right now!
Not sure I can spell my own name!"
Go to sleep! I'm off now anyway |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
"Your calculation that led me to 99.7% chance of failure. Is that a unique number each or not?
No, that's with them not knowing what anyone else chose and being able to duplicate combinations previously tried.
So if they did know the others' choices, would the probability be 3%?
Sorry, 3.333r%
Honestly... no idea right now!
Not sure I can spell my own name!
Go to sleep! I'm off now anyway "
Tried. Failed. |
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By (user no longer on site)
over a year ago
|
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings. "
Very high, I set the combination, and I'm having first choice |
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By (user no longer on site)
over a year ago
|
3,628,800 possible combinations.
So that's the odds on one person getting it.
If 300 people try then each have that same chance......so it should be 1:12,096 odds.
But even if 3,628,800 people tried, they may not get it as some may repeat numbers. Or several/many of them may get it. That's the way odds work! |
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By (user no longer on site)
over a year ago
|
"3,628,800 possible combinations.
I understand the first million ways of re-arranging 6 0-9 options, if you can name even a 1000001st I'll give you the money myself.
Mr ddc"
Just thought I'd make up some crap...but give it big numbers so it sounds feasible.
Of course the answer is 10 to the power of six combinations. But bet it made a few people consider it? |
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"
Mr ddc
Just thought I'd make up some crap...but give it big numbers so it sounds feasible.
"
Oh you bugger, you won't believe the length of time I spent trying to work out how you came up with that figure.
|
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
OK, I *think* this is right!
If nobody guessing knows the previous combinations tried, so combinations are not unique:
Each try is a win or lose scenario, (which is a binomial distribution problem), with a probability of winning of 1/1000000 = 0.000001, or 0.0001%
Assuming exactly 3000 attempts are made, the chance of exactly one of those being the correct combination is 0.2991%
I *think* if only unique codes were used each time, the chance of someone guessing correctly would be 0.3%, so not much better.
But I'm a chemist, not a mathematician! |
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By *ee VianteWoman
over a year ago
Somewhere in North Norfolk |
A binomial experiment is a statistical experiment that has the following properties:
-The experiment consists of n repeated trials.
-Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure.
-The probability of success, denoted by P, is the same on every trial.
-The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.
There is more info on this on a site called Stat Trek if you search for Binomial Probability Distribution.
They also have a Binomial Calculator into which you put the probability of success from one attempt, the number of attempts you make and the number of successes you are looking for the probability of having, (in the case above, 0.000001, 3000 and 1), and it'll work it out for you. |
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By (user no longer on site)
over a year ago
|
"A safe with a combination lock 0-9 with a 6 digit code.
3000 people have ONE chance each to guess the code and open the safe.
1) What are the odds of one person guessing the combination? (3 marks)
2) What are the odds that one of the 3000 people will guess the combination? Assuming none of them know which combinations have already been tried. (3 marks)
Show your workings. "
the odds are a lot less than me finding a sexy fb on here lol |
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By (user no longer on site)
over a year ago
|
"I don't need to be a mathematician to work out the answer is bring a sledge hammer "
Sledgehammers don't work. Most digital safes have a weak point in the production process, identify that and you can easily get inside.
Apparently that is. I read it somewhere  |
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