Lets say the phone book pages are numbered from 1 - 1000. And assume that the contact is definitely there in the book.
So when you randomly flip open a book there are 1 of 500 choices that it is one one of the two pages that appear before you.
So the probability is 1/500.
Next Question : How many attempts will it take for me to be certain that the contact is there?
Well you can NEVER be 100% sure .... for obvious reasons :)
Next Question : So how many attempts will it take for me to be 50% sure that the contact is found in one of the two pages that show up?
The 50% is called percent confidence in statistics lingo.
This is best solved by the difference from 1 for the probability of the contact NOT being on the page.
For each attempt to fail we need to multiply 499/500 . So for n attempts to fail it is (499/500)^n. So the probability of it being successful in n attempts is {1-(499/500)^n}
So you need to solve the equation :
{1-(499/500)^n} = .5
which is obviously :
(499/500)^n = .5
And taking log to the base 499/500 on both sides (using python) :
n = 347 attempts
double verify :
So when you randomly flip open a book there are 1 of 500 choices that it is one one of the two pages that appear before you.
So the probability is 1/500.
Next Question : How many attempts will it take for me to be certain that the contact is there?
Well you can NEVER be 100% sure .... for obvious reasons :)
Next Question : So how many attempts will it take for me to be 50% sure that the contact is found in one of the two pages that show up?
The 50% is called percent confidence in statistics lingo.
This is best solved by the difference from 1 for the probability of the contact NOT being on the page.
For each attempt to fail we need to multiply 499/500 . So for n attempts to fail it is (499/500)^n. So the probability of it being successful in n attempts is {1-(499/500)^n}
So you need to solve the equation :
{1-(499/500)^n} = .5
which is obviously :
(499/500)^n = .5
And taking log to the base 499/500 on both sides (using python) :
>>> math.log(.5,499/500.0) 346.22690104949118
n = 347 attempts
double verify :
>>> math.pow(499/500.0,346.22690104949118) 0.5
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