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Bill H
science forum beginner

Joined: 11 May 2006
Posts: 9

Posted: Thu May 11, 2006 2:16 pm    Post subject: probability of the same password

I work some probability problems each morning to wake my mind up. I'm
working through The Probability Tutoring Book: an Intuitive Course for
Engineers and Scientists by Carol Ash, 1993, because it has worked
solutions to all the problems. This morning I have to disagree with
one of the solutions. Here's the problem,

"In a certain computer system you must identify yourself with a
password consisting of a single letter or a letter followed by as many
as 6 symbols which may be letters or digits, for example, Z, ZZZZZZ6,
RUNNER, JIMBO, R2D2. Assuming that any password is as likely to be
chosen as any other, what is the probability that John and Mary choose

My approach was to consider the two independent events "John chooses a
password" and "Mary chooses a password", find the corresponding
probabilities, they should be the same, then multiply. But the
solution begins with the sentence,

"Assume John has chosen his password . . . ."

What??? That's an answer to a totally different problem, right?!
Michael Zedeler
science forum beginner

Joined: 29 Nov 2005
Posts: 17

Posted: Thu May 11, 2006 5:39 pm    Post subject: Re: probability of the same password

Bill H wrote:
 Quote: I work some probability problems each morning to wake my mind up. I'm working through The Probability Tutoring Book: an Intuitive Course for Engineers and Scientists by Carol Ash, 1993, because it has worked solutions to all the problems. This morning I have to disagree with one of the solutions. Here's the problem, "In a certain computer system you must identify yourself with a password consisting of a single letter or a letter followed by as many as 6 symbols which may be letters or digits, for example, Z, ZZZZZZ6, RUNNER, JIMBO, R2D2. Assuming that any password is as likely to be chosen as any other, what is the probability that John and Mary choose the same password?" My approach was to consider the two independent events "John chooses a password" and "Mary chooses a password", find the corresponding probabilities, they should be the same, then multiply. But the solution begins with the sentence, "Assume John has chosen his password . . . ." What??? That's an answer to a totally different problem, right?!

Try specifying exactly where you think there is an error in the proposed
solution.

Regards,

Michael.
--
Which is more dangerous? TV guided missiles or TV guided families?
Get my vcard at http://michael.zedeler.dk/vcard.vcf
Dan Akers

Joined: 19 Jul 2005
Posts: 56

Mike1170

Joined: 17 Sep 2005
Posts: 74

Posted: Fri May 12, 2006 4:34 am    Post subject: Re: probability of the same password

"D. Akers" wrote in message
 Quote: Yes, there are two different ways to approach this problem and two different answers that may be obtained. As a simplified example, with each choosing a one digit passnumber, with just 10 single digits, 0 through 9, to choose from, you could ask: "What is the probability that two individuals, A and B, will choose the same passnumber? OR, you could ask: "Given that A has passnumber X, what is the probability that B will choose the same passnumber? In the first case, it's 0.1^2 or 1%, in the second, it's 10%. I hope that helps with your problem...

Good point but for your math to hold the first question should be reworded
to
"... will choose the same given passnumber?" Your math suggests that you
want the probability that A and B both choose the same passnumber as chosen
by a 3rd party (or just a given passnumber) but your question as worded
merely
asks for the probability that their passnumbers match which is 10% and
equivalent
to your 2nd problem.
Mike1170

Joined: 17 Sep 2005
Posts: 74

Posted: Fri May 12, 2006 4:45 am    Post subject: Re: probability of the same password

"Bill H" <whowells@yahoo.com> wrote in message
 Quote: I work some probability problems each morning to wake my mind up. I'm working through The Probability Tutoring Book: an Intuitive Course for Engineers and Scientists by Carol Ash, 1993, because it has worked solutions to all the problems. This morning I have to disagree with one of the solutions. Here's the problem, "In a certain computer system you must identify yourself with a password consisting of a single letter or a letter followed by as many as 6 symbols which may be letters or digits, for example, Z, ZZZZZZ6, RUNNER, JIMBO, R2D2. Assuming that any password is as likely to be chosen as any other, what is the probability that John and Mary choose the same password?" My approach was to consider the two independent events "John chooses a password" and "Mary chooses a password", find the corresponding probabilities, they should be the same, then multiply. But the solution begins with the sentence, "Assume John has chosen his password . . . ." What??? That's an answer to a totally different problem, right?!

No, same problem.
Assume there are 10^7 different passwords. The joint sample space consists
of 10^7 x 10^7 sample point of which 10^7 consist of matching pairs.
Therefore, the probability their passwords match is 10^7 / 10^7 x 10^7 = 1
/ 10^7. You might as well assume that A or B picks first then ask what the
probability that the other's pick matches the first,
as Carol does. (You have to get up pretty early to catch Carol Ash in a
mistake. I know the book well. Love her visual explanation/proof of
Buffon's needle problem.)
Bill H
science forum beginner

Joined: 11 May 2006
Posts: 9

 Posted: Mon May 15, 2006 1:48 pm    Post subject: Re: probability of the same password Suppose that the probability of choosing a password is the same as the original problem, but the author presents the following two setups: 1a. "John and Mary attend the first computer lab of the year. They are instructed to create new passwords. What is the probability that John and Mary choose the same password?" 1b. "John and Mary attend the second computer lab of the year. Mary missed the first lab and must create a new password. What is the probability that John and Mary choose the same password?" Do you think these are the same probabilities?
Mike1170

Joined: 17 Sep 2005
Posts: 74

Posted: Tue May 16, 2006 2:55 am    Post subject: Re: probability of the same password

Bill H wrote:
 Quote: Suppose that the probability of choosing a password is the same as the original problem, but the author presents the following two setups: 1a. "John and Mary attend the first computer lab of the year. They are instructed to create new passwords. What is the probability that John and Mary choose the same password?" 1b. "John and Mary attend the second computer lab of the year. Mary missed the first lab and must create a new password. What is the probability that John and Mary choose the same password?" Do you think these are the same probabilities?

Yes.
Michael Zedeler
science forum beginner

Joined: 29 Nov 2005
Posts: 17

Posted: Tue May 16, 2006 10:04 am    Post subject: Re: probability of the same password

Bill H wrote:
 Quote: Suppose that the probability of choosing a password is the same as the original problem, but the author presents the following two setups: 1a. "John and Mary attend the first computer lab of the year. They are instructed to create new passwords. What is the probability that John and Mary choose the same password?"

Lets just write down a table. Assume there is four different passwords.

John | Mary
------+------
1 | 1
1 | 2
1 | 3
1 | 4
2 | 1
2 | 2
2 | 3
2 | 4
3 | 1
3 | 2
3 | 3
3 | 4
4 | 1
4 | 2
4 | 3
4 | 4

Now all you have to do is to count the number of equal pairs and divide
by the total count. The equal pairs are (1,1), (2,2), (3,3) and (4,4).
The total is 16. This gives 1/4 = 25%.

 Quote: 1b. "John and Mary attend the second computer lab of the year. Mary missed the first lab and must create a new password. What is the probability that John and Mary choose the same password?"

The time that the passwords are chosen has no influence on the outcome.
Assume there is four different passwords.

John | Mary
------+------
1 | 1
1 | 2
1 | 3
1 | 4
2 | 1
2 | 2
2 | 3
2 | 4
3 | 1
3 | 2
3 | 3
3 | 4
4 | 1
4 | 2
4 | 3
4 | 4

Now all you have to do is to count the number of equal pairs and divide
by the total count. The equal pairs are (1,1), (2,2), (3,3) and (4,4).
The total is 16. This gives 1/4 = 25%.

 Quote: Do you think these are the same probabilities?

Yeps.

Regards,

Michael.
--
Which is more dangerous? TV guided missiles or TV guided families?
Get my vcard at http://michael.zedeler.dk/vcard.vcf
Bill H
science forum beginner

Joined: 11 May 2006
Posts: 9

 Posted: Wed May 17, 2006 2:20 pm    Post subject: Re: probability of the same password You are correct. I was thinking that 1a is the joint probability Pr(M=x_i, J=x_j) and 1b the conditional Pr(M=x_i | J=x_j) where x_i and x_j are elements of the sample space of all possible passwords, say numbered 1 . . . N. Then Pr(M=x_i, J=x_j) = Pr(M=x_i)*Pr(J=x_j) by independence, and Pr(M=x_i | J=x_j) = Pr(M=x_i)*Pr(J=x_j) / Pr(J=x_j) by independence, which canceling equals Pr(M=x_i), which in general shouldn't equal the joint Pr(M=x_i)*Pr(J=x_j). But for the event "x_i equals x_j", the joint probability is the double sum from i=1 to N and j=1 to N of Pr(M=x_i)*Pr(J=x_j) = (1/N)*(1/N) times N "favorable" outcomes of x_i=x_j, just equals 1/N, same as Pr(M=x_i), assuming EQUALLY LIKELY PASSWORDS. However, with unequally likely passwords, this isn't true. For example, suppose John and Mary choose from the sample space {1, 2, 2, 4}. Then the Pr(M=x_i, J=x_j) = 6/16 = 3/8 (for x_i=x_j), while Pr(M=x_i | J=x_j) = 1/4 (for x_i=x_j). So Carol Ash was wise to say "assuming that any password is as likely to be chosen as any other". She definitely knows her stuff!

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