## RE: [skeptical] Digest Number 178

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• Hi, In many probability problems, it s easier to calculate the odds of an event NOT happening, and subtracting from 1: The probability of an astrologer
Message 1 of 4 , Feb 10, 2002
Hi,

In many probability problems, it's easier to calculate the odds of an event
NOT happening, and subtracting from 1:

The probability of an astrologer matching one person to one of 12 horoscopes
based on pure chance is 1 in 12, or 8.33%.

The probability of him getting one person wrong is therefore 11 in 12. The
probability of getting 2 people wrong is (11/12)^2, or roughly 84%.

The probability of him getting all 12 people wrong is (11/12)^12, which
works out to a bit over 35%. Therefore, the probability of him getting AT
LEAST one correct is the complement of that, i.e., 1 - (11/12)^12, that is,
65%.

So you're right in that his chances were better than 50-50 to get at least
one right, but they're not *that* much better.

Cheers,

Rony

-----Original Message-----
From: skeptical@yahoogroups.com [mailto:skeptical@yahoogroups.com]
Sent: Saturday, February 09, 2002 4:44 PM
To: skeptical@yahoogroups.com
Subject: [skeptical] Digest Number 178

There is 1 message in this issue.

Topics in this digest:

1. probability math
From: "pentatonika2000" <pentatonika@...>

________________________________________________________________________
________________________________________________________________________

Message: 1
Date: Sat, 09 Feb 2002 06:54:21 -0000
From: "pentatonika2000" <pentatonika@...>
Subject: probability math

for 12 unidentified people, meet those 12 people, and then guess
which was which.
The astrologer didn't make a single correct guess.

That surprised me, because I thought that the chances were greater
than 50% in favor of getting at least one correct guess.

But I think I was wrong. I began calculating the formula for the
chances of getting at least one correct match given the number of
people being matched. Here's what I came up with:

Given 1 person, the chances are 1 out of 1.
Given 2 persons, the chances are 1 out of 2.
Given 3 persons, the chances are 4 out of 6.
Given 4 persons, the chances are 6 out of 16.

And that's as far as I figured.
I can't discern a pattern from this data, and it would be quite
laborious to figure out the chances given 5 persons.
Does anything already know what the formula is?
Or does anything see a formula from this data?

Thomas Robertson
http://www.pentatonika.com

________________________________________________________________________
________________________________________________________________________

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
• ... The above statement is correct, but... ... I believe that this one is not. These guesses are not truly independent since you don t have 12 to choose from
Message 2 of 4 , Feb 10, 2002
> In many probability problems, it's easier to calculate the odds of an event
> NOT happening, and subtracting from 1:

The above statement is correct, but...

>
> The probability of an astrologer matching one person to one of 12 horoscopes
> based on pure chance is 1 in 12, or 8.33%.

>
> The probability of him getting one person wrong is therefore 11 in 12. The
> probability of getting 2 people wrong is (11/12)^2, or roughly 84%.

I believe that this one is not. These guesses are not truly independent since
you don't have 12 to choose from once you guess the first. You
now have only 11 left to guess from. Furthermore, the probability of
getting the second one right is influenced (albeit subtly and unknown to you)
by whether or not you got the first one right.

A counteraxample will prove the point: Suppose there were only
two people to guess, A and B.

The probablility of getting the first guess right is 1/2 = .5.

But if you get that one wrong, you *have* to get the other one
wrong too. In other words, if you guess A when it was B, then
necessarily you will guess B when it is A (for the second try).
So the probability of getting at least one right is the same as
the probability of getting either right = probability both right = .5.
Not .5^2 = .25.

It gets hairier as the numbers increase.

You might think that the sequence would be (11/12) x (10/11)
x (9/10) x ... but I don't think that quite works either.
The fact that you got the first one wrong changes the mix a bit.
(I tested this with 4 and was surprised at what came out).

I tried to think of a simple way to solve it, but haven't found one.
A Basic program to enumerate the possible sequences might be
the simplest approach. There are only 479 million orders....

Or perhaps just a simulation. Go into a spreadsheet and
enter the numbers 1 thru 12 in each of two columns.
In the third column put random numbers from a function.

Sort the second and third columns (not the first) by those
random numbers. Now count how many matches there
are. Doing this 10 or 20 times should give a good approximation.

I did this 20 times and got 9 with no matches, 6 with one
match, 2 with 2 matches and 3 with 3 matches. Thus 55%
had at least one match. There's a margin for error here, but
clearly it isn't hard to get one with or without a match.

Ed

>
> The probability of him getting all 12 people wrong is (11/12)^12, which
> works out to a bit over 35%. Therefore, the probability of him getting AT
> LEAST one correct is the complement of that, i.e., 1 - (11/12)^12, that is,
> 65%.
>
> So you're right in that his chances were better than 50-50 to get at least
> one right, but they're not *that* much better.
>
> Cheers,
>
> Rony
>
> -----Original Message-----
> From: skeptical@yahoogroups.com [mailto:skeptical@yahoogroups.com]
> Sent: Saturday, February 09, 2002 4:44 PM
> To: skeptical@yahoogroups.com
> Subject: [skeptical] Digest Number 178
>
> There is 1 message in this issue.
>
> Topics in this digest:
>
> 1. probability math
> From: "pentatonika2000" <pentatonika@...>
>
> ________________________________________________________________________
> ________________________________________________________________________
>
> Message: 1
> Date: Sat, 09 Feb 2002 06:54:21 -0000
> From: "pentatonika2000" <pentatonika@...>
> Subject: probability math
>
> for 12 unidentified people, meet those 12 people, and then guess
> which was which.
> The astrologer didn't make a single correct guess.
>
> That surprised me, because I thought that the chances were greater
> than 50% in favor of getting at least one correct guess.
>
> But I think I was wrong. I began calculating the formula for the
> chances of getting at least one correct match given the number of
> people being matched. Here's what I came up with:
>
> Given 1 person, the chances are 1 out of 1.
> Given 2 persons, the chances are 1 out of 2.
> Given 3 persons, the chances are 4 out of 6.
> Given 4 persons, the chances are 6 out of 16.
>
> And that's as far as I figured.
> I can't discern a pattern from this data, and it would be quite
> laborious to figure out the chances given 5 persons.
> Does anything already know what the formula is?
> Or does anything see a formula from this data?
> If so, please write back.
>
> Thomas Robertson
> http://www.pentatonika.com
>
> ________________________________________________________________________
> ________________________________________________________________________
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

--
-----------------------------------------------------------------
Ed Gracely

gracely@...
(For personal non-urgent e-mail and ALL weekend e-mail).

Edward.Gracely@...
(Mainly for work. Personal OK too, if you need to reach me
quickly during the work week.)
• Ed, Thanks for pointing this out - I gave the right answer to the wrong problem... Since I m too lazy to dig out my probability textbook, I wrote a quick and
Message 3 of 4 , Feb 11, 2002
Ed,

Thanks for pointing this out - I gave the right answer to the wrong
problem...

Since I'm too lazy to dig out my probability textbook, I wrote a quick and
dirty simulation.

It seems that the probability of guessing all wrong approaches a limit as N
increases (N=12 for astrology):

N P
3 2/6 (0.3333)
4 9/24 (0.3750)
5 11/30 (0.3667)
6 265/720 (0.3681)
7 1854/5040 (0.3679)
8 14833/40320 (0.3679)
9 133496/362880 (0.3679)
10 1334961/3628800 (0.3679)

Extrapolating to 12 is left as an exercise to the student...

Rony

-----Original Message-----
From: Ed Gracely [mailto:gracely@...]
Sent: Sunday, February 10, 2002 9:10 PM
To: Rony Shapiro
Cc: skeptical@yahoogroups.com; pentatonika@...
Subject: Re: [skeptical] Digest Number 178

> In many probability problems, it's easier to calculate the odds of an
event
> NOT happening, and subtracting from 1:

The above statement is correct, but...

>
> The probability of an astrologer matching one person to one of 12
horoscopes
> based on pure chance is 1 in 12, or 8.33%.

>
> The probability of him getting one person wrong is therefore 11 in 12. The
> probability of getting 2 people wrong is (11/12)^2, or roughly 84%.

I believe that this one is not. These guesses are not truly independent
since
you don't have 12 to choose from once you guess the first. You
now have only 11 left to guess from. Furthermore, the probability of
getting the second one right is influenced (albeit subtly and unknown to
you)
by whether or not you got the first one right.

A counteraxample will prove the point: Suppose there were only
two people to guess, A and B.

The probablility of getting the first guess right is 1/2 = .5.

But if you get that one wrong, you *have* to get the other one
wrong too. In other words, if you guess A when it was B, then
necessarily you will guess B when it is A (for the second try).
So the probability of getting at least one right is the same as
the probability of getting either right = probability both right = .5.
Not .5^2 = .25.

It gets hairier as the numbers increase.

You might think that the sequence would be (11/12) x (10/11)
x (9/10) x ... but I don't think that quite works either.
The fact that you got the first one wrong changes the mix a bit.
(I tested this with 4 and was surprised at what came out).

I tried to think of a simple way to solve it, but haven't found one.
A Basic program to enumerate the possible sequences might be
the simplest approach. There are only 479 million orders....

Or perhaps just a simulation. Go into a spreadsheet and
enter the numbers 1 thru 12 in each of two columns.
In the third column put random numbers from a function.

Sort the second and third columns (not the first) by those
random numbers. Now count how many matches there
are. Doing this 10 or 20 times should give a good approximation.

I did this 20 times and got 9 with no matches, 6 with one
match, 2 with 2 matches and 3 with 3 matches. Thus 55%
had at least one match. There's a margin for error here, but
clearly it isn't hard to get one with or without a match.

Ed

>
> The probability of him getting all 12 people wrong is (11/12)^12, which
> works out to a bit over 35%. Therefore, the probability of him getting AT
> LEAST one correct is the complement of that, i.e., 1 - (11/12)^12, that
is,
> 65%.
>
> So you're right in that his chances were better than 50-50 to get at least
> one right, but they're not *that* much better.
>
> Cheers,
>
> Rony
>
> -----Original Message-----
> From: skeptical@yahoogroups.com [mailto:skeptical@yahoogroups.com]
> Sent: Saturday, February 09, 2002 4:44 PM
> To: skeptical@yahoogroups.com
> Subject: [skeptical] Digest Number 178
>
> There is 1 message in this issue.
>
> Topics in this digest:
>
> 1. probability math
> From: "pentatonika2000" <pentatonika@...>
>
> ________________________________________________________________________
> ________________________________________________________________________
>
> Message: 1
> Date: Sat, 09 Feb 2002 06:54:21 -0000
> From: "pentatonika2000" <pentatonika@...>
> Subject: probability math
>
> for 12 unidentified people, meet those 12 people, and then guess
> which was which.
> The astrologer didn't make a single correct guess.
>
> That surprised me, because I thought that the chances were greater
> than 50% in favor of getting at least one correct guess.
>
> But I think I was wrong. I began calculating the formula for the
> chances of getting at least one correct match given the number of
> people being matched. Here's what I came up with:
>
> Given 1 person, the chances are 1 out of 1.
> Given 2 persons, the chances are 1 out of 2.
> Given 3 persons, the chances are 4 out of 6.
> Given 4 persons, the chances are 6 out of 16.
>
> And that's as far as I figured.
> I can't discern a pattern from this data, and it would be quite
> laborious to figure out the chances given 5 persons.
> Does anything already know what the formula is?
> Or does anything see a formula from this data?
> If so, please write back.
>
> Thomas Robertson
> http://www.pentatonika.com
>
> ________________________________________________________________________
> ________________________________________________________________________
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

--
-----------------------------------------------------------------
Ed Gracely

gracely@...
(For personal non-urgent e-mail and ALL weekend e-mail).

Edward.Gracely@...
(Mainly for work. Personal OK too, if you need to reach me
quickly during the work week.)
• Rony: I received this version, even though you got a return to sender on it. Hmm. Well, in any case, the number your sequence is going towards is probably1/e.
Message 4 of 4 , Feb 12, 2002
Rony:

on it. Hmm.

Well, in any case, the number your sequence is going towards
is probably1/e. This is the probability, in the limit, that all K
trials will be negative given that the probability of a positive
result on any particular trial is 1/K.

So, for 1000 trials, each with success probability = .001,
the probability that all 1,000 will be negative is .999^1000
= .36795...

I'm surprised that this is the same number for the probability
of no hits in the 12 guess situation, but it seems to be right.
Interesting!

Ed

Rony Shapiro wrote:

> Ed,
>
> Thanks for pointing this out - I gave the right answer to the wrong
> problem...
>
> Since I'm too lazy to dig out my probability textbook, I wrote a quick
> and
> dirty simulation.
>
> It seems that the probability of guessing all wrong approaches a limit
> as N
> increases (N=12 for astrology):
>
> N P
> 3 2/6 (0.3333)
> 4 9/24 (0.3750)
> 5 11/30 (0.3667)
> 6 265/720 (0.3681)
> 7 1854/5040 (0.3679)
> 8 14833/40320 (0.3679)
> 9 133496/362880 (0.3679)
> 10 1334961/3628800 (0.3679)
>
> Extrapolating to 12 is left as an exercise to the student...
>
> Rony
>
> -----Original Message-----
> From: Ed Gracely [mailto:gracely@...]
> Sent: Sunday, February 10, 2002 9:10 PM
> To: Rony Shapiro
> Cc: skeptical@yahoogroups.com; pentatonika@...
> Subject: Re: [skeptical] Digest Number 178
>
> > In many probability problems, it's easier to calculate the odds of
> an
> event
> > NOT happening, and subtracting from 1:
>
> The above statement is correct, but...
>
> >
> > The probability of an astrologer matching one person to one of 12
> horoscopes
> > based on pure chance is 1 in 12, or 8.33%.
>
> >
> > The probability of him getting one person wrong is therefore 11 in
> 12. The
> > probability of getting 2 people wrong is (11/12)^2, or roughly 84%.
>
> I believe that this one is not. These guesses are not truly
> independent
> since
> you don't have 12 to choose from once you guess the first. You
> now have only 11 left to guess from. Furthermore, the probability of
> getting the second one right is influenced (albeit subtly and unknown
> to
> you)
> by whether or not you got the first one right.
>
> A counteraxample will prove the point: Suppose there were only
> two people to guess, A and B.
>
> The probablility of getting the first guess right is 1/2 = .5.
>
> But if you get that one wrong, you *have* to get the other one
> wrong too. In other words, if you guess A when it was B, then
> necessarily you will guess B when it is A (for the second try).
> So the probability of getting at least one right is the same as
> the probability of getting either right = probability both right = .5.
>
> Not .5^2 = .25.
>
> It gets hairier as the numbers increase.
>
> You might think that the sequence would be (11/12) x (10/11)
> x (9/10) x ... but I don't think that quite works either.
> The fact that you got the first one wrong changes the mix a bit.
> (I tested this with 4 and was surprised at what came out).
>
> I tried to think of a simple way to solve it, but haven't found one.
> A Basic program to enumerate the possible sequences might be
> the simplest approach. There are only 479 million orders....
>
> Or perhaps just a simulation. Go into a spreadsheet and
> enter the numbers 1 thru 12 in each of two columns.
> In the third column put random numbers from a function.
>
> Sort the second and third columns (not the first) by those
> random numbers. Now count how many matches there
> are. Doing this 10 or 20 times should give a good approximation.
>
> I did this 20 times and got 9 with no matches, 6 with one
> match, 2 with 2 matches and 3 with 3 matches. Thus 55%
> had at least one match. There's a margin for error here, but
> clearly it isn't hard to get one with or without a match.
>
> Ed
>
> >
> > The probability of him getting all 12 people wrong is (11/12)^12,
> which
> > works out to a bit over 35%. Therefore, the probability of him
> getting AT
> > LEAST one correct is the complement of that, i.e., 1 - (11/12)^12,
> that
> is,
> > 65%.
> >
> > So you're right in that his chances were better than 50-50 to get at
> least
> > one right, but they're not *that* much better.
> >
> > Cheers,
> >
> > Rony
> >
> > -----Original Message-----
> > From: skeptical@yahoogroups.com [mailto:skeptical@yahoogroups.com]
> > Sent: Saturday, February 09, 2002 4:44 PM
> > To: skeptical@yahoogroups.com
> > Subject: [skeptical] Digest Number 178
> >
> > There is 1 message in this issue.
> >
> > Topics in this digest:
> >
> > 1. probability math
> > From: "pentatonika2000" <pentatonika@...>
> >
> >
> ________________________________________________________________________
>
> >
> ________________________________________________________________________
>
> >
> > Message: 1
> > Date: Sat, 09 Feb 2002 06:54:21 -0000
> > From: "pentatonika2000" <pentatonika@...>
> > Subject: probability math
> >
> horoscopes
> > for 12 unidentified people, meet those 12 people, and then guess
> > which was which.
> > The astrologer didn't make a single correct guess.
> >
> > That surprised me, because I thought that the chances were greater
> > than 50% in favor of getting at least one correct guess.
> >
> > But I think I was wrong. I began calculating the formula for the
> > chances of getting at least one correct match given the number of
> > people being matched. Here's what I came up with:
> >
> > Given 1 person, the chances are 1 out of 1.
> > Given 2 persons, the chances are 1 out of 2.
> > Given 3 persons, the chances are 4 out of 6.
> > Given 4 persons, the chances are 6 out of 16.
> >
> > And that's as far as I figured.
> > I can't discern a pattern from this data, and it would be quite
> > laborious to figure out the chances given 5 persons.
> > Does anything already know what the formula is?
> > Or does anything see a formula from this data?
> > If so, please write back.
> >
> > Thomas Robertson
> > http://www.pentatonika.com
> >
> >
> ________________________________________________________________________
>
> >
> ________________________________________________________________________
>
> >
> > Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
>
> --
> -----------------------------------------------------------------
> Ed Gracely
>
> gracely@...
> (For personal non-urgent e-mail and ALL weekend e-mail).
>
> Edward.Gracely@...
> (Mainly for work. Personal OK too, if you need to reach me
> quickly during the work week.)
>
>
>
>
>
> Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/

--
-----------------------------------------------------------------
Ed Gracely

gracely@...
(For personal non-urgent e-mail and ALL weekend e-mail).

Edward.Gracely@...
(Mainly for work. Personal OK too, if you need to reach me
quickly during the work week.)

--
-----------------------------------------------------------------
Ed Gracely

gracely@...
(For personal non-urgent e-mail and ALL weekend e-mail).

Edward.Gracely@...
(Mainly for work. Personal OK too, if you need to reach me
quickly during the work week.)
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