- 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

I recently read about an astrologer who was asked to write 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/ > In many probability problems, it's easier to calculate the odds of an event

The above statement is correct, but...

> NOT happening, and subtracting from 1:

>

I believe that this one is not. These guesses are not truly independent since

> 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%.

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

>

> I recently read about an astrologer who was asked to write 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.)- 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...

>

horoscopes

> The probability of an astrologer matching one person to one of 12

> based on pure chance is 1 in 12, or 8.33%.

I believe that this one is not. These guesses are not truly independent

>

> 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%.

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

>

is,

> 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

> 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

>

> I recently read about an astrologer who was asked to write 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.) - 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. 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

> >

> > I recently read about an astrologer who was asked to write

> 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.)

>

> ------------------------ Yahoo! Groups Sponsor

>

>

>

>

> 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.)