--- On Mon, 10/26/09, ujjwalkotdiya <ujjwalkotdiya@...> wrote:



From: ujjwalkotdiya <ujjwalkotdiya@...>
Subject: [MATH for FUN] Re: maths project
To: mathforfun@yahoogroups.com
Date: Monday, October 26, 2009, 6:15 AM
just work onit theoritically i.e do some question on this from chapters height
and distances
based on that use a single string that is connecting edge of pole ,edge of hadow
& light souurce .
just use another pole &keep it parallel with another and from this measure angle
b/w shadow and string 
by using that angle and length of shadow calculate length of pole   

 




Recent Activity


 1
New MembersVisit Your Group



Give Back
Yahoo! for Good
Get inspired
by a good cause.

Y! Toolbar
Get it Free!
easy 1-click access
to your groups.

Yahoo! Groups
Start a group
in 3 easy steps.
Connect with others.
.


















[Non-text portions of this message have been removed]

--- On Mon, 26/10/09, ujjwal kotdiya <ujjwalkotdiya@...> wrote:

From: ujjwal kotdiya <ujjwalkotdiya@...>
Subject: Re: [MATH for FUN] Re: maths project
To: mathforfun@yahoogroups.com
Date: Monday, 26 October, 2009, 6:54 PM






 









--- On Mon, 10/26/09, ujjwalkotdiya <ujjwalkotdiya@ yahoo.com> wrote:



From: ujjwalkotdiya <ujjwalkotdiya@ yahoo.com>

Subject: [MATH for FUN] Re: maths project

To: mathforfun@yahoogro ups.com

Date: Monday, October 26, 2009, 6:15 AM

just work onit theoritically i.e do some question on this from chapters height
and distances

based on that use a single string that is connecting edge of pole ,edge of hadow
& light souurce .

just use another pole &keep it parallel with another and from this measure angle
b/w shadow and string 

by using that angle and length of shadow calculate length of pole




 



Recent Activity



 1

New MembersVisit Your Group



Give Back

Yahoo! for Good

Get inspired

by a good cause.



Y! Toolbar

Get it Free!

easy 1-click access

to your groups.



Yahoo! Groups

Start a group

in 3 easy steps.

Connect with others.

.



[Non-text portions of this message have been removed]





























       Add whatever you love to the Yahoo! India homepage. Try now!
http://in.yahoo.com/trynew

[Non-text portions of this message have been removed]

any project or working model on probability for school level

Give a step by step proof that \sum_{i=0}^n (-a-b+1)^{n-i} *\binom{{n}{i}}
*(a^i+b^i)\neq 1 for non zero  integers a and b where n>2


Have a safe and happy Halloween and All Saints Day.
bq

Please disregard my last post. I noticed a huge flaw when I got home. I'll
re-post the problem using shift operator notation. A few definitions first.
x^nE^i=E^ix^n E is the shift operator. E^ix^n*k^j=x^n*k^(i+j). The problem is
FLT in terms of Jackson's difference fan (the "J-transform").
(E-((a^n+b^n)^(1/n))+1)^n(a^0+b^0) not= to 1 where a and b are non zero integers
and n>2 And that's it.

Welcome back Clooneman!

Thank you!

--- In mathforfun@yahoogroups.com, bqllpd <no_reply@...> wrote:
>
> Please disregard my last post. I noticed a huge flaw when I got home. I'll
re-post the problem using shift operator notation. A few definitions first.
x^nE^i=E^ix^n E is the shift operator. E^ix^n*k^j=x^n*k^(i+j). The problem is
FLT in terms of Jackson's difference fan (the "J-transform").
> (E-((a^n+b^n)^(1/n))+1)^n(a^0+b^0) not= to 1 where a and b are non zero
integers and n>2 And that's it.
>
> Welcome back Clooneman!
>

Why is this guy targeting us?  Does he think we have trouble attracting women?

Most women light right up when I start talking about the differences between
aleph-null and aleph-one :)

Ed




________________________________
From: ensolution <no_reply@yahoogroups.com>
To: mathforfun@yahoogroups.com
Sent: Wed, November 4, 2009 12:15:18 PM
Subject: [MATH for FUN] How To Attract Women

 
When are women the most receptive to meeting a guy?
Do you want to know the answer... Yeh.. Why not Follow the link below.
http://buildournation.ning.com/profiles/blogs/how-to-attract-women-by-paul

Yours most friendly,
Don Sugath







[Non-text portions of this message have been removed]

--- In mathforfun@yahoogroups.com, Edward Moore <emoore06905@...> wrote:
>
> Why is this guy targeting us?  Does he think we have trouble attracting women?
>
> Most women light right up when I start talking about the differences between
aleph-null and aleph-one :)
>
> Ed
>
>
>
>
> ________________________________
> From: ensolution <no_reply@yahoogroups.com>
> To: mathforfun@yahoogroups.com
> Sent: Wed, November 4, 2009 12:15:18 PM
> Subject: [MATH for FUN] How To Attract Women
>
>  
> When are women the most receptive to meeting a guy?
> Do you want to know the answer... Yeh.. Why not Follow the link below.
> http://buildournation.ning.com/profiles/blogs/how-to-attract-women-by-paul
>
> Yours most friendly,
> Don Sugath
>
>
>
>
>
>
>
> [Non-text portions of this message have been removed]
>

Yes he does! One hundred and one percent! :)

--- In mathforfun@yahoogroups.com, Edward Moore <emoore06905@...> wrote:
>
> Why is this guy targeting us?  Does he think we have trouble attracting women?
>
> Most women light right up when I start talking about the differences between
aleph-null and aleph-one :)
>
> Ed
>
>
>
>
> ________________________________
> From: ensolution <no_reply@yahoogroups.com>
> To: mathforfun@yahoogroups.com
> Sent: Wed, November 4, 2009 12:15:18 PM
> Subject: [MATH for FUN] How To Attract Women
>
>  
> When are women the most receptive to meeting a guy?
> Do you want to know the answer... Yeh.. Why not Follow the link below.
> http://buildournation.ning.com/profiles/blogs/how-to-attract-women-by-paul
>
> Yours most friendly,
> Don Sugath
>
>
>
>
>
>
>
> [Non-text portions of this message have been removed]
>
Special, wheneveer some of them realise what number is between! :):)

Here's one for you all to put your brains together about. A number is chosen
between 1 and... say, 10, *inclusive*. Two players then take it in turns to
guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.

Example (correct answer: 6)
A: 3
"Higher"
B: 8
"Lower"
A: 5
"Higher"
B: 6
Correct, B wins

So... Is there an optimal strategy to improve a player's chances of winning?

Yes, there is. It is well-known in computer science. See
http://en.wikipedia.org/wiki/Binary_search_algorithm#Number_guessing_game

-- Rick

--- On Thu, 11/12/09, clooneman2000 <clooneman2000@...> wrote:

From: clooneman2000 <clooneman2000@...>
Subject: [MATH for FUN] Optimal strategy for number-guessing game
To: mathforfun@yahoogroups.com
Date: Thursday, November 12, 2009, 11:45 AM

Here's one for you all to put your brains together about. A number is chosen
between 1 and... say, 10, *inclusive*. Two players then take it in turns to
guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.

Example (correct answer: 6)
A: 3
"Higher"
B: 8
"Lower"
A: 5
"Higher"
B: 6
Correct, B wins

So... Is there an optimal strategy to improve a player's chances of winning?



------------------------------------

Yahoo! Groups Links





[Non-text portions of this message have been removed]

I only quickly scanned through that site, but this looks slightly different than
the well-known puzzle - this looks like there are three people, one who picks a
number, and the other two taking turns to see who gets the correct number
first.  Is the strategy for playing this variant the same as the two-player
version?

Ed




________________________________
From: Rick <rcastrap@...>
To: mathforfun@yahoogroups.com
Sent: Thu, November 12, 2009 2:53:09 PM
Subject: Re: [MATH for FUN] Optimal strategy for number-guessing game

 
Yes, there is. It is well-known in computer science. See
http://en.wikipedia.org/wiki/Binary_search_algorithm#Number_guessing_game

-- Rick

--- On Thu, 11/12/09, clooneman2000 <clooneman2000@ yahoo.com> wrote:

From: clooneman2000 <clooneman2000@ yahoo.com>
Subject: [MATH for FUN] Optimal strategy for number-guessing game
To: mathforfun@yahoogro ups.com
Date: Thursday, November 12, 2009, 11:45 AM

Here's one for you all to put your brains together about. A number is chosen
between 1 and... say, 10, *inclusive*. Two players then take it in turns to
guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.

Example (correct answer: 6)
A: 3
"Higher"
B: 8
"Lower"
A: 5
"Higher"
B: 6
Correct, B wins

So... Is there an optimal strategy to improve a player's chances of winning?

------------ --------- --------- ------

Yahoo! Groups Links

[Non-text portions of this message have been removed]







[Non-text portions of this message have been removed]

hi,

I need to take a math 20 placement test at my home community collage, I took
math back home in Lebanon and was instructed in French and therefore I am not
quite sure what math 20 constitutes ...please help



________________________________
From: Rick <rcastrap@...>
To: mathforfun@yahoogroups.com
Sent: Thu, November 12, 2009 11:53:09 AM
Subject: Re: [MATH for FUN] Optimal strategy for number-guessing game


Yes, there is. It is well-known in computer science. See http://en.wikipedia
.org/wiki/ Binary_search_ algorithm# Number_guessing_ game

-- Rick

--- On Thu, 11/12/09, clooneman2000 <clooneman2000@ yahoo.com> wrote:

From: clooneman2000 <clooneman2000@ yahoo.com>
Subject: [MATH for FUN] Optimal strategy for number-guessing game
To: mathforfun@yahoogro ups.com
Date: Thursday, November 12, 2009, 11:45 AM

Here's one for you all to put your brains together about. A number is chosen
between 1 and... say, 10, *inclusive*. Two players then take it in turns to
guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.

Example (correct answer: 6)
A: 3
"Higher"
B: 8
"Lower"
A: 5
"Higher"
B: 6
Correct, B wins

So... Is there an optimal strategy to improve a player's chances of winning?

------------ --------- --------- ------

Yahoo! Groups Links

[Non-text portions of this message have been removed]







[Non-text portions of this message have been removed]

The course numbering system is different at every college, so "Math 20" is
meaningless outside of your college. Look up the course outline on the college
web site, or go to the math dept. office and ask for the course outline.

-- Rick

--- On Thu, 11/12/09, Amal Abdallah <amalchehab@...> wrote:

From: Amal Abdallah <amalchehab@...>
Subject: Re: [MATH for FUN] Optimal strategy for number-guessing game
To: mathforfun@yahoogroups.com
Date: Thursday, November 12, 2009, 12:05 PM

hi,

I need to take a math 20 placement test at my home community collage, I took
math back home in Lebanon and was instructed in French and therefore I am not
quite sure what math 20 constitutes ...please help



________________________________
From: Rick <rcastrap@...>
To: mathforfun@yahoogroups.com
Sent: Thu, November 12, 2009 11:53:09 AM
Subject: Re: [MATH for FUN] Optimal strategy for number-guessing game

 
Yes, there is. It is well-known in computer science. See http://en.wikipedia
.org/wiki/ Binary_search_ algorithm# Number_guessing_ game

-- Rick

--- On Thu, 11/12/09, clooneman2000 <clooneman2000@ yahoo.com> wrote:

From: clooneman2000 <clooneman2000@ yahoo.com>
Subject: [MATH for FUN] Optimal strategy for number-guessing game
To: mathforfun@yahoogro ups.com
Date: Thursday, November 12, 2009, 11:45 AM

Here's one for you all to put your brains together about. A number is chosen
between 1 and... say, 10, *inclusive*. Two players then take it in turns to
guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.

Example (correct answer: 6)
A: 3
"Higher"
B: 8
"Lower"
A: 5
"Higher"
B: 6
Correct, B wins

So... Is there an optimal strategy to improve a player's chances of winning?

------------ --------- --------- ------

Yahoo! Groups Links

[Non-text portions of this message have been removed]





     

[Non-text portions of this message have been removed]



------------------------------------

Yahoo! Groups Links





[Non-text portions of this message have been removed]

Yes this looks like a more difficult and interesting problem - thanks Clooneman.

I haven't solved it yet but consider that after the first player chooses, the
game is formally similar except for 10 being replaced by a lower number. For
example in the above example if 3 isn't the correct number the rest of the game
is formally identical to the original game except with 1,2...,10 being replaced
by 1,2 or by 4,5,...,10 and the players reversing roles. If there are N numbers
left the probability of each being the correct one is 1/N.

So define P(N,n) to be the probability that the player whose turn it is to
choose will ultimately win if there are N consecutive numbers and he decides to
pick the nth of them (n being an ordinal number between 1 and N inclusive)
assuming both players choose optimally thereafter.

It should be possible to write a recursion relation for P(N,n) in terms of lower
values of N then use it to work up to 10.



--- In mathforfun@yahoogroups.com, Edward Moore <emoore06905@...> wrote:
>
> I only quickly scanned through that site, but this looks slightly different
than the well-known puzzle - this looks like there are three people, one who
picks a number, and the other two taking turns to see who gets the correct
number first.  Is the strategy for playing this variant the same as the
two-player version?
>
> Ed
>
>
>
>
> ________________________________
> From: Rick <rcastrap@...>
> To: mathforfun@yahoogroups.com
> Sent: Thu, November 12, 2009 2:53:09 PM
> Subject: Re: [MATH for FUN] Optimal strategy for number-guessing game
>
>  
> Yes, there is. It is well-known in computer science. See
http://en.wikipedia.org/wiki/Binary_search_algorithm#Number_guessing_game
>
> -- Rick
>
> --- On Thu, 11/12/09, clooneman2000 <clooneman2000@ yahoo.com> wrote:
>
> From: clooneman2000 <clooneman2000@ yahoo.com>
> Subject: [MATH for FUN] Optimal strategy for number-guessing game
> To: mathforfun@yahoogro ups.com
> Date: Thursday, November 12, 2009, 11:45 AM
>
> Here's one for you all to put your brains together about. A number is chosen
between 1 and... say, 10, *inclusive*. Two players then take it in turns to
guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.
>
> Example (correct answer: 6)
> A: 3
> "Higher"
> B: 8
> "Lower"
> A: 5
> "Higher"
> B: 6
> Correct, B wins
>
> So... Is there an optimal strategy to improve a player's chances of winning?
>
> ------------ --------- --------- ------
>
> Yahoo! Groups Links
>
> [Non-text portions of this message have been removed]
>
>
>
>
>
>
>
> [Non-text portions of this message have been removed]
>

That's exactly the approach I was thinking of.

When there are two numbers to choose from, it doesn't matter which number you
choose; you have a 50-50 chance of winning or losing the game.

When there are three consecutive numbers to choose from, things get interesting.
If you choose the number in the middle, you have a 33.3% chance of winning the
game; if the number is wrong, your opponent will guess correctly on his/her next
turn, barring a monumental error. But if you choose either of the other numbers,
your success probability climbs to 66.7%;
- there is a 33.3% probability your guess is correct;
- there is a 33.3% probability your guess is incorrect and your opponent guesses
wrong.
(And of course there is a 33.3% probability your guess is incorrect and your
opponent guesses right.)

When there are four consecutive numbers to choose from, if you choose one of the
numbers at the ends of the selection,
- there is a 25% probability your guess is correct - SUCCESS;
- there is a 75% probability your guess is incorrect, leaving your opponent with
the same options you would have in a 3-number game, and at that point, he has a
66.7% probability of winning.
-- there is a 25% (75% x 33.3%) probability your guess is incorrect and your
opponent guesses correctly - FAILURE;
-- there is a 25% probability your guess is incorrect, your opponent guesses
incorrectly, and you guess correctly from the remaining two numbers - SUCCESS;
-- there is a 25% probability your guess is incorrect, your opponent guesses
incorrectly, and you guess incorrectly from the remaining two numbers, handing
victory to your opponent - FAILURE.

So that particular strategy gives you a 50-50 chance of winning.

When there are four consecutive numbers to choose from, if you don't choose one
of the numbers at the ends of the selection (e.g. if the numbers are 4, 5, 6 and
7 and you choose either 5 or 6),
- there is a 25% probability your guess is correct - SUCCESS;
- there is a 25% probability your guess is incorrect and your opponent has only
one number to choose from (e.g. following on from above, you guess 5 and the
correct answer is 4), giving him/her the win - FAILURE;
- there is a 25% probability your guess is incorrect and your opponent guesses
correctly from the remaining two numbers - FAILURE;
- there is a 25% probability your guess is incorrect and your opponent guesses
incorrectly from the remaining two numbers, giving you victory - SUCCESS.

Again, 50-50.

Now, my head is uneasy from last night. Who wants to have a go at a 5-number
game?

--- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@...> wrote:
>
> Yes this looks like a more difficult and interesting problem - thanks
Clooneman.
>
> I haven't solved it yet but consider that after the first player chooses, the
game is formally similar except for 10 being replaced by a lower number. For
example in the above example if 3 isn't the correct number the rest of the game
is formally identical to the original game except with 1,2...,10 being replaced
by 1,2 or by 4,5,...,10 and the players reversing roles. If there are N numbers
left the probability of each being the correct one is 1/N.
>
> So define P(N,n) to be the probability that the player whose turn it is to
choose will ultimately win if there are N consecutive numbers and he decides to
pick the nth of them (n being an ordinal number between 1 and N inclusive)
assuming both players choose optimally thereafter.
>
> It should be possible to write a recursion relation for P(N,n) in terms of
lower values of N then use it to work up to 10.
>
>
>
> --- In mathforfun@yahoogroups.com, Edward Moore <emoore06905@> wrote:
> >
> > I only quickly scanned through that site, but this looks slightly different
than the well-known puzzle - this looks like there are three people, one who
picks a number, and the other two taking turns to see who gets the correct
number first.  Is the strategy for playing this variant the same as the
two-player version?
> >
> > Ed
> >
> >
> >
> >
> > ________________________________
> > From: Rick <rcastrap@>
> > To: mathforfun@yahoogroups.com
> > Sent: Thu, November 12, 2009 2:53:09 PM
> > Subject: Re: [MATH for FUN] Optimal strategy for number-guessing game
> >
> >  
> > Yes, there is. It is well-known in computer science. See
http://en.wikipedia.org/wiki/Binary_search_algorithm#Number_guessing_game
> >
> > -- Rick
> >
> > --- On Thu, 11/12/09, clooneman2000 <clooneman2000@ yahoo.com> wrote:
> >
> > From: clooneman2000 <clooneman2000@ yahoo.com>
> > Subject: [MATH for FUN] Optimal strategy for number-guessing game
> > To: mathforfun@yahoogro ups.com
> > Date: Thursday, November 12, 2009, 11:45 AM
> >
> > Here's one for you all to put your brains together about. A number is chosen
between 1 and... say, 10, *inclusive*. Two players then take it in turns to
guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.
> >
> > Example (correct answer: 6)
> > A: 3
> > "Higher"
> > B: 8
> > "Lower"
> > A: 5
> > "Higher"
> > B: 6
> > Correct, B wins
> >
> > So... Is there an optimal strategy to improve a player's chances of winning?
> >
> > ------------ --------- --------- ------
> >
> > Yahoo! Groups Links
> >
> > [Non-text portions of this message have been removed]
> >
> >
> >
> >
> >
> >
> >
> > [Non-text portions of this message have been removed]
> >
>

Nope; what I have in mind is two players in competition with one another, not
one person taking repeated guesses.

--- In mathforfun@yahoogroups.com, Rick <rcastrap@...> wrote:
>
> Yes, there is. It is well-known in computer science. See
http://en.wikipedia.org/wiki/Binary_search_algorithm#Number_guessing_game
>
> -- Rick
>
> --- On Thu, 11/12/09, clooneman2000 <clooneman2000@...> wrote:
>
> From: clooneman2000 <clooneman2000@...>
> Subject: [MATH for FUN] Optimal strategy for number-guessing game
> To: mathforfun@yahoogroups.com
> Date: Thursday, November 12, 2009, 11:45 AM
>
> Here's one for you all to put your brains together about. A number is chosen
between 1 and... say, 10, *inclusive*. Two players then take it in turns to
guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.
>
> Example (correct answer: 6)
> A: 3
> "Higher"
> B: 8
> "Lower"
> A: 5
> "Higher"
> B: 6
> Correct, B wins
>
> So... Is there an optimal strategy to improve a player's chances of winning?
>
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>
>
>
> [Non-text portions of this message have been removed]
>

Claim (I have a proof but maybe someone else wants to try to prove or disprove
it):

Say there are N numbers left, this is a generalization from those cases N=2,3,4:

If N is even there's no advantage in choosing any number over any other. The
probability of ultimately winning is 1/2.

If N is odd it's best to choose one of: 1st,3rd,5th,...,N. In that case the
probability of winning is: (N+1)/(2N). If one of the other numbers is chosen the
probability of winning is (N-1)/(2N). This is assuming both players choose
optimally thereafter.

****

So a simple strategy optimal for all values of N would be just to choose the
lowest number possible.

Note the game is fair only if the starting N is even (like 10). If N is odd the
first player has an advantage. When N = 3,7,11,..., it's a mistake to choose the
middle of the range, for example choosing 4 out of {1,2,3,4,5,6,7}. That would
reduce the probability of winning to 3/7.

--- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@...> wrote:
>
> That's exactly the approach I was thinking of.
>
> When there are two numbers to choose from, it doesn't matter which number you
choose; you have a 50-50 chance of winning or losing the game.
>
> When there are three consecutive numbers to choose from, things get
interesting. If you choose the number in the middle, you have a 33.3% chance of
winning the game; if the number is wrong, your opponent will guess correctly on
his/her next turn, barring a monumental error. But if you choose either of the
other numbers, your success probability climbs to 66.7%;
> - there is a 33.3% probability your guess is correct;
> - there is a 33.3% probability your guess is incorrect and your opponent
guesses wrong.
> (And of course there is a 33.3% probability your guess is incorrect and your
opponent guesses right.)
>
> When there are four consecutive numbers to choose from, if you choose one of
the numbers at the ends of the selection,
> - there is a 25% probability your guess is correct - SUCCESS;
> - there is a 75% probability your guess is incorrect, leaving your opponent
with the same options you would have in a 3-number game, and at that point, he
has a 66.7% probability of winning.
> -- there is a 25% (75% x 33.3%) probability your guess is incorrect and your
opponent guesses correctly - FAILURE;
> -- there is a 25% probability your guess is incorrect, your opponent guesses
incorrectly, and you guess correctly from the remaining two numbers - SUCCESS;
> -- there is a 25% probability your guess is incorrect, your opponent guesses
incorrectly, and you guess incorrectly from the remaining two numbers, handing
victory to your opponent - FAILURE.
>
> So that particular strategy gives you a 50-50 chance of winning.
>
> When there are four consecutive numbers to choose from, if you don't choose
one of the numbers at the ends of the selection (e.g. if the numbers are 4, 5, 6
and 7 and you choose either 5 or 6),
> - there is a 25% probability your guess is correct - SUCCESS;
> - there is a 25% probability your guess is incorrect and your opponent has
only one number to choose from (e.g. following on from above, you guess 5 and
the correct answer is 4), giving him/her the win - FAILURE;
> - there is a 25% probability your guess is incorrect and your opponent guesses
correctly from the remaining two numbers - FAILURE;
> - there is a 25% probability your guess is incorrect and your opponent guesses
incorrectly from the remaining two numbers, giving you victory - SUCCESS.
>
> Again, 50-50.
>
> Now, my head is uneasy from last night. Who wants to have a go at a 5-number
game?
>
> --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:
> >
> > Yes this looks like a more difficult and interesting problem - thanks
Clooneman.
> >
> > I haven't solved it yet but consider that after the first player chooses,
the game is formally similar except for 10 being replaced by a lower number. For
example in the above example if 3 isn't the correct number the rest of the game
is formally identical to the original game except with 1,2...,10 being replaced
by 1,2 or by 4,5,...,10 and the players reversing roles. If there are N numbers
left the probability of each being the correct one is 1/N.
> >
> > So define P(N,n) to be the probability that the player whose turn it is to
choose will ultimately win if there are N consecutive numbers and he decides to
pick the nth of them (n being an ordinal number between 1 and N inclusive)
assuming both players choose optimally thereafter.
> >
> > It should be possible to write a recursion relation for P(N,n) in terms of
lower values of N then use it to work up to 10.
> >
> >
> >
> > --- In mathforfun@yahoogroups.com, Edward Moore <emoore06905@> wrote:
> > >
> > > I only quickly scanned through that site, but this looks slightly
different than the well-known puzzle - this looks like there are three people,
one who picks a number, and the other two taking turns to see who gets the
correct number first.  Is the strategy for playing this variant the same as the
two-player version?
> > >
> > > Ed
> > >
> > >
> > >
> > >
> > > ________________________________
> > > From: Rick <rcastrap@>
> > > To: mathforfun@yahoogroups.com
> > > Sent: Thu, November 12, 2009 2:53:09 PM
> > > Subject: Re: [MATH for FUN] Optimal strategy for number-guessing game
> > >
> > >  
> > > Yes, there is. It is well-known in computer science. See
http://en.wikipedia.org/wiki/Binary_search_algorithm#Number_guessing_game
> > >
> > > -- Rick
> > >
> > > --- On Thu, 11/12/09, clooneman2000 <clooneman2000@ yahoo.com> wrote:
> > >
> > > From: clooneman2000 <clooneman2000@ yahoo.com>
> > > Subject: [MATH for FUN] Optimal strategy for number-guessing game
> > > To: mathforfun@yahoogro ups.com
> > > Date: Thursday, November 12, 2009, 11:45 AM
> > >
> > > Here's one for you all to put your brains together about. A number is
chosen between 1 and... say, 10, *inclusive*. Two players then take it in turns
to guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.
> > >
> > > Example (correct answer: 6)
> > > A: 3
> > > "Higher"
> > > B: 8
> > > "Lower"
> > > A: 5
> > > "Higher"
> > > B: 6
> > > Correct, B wins
> > >
> > > So... Is there an optimal strategy to improve a player's chances of
winning?
> > >
> > > ------------ --------- --------- ------
> > >
> > > Yahoo! Groups Links
> > >
> > > [Non-text portions of this message have been removed]
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > [Non-text portions of this message have been removed]
> > >
> >
>

Hmmm... I like it. Even when I was wondering to myself that perhaps choosing
from the extremities of the selection was the best way to go. I won't be
attmpting to prove or disprove it, but I'd be interested in seeing yours!

--- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@...> wrote:
>
> Claim (I have a proof but maybe someone else wants to try to prove or disprove
it):
>
> Say there are N numbers left, this is a generalization from those cases
N=2,3,4:
>
> If N is even there's no advantage in choosing any number over any other. The
probability of ultimately winning is 1/2.
>
> If N is odd it's best to choose one of: 1st,3rd,5th,...,N. In that case the
probability of winning is: (N+1)/(2N). If one of the other numbers is chosen the
probability of winning is (N-1)/(2N). This is assuming both players choose
optimally thereafter.
>
> ****
>
> So a simple strategy optimal for all values of N would be just to choose the
lowest number possible.
>
> Note the game is fair only if the starting N is even (like 10). If N is odd
the first player has an advantage. When N = 3,7,11,..., it's a mistake to choose
the middle of the range, for example choosing 4 out of {1,2,3,4,5,6,7}. That
would reduce the probability of winning to 3/7.
>
> --- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@> wrote:
> >
> > That's exactly the approach I was thinking of.
> >
> > When there are two numbers to choose from, it doesn't matter which number
you choose; you have a 50-50 chance of winning or losing the game.
> >
> > When there are three consecutive numbers to choose from, things get
interesting. If you choose the number in the middle, you have a 33.3% chance of
winning the game; if the number is wrong, your opponent will guess correctly on
his/her next turn, barring a monumental error. But if you choose either of the
other numbers, your success probability climbs to 66.7%;
> > - there is a 33.3% probability your guess is correct;
> > - there is a 33.3% probability your guess is incorrect and your opponent
guesses wrong.
> > (And of course there is a 33.3% probability your guess is incorrect and your
opponent guesses right.)
> >
> > When there are four consecutive numbers to choose from, if you choose one of
the numbers at the ends of the selection,
> > - there is a 25% probability your guess is correct - SUCCESS;
> > - there is a 75% probability your guess is incorrect, leaving your opponent
with the same options you would have in a 3-number game, and at that point, he
has a 66.7% probability of winning.
> > -- there is a 25% (75% x 33.3%) probability your guess is incorrect and your
opponent guesses correctly - FAILURE;
> > -- there is a 25% probability your guess is incorrect, your opponent guesses
incorrectly, and you guess correctly from the remaining two numbers - SUCCESS;
> > -- there is a 25% probability your guess is incorrect, your opponent guesses
incorrectly, and you guess incorrectly from the remaining two numbers, handing
victory to your opponent - FAILURE.
> >
> > So that particular strategy gives you a 50-50 chance of winning.
> >
> > When there are four consecutive numbers to choose from, if you don't choose
one of the numbers at the ends of the selection (e.g. if the numbers are 4, 5, 6
and 7 and you choose either 5 or 6),
> > - there is a 25% probability your guess is correct - SUCCESS;
> > - there is a 25% probability your guess is incorrect and your opponent has
only one number to choose from (e.g. following on from above, you guess 5 and
the correct answer is 4), giving him/her the win - FAILURE;
> > - there is a 25% probability your guess is incorrect and your opponent
guesses correctly from the remaining two numbers - FAILURE;
> > - there is a 25% probability your guess is incorrect and your opponent
guesses incorrectly from the remaining two numbers, giving you victory -
SUCCESS.
> >
> > Again, 50-50.
> >
> > Now, my head is uneasy from last night. Who wants to have a go at a 5-number
game?
> >
> > --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:
> > >
> > > Yes this looks like a more difficult and interesting problem - thanks
Clooneman.
> > >
> > > I haven't solved it yet but consider that after the first player chooses,
the game is formally similar except for 10 being replaced by a lower number. For
example in the above example if 3 isn't the correct number the rest of the game
is formally identical to the original game except with 1,2...,10 being replaced
by 1,2 or by 4,5,...,10 and the players reversing roles. If there are N numbers
left the probability of each being the correct one is 1/N.
> > >
> > > So define P(N,n) to be the probability that the player whose turn it is to
choose will ultimately win if there are N consecutive numbers and he decides to
pick the nth of them (n being an ordinal number between 1 and N inclusive)
assuming both players choose optimally thereafter.
> > >
> > > It should be possible to write a recursion relation for P(N,n) in terms of
lower values of N then use it to work up to 10.
> > >
> > >
> > >
> > > --- In mathforfun@yahoogroups.com, Edward Moore <emoore06905@> wrote:
> > > >
> > > > I only quickly scanned through that site, but this looks slightly
different than the well-known puzzle - this looks like there are three people,
one who picks a number, and the other two taking turns to see who gets the
correct number first.  Is the strategy for playing this variant the same as the
two-player version?
> > > >
> > > > Ed
> > > >
> > > >
> > > >
> > > >
> > > > ________________________________
> > > > From: Rick <rcastrap@>
> > > > To: mathforfun@yahoogroups.com
> > > > Sent: Thu, November 12, 2009 2:53:09 PM
> > > > Subject: Re: [MATH for FUN] Optimal strategy for number-guessing game
> > > >
> > > >  
> > > > Yes, there is. It is well-known in computer science. See
http://en.wikipedia.org/wiki/Binary_search_algorithm#Number_guessing_game
> > > >
> > > > -- Rick
> > > >
> > > > --- On Thu, 11/12/09, clooneman2000 <clooneman2000@ yahoo.com> wrote:
> > > >
> > > > From: clooneman2000 <clooneman2000@ yahoo.com>
> > > > Subject: [MATH for FUN] Optimal strategy for number-guessing game
> > > > To: mathforfun@yahoogro ups.com
> > > > Date: Thursday, November 12, 2009, 11:45 AM
> > > >
> > > > Here's one for you all to put your brains together about. A number is
chosen between 1 and... say, 10, *inclusive*. Two players then take it in turns
to guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.
> > > >
> > > > Example (correct answer: 6)
> > > > A: 3
> > > > "Higher"
> > > > B: 8
> > > > "Lower"
> > > > A: 5
> > > > "Higher"
> > > > B: 6
> > > > Correct, B wins
> > > >
> > > > So... Is there an optimal strategy to improve a player's chances of
winning?
> > > >
> > > > ------------ --------- --------- ------
> > > >
> > > > Yahoo! Groups Links
> > > >
> > > > [Non-text portions of this message have been removed]
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > [Non-text portions of this message have been removed]
> > > >
> > >
> >
>

I am thinking that it is either you or your opponent who wins.
Since any number is equally likely to occur, it doesn't matter which number you
choose - your chances of winning are the same.

Your strategy should thus be to reduce the  probability of your opponent winning
if you guess wrong so they get a chance to guess.

The way to minimise their chances, is to maximise the number of choices they
have.

You should thus choose/guess either the smallest or largest existing number.

Does that sound like good logic??

The original question [see below] asked "So... Is there an optimal strategy to
improve a player's chances of winning?"

Now if the player, whose chances of winning you wish to improve, is in fact your
opponent - then you should choose the middle number, thus minimising the number
of choices they will have.

Peter


--- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@...> wrote:
>
> Here's one for you all to put your brains together about. A number is chosen
between 1 and... say, 10, *inclusive*. Two players then take it in turns to
guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.
>
> Example (correct answer: 6)
> A: 3
> "Higher"
> B: 8
> "Lower"
> A: 5
> "Higher"
> B: 6
> Correct, B wins
>
> So... Is there an optimal strategy to improve a player's chances of winning?
>

Peter's explanation for choosing a number on one of the extremes looks
reasonable - normally after an incorrect guess there's some helpful information
saying it was too high or too low but the advantage of that information all goes
to the opponent so it would be best to deprive him of it.

Anyway, a proof of:

Let P be the probability of ultimately winning by choosing the kth number out of
N consecutive numbers. k being one of the ordinals 1,2,...,N. Assume optimal
play by both sides thereafter.

(1) If N is even:
P = 1/2

(2) If N is odd:
P = (N+1)/(2N) for k odd
P = (N-1)/(2N) for k even

This is proved by induction on N. First, for N=1,P=1 and for N=2,P=1/2 so (1)
and (2) hold for N <= 2.

Now assume both (1) and (2) hold for any set of n < N consecutive numbers and
try to show they hold for N.

The assumption of optimal play comes in when the opponent is presented with an
odd number n of choices. He'll pick optimally (odd) in that case making his
probability of winning (n+1)/2n and the original player's probability of winning
(n-1)/2n.


First suppose N is even. If k is even counting from one end of the range, it's
odd counting from the other, so assume without loss of generality k is even.

There are three possibilities: k correct (Probability 1/N), "lower" (Prob.
(k-1)/N) and "higher" (Probability (N-k)/N). In the second case the opponent is
presented with k-1 (odd) choices, so the probability of the original player
going on to win is ((k-1)-1)/(2(k-1)). In the third case the probability of
winning is 1/2. So in total:

P = 1/N + ((k-1)/N)*((k-1)-1)/(2(k-1)) + ((N-k)/N)*(1/2) = 1/2

QED for (1)


Now suppose N is odd. This is very similar to the previous argument except that
(k-1) and (N-k) are either both even (k odd) or both odd (k even). If they're
both even, the probability of winning is:

1/N + ((k-1)/N)*(1/2) + ((N-k)/N)*(1/2) = (N+1)/(2N)

If they're both odd:

1/N + ((k-1)/N)*((k-1)-1)/(2(k-1)) + ((N-k)/N)*((N-k)-1)/(2(N-k)) = (N-1)/(2N)

QED for (2)




--- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@...> wrote:
>
> Hmmm... I like it. Even when I was wondering to myself that perhaps choosing
from the extremities of the selection was the best way to go. I won't be
attmpting to prove or disprove it, but I'd be interested in seeing yours!
>
> --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:
> >
> > Claim (I have a proof but maybe someone else wants to try to prove or
disprove it):
> >
> > Say there are N numbers left, this is a generalization from those cases
N=2,3,4:
> >
> > If N is even there's no advantage in choosing any number over any other. The
probability of ultimately winning is 1/2.
> >
> > If N is odd it's best to choose one of: 1st,3rd,5th,...,N. In that case the
probability of winning is: (N+1)/(2N). If one of the other numbers is chosen the
probability of winning is (N-1)/(2N). This is assuming both players choose
optimally thereafter.
> >
> > ****
> >
> > So a simple strategy optimal for all values of N would be just to choose the
lowest number possible.
> >
> > Note the game is fair only if the starting N is even (like 10). If N is odd
the first player has an advantage. When N = 3,7,11,..., it's a mistake to choose
the middle of the range, for example choosing 4 out of {1,2,3,4,5,6,7}. That
would reduce the probability of winning to 3/7.
> >
> > --- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@> wrote:
> > >
> > > That's exactly the approach I was thinking of.
> > >
> > > When there are two numbers to choose from, it doesn't matter which number
you choose; you have a 50-50 chance of winning or losing the game.
> > >
> > > When there are three consecutive numbers to choose from, things get
interesting. If you choose the number in the middle, you have a 33.3% chance of
winning the game; if the number is wrong, your opponent will guess correctly on
his/her next turn, barring a monumental error. But if you choose either of the
other numbers, your success probability climbs to 66.7%;
> > > - there is a 33.3% probability your guess is correct;
> > > - there is a 33.3% probability your guess is incorrect and your opponent
guesses wrong.
> > > (And of course there is a 33.3% probability your guess is incorrect and
your opponent guesses right.)
> > >
> > > When there are four consecutive numbers to choose from, if you choose one
of the numbers at the ends of the selection,
> > > - there is a 25% probability your guess is correct - SUCCESS;
> > > - there is a 75% probability your guess is incorrect, leaving your
opponent with the same options you would have in a 3-number game, and at that
point, he has a 66.7% probability of winning.
> > > -- there is a 25% (75% x 33.3%) probability your guess is incorrect and
your opponent guesses correctly - FAILURE;
> > > -- there is a 25% probability your guess is incorrect, your opponent
guesses incorrectly, and you guess correctly from the remaining two numbers -
SUCCESS;
> > > -- there is a 25% probability your guess is incorrect, your opponent
guesses incorrectly, and you guess incorrectly from the remaining two numbers,
handing victory to your opponent - FAILURE.
> > >
> > > So that particular strategy gives you a 50-50 chance of winning.
> > >
> > > When there are four consecutive numbers to choose from, if you don't
choose one of the numbers at the ends of the selection (e.g. if the numbers are
4, 5, 6 and 7 and you choose either 5 or 6),
> > > - there is a 25% probability your guess is correct - SUCCESS;
> > > - there is a 25% probability your guess is incorrect and your opponent has
only one number to choose from (e.g. following on from above, you guess 5 and
the correct answer is 4), giving him/her the win - FAILURE;
> > > - there is a 25% probability your guess is incorrect and your opponent
guesses correctly from the remaining two numbers - FAILURE;
> > > - there is a 25% probability your guess is incorrect and your opponent
guesses incorrectly from the remaining two numbers, giving you victory -
SUCCESS.
> > >
> > > Again, 50-50.
> > >
> > > Now, my head is uneasy from last night. Who wants to have a go at a
5-number game?
> > >
> > > --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:
> > > >
> > > > Yes this looks like a more difficult and interesting problem - thanks
Clooneman.
> > > >
> > > > I haven't solved it yet but consider that after the first player
chooses, the game is formally similar except for 10 being replaced by a lower
number. For example in the above example if 3 isn't the correct number the rest
of the game is formally identical to the original game except with 1,2...,10
being replaced by 1,2 or by 4,5,...,10 and the players reversing roles. If there
are N numbers left the probability of each being the correct one is 1/N.
> > > >
> > > > So define P(N,n) to be the probability that the player whose turn it is
to choose will ultimately win if there are N consecutive numbers and he decides
to pick the nth of them (n being an ordinal number between 1 and N inclusive)
assuming both players choose optimally thereafter.
> > > >
> > > > It should be possible to write a recursion relation for P(N,n) in terms
of lower values of N then use it to work up to 10.
> > > >
> > > >
> > > >
> > > > --- In mathforfun@yahoogroups.com, Edward Moore <emoore06905@> wrote:
> > > > >
> > > > > I only quickly scanned through that site, but this looks slightly
different than the well-known puzzle - this looks like there are three people,
one who picks a number, and the other two taking turns to see who gets the
correct number first.  Is the strategy for playing this variant the same as the
two-player version?
> > > > >
> > > > > Ed
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > ________________________________
> > > > > From: Rick <rcastrap@>
> > > > > To: mathforfun@yahoogroups.com
> > > > > Sent: Thu, November 12, 2009 2:53:09 PM
> > > > > Subject: Re: [MATH for FUN] Optimal strategy for number-guessing game
> > > > >
> > > > >  
> > > > > Yes, there is. It is well-known in computer science. See
http://en.wikipedia.org/wiki/Binary_search_algorithm#Number_guessing_game
> > > > >
> > > > > -- Rick
> > > > >
> > > > > --- On Thu, 11/12/09, clooneman2000 <clooneman2000@ yahoo.com> wrote:
> > > > >
> > > > > From: clooneman2000 <clooneman2000@ yahoo.com>
> > > > > Subject: [MATH for FUN] Optimal strategy for number-guessing game
> > > > > To: mathforfun@yahoogro ups.com
> > > > > Date: Thursday, November 12, 2009, 11:45 AM
> > > > >
> > > > > Here's one for you all to put your brains together about. A number is
chosen between 1 and... say, 10, *inclusive*. Two players then take it in turns
to guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.
> > > > >
> > > > > Example (correct answer: 6)
> > > > > A: 3
> > > > > "Higher"
> > > > > B: 8
> > > > > "Lower"
> > > > > A: 5
> > > > > "Higher"
> > > > > B: 6
> > > > > Correct, B wins
> > > > >
> > > > > So... Is there an optimal strategy to improve a player's chances of
winning?
> > > > >
> > > > > ------------ --------- --------- ------
> > > > >
> > > > > Yahoo! Groups Links
> > > > >
> > > > > [Non-text portions of this message have been removed]
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > [Non-text portions of this message have been removed]
> > > > >
> > > >
> > >
> >
>

The game described is one between two players - you and your opponent, for
example. So yes, it is a scenario where, by definition, one player will win and
one will lose.

The question "Is there an optimal strategy to improve a player's chances of
winning?" refers to an optimal strategy to be adopted by a player for that same
player's victory.

--- In mathforfun@yahoogroups.com, "bogaduck" <pmaxotzen@...> wrote:
>
> I am thinking that it is either you or your opponent who wins.
> Since any number is equally likely to occur, it doesn't matter which number
you choose - your chances of winning are the same.
>
> Your strategy should thus be to reduce the  probability of your opponent
winning if you guess wrong so they get a chance to guess.
>
> The way to minimise their chances, is to maximise the number of choices they
have.
>
> You should thus choose/guess either the smallest or largest existing number.
>
> Does that sound like good logic??
>
> The original question [see below] asked "So... Is there an optimal strategy to
improve a player's chances of winning?"
>
> Now if the player, whose chances of winning you wish to improve, is in fact
your opponent - then you should choose the middle number, thus minimising the
number of choices they will have.
>
> Peter
>
>
> --- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@> wrote:
> >
> > Here's one for you all to put your brains together about. A number is chosen
between 1 and... say, 10, *inclusive*. Two players then take it in turns to
guess the number. When a player guesses correctly, that player wins. When a
player guesses incorrectly, the players are told "higher" or "lower" depending
on whether the correct number is higher or lower than the guess.
> >
> > Example (correct answer: 6)
> > A: 3
> > "Higher"
> > B: 8
> > "Lower"
> > A: 5
> > "Higher"
> > B: 6
> > Correct, B wins
> >
> > So... Is there an optimal strategy to improve a player's chances of winning?
> >
>

Brilliant, sir.

Of course, the disclaimer is that the optimal strategy doesn't guarantee
victory, but it does increase your chances. And, if you have to choose between,
say, 1 and 21, you can choose any odd number in the selection to begin with
(e.g. 9 or 11) to reduce the playing time without reducing the probability of
winning. --Am I right?-- In practice, though, if such a game were to be played
by one of us against someone not among us (readers of this topic), the opponent
would not have an optimal strategy in mind, and we could just pick off the
numbers at the extremities of the selection and let the opponent do all the hard
work, particularly if the length of the selection open to us is even and the
opponent then errs and leaves us with an odd selection length.

--- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@...> wrote:
>
> Peter's explanation for choosing a number on one of the extremes looks
reasonable - normally after an incorrect guess there's some helpful information
saying it was too high or too low but the advantage of that information all goes
to the opponent so it would be best to deprive him of it.
>
> Anyway, a proof of:
>
> Let P be the probability of ultimately winning by choosing the kth number out
of N consecutive numbers. k being one of the ordinals 1,2,...,N. Assume optimal
play by both sides thereafter.
>
> (1) If N is even:
> P = 1/2
>
> (2) If N is odd:
> P = (N+1)/(2N) for k odd
> P = (N-1)/(2N) for k even
>
> This is proved by induction on N. First, for N=1,P=1 and for N=2,P=1/2 so (1)
and (2) hold for N <= 2.
>
> Now assume both (1) and (2) hold for any set of n < N consecutive numbers and
try to show they hold for N.
>
> The assumption of optimal play comes in when the opponent is presented with an
odd number n of choices. He'll pick optimally (odd) in that case making his
probability of winning (n+1)/2n and the original player's probability of winning
(n-1)/2n.
>
>
> First suppose N is even. If k is even counting from one end of the range, it's
odd counting from the other, so assume without loss of generality k is even.
>
> There are three possibilities: k correct (Probability 1/N), "lower" (Prob.
(k-1)/N) and "higher" (Probability (N-k)/N). In the second case the opponent is
presented with k-1 (odd) choices, so the probability of the original player
going on to win is ((k-1)-1)/(2(k-1)). In the third case the probability of
winning is 1/2. So in total:
>
> P = 1/N + ((k-1)/N)*((k-1)-1)/(2(k-1)) + ((N-k)/N)*(1/2) = 1/2
>
> QED for (1)
>
>
> Now suppose N is odd. This is very similar to the previous argument except
that (k-1) and (N-k) are either both even (k odd) or both odd (k even). If
they're both even, the probability of winning is:
>
> 1/N + ((k-1)/N)*(1/2) + ((N-k)/N)*(1/2) = (N+1)/(2N)
>
> If they're both odd:
>
> 1/N + ((k-1)/N)*((k-1)-1)/(2(k-1)) + ((N-k)/N)*((N-k)-1)/(2(N-k)) = (N-1)/(2N)
>
> QED for (2)
>
>
>
>
> --- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@> wrote:
> >
> > Hmmm... I like it. Even when I was wondering to myself that perhaps choosing
from the extremities of the selection was the best way to go. I won't be
attmpting to prove or disprove it, but I'd be interested in seeing yours!
> >
> > --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:
> > >
> > > Claim (I have a proof but maybe someone else wants to try to prove or
disprove it):
> > >
> > > Say there are N numbers left, this is a generalization from those cases
N=2,3,4:
> > >
> > > If N is even there's no advantage in choosing any number over any other.
The probability of ultimately winning is 1/2.
> > >
> > > If N is odd it's best to choose one of: 1st,3rd,5th,...,N. In that case
the probability of winning is: (N+1)/(2N). If one of the other numbers is chosen
the probability of winning is (N-1)/(2N). This is assuming both players choose
optimally thereafter.
> > >
> > > ****
> > >
> > > So a simple strategy optimal for all values of N would be just to choose
the lowest number possible.
> > >
> > > Note the game is fair only if the starting N is even (like 10). If N is
odd the first player has an advantage. When N = 3,7,11,..., it's a mistake to
choose the middle of the range, for example choosing 4 out of {1,2,3,4,5,6,7}.
That would reduce the probability of winning to 3/7.
> > >
> > > --- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@> wrote:
> > > >
> > > > That's exactly the approach I was thinking of.
> > > >
> > > > When there are two numbers to choose from, it doesn't matter which
number you choose; you have a 50-50 chance of winning or losing the game.
> > > >
> > > > When there are three consecutive numbers to choose from, things get
interesting. If you choose the number in the middle, you have a 33.3% chance of
winning the game; if the number is wrong, your opponent will guess correctly on
his/her next turn, barring a monumental error. But if you choose either of the
other numbers, your success probability climbs to 66.7%;
> > > > - there is a 33.3% probability your guess is correct;
> > > > - there is a 33.3% probability your guess is incorrect and your opponent
guesses wrong.
> > > > (And of course there is a 33.3% probability your guess is incorrect and
your opponent guesses right.)
> > > >
> > > > When there are four consecutive numbers to choose from, if you choose
one of the numbers at the ends of the selection,
> > > > - there is a 25% probability your guess is correct - SUCCESS;
> > > > - there is a 75% probability your guess is incorrect, leaving your
opponent with the same options you would have in a 3-number game, and at that
point, he has a 66.7% probability of winning.
> > > > -- there is a 25% (75% x 33.3%) probability your guess is incorrect and
your opponent guesses correctly - FAILURE;
> > > > -- there is a 25% probability your guess is incorrect, your opponent
guesses incorrectly, and you guess correctly from the remaining two numbers -
SUCCESS;
> > > > -- there is a 25% probability your guess is incorrect, your opponent
guesses incorrectly, and you guess incorrectly from the remaining two numbers,
handing victory to your opponent - FAILURE.
> > > >
> > > > So that particular strategy gives you a 50-50 chance of winning.
> > > >
> > > > When there are four consecutive numbers to choose from, if you don't
choose one of the numbers at the ends of the selection (e.g. if the numbers are
4, 5, 6 and 7 and you choose either 5 or 6),
> > > > - there is a 25% probability your guess is correct - SUCCESS;
> > > > - there is a 25% probability your guess is incorrect and your opponent
has only one number to choose from (e.g. following on from above, you guess 5
and the correct answer is 4), giving him/her the win - FAILURE;
> > > > - there is a 25% probability your guess is incorrect and your opponent
guesses correctly from the remaining two numbers - FAILURE;
> > > > - there is a 25% probability your guess is incorrect and your opponent
guesses incorrectly from the remaining two numbers, giving you victory -
SUCCESS.
> > > >
> > > > Again, 50-50.
> > > >
> > > > Now, my head is uneasy from last night. Who wants to have a go at a
5-number game?
> > > >
> > > > --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:
> > > > >
> > > > > Yes this looks like a more difficult and interesting problem - thanks
Clooneman.
> > > > >
> > > > > I haven't solved it yet but consider that after the first player
chooses, the game is formally similar except for 10 being replaced by a lower
number. For example in the above example if 3 isn't the correct number the rest
of the game is formally identical to the original game except with 1,2...,10
being replaced by 1,2 or by 4,5,...,10 and the players reversing roles. If there
are N numbers left the probability of each being the correct one is 1/N.
> > > > >
> > > > > So define P(N,n) to be the probability that the player whose turn it
is to choose will ultimately win if there are N consecutive numbers and he
decides to pick the nth of them (n being an ordinal number between 1 and N
inclusive) assuming both players choose optimally thereafter.
> > > > >
> > > > > It should be possible to write a recursion relation for P(N,n) in
terms of lower values of N then use it to work up to 10.
> > > > >
> > > > >
> > > > >
> > > > > --- In mathforfun@yahoogroups.com, Edward Moore <emoore06905@> wrote:
> > > > > >
> > > > > > I only quickly scanned through that site, but this looks slightly
different than the well-known puzzle - this looks like there are three people,
one who picks a number, and the other two taking turns to see who gets the
correct number first.  Is the strategy for playing this variant the same as the
two-player version?
> > > > > >
> > > > > > Ed
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > > ________________________________
> > > > > > From: Rick <rcastrap@>
> > > > > > To: mathforfun@yahoogroups.com
> > > > > > Sent: Thu, November 12, 2009 2:53:09 PM
> > > > > > Subject: Re: [MATH for FUN] Optimal strategy for number-guessing
game
> > > > > >
> > > > > >  
> > > > > > Yes, there is. It is well-known in computer science. See
http://en.wikipedia.org/wiki/Binary_search_algorithm#Number_guessing_game
> > > > > >
> > > > > > -- Rick
> > > > > >
> > > > > > --- On Thu, 11/12/09, clooneman2000 <clooneman2000@ yahoo.com>
wrote:
> > > > > >
> > > > > > From: clooneman2000 <clooneman2000@ yahoo.com>
> > > > > > Subject: [MATH for FUN] Optimal strategy for number-guessing game
> > > > > > To: mathforfun@yahoogro ups.com
> > > > > > Date: Thursday, November 12, 2009, 11:45 AM
> > > > > >
> > > > > > Here's one for you all to put your brains together about. A number
is chosen between 1 and... say, 10, *inclusive*. Two players then take it in
turns to guess the number. When a player guesses correctly, that player wins.
When a player guesses incorrectly, the players are told "higher" or "lower"
depending on whether the correct number is higher or lower than the guess.
> > > > > >
> > > > > > Example (correct answer: 6)
> > > > > > A: 3
> > > > > > "Higher"
> > > > > > B: 8
> > > > > > "Lower"
> > > > > > A: 5
> > > > > > "Higher"
> > > > > > B: 6
> > > > > > Correct, B wins
> > > > > >
> > > > > > So... Is there an optimal strategy to improve a player's chances of
winning?
> > > > > >
> > > > > > ------------ --------- --------- ------
> > > > > >
> > > > > > Yahoo! Groups Links
> > > > > >
> > > > > > [Non-text portions of this message have been removed]
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > > [Non-text portions of this message have been removed]
> > > > > >
> > > > >
> > > >
> > >
> >
>

--- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@...> wrote:
>
> Brilliant, sir.
>
> Of course, the disclaimer is that the optimal strategy doesn't guarantee
victory, but it does increase your chances. And, if you have to choose between,
say, 1 and 21, you can choose any odd number in the selection to begin with
(e.g. 9 or 11) to reduce the playing time without reducing the probability of
winning. --Am I right?-- In practice, though, if such a game were to be played
by one of us against someone not among us (readers of this topic), the opponent
would not have an optimal strategy in mind, and we could just pick off the
numbers at the extremities of the selection and let the opponent do all the hard
work, particularly if the length of the selection open to us is even and the
opponent then errs and leaves us with an odd selection length.
>

Thanks - yes picking an odd number near the center or even at the exact center
(if it's odd) would be the thing to do to get the game over with quickly without
a disadvantage.

In your original example picking 3 first seems like a good choice since if the
other guy hadn't analyzed the game he might have instinctively picked the center
of the range (4,5,6,7,8,9,10) (i.e. 7) which would have given a slight advantage
to the first player.
(But as a math problem you kind of have to assume the other player will always
play optimally).



> --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:
> >
> > Peter's explanation for choosing a number on one of the extremes looks
reasonable - normally after an incorrect guess there's some helpful information
saying it was too high or too low but the advantage of that information all goes
to the opponent so it would be best to deprive him of it.
> >
> > Anyway, a proof of:
> >
> > Let P be the probability of ultimately winning by choosing the kth number
out of N consecutive numbers. k being one of the ordinals 1,2,...,N. Assume
optimal play by both sides thereafter.
> >
> > (1) If N is even:
> > P = 1/2
> >
> > (2) If N is odd:
> > P = (N+1)/(2N) for k odd
> > P = (N-1)/(2N) for k even
> >
> > This is proved by induction on N. First, for N=1,P=1 and for N=2,P=1/2 so
(1) and (2) hold for N <= 2.
> >
> > Now assume both (1) and (2) hold for any set of n < N consecutive numbers
and try to show they hold for N.
> >
> > The assumption of optimal play comes in when the opponent is presented with
an odd number n of choices. He'll pick optimally (odd) in that case making his
probability of winning (n+1)/2n and the original player's probability of winning
(n-1)/2n.
> >
> >
> > First suppose N is even. If k is even counting from one end of the range,
it's odd counting from the other, so assume without loss of generality k is
even.
> >
> > There are three possibilities: k correct (Probability 1/N), "lower" (Prob.
(k-1)/N) and "higher" (Probability (N-k)/N). In the second case the opponent is
presented with k-1 (odd) choices, so the probability of the original player
going on to win is ((k-1)-1)/(2(k-1)). In the third case the probability of
winning is 1/2. So in total:
> >
> > P = 1/N + ((k-1)/N)*((k-1)-1)/(2(k-1)) + ((N-k)/N)*(1/2) = 1/2
> >
> > QED for (1)
> >
> >
> > Now suppose N is odd. This is very similar to the previous argument except
that (k-1) and (N-k) are either both even (k odd) or both odd (k even). If
they're both even, the probability of winning is:
> >
> > 1/N + ((k-1)/N)*(1/2) + ((N-k)/N)*(1/2) = (N+1)/(2N)
> >
> > If they're both odd:
> >
> > 1/N + ((k-1)/N)*((k-1)-1)/(2(k-1)) + ((N-k)/N)*((N-k)-1)/(2(N-k)) =
(N-1)/(2N)
> >
> > QED for (2)
> >
> >
> >
> >
> > --- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@> wrote:
> > >
> > > Hmmm... I like it. Even when I was wondering to myself that perhaps
choosing from the extremities of the selection was the best way to go. I won't
be attmpting to prove or disprove it, but I'd be interested in seeing yours!
> > >
> > > --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:
> > > >
> > > > Claim (I have a proof but maybe someone else wants to try to prove or
disprove it):
> > > >
> > > > Say there are N numbers left, this is a generalization from those cases
N=2,3,4:
> > > >
> > > > If N is even there's no advantage in choosing any number over any other.
The probability of ultimately winning is 1/2.
> > > >
> > > > If N is odd it's best to choose one of: 1st,3rd,5th,...,N. In that case
the probability of winning is: (N+1)/(2N). If one of the other numbers is chosen
the probability of winning is (N-1)/(2N). This is assuming both players choose
optimally thereafter.
> > > >
> > > > ****
> > > >
> > > > So a simple strategy optimal for all values of N would be just to choose
the lowest number possible.
> > > >
> > > > Note the game is fair only if the starting N is even (like 10). If N is
odd the first player has an advantage. When N = 3,7,11,..., it's a mistake to
choose the middle of the range, for example choosing 4 out of {1,2,3,4,5,6,7}.
That would reduce the probability of winning to 3/7.
> > > >
> > > > --- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@>
wrote:
> > > > >
> > > > > That's exactly the approach I was thinking of.
> > > > >
> > > > > When there are two numbers to choose from, it doesn't matter which
number you choose; you have a 50-50 chance of winning or losing the game.
> > > > >
> > > > > When there are three consecutive numbers to choose from, things get
interesting. If you choose the number in the middle, you have a 33.3% chance of
winning the game; if the number is wrong, your opponent will guess correctly on
his/her next turn, barring a monumental error. But if you choose either of the
other numbers, your success probability climbs to 66.7%;
> > > > > - there is a 33.3% probability your guess is correct;
> > > > > - there is a 33.3% probability your guess is incorrect and your
opponent guesses wrong.
> > > > > (And of course there is a 33.3% probability your guess is incorrect
and your opponent guesses right.)
> > > > >
> > > > > When there are four consecutive numbers to choose from, if you choose
one of the numbers at the ends of the selection,
> > > > > - there is a 25% probability your guess is correct - SUCCESS;
> > > > > - there is a 75% probability your guess is incorrect, leaving your
opponent with the same options you would have in a 3-number game, and at that
point, he has a 66.7% probability of winning.
> > > > > -- there is a 25% (75% x 33.3%) probability your guess is incorrect
and your opponent guesses correctly - FAILURE;
> > > > > -- there is a 25% probability your guess is incorrect, your opponent
guesses incorrectly, and you guess correctly from the remaining two numbers -
SUCCESS;
> > > > > -- there is a 25% probability your guess is incorrect, your opponent
guesses incorrectly, and you guess incorrectly from the remaining two numbers,
handing victory to your opponent - FAILURE.
> > > > >
> > > > > So that particular strategy gives you a 50-50 chance of winning.
> > > > >
> > > > > When there are four consecutive numbers to choose from, if you don't
choose one of the numbers at the ends of the selection (e.g. if the numbers are
4, 5, 6 and 7 and you choose either 5 or 6),
> > > > > - there is a 25% probability your guess is correct - SUCCESS;
> > > > > - there is a 25% probability your guess is incorrect and your opponent
has only one number to choose from (e.g. following on from above, you guess 5
and the correct answer is 4), giving him/her the win - FAILURE;
> > > > > - there is a 25% probability your guess is incorrect and your opponent
guesses correctly from the remaining two numbers - FAILURE;
> > > > > - there is a 25% probability your guess is incorrect and your opponent
guesses incorrectly from the remaining two numbers, giving you victory -
SUCCESS.
> > > > >
> > > > > Again, 50-50.
> > > > >
> > > > > Now, my head is uneasy from last night. Who wants to have a go at a
5-number game?
> > > > >
> > > > > --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@>
wrote:
> > > > > >
> > > > > > Yes this looks like a more difficult and interesting problem -
thanks Clooneman.
> > > > > >
> > > > > > I haven't solved it yet but consider that after the first player
chooses, the game is formally similar except for 10 being replaced by a lower
number. For example in the above example if 3 isn't the correct number the rest
of the game is formally identical to the original game except with 1,2...,10
being replaced by 1,2 or by 4,5,...,10 and the players reversing roles. If there
are N numbers left the probability of each being the correct one is 1/N.
> > > > > >
> > > > > > So define P(N,n) to be the probability that the player whose turn it
is to choose will ultimately win if there are N consecutive numbers and he
decides to pick the nth of them (n being an ordinal number between 1 and N
inclusive) assuming both players choose optimally thereafter.
> > > > > >
> > > > > > It should be possible to write a recursion relation for P(N,n) in
terms of lower values of N then use it to work up to 10.
> > > > > >
> > > > > >
> > > > > >
> > > > > > --- In mathforfun@yahoogroups.com, Edward Moore <emoore06905@>
wrote:
> > > > > > >
> > > > > > > I only quickly scanned through that site, but this looks slightly
different than the well-known puzzle - this looks like there are three people,
one who picks a number, and the other two taking turns to see who gets the
correct number first.  Is the strategy for playing this variant the same as the
two-player version?
> > > > > > >
> > > > > > > Ed
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > ________________________________
> > > > > > > From: Rick <rcastrap@>
> > > > > > > To: mathforfun@yahoogroups.com
> > > > > > > Sent: Thu, November 12, 2009 2:53:09 PM
> > > > > > > Subject: Re: [MATH for FUN] Optimal strategy for number-guessing
game
> > > > > > >
> > > > > > >  
> > > > > > > Yes, there is. It is well-known in computer science. See
http://en.wikipedia.org/wiki/Binary_search_algorithm#Number_guessing_game
> > > > > > >
> > > > > > > -- Rick
> > > > > > >
> > > > > > > --- On Thu, 11/12/09, clooneman2000 <clooneman2000@ yahoo.com>
wrote:
> > > > > > >
> > > > > > > From: clooneman2000 <clooneman2000@ yahoo.com>
> > > > > > > Subject: [MATH for FUN] Optimal strategy for number-guessing game
> > > > > > > To: mathforfun@yahoogro ups.com
> > > > > > > Date: Thursday, November 12, 2009, 11:45 AM
> > > > > > >
> > > > > > > Here's one for you all to put your brains together about. A number
is chosen between 1 and... say, 10, *inclusive*. Two players then take it in
turns to guess the number. When a player guesses correctly, that player wins.
When a player guesses incorrectly, the players are told "higher" or "lower"
depending on whether the correct number is higher or lower than the guess.
> > > > > > >
> > > > > > > Example (correct answer: 6)
> > > > > > > A: 3
> > > > > > > "Higher"
> > > > > > > B: 8
> > > > > > > "Lower"
> > > > > > > A: 5
> > > > > > > "Higher"
> > > > > > > B: 6
> > > > > > > Correct, B wins
> > > > > > >
> > > > > > > So... Is there an optimal strategy to improve a player's chances
of winning?
> > > > > > >
> > > > > > > ------------ --------- --------- ------
> > > > > > >
> > > > > > > Yahoo! Groups Links
> > > > > > >
> > > > > > > [Non-text portions of this message have been removed]
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > [Non-text portions of this message have been removed]
> > > > > > >
> > > > > >
> > > > >
> > > >
> > >
> >
>

The 'numbertheory' Yahoo group was very well-maintained, had a membership of
1200 and the discussion was of a high level of sophistication.

Unfortunately the list owner, Dennis Kluk, died in 2007 and messages ceased to
be forwarded to the list.

The remaining moderator, an elderly friend of Mr Kluk, went to a great deal of
trouble and some expense to obtain the death certficate of Dennis Kluk with a
view to appointing a new list owner so that the group could continue. Ultimately
this proved unsatisfactory for Yahoo, so instead a new group,
Number_Theory_group, has been set up and has already inherited many of the best
contributors from Dennis' group.

To join, please visit
http://tech.groups.yahoo.com/group/Number_Theory_group/ .

List owner, Number_Theory_group

So despite a death certificate being obtained, Yahoo were still not for turning?
Doesn't surprise me, considering the problems I've had myself, although not
remotely as complicated as those endured by Mr Kluk's friend. Best of luck and
evvery success with the group.

--- In mathforfun@yahoogroups.com, "primorial3" <outfinite@...> wrote:
>
> The 'numbertheory' Yahoo group was very well-maintained, had a membership of
1200 and the discussion was of a high level of sophistication.
>
> Unfortunately the list owner, Dennis Kluk, died in 2007 and messages ceased to
be forwarded to the list.
>
> The remaining moderator, an elderly friend of Mr Kluk, went to a great deal of
trouble and some expense to obtain the death certficate of Dennis Kluk with a
view to appointing a new list owner so that the group could continue. Ultimately
this proved unsatisfactory for Yahoo, so instead a new group,
Number_Theory_group, has been set up and has already inherited many of the best
contributors from Dennis' group.
>
> To join, please visit
> http://tech.groups.yahoo.com/group/Number_Theory_group/ .
>
> List owner, Number_Theory_group
>

find free cours and exercies of mathematica and informatica and physice and
chimistry for university student don't mis it go see it now

http://tinyurl.com/yjdqhqu

The worst thing was that they originally suggested a death certificate would be
satisfactory for the new appointment to be feasible.



--- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@...> wrote:
>
> So despite a death certificate being obtained, Yahoo were still not for
turning? Doesn't surprise me, considering the problems I've had myself, although
not remotely as complicated as those endured by Mr Kluk's friend. Best of luck
and evvery success with the group.
>
> --- In mathforfun@yahoogroups.com, "primorial3" <outfinite@> wrote:
> >
> > The 'numbertheory' Yahoo group was very well-maintained, had a membership of
1200 and the discussion was of a high level of sophistication.
> >
> > Unfortunately the list owner, Dennis Kluk, died in 2007 and messages ceased
to be forwarded to the list.
> >
> > The remaining moderator, an elderly friend of Mr Kluk, went to a great deal
of trouble and some expense to obtain the death certficate of Dennis Kluk with a
view to appointing a new list owner so that the group could continue. Ultimately
this proved unsatisfactory for Yahoo, so instead a new group,
Number_Theory_group, has been set up and has already inherited many of the best
contributors from Dennis' group.
> >
> > To join, please visit
> > http://tech.groups.yahoo.com/group/Number_Theory_group/ .
> >
> > List owner, Number_Theory_group
> >
>

Hi everyone..

I have a new blog up for a couple of weeks-ish. It is mainly about my ideas on
philosophy (some contains math related stuff), history, politics, and even
culinary :)

Not much people in my environment, apparently, has noticeable interest in it...
Most of them dont understand >,<

So, please check it out and leave comments... I hope this could be open to
debate :)

The address:
fajrithinkingoutloud.wordpress.com

Thank you all :D

That's horrendous, moving the goalposts like that as we say on this side of the
Atlantic. For shame...

--- In mathforfun@yahoogroups.com, "primorial3" <outfinite@...> wrote:
>
> The worst thing was that they originally suggested a death certificate would
be satisfactory for the new appointment to be feasible.
>
>
>
> --- In mathforfun@yahoogroups.com, "clooneman2000" <clooneman2000@> wrote:
> >
> > So despite a death certificate being obtained, Yahoo were still not for
turning? Doesn't surprise me, considering the problems I've had myself, although
not remotely as complicated as those endured by Mr Kluk's friend. Best of luck
and evvery success with the group.
> >
> > --- In mathforfun@yahoogroups.com, "primorial3" <outfinite@> wrote:
> > >
> > > The 'numbertheory' Yahoo group was very well-maintained, had a membership
of 1200 and the discussion was of a high level of sophistication.
> > >
> > > Unfortunately the list owner, Dennis Kluk, died in 2007 and messages
ceased to be forwarded to the list.
> > >
> > > The remaining moderator, an elderly friend of Mr Kluk, went to a great
deal of trouble and some expense to obtain the death certficate of Dennis Kluk
with a view to appointing a new list owner so that the group could continue.
Ultimately this proved unsatisfactory for Yahoo, so instead a new group,
Number_Theory_group, has been set up and has already inherited many of the best
contributors from Dennis' group.
> > >
> > > To join, please visit
> > > http://tech.groups.yahoo.com/group/Number_Theory_group/ .
> > >
> > > List owner, Number_Theory_group
> > >
> >
>