# Bharat Rakshak

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 Post subject: BR Maths Corner-1Posted: 22 Jul 2008 00:24
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Please use this thread to discuss mathematics, math history, post puzzles and brain teasers. I not there is some interest in knowing problem solving techniques in maths.

Thanks, ramana
--------------
ramana wrote:
AmberG, Considering that the more cerebral folks would like to be given a chance to exercise their math aptitude, would you consider a request to host a maths thread?

Vina's above reply convices me of a need for that. It might have infrequent visitors but would serve a purpose.

ramana

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 00:27
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vina wrote:
Amber G. wrote:
Kalva's page is good. (though, this being this year's problem, will not be there for a while).
Would you like to give brief way to do it?
R- I knew what you meant (that's why -(BTW, the youngest one graduated from MIT this year, so pretty proud etc) ..
But it may amuse you, that the best of madarsa pupils (from the land of the pure - and best of the best math students there) generally score zeros (or number in single digit ) for problems like these...

Here is one solution of 2(b) given by someone -
For any integer k, x=-k(k+1), y= -(k+1)/k^2, z= k/(k+1)^2, one can see that xyz=1 and x^2/(x-1)^2+ ... = 1, and taking k =0,1, etc.. one indeed has infinite rational solutions.

J. You may already know that, but I would say most of the college math profs (say 90%) will have difficulty with IMO problems. FWIW , that applies to best of the best students/ participants.. (that includes ranking IIT-JEE type kids).. that's the beauty of math.

Ah Amber G. you are so cruel. I am no "nuke scientist" who can do Cauchy and Riemann Integral and talk about Bessels Functions and Gamma functions and all that (at least not in any shape over the past decade after defecting to the dark side.. all I can do now is Excel Spreadheets for Finance and "analysis" for "strategic and corporate planning" (whatever that means)), but still your problems draw me like a moth to a flame and against my better judgment put pencil on paper and try to do it.

Well , I did do the problem you posted , of course no way was my solution as brilliant and as elegant as the one you posted.

Mine was a= x +1 , b = y+1 and c = y+1,; and then plugging it into the problem you gave and using the identity xyz =1 , I got 3 + (sum of squares of ) -2 (some terms) . I was able to show that the last two terms were >= -2 and hence the identity >=1 held true for all real values and sort of finessed 2nd part by saying infinite real values also implies infinite rational and irrational solutions, don't know how acceptable that 2nd part would be, my guess it wouldn't be. but hey it was an honest attempt in my middle age , when the edge is gone and rusted from disuse and neglect. (btw , if you plug in k = sqrt 2, sqrt 3 etc, you do get infinite irrational solutions as well , so maybe someone could have bought that ! )

Math might be beautiful, but it is scary and I will run as far away from it as possible , but the "problems " you post exert an inexorable pull like a black hole.

and
Amber G. wrote:
Vina – Here is one method: :
Let p=x/(1-x), q=y/(1-y),r=z/(1-z)
One can see that p+1 = 1/(1-x) etc, and noticing \$xyz=1\$ we have:
pqr = (p+1)(q+1)(r+1) (because both are equal to 1/((1-x)(1-y)(1-z))
= pqr + (pq+qr+rs) + (p+q+r) +1

That is (pq+qr+rs) + 1 = - (p+q+r)
Squaring both sides:
(pq+qr+rs)^2 +2 (pq+qr+rs) + 1 = p^2+q^2+r^2 + 2(pq+qr+rs)
or (pq+qr+rs)^2 +1 = p^2+q^2+r^2
Since (pq+qr+rs)^2 or (any_thing)^2 is always >=0 we have p^2+q^2+r^2 >=1
QED.
Eqaulity holds only when pq+qr+rs=0,
And it is fairy easy then to deduce the part (b) as the equation is same as saying (xy+yz+zx) = 3 or xy+ (1/x)+(1/y)=3, (all you have to do solve the quadratic eq in say y, see if one can make the discriminate a perfect square some how, and you done)
Of course, you are right that there are infinite solutions (Infinite rational, triplets and infinite irrational triplets too) the question only asked to show that there are infinite rational ones...(showing infinite solutions are there, as you say, is quite trivial but infinite real solutions doe NOT automatically imply infinite rational solutions, hence the part (b) )

The problem set, is up on the official IMO web, btw.. solutions are not going to be up for a while.. Indian team, according to rumors, though did better than before may be looking at 1-2 silver medals, and 1-3 bronze.. but one can be wrong.. the results will be out soon.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 02:43
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Ramana – Thanks for starting the thread. I do understand that there are many forums, formal classes , and magazines which have math problems etc… but we have shero shayree thread even though there may be dedicated magazines for that – so why not .. see if any one is interested.

Here I start:

If you post a solution , please make the size tiny Like tiny ) (IOW use size=50 html tag) so that, others, if they don’t like to peek at the solution may skip it. If you have seen the problem before, give it a day or before posting the solution,

The problems here are, for middle school to high-school level, so all can enjoy. (Does not require calculators etc, but feel free to use it if it gives you a hint, same way use calculus or higher math etc.. if that gives you a hint…)

1. Prove that (1/2)(3/4)(5/6)………..(99/100)
is less than (1/10) and greater than (1/15)
For those who have calculators change the last term to say (99999999/100000000) and then show the result to say 4 significant figures.

2. Find at (at least ) 3 proper factors of
2^58+1

3 prove
(cos 20 ) (cos 40) (cos 80) = 1/8

(Cos 20) means cosine of 20 degrees)

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 03:16
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why not find at least 2 proper factors of 2^16 + 1 for a start ?

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 03:33
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Why not indeed, but I think pepins test (http://en.wikipedia.org/wiki/P%C3%A9pin's_test) will find the number prime.

(For those, who don't know, 2^(2^n)+1 are called fermat's primes, which are prime for n=0,1,2,3,4 (3,17,257, 2^16+1=65537) ... when n>4 most such numbers are not prime (actually no one has found any prime of this kind whne n>4 yet .. Euler found that 2^32+1 was not a prime.
Pepins test is a fast way to check )

Last edited by Amber G. on 22 Jul 2008 03:37, edited 1 time in total.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 03:35
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btw, has it been proved that there are infinite number of primes ?
remember reading something on this long back, don't know what the situation is now.
thanks.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 03:41
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Yes, and the proof is really very simple, suppose if (p1,p2,p3,.... p_n) are the only prime numbers.. take \$(p1)(p2)(p3)..(p_n)+1\$ is either prime, or not divisible by any of the primes above (it will leave a remainder or 1) .. in any case you found a prime larger than p_n ...so you found 1 more... QED.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 03:52
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Also x-post from the other thread:
Najunamar wrote:
Amber G. wrote:
Okay, at the risk of being accused of torturer - (But we do have nuclear weapons designers here in this Forum) Today was IMO (Math Olympiad) 'Spain first day of problems This is Problem number 2.

(This is HighSchool Level math, No calculus needed)

2)given xyz=1 and None of x,y,z equal to 1 .

Prove that (x^2)/(x-1)^2+ Y^2/(y-1)^2+z^2/(z-1)^2 >= 1

b) Show that there are infinite number of rational triplets ( x,y,z) which will satisfy the equality.

Try that out, its not that hard.
(The problem wording is not exact --- more from the memory etc..)

This part of Contest ends (problems sets part II) tomorrow. results will be known in a few days.. so wish India team good luck.

Yes, it is not hard - for a good list of IMO type problems and solutions see Kalva's page. The section a - you can split into two cases; in both you'll see there is atleast 1 term >1 and all terms positive. Fairly elementary or ailimentary as Ramana puts it.

Najunamar: would you like to post, how you split this inot two cases etc...?
Thanks.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 04:24
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I would suggest that when someone posts an answer to any problem they explicitly state that "THIS IS AN ANSWER POST" so that aam abduls who intend to solve the problem later with honesty do not accidentally read the answers. Thank you AmberG for this thread, I have a chance to save my H&D in my eyes by trying some problems now.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 04:34
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archan wrote:
I would suggest that when someone posts an answer to any problem they explicitly state that "THIS IS AN ANSWER POST" so that aam abduls who intend to solve the problem later with honesty do not accidentally read the answers.

Yes, also as I suggested also try to make the answer tiny Like below: (those who want to peek can alway make it bigger by using the quote button and remove the size tag and preview and peek :
Here is an example
(Lines below this can not be seen easily but if you use "quote" you can see it.

All this solution business can go here... for
etc...\
etc...

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 04:45
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I suppose he means that xyz =1 with x,y,z /= 1
--->
case 1) at least 1 of the three > 1 (other two < 1)
case 2) at most 2 of the three > 1 (other one < 1)

either case, at least one term is > 1 with the others +ve ---> the sum > 1 .
(this is the first thing that came to my mind, too.) it works for +ve numbers but can't satisfy the equality for any value of +ve only x,y,z.

problem is there may be negative numbers in pairs among the three and the phrase "modulus of"
has to flag off both the cases. this would mean that corresponding terms may not be > 1 and this way of proof will fail.

btw, amber g, the solution of b you posted in nukkad, will not work for k=0.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 05:05
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Quote:
btw, amber g, the solution of b you posted in nukkad, will not work for k=0.

You are correct!
(but 1,2,3,4 ..... and you still have infinity even without a zero

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 05:42
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Here is a well known and famous puzzle for the Rakshaks, dont cheat.

There are 3 doors (A,B,C) ahead of you, two of them have goats behind them, one of them has a car behind it. The host knows where the car and the goats are and allows you to choose a door. You choose it, but do not open it yet. Let us assume you chose door A.

The host walks over, and opens door C and shows you a goat. He now gives you a choice between switching to door B or staying at door A. What would you do ? Do your chances of winning a car

1. Go up if you switch
2. Go down if you switch
3. Remain the same if you switch

Why ?

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 05:51
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I know the supposed solution from a friend but was never convinced by his logic.
let's see if BRF does any better !

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 05:58
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Quote:
1. Go up if you switch
2. Go down if you switch
3. Remain the same if you switch

You switch. your prob goes up from 1/3 to 2/3 - assuming host knows where the car is..
If you have ever played cards eg bridge = this is called principle of restricted choice every bridge expert knows what to do.

Easiest way to see is your chances before (if you don't switch is 1/3)
No new information given (because host can always open a goat door) so it remains the same (1/3)
So obviously remaining case has possibility = 1-1/3 = 2/3 ===> you switch.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 06:55
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Rahul M wrote:
I suppose he means that xyz =1 with x,y,z /= 1
--->
case 1) at least 1 of the three > 1 (other two < 1)
case 2) at most 2 of the three > 1 (other one < 1)

either case, at least one term is > 1 with the others +ve ---> the sum > 1 .
.....

Rahul I really can't understand, how splitting in these cases help, x=1/4,y=-2,z=-2 with xyz=1
and ALL of them are less then 1 ( don't know where they fit?). Problem nowhere says that x,y.z are positive.. so I don't know how that helps.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 09:17
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if the goat is seen, then doesn't probability stay at 1/2? pl explain.
/math moorkh

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 09:40
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Buy some of the Vedic math books
They are very cheap and easy to understand

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 11:41
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SaiK wrote:
if the goat is seen, then doesn't probability stay at 1/2? pl explain.
/math moorkh

SaiK-saar, Rahul M-saar,

Let us assume you dont switch. 1/3 of the time, you choose a car and win.

Let us assume you switch.
1/3 of the time, you choose a Car. In this case, switching gives you a goat and you lose.
2/3 of the time, you choose a Goat. In this case, switching gives you a car and you win.
Chance of winning = 2/3 if you switch.

Another way of thinking about it is as follows: One third of the time, (when you have chosen a car), the host is trying to make an idiot out of you and you should discard his suggestion. Two thirds of the time (when you have chosen a goat), he is trying to help you by indicating which door has the car. So it is better to follow his suggestion.

 Last edited by Rahul M on 22 Jul 2008 14:48, edited 1 time in total. Edited to tiny font size.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 12:28
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Not answer, but a hint sort of, re: goat/car behind door.

Imagine the same problem with a 100 doors. 99 has goats, one has car. Host knows where the car is. You choose a door. Host goes and opens 98 of the remaining doors, revealing goats. Should you switch?

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 12:34
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Rahul M wrote:
btw, has it been proved that there are infinite number of primes ?
remember reading something on this long back, don't know what the situation is now.
thanks.

This was actually proved eons ago -- the proof appears in Euclid's Elements. While Euclid wrote books based on other people's discoveries, this one is supposed to have been proved by Euclid himself.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 12:36
 BRFite

Joined: 08 Jan 2002 12:31
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A seemingly simple probability problem with application to real life.

Doctors have developed an early test for a fatal illness X, which affects one in 1,000 people. The test is fairly accurate, giving only 5% false positives.

You took the test. It came back positive. What is the probability you are afflicted with X?

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 12:44
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Ah.. Goats. Nayakuddin should be reminded of a defunct thread and should now jump right in..

I dont know how the others got the 2/3rd no. I think that is wrong, it disregards value of information and the value of real options (economics geeks jump in here please and go on and on about why and how forecast creates economic value and decision making under uncertainty) . But let me throw my hat in the ring.

This is a case of conditional probability , with value of information thrown in.

Assuming that you will have to open the door you choose and the goat/car dealer doesn't give you the option of opening a goat door and showing the goat , ie it is a one shot "blind game"

We have total 3 doors, with 2 of them goats and 1 with a car.

So probability you chose a goat is 2/3 and that you choose a car is 1/3.

Now coming back to the game where the nice goat /car dealer allows you options of holding your hand at the chosen door , shows you a goat and gives you an option to switch. .

Now upto the time you put your hand on the door, the probabilities are like the "blind game"

Probability that what is behind the door you touched is a goat is 2/3
Probability that what is behind the door you touched is a car is 1/3.

Now the dealer "reveals" further information and shows another goat.

So the problem is now this.. There are two doors. The one you are touching and the other unopened. One contains a car and another a goat. each has a probability of 1/2The question now is which door to open, the door you are touching or the other one ?.

You will now have to evaluate conditional probability.

What is the probability that the door you are touching has a goat ?

That probability is 2/3* 1/2 = 1/3

What is the probability that the door you are touching has a car ?
That is 1/3 * 1/2 = 1/6

Now what is the probability that the other door has a goat
That is 2/3 * 1/2 = 1/3

What is the probability that the other door has a car ?

That is 2/3 * 1/2 = 1/3 .

So it is easy to see that while the door you are touching has a car behind it is has probability 1/6 , the probability that that the door you are NOT touching has a car behind it is 1/3. Obviously you would switch and choose the other door, because the odds of you winning are twice if you switch than if you stuck to your first choice.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 14:47
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Amber G. wrote:
Rahul M wrote:
I suppose he means that xyz =1 with x,y,z /= 1
--->
case 1) at least 1 of the three > 1 (other two < 1)
case 2) at most 2 of the three > 1 (other one < 1)

either case, at least one term is > 1 with the others +ve ---> the sum > 1 .
.....

Rahul I really can't understand, how splitting in these cases help, x=1/4,y=-2,z=-2 with xyz=1
and ALL of them are less then 1 ( don't know where they fit?). Problem nowhere says that x,y.z are positive.. so I don't know how that helps.

amber saab, I didn't mean to say that this was a correct solution !!
I think I have made that clear in the subsequent part of my post.

at first glance I thought in this way but that only satisfies the conditions
xyz =1 and f(x,y,z) > 1.
the equality can't be satisfied if all x,y,z are > 0.

I was trying to show how it was possible to think in terms of "splitting in 2" and go wrong !
regards.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 14:51
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ArmenT wrote:
Rahul M wrote:
btw, has it been proved that there are infinite number of primes ?
remember reading something on this long back, don't know what the situation is now.
thanks.

This was actually proved eons ago -- the proof appears in Euclid's Elements. While Euclid wrote books based on other people's discoveries, this one is supposed to have been proved by Euclid himself.

wasn't there another theory somewhere that the number of prime numbers decrease as you go along to higher numbers (obvious) and that may lead to a final drying up of primes as you crossed a certain number ? this is from childhood memory, so please forgive me if I the details aren't right !

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 15:18
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Quote:
3 prove
(cos 20 ) (cos 40) (cos 80) = 1/8

Bwahaaha, I actually managed something...

cos 20 = cos x

(cos x)(cos [60-x])(cos[60+x])
Expanding this out, we get
(1/4)(cosx){cos^2 x - 3sin^2x}
But sin^2 x = 1 - cos^2 x
So above reduces to cos^3 x - (3/4)cosx
which can be further simplified to cos3x / 4
cos3x = cos 60 = 1/2
so above is 1/8

Proved

The first few steps were easy, getting it to cos3x format proved a bit difficult, since I didnt know the formulae.

Hey, I can do maths!

BTW I took the same approach as vinas for the goat solution...

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 15:39
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Nandu wrote:
A seemingly simple probability problem with application to real life.

Doctors have developed an early test for a fatal illness X, which affects one in 1,000 people. The test is fairly accurate, giving only 5% false positives.

You took the test. It came back positive. What is the probability you are afflicted with X?

Isnt this simple conditional probability? P(having X) = 1/1000
P(having x/given test positive) = 95/100
P(given test positive/having X) = 1/1000 * 0.95 = 0.95/1000

Of course it cant be that simple, so I am wrong onleee.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 16:23
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The goat and car problem reminds me a joke. Of course you heard that in other versions.

Someone had to take a flight. He was afraid that someone will bring a bomb on board and blow it up. So, he asked the statistician what is the probability of that happening.

The answer was 1 in 24000.

He wasnt happy about it, so he asked how can that be improved.

"YOU carry a bomb on board. TWO independent people carrying a bomb aboard is 1 in 1 billion", said the statistician.

(Both numbers are indicative onlee. I NEVER had a proper education in statistics myself)

Now, if you ask me, You forget everything happened till the host revealed the goat. That doesn't exist. Then I take the available data.

Two doors. One with car, the other with goat. No REAL advantage in switching. It is like carrying a bomb on board.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 16:37
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Dileep wrote:
Now, if you ask me, You forget everything happened till the host revealed the goat. That doesn't exist. Then I take the available data.

which is precisely why I can't seem to understand the other solution !

3 doors A,B,C. you choose A but don't open. host opens C to show a goat.

problem reduces to 2 doors with one favourable outcome. probability does not change on switch. where is the fallacy ?

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 16:53
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The difference is you are looking at the probability at two different points of time.

The official solution (ie, switch is better) increases the probability looking from the beginning. The pragmatic solution (ie equal chance, so no switch) gives you the probability at the CURRENT TIME, which makes sense.

I would argue that the question is not completely defined. The question should specify WHICH probability. The probability looking from the beginning, or the probability NOW.

But if you do, it is no fun right?

The teacher of my 11th grade started the statistics class by telling us that it sometimes gives absurd results. The example she gave was not as funny as the bomb case. She said, statistics show that vehicles hit pedestrians more at the sides of the roads and less at the middle. So, statistically, it is safe to walk in the middle of the road.

Didn't I say I never had a formal statistics education? That is because I fell sick after that class and missed the whole thing. When I got into to engg, while we were in 1st year, it was in 5th sem. Then the syllabus changed and when we reached 5th, it went back to 1st year. So, my batch never had statistics!!

Hmm, like that matters much anyway!!

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 16:58
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there was a similar problem (well not quite !) that may be interesting in this case.
you toss an unbiased coin 6 times and it comes up head 4times.

what is the probability that the 7th toss will be a head ?

BTW, Dileep ji, ever thought of penning another of those 'needle in a haystack' type scenarios ? I was a big fan !

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 17:28
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Rahul, it is EXACTLY the same kind of problem.

Thanx for the OT. Who knows?!!

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 17:39
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so we have three different answers to the problem !!
P(car on switch) = 1/2,2/3,1/3 !!

now, who wants a poll to find out which is correct ?

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 18:16
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Rahul M wrote:
there was a similar problem (well not quite !) that may be interesting in this case.
you toss an unbiased coin 6 times and it comes up head 4times.

what is the probability that the 7th toss will be a head ?

BTW, Dileep ji, ever thought of penning another of those 'needle in a haystack' type scenarios ? I was a big fan !

Eh, arent the repeated coin tosses mutually independent events? So in this case, answer every time is 1/2 since you have stated the coin ins unbiased.

As far as the goat thing, as Dileep has said, it depends on what knowledge or instant in time you are measuring. The 2/3 solution assumes prior knowledge of the 2-stage experiment. The 1/2 solution experiment assumes that these are 2 distinct and independent experiments, with one part not affecting the other. Hence the different answers.

For the walking in the middle of the road part, statistics is of course right and shows lower values for that. But is it safer? Statistics cant answer that question, it can only show the measures. Deciding whether it is safe or not is an inference and a subjective one at that, which statistics cant answer.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 18:18
 BRF Oldie

Joined: 27 May 2007 03:55
Posts: 5059
vina wrote:
So it is easy to see that while the door you are touching has a car behind it is has probability 1/6 , the probability that that the door you are NOT touching has a car behind it is 1/3.

vina-saar, given that there are only two doors left and one of them has a car, shouldn't these probabilities add up to 1 instead of 1/3+1/6=1/2?

Dileep wrote:
Two doors. One with car, the other with goat. No REAL advantage in switching. It is like carrying a bomb on board.

Dileep-saar, that is not true. there are two doors. One has a car and one has a goat. But what is the probability that you have your hand on the door with the goat (apriori, irrespective of what the host does) ? There is another way of thinking about it.

Game 1: There are two contestants, Dilbullah and Vamanullah. None of them know what is behind the closed doors. D picks a door, does not open it yet. V picks one of the doors and opens it. Now irrespective of what V finds behind the door, is it advantageous to switch ?

Game 2: There is one contestant and one host, Dilbullah and Lalprafessar. D has no idea what is behind closed doors, Lalprafessar knows everything . D picks a door and does not open it yet. L picks a door and opens it and shows a goat. Now is it advantageous to switch ?

Are Game 1 and Game 2 similar ? (Hint, the are not.) What is the difference ?

Consider a modification of Game 1.
Game 1(b) If V finds a car, is it advantageous to switch ? (if V finds a goat, the game is called off)
Game 1(c) If V finds a goat, is it advantageous to switch ? (if V finds a car, the game is called off) Hint: Does the fact that we have removed those instances that V finds a car and replay the game, have any effect ?

Is Game 2 similar to Game 1(b) or Game 1(c) ? What are the sample spaces (list of possible outcomes) for Game 1(b) and Game 1(c) ? So is it advantageous to switch in Game 2 ?

Dileep wrote:
"YOU carry a bomb on board. TWO independent people carrying a bomb aboard is 1 in 1 billion", said the statistician.

The logical flaw being that the experiment has been influenced and is not the previous experiment anymore. Take for example this experiment.

I have a fair coin, I toss it, what is the probability I will have heads ? It is 1/2

You and I have a fair coin each, we toss both them. What is the probability we will get two heads ? It is 1/4. (Dont drag in Mr Bose here ).

You and I have fair coins, you cheat and show me a head always. I toss. What is the probability we will get two heads ? It is 1/2 and not 1/4.

The reduction in probability comes only if you toss and not if you cheat. Similarly, replace coin with Bomb. Bomb on the plane is 1 in 24000 (chance of tossing a heads). If you always carry a bomb (always show me heads), probability that someone else has a bomb too is 1 in 24000 and does not change.

Dileep wrote:
The teacher of my 11th grade started the statistics class by telling us that it sometimes gives absurd results. The example she gave was not as funny as the bomb case. She said, statistics show that vehicles hit pedestrians more at the sides of the roads and less at the middle. So, statistically, it is safe to walk in the middle of the road.

Dileep-saar,
That is also not true. This is a subtle wordplay on the conclusion. Let me restate using another example.
Number of people who get killed by snakes is less than number of people who die from complications from common cold. So it is safer to be bitten by a snake than to catch a cold ? Obviously it is safer to be bitten by a snake isnt it ?
The flaw in this line of reasoning is as follows: We should be addressing the question: If you get bitten by a snake, what is the chance that you die of snakebite ? If you get a cold, what is the chance you die of a cold ? Now, given a choice of one of the two (getting a cold vs snakebite), wont you pick having a cold over getting bitten by a snake ?

Similarly, in the road experiment, if you walk in the middle of the road, what is the chance that you die ? If you walk in the side of the road what is the chance that you die ? Given that you have to walk, is it better to walk on the side or the middle ?

Tanaji wrote:
For the walking in the middle of the road part, statistics is of course right and shows lower values for that. But is it safer? Statistics cant answer that question, it can only show the measures. Deciding whether it is safe or not is an inference and a subjective one at that, which statistics cant answer.

Tanaji-saar,
That is actually not true. Statistics does measure propensity. If the question is framed properly, statistics does help out. (a) If you compute statistics of one occurrence, and try to answer a totally unrelated question, like any other scientific method, statistics does not help. (b) If you do not factor in all information, like any other scientific method, statistics does not help.

There is a usual misunderstanding of (frequentists vs Bayesian), but both of them agree on the propensity, they simply dont agree if propensity can be explained....

Last edited by Anujan on 22 Jul 2008 19:10, edited 3 times in total.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 18:48
 Forum Moderator

Joined: 17 Aug 2005 21:09
Posts: 12427
Location: racetrack pattern over BRFATA.
Tanaji wrote:
Rahul M wrote:
there was a similar problem (well not quite !) that may be interesting in this case.
you toss an unbiased coin 6 times and it comes up head 4times.

what is the probability that the 7th toss will be a head ?

BTW, Dileep ji, ever thought of penning another of those 'needle in a haystack' type scenarios ? I was a big fan !

Eh, arent the repeated coin tosses mutually independent events? So in this case, answer every time is 1/2 since you have stated the coin ins unbiased.

Tanaji, that is precisely the point I am trying to make. once you know that the host has removed one door with goat behind it from the problem, the choice has to take that fact into account and this is in fact a new problem with just 2 doors and one car and one goat !

therefore, the three door problem has no relation to this problem, IMHO.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 19:26
 BRFite

Joined: 08 Jan 2002 12:31
Posts: 1343
So Dileep, what say you to the hundred door problem?

Tanaji, sorry, the conditional prob approach gives wrong answer.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 19:38
 BRF Oldie

Joined: 17 Dec 2002 12:31
Posts: 3321
Location: Ohio, USA
Vina et all -
The problems is fairly well discussed, at:
http://en.wikipedia.org/wiki/Monty_Hall_problem
As I said before, Your initial chances of success was (1/3) opening of the door by the host gives absolutely NO other information and the probability so it still remains same (1/3) that it was in the original door. Hence the other remaining door must have prob = 1-(1/3) = 2/3

The problem is very old. As I said, in Card game, bridge for example, the same type of logic (check out google/wiki for "principle of restricted choice" ) is helpful at the card table. (Even there, most of the card players refuse to believe that is so ... there was an article in mid 1940's by world champ Reese in Bridge world, and it sort of caused a major sensation, because most of the ordinary folks thought it was counter intuitive)

Basically, in similar problems, you don't go wrong if you assume the host had "no choice" to gauge which option is better. (Host has no choice when your original choice was wrong - Hence switch) .. You arrive at the correct answer without much math.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 19:57
 BRF Oldie

Joined: 17 Dec 2002 12:31
Posts: 3321
Location: Ohio, USA
BTW has anyone has solution to the three problems I posted?

Added later (after reading the tanaji's post): I one is done. thanks.

Last edited by Amber G. on 22 Jul 2008 20:20, edited 1 time in total.

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 Post subject: Re: BR Maths Corner-1Posted: 22 Jul 2008 20:06
 BRF Oldie

Joined: 21 Jun 2000 11:31
Posts: 3020
I got the easiest one, see earlier post.

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