GPrep 2

[sanyalpritish]

plz read with patience
So many Explanations and so many formula..

Simple Draw 4 dash _ _ _ _
Now lets figure out in how many ways can we just put A in right place.
If A is in the right place we are left with B,C,D in the
2nd _ we can't put B so we have C,D hence 2 options
3rd _ we can only put D hence 1 options
4th _ we can only put C hence 1 options
We multiply all the options 1x2x1x1=2
If to put A in the right place its 2options similarly for B,C,D we have 2,2,2 you may want to try it out.
Add all the chances 2+2+2+2=8 Magic Figure
All possible combination is =24
P=8/24

[mschwrtz]

That's correct sanyalpritish. Note that your approach, although it isn't laden with formulas, avoids the error of at least two earlier explanations, which explanations Ron correctly described as "fundamentally flawed."

Unlike those explanations, yours asks what's possible for the second envelope given what we've stipulated of the first, and then what's possible for the third given what we've stipulated of the first and second, etc. I encourage you to use the approach you've described.

I would submit, though, that the incidence of error throughout this thread is a good argument that most should use Ron's brute-force approach.

[singh.ambesh]

The 'brute force way' proves the distibution of possibilities and is indeed a practical approach. If Tanya had more than 4 letters and envelopes, things will turn very 'brutal' this way.

Now the math involved is a bit difficullt to explain on this form, as it involves Venn Diagrams, Baye's theorem(Conditional Probability) and combinotrics. Appling these will enable you to solve it in a single line and would not take even 45 secs.

in this prob, it goes like this:

Prob(only1 in correct envelope)=(1/2+1/6-1/24)-(1/2+1/6+1/24)=1/3

Hi ..I could think of Rons's brute force approach ,but I want to understand the other way of solving this too.I found it difficult to comprehend.
Could some one explain please. Thank you.

[jnelson0612]

Hi Singh,
Could you please be more specific about what is confusing you? That will help us provide better explanations.

Thank you,

[david.khoy]

you have to be extremely careful whenever you multiply consecutive probabilities for statements that encompass multiple possibilities (especially negative statements, like the ones here). let me point out what can go wrong, using your first group of calculations as an example:

Hi friends,

My above post in summary:

1. The probability that ONLY A matches
The probability of Letter A going into Envelope A = 1/4
The probability of Letter B NOT going into Envelope B = 2/3
The probability of Letter C NOT going into Envelope C = 1/2
The probability of Letter D NOT going into Envelope D = 1
ONLY A matches is 1/4 x 2/3 x 1/2 x 1 = 2/24 = 1/12

Since there are 4 sets = 4 x 1/12 = 1/3.

there are issues with this calculation. it happens to hit upon the correct value at the end, but that's a total coincidence.

i agree with the first two probabilities: the probability that letter a goes into envelope a is indeed 1/4, and the probability if that happens that the letter b goes into an envelope other than b is 2/3.
however, it's downhill from there: if letter b actually went into envelope c, then the probability of letter c not going into envelope c is 1. the probability is only 1/2 (as you've stated) if letter b winds up in envelope d.
similarly, the final probability is either 0 or 1, depending on whether the last envelope remaining is envelope d or not. if letter b goes in envelope c and letter c goes in envelope b (fulfilling all of your conditions), then letter d is stuck going into envelope d, making that last probability 0.

so, if you're going to go this route, you're stuck with doing the following:
* first 2 steps = same as you have them now
* 3rd step = 2 branches of a probability tree, depending on whether envelope c is still available (vs. whether it was used for letter b)
* 4th step = 2 branches off EACH of those prior 2 branches, depending on whether envelope d is still available (vs. whether it was used for letter b or c)

that really, really sucks.

in general, if you have a COMBINATION problem with a SMALL # OF COMBINATIONS, you are much better off just counting possibilities than trying to employ fancy probability tricks.

hth!

> Ron:

Well, I really think you are wrong.

Is that really a coincindence that the consecutive probabilities method hit upon the correct value at the end?

Let's try with 3 enveloppes.

-Brutal force method:
We know there is a total of 3*2 = 6 possibilites:
abc: no good
acb: ok
bac: ok
bca: no good
cab: no good
cba: ok
So the probability is 3/6 = 1/2

-Consecutive probabilities method:
3 * 1/3 * 1/2 = 1/2

Same result again.

I do not agree with this part of your explanation:

however, it's downhill from there: if letter b actually went into envelope c, then the probability of letter c not going into envelope c is 1. the probability is only 1/2 (as you've stated) if letter b winds up in envelope d.

similarly, the final probability is either 0 or 1, depending on whether the last envelope remaining is envelope d or not. if letter b goes in envelope c and letter c goes in envelope b (fulfilling all of your conditions), then letter d is stuck going into envelope d, making that last probability 0.

I think this line of reasoning is wrong.

Letter A has a probability of 1/4 to get into envelope A.
Letter B has a probability of 2/3 not to get into envelope B.

Now let's pretend that letter B got into enveloppe D.
Letter C has a probability of 1/2 not to get into envelope C.
(At this point, C could get into enveloppe B or C)

Now let's pretend that letter B got into enveloppe C.
Now we seem to be stuck because the probability of letter C not going into envelope C is 1. However, we have to think of letter D first.

Letter D has also a probability of 1/2 not to get into envelope D. (At this point, D could get into enveloppe B or D)

Whether B gets into enveloppe C or D, we always have a letter (C or D) that has a probability of 1/2 of not going into the right envelope.

Thus, the consecutive probabilities method will always yield:
4 * 1/4 * 2/3 * 1/2 = 1/3.

I think we do not have to make trees. We have to assume the events x, y and z are independent. We have to assume that x and y occured before thinking of z, and then multiply the three probabilities together.

What would be the result with 5 letters and 5 envelopes? I am pretty sure the answer would be 1/4.

5 * 1/5 * 3/4 * 2/3 * 1/2 = 1/4

[RonPurewal]

Ron:
Well, I really think you are wrong.

Is that really a coincindence that the consecutive probabilities method hit upon the correct value at the end?

yep, it's a coincidence.

I think we do not have to make trees. We have to assume the events x, y and z are independent. We have to assume that x and y occured before thinking of z, and then multiply the three probabilities together.
What would be the result with 5 letters and 5 envelopes? I am pretty sure the answer would be 1/4.

5 * 1/5 * 3/4 * 2/3 * 1/2 = 1/4

nope, it's 3/8. so, i guess, there's proof that the consecutive probability approach DOESN'T work.

here's a list of the 9 ways in which you can match only envelope 'E' in this case:
BADCE
BCDAE
BDACE
CADBE
CDABE
CDBAE
DABCE
DCABE
DCBAE
there's nothing special about envelope 'E', so similar lists of 9 possibilities will exist for each other envelope.
these lists are mutually exclusive, so there are 45 different successful possibilities, out of 5! = 120 total.
so, the probability is 45/120 = 3/8.

this is not 1/4, so it's not worth investigating why this other proposed method doesn't work; it just doesn't work, for the reasons i mentioned above.
the coincidence for 3 or 4 envelopes is just that -- a coincidence.

[david.khoy]

Ok, thanks Ron.

The consecutive probability approach does not work with 5 enveloppes indeed.

However I still think that the fact that it worked for 3 and 4 enveloppes was not a mere coincindence.

With 3 enveloppes, whatever happens you always have a 1/3 probability of getting the first letter into the right enveloppe, and then you always have a 1/2 probability of getting the second letter into the wrong enveloppe (only two letters and two enveloppes remain). The consecutive probability approach really works with 3 enveloppes, because there are not many possibilites.

I believe that making the trees you were talking about only becomes necessary with 5 enveloppes, meaning the consecutive probability approach does not work anymore. That's because with 5 enveloppes, there are many possible scenarios.

That's probably why the GMAC sticked with 4 enveloppes.

[RonPurewal]

Ok, thanks Ron.

The consecutive probability approach does not work with 5 enveloppes indeed.

However I still think that the fact that it worked for 3 and 4 enveloppes was not a mere coincindence.

... except it was. if it weren't, then the same approach would also work for N envelopes.

by the way, fyi, there's only one "p" in "envelope".
(in french there are two "p"s, but not in english.)

With 3 enveloppes, whatever happens you always have a 1/3 probability of getting the first letter into the right enveloppe, and then you always have a 1/2 probability of getting the second letter into the wrong enveloppe (only two letters and two enveloppes remain).

no, this is incorrect, and the answer really does come from pure coincidence, as i've explained above.

this is the beauty of mathematics: there's no such thing as an approach that "just stops working". if it works for a class of problems, then it will work for *every* problem in that class, unless there is an actual, material difference between the problems (such as introducing 2 letters that are indistinguishable from each other, etc.).
otherwise, a fomula that works for *some* items in a class, but not all of them, is ... well ... a wrong formula.

here's another, unrelated example of the same thing:
take this polynomial
x^2 - x + 41
and start plugging positive integers into it, and see whether the result is prime.
you'll find that the result is prime for EVERY single integer from 1 to 40 ... but then it breaks down at 41.
does this mean that this formula "works" as a prime-generator until then? nope -- it's a pure, unadulterated coincidence, this time for forty cases in a row rather than just a couple.

That's probably why the GMAC sticked with 4 enveloppes.

it's much more likely that they were trying to write a problem on which you just have to break down and count things, but that the problem happened to have an accidental correct solution with an incorrect method.

one of the major hallmarks of this test is the inclusion of problems that absolutely can't be solved with "traditional" methods, so it wouldn't surprise me at all if that were their intention.

see the NOV. 4, 2010 session here for a whole hour and a half of problems that can't be solved with traditional methods:
http://www.manhattangmat.com/thursdays-with-ron.cfm

[david.khoy]

Ok Ron, you convinced me. Thanks.

I really do not like the brutal force method (I think it is time consuming and error-prone), but I guess it is the only way to get this question right without relying on coincidences.

Thank you for the envelope remark. (I am french. At first I wrote "envelope" in my messages, but after a while, I wrote "enveloppe" without thinking.)

By the way, in the solution you wrote back in 2007, you used a really brutal method by writing all the 24 possibilities. I prefer the method you used to prove that the consecutive probabilities method does not work with 5 envelopes. When it is applied to the 4 envelopes case, it's really quick and safe:

There are 4 * 3 *2 = 24 possible arrangements.

There are only two ways to match only letter A:

ACDB
ADBC

Since there are 4 letters, there are 2 * 4 = 8 ways to match only one letter.

8/24 = 1/3

[tim]

glad we got this resolved.. :)

[me.parashar]

Other than the brute force method of Ron, I really liked the prude_sb's method. I hope it's right. I didn't see any remark from Ron on that.

[RonPurewal]

Other than the brute force method of Ron, I really liked the prude_sb's method. I hope it's right. I didn't see any remark from Ron on that.

prude_sb's method (previous page) does work, yes. the drawback to that method is that it requires a tremendous amount of intuition, which can't be picked up just by rote studying.
the good thing about just listing all 24 possibilities is that anyone can do it -- you just have to get started early!

[me.parashar]

Other than the brute force method of Ron, I really liked the prude_sb's method. I hope it's right. I didn't see any remark from Ron on that.

prude_sb's method (previous page) does work, yes. the drawback to that method is that it requires a tremendous amount of intuition, which can't be picked up just by rote studying.
the good thing about just listing all 24 possibilities is that anyone can do it -- you just have to get started early!

Great... actually when I was answering the question in official test conditions, i did do the question with his method but didn't multiply the 2 possibilities by 4 (for each letter) hence ended up getting 2/24 instead of 8/24. So when i started to search the question on google and came on the forum, i was specifically looking for that method and the mistake I made during test.

thanks to you I now have something to rely on if my intuition doesn't work (which happens sometimes with me under stress) :)

Oh.. and before I forget... thanks for the video about unconventional methods to solve problems (where algebra doesn't work)

[RonPurewal]

Oh.. and before I forget... thanks for the video about unconventional methods to solve problems (where algebra doesn't work)

that's an important video; way too many people depend solely on algebra or "textbook" methods.
along similar lines, i would also recommend the february 4 video (backup methods).

[parthian7]

I personally dislike the brute force method, even though I agree with Ron that it may be ideal for some people, since like he says anyone can do it.

I would use the following approach and thought I'd share it as it may be useful to some other people as well.

so we have letters: A, B, C and D
and envelopes: Ea, Eb, Ec, Ed

total # of possibilities: 4! = 24

for desired # of possibilities:
let's say we want A to end up in Ea, but B not in Eb, C not in Ec and D not in Ed:

I would use the following tree shaped diagram:

-----------/ B
--------/ D
----/ C
A
----\ D
--------\ B
-----------\ C


Notice we have 2 possibilities for this case.
Pretty much the only subtlety here is recognizing that we have 2 choices for the 2nd spot (C and D), but 1 for 3rd and 4th.
At first glance, it may look like we have 2 choices for the 3rd spot as well: i.e. after placing C in 2nd spot, we may think we can put either B or D in Ec. However, notice that D CANNOT go to Ed, therefore, D has to go to Ec (1 choice) and B has to go to Ed (1 choice). It's the same logic for when D goes into Eb. Afterwards, we are left with 1 choice for Ec: B and 1 choice for Ed: C.

We have 4 such cases, hence 4*2 -> 8 possibilities altogether

desired possibilities: 8
total possibilities: 24

8/24 = 1/3

Also notice that this way there's no need to write down all possibilities, which would be brutal if we had more than 4 letters. We only draw it out for 1 case and multiply by 4, since they are identical.

hope that helps