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

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:



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!

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.

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?

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

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).

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

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!

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