Wayxi wrote:
You can solve this the traditional way:
Probability that first letter in the right envelope= 1/4
Probability that second letter in wrong envelope = 2/3
Probability that third letter in wrong envelope = 1/2
Probability that forth letter in wrong envelope = 1
1/4 * 2/3 * 1/2 = 1/12
Multiply by 4, representing the four letters in the correct envelope:
1/12 * 4 = 4/12 = 1/3
For anyone wondering about this solution, it is actually correct only by coincidence. This method will fail when the number of letters increased to 5, and is therefore an incorrect formula.
"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)"
If you would like to read further, see
Manhattan Prep's forum post on this question.
This is a quote from Ron Purewal