A mailman has thirty letters that each have a corresponding mailbox (all situated neatly together) in which each letter belongs. The mailman randomly places each letter in a mailbox. a) What are the odds that one letter was placed in the correct mailbox? b) What are the odds that AT LEAST one letter was placed in the correct mailbox? c) What are the odds that NO letters were placed in ANY of the correct mailboxes? d) How many possible permutations are there?

A:If you mean ONLY one letter, than extremely low. B: Almost a certainty C: Almost as unlikely as hitting the lottery. And getting stuck by lightening on the same day. D: A lot (just guesses off the top of my head...way too late to do the math...besides, you talked me out of it with .1% being able to solve it) An easier version of a similar problem. 20 people in a room, what are the odds of 2 having the same birthday? Your problem reminds me of a sort of perverse reverse Keno. And I can tell you that no one even bothers figuring out odds on keno. They are so incredibly astronomical as to be a waste of time to try and calculate. Of course it can be done, but why bother? The best bet, (the fairest payoff odds)is picking one number and getting ripped for something like 25% on your payoff. Then it just gets worse from there. http://conjelco.com/faq/keno-odds.html

I am too lazy to do the math too but my estimations are different: a) about 1/3 b) about 2/3 c) about 1/3 d) 30! I am pretty sure that this problem has no similarities with "the same birthday" problem. There must be a lot of factorials involved in the solution of this problem. Because of this, I am guessing that if I change 3's to e's I might get closer.

a) don't know b) don't know c) 1 out of {the reciprocal of (29/30 X 28/29 X 27/28 ... X 1/2)} The odds of misplacing the first letter are 29/30. If that happens, the odds of misplacing the second letter are 28/29. If that happens, the odds of misplacing the third letter are 27/28. Then you keep going down until you get to 1/2. The odds of more than one event ALL happening is found by multiplying the odds together...so multiply all those fractions together, you'll get some really small decimal. Take the reciprocal of that number, and 1 over that number is the odds of misplacing all letters. d) 30! (30X29X28...X1)

Wrong. The odds of misplacing the second letter are 28/29 only if you placed the first letter correctly.

What a way to spend a Friday night. I took statistics 101, twice, and it was a while ago but here goes. The probability of any letter being in the wrong mailbox is 29/30. The probability of 29 of them being in the wrong mailbox is 29/30 to the 29th power or .37413, which is the same as one and only one being in the right mailbox. The probability of 30 of them being in the wrong mailbox is 29/30 to the 30th power or .36166 which is the same case as none being in the right mailbox. There are 30 ways to place a letter in the first box, 29 ways to place a letter in the second box, etc. The permutations are 30 factorial, or 2.65 to the 32nd power. The calculation of the probability of one or more being in the correct mailbox is way beyond discussing on a stock trading forum (in other words I have not a clue) Now regarding the hat question, the card on the floor is not the white-white one, therefore it is a coin toss between being red-red or red- white, 50%. Max

This is the easy part - the answer to (c) is simply 1 minus the answer to (b). In fact, as the number of letters and mailboxes increases, the answer to (c) approaches 1/e (0.367879...). Another interesting fact - the average number of correct matches is 1. In other words, if the mailman repeats this process daily, then on average there will be 1 letter in the correct mailbox - and this is true whether there are 30 letters (and mailboxes) or 300 letters (and mailboxes). See the thread for enlightenment.

Wrong. The probability of any letter being in the wrong mailbox is 29/30 only if you decide to count as possible outcome such outcomes where the letter which is supposed to go into mailbox number 1 is in all 30 mailboxes. Here is an illustration for 3 mailboxes. The possible outcomes are: 123 132 ... etc while you count such outcomes as 111 112 ... etc Since your assumption is wrong, your conclusions are wrong too.

30! equals about 266 million trillion trillion (~2.65528598*10^32), or about how much money I have lost in the market on my worst days . Somebody else will have to generate all those combinations and see which ones have letters in the right boxes.