# Letters and Envelops problem

Consider a machine whose job is to place 100 letters into 100 envelops.The machine is defective and makes mistakes.What is the probability that in a group of 100 letters no letter is put into the correct envelope?

I did like this.

Total ways of putting letters into envelops=$$100!$$

And there is only 1 way by which all letter go into correct envelope.

So, I thought my answer must be $$\frac{100!-1}{100!}$$

But the answer is given to be $$\frac{D_{100}}{100!}$$

Where $$D_{100}$$ represents derangement of 100 letters.

• Consider $3$ letters. The choices are $\color{green}{ABC}, \color{blue}{ACB}, \color{blue}{BAC}, \color{red}{BCA}, \color{red}{CAB}, \color{blue}{CBA}$. $\color{green}{Green} -$ all correct, $\color{blue}{blue} -$ one correct and $\color{red}{red} -$ all incorrect. Note that there is no "two correct" (why?) Your method gives $\frac{3!-1}{3!}=\frac56$, while the given answer gives $\frac{!3}{3!}=\frac26=\frac13$. – farruhota Dec 4 '18 at 8:28
• Nitpicking: It appears somewhat bold to conclude that every permutation is equally likely when we are only told that "the machine makes mistakes". (Not to mention that other kinds of mistakes could be that multiple letters end up in the same envelope or some letters are destroyed etc.) – Hagen von Eitzen Dec 4 '18 at 18:54

Why is the Derangement Probability so Close to $$\frac{1}{e}$$?