Probability from hell: Calculate probability of drawing scrabble tiles in alphabetical order?

Question

I have 4 scrabble tiles: "A", "B", "C", and "D". I draw a letter at random and place it on the table, then draw a second and then a third, placing them down next to the previously drawn letter.

NOTICE! I'm only drawing 3 tiles, not four!

What is the probability that I draw the letters in alphabetical order?

The answer my book (in which, the probability of finding an honest answer is 0.1) is $\frac{1}{24}$.

The answer I through solving is $\frac{1}{6}$, and I, in no way, can figure whether I'm wrong or my book's answer section should surrender.

Answer 1

If you're only picking $3$, pick any tile first and throw it away (not literally!). You now have a $1:6$ chance.

Otherwise the book is asking a very trivial question.

Answer 2

What we are counting is the number of ways to choose four objects without replacement. Since only one of these will be in alphabetical order, it will be 1 divided by that.

The number of ways to choose four items without replacement is given by $4!=24$, and so the probability is indeed $\frac{1}{24}$.

An explanation of drawing without replacement: We are choosing tiles one at a time. For our first choice, we can choose any of the four. For the second choice, we must choose one of the remaining three, and our third choice we will have to choose from two, and our final choice will be a single one. The number of total possibilities of this is $4*3*2*1=4!=24$. Only one of these possibilities is ABCD, and so the probability is $\frac{1}{24}$.

Answer 3

• The probability you draw A first from A, B, C, D is $\frac14$
• The probability you draw B next from B, C, D is $\frac13$
• The probability you draw C next from C, D is $\frac12$

Then $$\frac14 \times \frac13 \times \frac12 = \frac1{24}$$