Suppose a player draws 13 cards from a deck. What is the probability that their hand does not contain any spades?
I've found out that the answer might be easier than my initial approach:
There are ${52}\choose{13}$ ways of picking a hand from the deck, ${39}\choose{13}$ of which are without any spades. That leaves for a total probability of ${39}\choose{13}$$/$${52}\choose{13}$ $= 0.0127\dots$ though I'm not certain on this.
My initial approach was:
The chance of drawing a non-spade card is at first $\frac{39}{52}$, and for the second card it's $\frac{38}{51}$ and so on. This gives a total probability of $\frac{39\times38\times\dots \times 27}{52\times51\times\dots\times40}=0.0127\dots$
Are both of these methods correct?
Best Answer
They're both correct because they're equivalent
$$\frac{39\choose13}{52\choose13}=\frac{{39\choose13}\cdot13!}{{52\choose13 }\cdot13!}=\frac{39\cdot38\cdot{\dots}\cdot27}{52\cdot51\cdot{\dots}\cdot40}.$$