# Probability of drawing 13-card hand from a deck of cards without any spades

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?

• Your fraction is upside down, but otherwise the first method is correct. As is the second method (though you miscomputed the product). – lulu May 1 '18 at 11:50
• @lulu thank you for the quick answer. I recomputed it and noticed I put some extra multiplications that gave the wrong answer. But as they now give the same answer I see that both methods are probably correct, as you pointed out! – Marc May 1 '18 at 11:53

$$\frac{39\choose13}{52\choose13}=\frac{{39\choose13}\cdot13!}{{52\choose13 }\cdot13!}=\frac{39\cdot38\cdot{\dots}\cdot27}{52\cdot51\cdot{\dots}\cdot40}.$$
You should have: $\left.\dbinom{39}{13}\middle/\dbinom{52}{13}\right.=\dfrac{17063919}{1334062100}\approx0.012790948037576361700103765784216491870955632425...$