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

Question

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?

Answer 1

You should have: $\left.\dbinom{39}{13}\middle/\dbinom{52}{13}\right.=\dfrac{17063919}{1334062100}\approx0.012790948037576361700103765784216491870955632425...$

Your original method was not wrong, so should also have given this result.   So you simply made a calculation error.

Answer 2

They're both correct because they're equivalent

$$\frac{39\choose13}{52\choose13}=\frac{{39\choose13}\cdot13!}{{52\choose13 }\cdot13!}=\frac{39\cdot38\cdot{\dots}\cdot27}{52\cdot51\cdot{\dots}\cdot40}.$$

Answer 3

There are 52 cards in the deck, 39 of them "not spades". So the probability of drawing a "not spade" the first time is 39/52= 3/4. There are then 51 cards left, 38 of them "not spades". The probability of drawing a "not spades" the second time is 38/51. There are then 50 cards in the deck, 37 of the "not spades". The probability of drawing a 'not spades" is 37/50. Continuing in that way, the probability of drawing 13 "not spades" in a row is (39/52)(38/51)(37/50)...(27/40)= ((39)(38)(37)...(29)(28)(27))/((52)(51)(50)...(42)(41)(40)) which we can write as (39!/26!)(39!/52!)