Question

What is the probability that a hand of 13 cards dealt from a standard deck will contain no hearts?

Answer 1

The situation boils down to taking 13 cards out of a selection of 39 non-hearts from a total of 52.

1st card: P(non-heart)=39/52
2nd card: P(non-heart)=38/51
3rd card: P(non-heart)=37/50
And so on.

After that you multiply these probabilities, because it's an AND-situation.

GC:
(39 nCr 13)/(52 nCr 13)

Answer 2

Just another way of thinking about it:

There are `C(39,13)` ("`39` Choose `13`" or "`39C13`" depending upon the notation you prefer) ways of dealing a hand with 13 non-Heart cards.

There are a total of `C(52,13)` ways of dealing 13 cards with no restrictions.

The probability of dealing a non-Heart hand is
(the number of ways of dealing a non-Heart hand) divided by (the total number of ways of dealing a non-restricted hand)
`= (C(39,13)) / (C(52,13))`

`= (39!)/(13! 26!)` / `(52!)/(13! 39!)`

`= (39!)^2/(26! 52!)`

Answer 3

`P("no hearts") = 1 - P("all hearts")`

=`(52!-13!xx39!)/(52!)`

Explanation:

We could approach this by considering the other alternative .. that a player is dealt ALL the hearts.

The probability of 13 hearts can be determined as

1st card: `P(H) = 13/52,`

2nd card `P(H) = 12/51` and so on.

`P(HH......HHH) = 13/52 xx12/51 xx 11/50 xx ...... xx 1/40`

`13/52 xx12/51 xx 11/50 xx ...... xx 1/40 = (13! xx39!)/(52!)`

However, we want the probability that there are NO hearts in a hand..

`P("no hearts") = 1 - P("all hearts")`

`1- (13! xx39!)/(52!)`

=`(52!)/(52!) - (13! xx39!)/(52!)`

=`(52!-13!xx39!)/(52!)`

Is this thinking flawed?