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January 28, 2015

January 28, 2015

Posted by **Punkie** on Saturday, November 28, 2009 at 9:26pm.

Consider the experiment of drawing two cards without replacement from an ordinary deck of 52 playing cards.

What are the odds in favor of drawing a spade and a heart?

There are 52 playing cards in a deck

There are 13 spades and 13 hearts in deck

13 + 13 = 26

P(spade and heart) = 26/52

P(not a spade or heart) = 26/52

Odds in favor of A = Number of ways that A could occur/Number of ways that A could not occur

Odds in favor of A = 26/52 / 1- 26/52 = 0

- Math -
**Punkie**, Saturday, November 28, 2009 at 9:34pmOr would it be

Odds in favor of A = 26/52 / 1 - 26/52 = 26 / 51

- Math -
**MathMate**, Saturday, November 28, 2009 at 9:37pm"Odds in favor of A = 26/52 / (1- 26/52)= 0"

A in the above expression actually evaluates to 26/52 / (1-26/52) = (1/2) / (1/2) =1:1

However, I would reason it this way:

The first card can be either a spade or a heart, so the probability is 26/52.

The second card has to the complementary suit, namely a spade if the first one was a heart, and vice versa.

Probability of the second card is therefore limited to one suit out of 51 cards, or 13/51.

Probability of both events happening is

(26/52)*(13/51)=(1/2)*(13/51) = 13/102

Odds are

(13/102) / (1-13/102)

= 13/89

13:89

- Math -
**Punkie**, Saturday, November 28, 2009 at 10:03pmI have a question about the answer, how did you get 13/89, I don't understand how you got 89

- Math -
**MathMate**, Saturday, November 28, 2009 at 11:16pmHere it is, in a little more detail:

(13/102) / (1-13/102)

= (13/102) / ((102-13)/102)

=(13/102) / (89/102)

=13/89

=13:89

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