1.18 Calculating Odds. If we know the probabilities of a certain event, another way we can express these probabilities is through the odds – that is, the return given to betting on the event. To give an example, we can easily calculate the probability of the next card in a deck of 52 cards being a red card. So the odds offered on the next card being a red card are 1:1; that is, for every dollar bet, I give back 1 more dollar if the bet is a win. The probability is 0.5 = 1 1+1 . In the same spirit, the odds given to draw a club are 3:1 or 3$ for every 1$ bet. (In fact, the return is 4 if you count the dollar initially bet.) The probability of course is 0.25 = 1 1+3 . This idea will be seen in much more detail later when we talk about martingales. What happens if the probabilities are not of the form 1/n. For example, it is easy to calculate that the probability of drawing a black ball from an urn containing three red and four black balls is 4/7. No problem, the correct odds for betting on black are 3:4; that is, for every four dollars bet, you receive back three more. That is easy to see because the event has probability greater than 1/2. So the next questions are all asking you to formulate the correct odds (in a fair game) of the following events:

a) A card chosen at random from a deck of 52 is an ace.

b) A card chosen at random is either a Queen or a Club.

c) In 4 cards drawn from a deck of 52, there are at least 2 Hearts.

d) A hand of bridge (13 cards) is missing a suit (i.e., does not contain all four clubs, hearts, diamond, and spades).

e) You see two heads when tossing a fair coin twice.

f) You see either a sum of 7 or a sum of 11 when rolling two six-sided dies.