Chapter 3: Q22E (page 171)

Jai-alai bets. The Quinella bet at the paramutual game of jai-alai consists of picking the jai-alai players that will place first and second in a game irrespective of order. In jai-alai, eight players (numbered 1, 2, 3, . . . , 8) compete in every game.
a. How many different Quinella bets are possible?
b. Suppose you bet the Quinella combination of 2—7. If the players are of equal ability, what is the probability that you win the bet?

Short Answer

Expert verified

28
45/7

Step by step solution

Step-by-Step SolutionStep 1: Identify the different possible Quinella bets

The possibility of an event occurring is defined by probability. We may be required to forecast the result of an event in various real-life situations. To us, the result of an event might be certain or unknown.

Here, we have 8 players in the game, and we are selecting a person who has placed first and second in this game.

Therefore, we were only interested in two players and did not consider an order.

$Probability = {}^{8}C_{2} = \frac{8!}{2! \times (8 - 2)!} = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 28$

Hence, the different possible Quinella bets are 28.

Identify the probability of winning the bet

The possibility that wins the bet is:

$P (win the bet) = \frac{No . of possible outcomes}{Total outcomes} = \frac{{}^{6}C_{2}}{28} = \frac{\frac{6!}{2! \times 2!}}{28} = \frac{\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1}}{28} = \frac{180}{28} = \frac{45}{7}$