Solution Found!
Jai-alai bets. The Quinella bet at the paramutual game of
Chapter 3, Problem 20E(choose chapter or problem)
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?
Questions & Answers
QUESTION:
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?
ANSWER:Step 1 of 2
(a)
Jai-alai bets. The Quinella bet at the para mutual game of jai-alai consists of picking the jai-alai players that will place first and second in a game respective of order.
In jai-alai, eight players (numbered 1, 2, 3, ……….,8) compete in every game.
How many different Quinella bets are possible?
In the question, we have given that the order does not matter, so we can use the combinations theorem here.
Hence the number of different bets would be a combination of out of 8 players, 2 players will place a first and second bet.
Therefore number of ways would be,
\(\left(\begin{array}{l}
8 \\
2
\end{array}\right)=C_{2}^{8}=\frac{n !}{r ! \times(n-r) !}\)
[here n = 8 and r = 2]
\(\left(\begin{array}{l}
8 \\
2
\end{array}\right)=C_{2}^{8}=\frac{8 !}{2 ! \times(8-2) !}=28\)
Hence 28 different Quinella bets are possible.