Q.4.80 - A First Course in Probability

Question

A game popular in Nevada gambling casinos is Keno, which is played as follows: Twenty numbers are selected at random by the casino from the set of numbers $1$ through $80$. A player can select from$1$to $15$ numbers; a win occurs if some fraction of the player's chosen subset matches any of the $20$ numbers drawn by the house. The payoff is a function of the number of elements in the player's selection and the number of matches. For instance, if the player selects only 1 number, then he or she wins if this number is among the set of $20$ , and the payoff is $\backslash (2.20$ won for every dollar bet. (As the player's probability of winning in this case is $\frac{1}{4}$, it is clear that the "fair" payoff should be $\backslash )3$won for every $\backslash (1$ bet.) When the player selects 2 numbers, a payoff (of odds) of  won for every \)1 bet is made when both numbers are among the $20 .$

(a) What would be the fair payoff in this case?

Let $P_{n,k}$ denote the probability that exactly $k$ of the $n$ numbers chosen by the player are among the 20 selected by the house.

(b) Compute $P_{n,k}$

(c) The most typical wager at Keno consists of selecting 10 numbers. For such a bet, the casino pays off as shown in the following table. Compute the expected payoff:

Step-by-Step Solution

Answer

(a) $\$15.64$ won for every $\$1$ bet made.

$P_{n,k}=\frac{\left(\begin{array}{c}20\\ k\end{array}\right)\left(\begin{array}{c}60\\ n-k\end{array}\right)}{\left(\begin{array}{c}80\\ n\end{array}\right)}$

(c)$-\$0.25$

Step 1:Given information(part a)

A game popular in Nevada gambling casinos is Keno, which is played as follows: Twenty numbers are selected at random by the casino from the set of numbers $1$ through $80$. A player can select from $1$ to $15$numbers; a win occurs if some fraction of the player's chosen subset matches any of the$20$numbers drawn by the house. The payoff is a function of the number of elements in the player's selection and the number of matches. For instance, if the player selects only 1 number, then he or she wins if this number is among the set of $20$ , and the payoff is $\$2.20$ won for every dollar bet. (As the player's probability of winning in this case is $\frac{1}{4}$, it is clear that the "fair" payoff should be $\$3$ won for every $\$1$bet.) When the player selects $2$ numbers, a payoff (of odds) of $\$12$ won for every $\$1$ bet is made when both numbers are among the $20$ .

Step 2:Explanation(part a)

Twenty numbers are randomly 

picked from $1$ to $80$. A player can select 1 to $15$ numbers and wins when a ratio of his numbers matches the twenty numbers.

We need to select the fair payoff when the player selected $2$ numbers.

Let $X$ denote the number of matching numbers for the player. The number of successes among random draws from a finite population with two possible outcomes follows a hypergeometric distribution.

$N=$ Population size$=80$

$n=$ Number of draws $=2$

$m=$ Number of observed successes $=20$

Formula hypergeometric probability:

$P(X=i)=\frac{\left(\begin{array}{c}m\\ i\end{array}\right)\left(\begin{array}{c}N-m\\ n-i\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$

Evaluate the definition of hypergeometric probability at $i=2$ (which is when the player will win).

$P(X=2)=\frac{\left(\begin{array}{c}20\\ 2\end{array}\right)\left(\begin{array}{c}80-20\\ 2-2\end{array}\right)}{\left(\begin{array}{c}80\\ 2\end{array}\right)}$

$=\frac{\left(\begin{array}{c}20\\ 2\end{array}\right)\left(\begin{array}{c}60\\ 0\end{array}\right)}{\left(\begin{array}{c}80\\ 2\end{array}\right)}$

$=\frac{19}{316}$

$\approx 0.0601$

Since the probability of winning is about $0.0601$, the pay would be fair when you win $\frac{1}{0.0601}\approx \$16.64$and thus the fair payoff is then $\$15.64$won for every $\$1$ bet made (as the $\$1$ bet will also be returned resulting in a total gains of$\$16.64)$

Step 3:Final answer(part a)

(a) $\$15.64$ won for every $\$1$ bet made.

Step 4:Given information (part b)

Given in the question that a game popular in Nevada gambling casinos is Keno, which is played as follows: Twenty numbers are selected at random by the casino from the set of numbers$1$ through $80$. A player can select from $1$ to $15$ numbers; a win occurs if some fraction of the player’s chosen subset matches any of the $20$numbers drawn by the house. The payoff is a function of the number of elements in the player’s selection and the number of matches. For instance, if the player selects only $1$ number, then he or she wins if this number is among the set of $20$, and the payoff is $\$2.20$ won for every dollar bet. (As the player’s probability of winning in this case is$1 4$ , it is clear that the “fair” payoff should be$\$3$won for every $\$1$ bet.) When the player selects 2 numbers, a payoff (of odds) of$\$12$ won for every $\$1$ bet is made when both numbers are among the $20.$ 

Step 5:Explanation(part b)

 Probability that exactly $k$ of the $n$ numbers chosen are among the $20$ selected by the house.

Let $X$ represent the number of matching numbers for the player. The number of successes among random draws from a finite population with two possible outcomes follows a hypergeometric distribution.

$N=$ Population size $=80$

$n=$ Number of draws

$m=$Number of observed successes

Formula hypergeometric probability:

$P(X=i)=\frac{\left(\begin{array}{c}m\\ i\end{array}\right)\left(\begin{array}{c}N-m\\ n-i\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$

Evaluate the definition of hypergeometric probability at $i=k$.

$P_{n,k}=P(X=k)$

$=\frac{\left(\begin{array}{c}20\\ k\end{array}\right)\left(\begin{array}{c}80-20\\ n-k\end{array}\right)}{\left(\begin{array}{c}80\\ n\end{array}\right)}$

$=\frac{\left(\begin{array}{c}20\\ k\end{array}\right)\left(\begin{array}{c}60\\ n-k\end{array}\right)}{\left(\begin{array}{c}80\\ n\end{array}\right)}$

Since the probability of winning is $\frac{1}{0.0601}\approx \$16.64$  16.64$about $0.0601$, the pay would be fair when you win  16.64$ and thus the fair payoff is then $\$15.64$ won for every $\$1$ bet made (as the $\$1$ bet will also be returned resulting in a total gains of $\$16.64$).

Step 6:Final answer(part b)

$P_{n,k}=\frac{\left(\begin{array}{c}20\\ k\end{array}\right)\left(\begin{array}{c}60\\ n-k\end{array}\right)}{\left(\begin{array}{c}80\\ n\end{array}\right)}$

Step 7:Given information(part c)

Given in the question that,a game popular in Nevada gambling casinos is Keno, which is played as follows: Twenty numbers are selected at random by the casino from the set of numbers $1$ through $80$. A player can select from $1$to $15$numbers; a win occurs if some fraction of the player’s chosen subset matches any of the $20$ numbers drawn by the house. The payoff is a function of the number of elements in the player’s selection and the number of matches. For instance, if the player selects only $1$number, then he or she wins if this number is among the set of $20$, and the payoff is $\$2.20$ won for every dollar bet. (As the player’s probability of winning in this case is $14$ , it is clear that the “fair” payoff should be $3 won for every $1 bet.) When the player selects 2 numbers, a payoff (of odds) of $\$12$ won for every $\$1$ bet is made when both numbers are among the $20$ 

Step 8:Explanation(part c)

Let $X$ represent the number of matching numbers for the player. The number of successes among random draws from a finite population with two possible outcomes follows a hypergeometric distribution.

$N=$ Population size $=80$

$n=$Number of draws $=10$

$m=$Number of observed successes $=20$

Formula hypergeometric probability:

$P(X=i)=\frac{\left(\begin{array}{c}m\\ i\end{array}\right)\left(\begin{array}{c}N-m\\ n-i\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$

Evaluate the definition of hypergeometric probability at

$i=0,1,2,3,4,5,6,7,8,9,10\text{. }$

Step 9:Explanation(part c)

$P(X=0)=\frac{\left(\begin{array}{c}20\\ 0\end{array}\right)\left(\begin{array}{c}80-20\\ 10-0\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}=\frac{\left(\begin{array}{c}20\\ 0\end{array}\right)\left(\begin{array}{c}60\\ 10\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}\approx 0.0458$

$P(X=1)=\frac{\left(\begin{array}{c}20\\ 1\end{array}\right)\left(\begin{array}{c}80-20\\ 10-1\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}=\frac{\left(\begin{array}{c}20\\ 1\end{array}\right)\left(\begin{array}{c}60\\ 9\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}\approx 0.1796$

$P(X=2)=\frac{\left(\begin{array}{c}20\\ 2\end{array}\right)\left(\begin{array}{c}80-20\\ 10-2\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}=\frac{\left(\begin{array}{c}20\\ 2\end{array}\right)\left(\begin{array}{c}60\\ 8\end{array}\right)}{\left(\begin{array}{l}80\\ 10\end{array}\right)}\approx 0.2953$

$P(X=3)=\frac{\left(\begin{array}{c}20\\ 3\end{array}\right)\left(\begin{array}{c}80-20\\ 10-3\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}=\frac{\left(\begin{array}{c}20\\ 3\end{array}\right)\left(\begin{array}{c}60\\ 7\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}\approx 0.2674$

$P(X=4)=\frac{\left(\begin{array}{c}20\\ 4\end{array}\right)\left(\begin{array}{c}80-20\\ 10-4\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}=\frac{\left(\begin{array}{c}20\\ 4\end{array}\right)\left(\begin{array}{c}60\\ 6\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}\approx 0.1473$

$P(X=5)=\frac{\left(\begin{array}{c}20\\ 5\end{array}\right)\left(\begin{array}{c}80-20\\ 10-5\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}=\frac{\left(\begin{array}{c}20\\ 5\end{array}\right)\left(\begin{array}{c}60\\ 5\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}\approx 0.0514$

$P(X=6)=\frac{\left(\begin{array}{c}20\\ 6\end{array}\right)\left(\begin{array}{c}80-20\\ 10-6\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}=\frac{\left(\begin{array}{c}20\\ 6\end{array}\right)\left(\begin{array}{c}60\\ 4\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}\approx 0.0115$

$P(X=7)=\frac{\left(\begin{array}{c}20\\ 7\end{array}\right)\left(\begin{array}{c}80-20\\ 10-7\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}=\frac{\left(\begin{array}{c}20\\ 7\end{array}\right)\left(\begin{array}{c}60\\ 3\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}\approx 0.0016$

$P(X=8)=\frac{\left(\begin{array}{c}20\\ 8\end{array}\right)\left(\begin{array}{c}80-20\\ 10-8\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}=\frac{\left(\begin{array}{c}20\\ 8\end{array}\right)\left(\begin{array}{c}60\\ 2\end{array}\right)}{\left(\begin{array}{l}80\\ 10\end{array}\right)}\approx 0.0001$

$P(X=9)=\frac{\left(\begin{array}{c}20\\ 9\end{array}\right)\left(\begin{array}{c}80-20\\ 10-9\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}=\frac{\left(\begin{array}{c}20\\ 9\end{array}\right)\left(\begin{array}{c}60\\ 1\end{array}\right)}{\left(\begin{array}{c}80\\ 10\end{array}\right)}\approx 0.000006$

$P(X=10)=\frac{\left(\begin{array}{l}20\\ 10\end{array}\right)\left(\begin{array}{l}80-20\\ 10-10\end{array}\right)}{\left(\begin{array}{l}80\\ 10\end{array}\right)}=\frac{\left(\begin{array}{c}20\\ 10\end{array}\right)\left(\begin{array}{c}60\\ 0\end{array}\right)}{\left(\begin{array}{l}80\\ 10\end{array}\right)}\approx 0.0000001$

Step 10:Explanation(part c)

The expected value (or mean) is the sum of the product of each possibility $x$ (dollars won for each $\$1$ bet) with its probability

$\mu =\sum xP(X=x)$

$=(-1)\times 0.0458+(-1)\times 0.1796+(-1)\times 0.2953$

$+(-1)\times 0.2674+(-1)\times 0.1473+1\times 0.0514$

$+17\times 0.0115+179\times 0.0016+1299\times 0.0001$

$+2599\times 0.000006+24999\times 0.0000001$

$\approx -0.25$

$-\$0.25\$$ (a loss of $\$0.25$).

Step 11:Final answer(part c)

$-\$0.25$