you pick a card at random from an ordinary deck of 52 cards. if the card is an ace, you get 9 points. if…

you pick a card at random from an ordinary deck of 52 cards. if the card is an ace, you get 9 points. if not, you lose 1 point. write the equation for the expected - value $e(x)=\\frac{4}{52}(a)+\\frac{b}{52}(c)$. what is the expected value? should you play the game? for what expected value is a game fair? $e(x)=0$, $e(x)=1$, $e(x)>0$

you pick a card at random from an ordinary deck of 52 cards. if the card is an ace, you get 9 points. if not, you lose 1 point. write the equation for the expected - value $e(x)=\\frac{4}{52}(a)+\\frac{b}{52}(c)$. what is the expected value? should you play the game? for what expected value is a game fair? $e(x)=0$, $e(x)=1$, $e(x)>0$

Answer

Explanation:

Step1: Calculate probability of getting an ace

There are 4 aces in a 52 - card deck. So the probability of getting an ace, $P(\text{ace})=\frac{4}{52}$. When you get an ace, you get 9 points, so the contribution to the expected - value from getting an ace is $\frac{4}{52}\times9$.

Step2: Calculate probability of not getting an ace

The number of non - ace cards is $52 - 4=48$. So the probability of not getting an ace, $P(\text{not ace})=\frac{48}{52}$. When you don't get an ace, you lose 1 point, so the contribution to the expected - value from not getting an ace is $\frac{48}{52}\times(- 1)$.

Step3: Calculate the expected value

The formula for the expected value $E(X)$ of a discrete random variable is $E(X)=\sum_{i}x_ip_i$. Here, $E(X)=\frac{4}{52}\times9+\frac{48}{52}\times(-1)$. [ \begin{align*} E(X)&=\frac{4\times9}{52}-\frac{48}{52}\ &=\frac{36 - 48}{52}\ &=-\frac{12}{52}\ &=-\frac{3}{13} \end{align*} ] A game is fair when the expected value $E(X) = 0$. Since $E(X)=-\frac{3}{13}<0$, you should not play the game.

Answer:

$a = 9$, $b=-1$, and the expected value of the game is $-\frac{3}{13}$ points. You should not play the game. A game is fair when $E(X) = 0$.