You pick a card at random from an ordinary deck of 52 cards. If the card is an ace, you get a points; if not, you lose 1 point.
Write the equation for the expected value.
[tex]
\begin{aligned}
E(V) & =\frac{1}{52}(a)+\frac{b}{52}(c) \\
a & =\square c=\square
\end{aligned}
[/tex]
Answer (1)
The given equation is E ( V ) = 52 1 ( a ) + 52 b ( c ) .
a represents the points gained when drawing an ace.
c represents the points lost when not drawing an ace, which is -1.
Therefore, a = a and c = − 1 .
Explanation
Analyze the problem and data We are given the equation for the expected value of drawing a card from a deck: E ( V ) = 52 1 ( a ) + 52 b ( c ) We need to determine the values of a , b , and c .
Determine the value of a In a standard deck of 52 cards, there are 4 aces. If you draw an ace, you get a points. So, the probability of drawing an ace is 52 4 . The problem states the equation uses 52 1 ( a ) , which means that a represents the points you get if you draw an ace.
Determine the values of b and c If the card is not an ace, you lose 1 point. The number of non-ace cards is 52 − 4 = 48 . So, b represents the number of non-ace cards, which is 48. The points lost when a non-ace card is drawn is represented by c , which is -1.
Conclude the values of a and c Therefore, a is the number of points you get if you draw an ace, b is the number of non-ace cards, and c is the number of points lost if you draw a non-ace card. So, a is the points gained when drawing an ace, and c is the points lost when not drawing an ace, which is -1.
State the final answer Thus, we have a = a and c = − 1 .
Examples
This type of expected value calculation can be used in various games of chance, such as card games or lotteries, to determine the average outcome of playing the game over a long period. It helps in assessing whether the game is favorable or unfavorable in the long run.
