Expected value HW question

Question

Suppose a student is offered a chance to draw a card from an ordinary deck of 52 playing cards and win $26 for an ace, $10 for a king, and $3 for a queen. If the student pays $6 to play the game, what is the amount the student expects to win each time?

$___

Answer 1

26(1/13) + 10(1/13) + 3(1/13)

= (26+10+3)/13

= 39/13

=$3

if it costs $6 to play, the expected "winnings" are 3-6 = -$3

he expects to lose, on average, $3 each time he plays

but 10 out of 13 times, he'll lose $6. He'll lose $6 each time, 10 out of 13 times. He'll break even 1 out of 13 times, win $20 1 out of 13 times and win $4 1 out of 13 times.

the mean "winnings" is a $3 loss. The median and mode "winnings" is a $6 loss.

20+4+0 -10(6) = 24-60 = -36

-36 divided by 13 = -36/13 = -$3 on average. Although on no draw will he ever lose exactly $3. It's just an average = "expected" outcome. It's like saying the average US family has 2.3 children, yet there is no family that has 2.3 children. It's "just" an arithmetic average.