A deck of cards has the aces and face cards removed, so that the numbered cards 2-10 remain. The player draws a card and is paid the value of the card. Each play costs $5. If you play the game 20 times, how much money would you expect to win or lose overall?

Question

I started to make a probability table with the x and p(x) labeled but i don't know how to continue from here.

Answer 1

There are 9x4 cards in the deck. All are equally probable.
if you get a 2, you lose 3 dollars
3...lose 2 dollars
4..lose 1 dollar
5 break even
6 win 1
7 win 2
8 win 3
9 win 4
10 win 5
The mean win is 1.50 per play, in 20 plays, that is 30 dollars.

Answer 2

The prob of drawing any specific number is 4/36 = 1/9
expected return on one game
= expectation(2) + expectation(3) + ... + expectation(10)
= (1/9)(2) + (1/9)(3) + ... + (1/9)(10)
= (1/9)(2+3+4+...+10)
= (1/9)(45) = 5

so it will cost 20($5) or $100 to play the 20 games
the expected return is 20($5).
So you will break even.

Answer 3

2+3+4+..+10 = 54 , not 45

so the expected return is (1/9)(54) = $6

so cost to play 20 games = 100
return = 20(6) = 120
so you would expect a profit of $20 to play 20 games