Question

Manuel has a deck of 10 cards numbered 1 through 10. He is playing a game of chance.
This game is this: Manuel chooses one card from the deck at random. He wins an amount of money equal to the value of the card if an odd numbered card is drawn. He loses $4.20 if an even numbered card is drawn.

What can Manuel expect in the long run, after playing the game many times?
(He replaces the card in the deck each time.)

Manuel can expect to gain money.

He can expect to gain __ dollars per draw

Manuel can expect to lose money.
He can expect to lose __ dollars per draw

Manuel can expect to break even (neither gain nor lose money).

Answer 1

prob(to pick a specific card) = 1/10
expectation of win
= (1/10)(1+3+5+7+9) = 25/10 = 2.50
expectation of loss
= (1/10)(2+4+6+8+10) = 30/10 = 3.00

bad game to play!