It is a game you are playing with toss a coin and roll a die. In each game, you pay $2. If the coin shows a tail and die shows at least a five, you win $10..

Question

If you played this game 50 times, how much would you expect to win /loose? (Please write amount)

and what is the expected value ?

Answer 1

My approach is to calculate the probability of winning one game, then multiply expected win/loss by $50.

Probability of Winning One Game

P(coin_win) = probability of coin win condition (tail) = 1/2

Since coins have a 50/50 chance of heads and tails, the probability of getting tail is 1/2.

P(die_win) = probability of die win condition (5 and 6) = 1/6 + 1/6 = 2/6 = 1/3

For the die you will meet the winning condition if you roll a 5 or a 6, since probability of each number is 1/6, total probability for rolling 5 or 6 is 1/3.

P(win) = P(coin_win, die_win) = P(coin_win) * P(die_win) = 1/2 * 1/3 = 1/6

To win you need to meet BOTH coin and die conditions, this is a joint probability (probability of one event occurring at the same time that another event occurs). Joint probabilities are calculated by multiplying the probability of each event.

Win/Loss of One Game

Win = P(win) * Amount_win - cost = 1/6 * $10 - $2 = - $1/3

Expected win is probability of win (1/6) multiplied by amount you would win ($10) minus the cost to play ($2). You expect to lose $1/3 of a dollar every game.

Total Loss

Multiply one game's loss by number of times played:

(-$1/3) * 50 = -$16.67

Therefore you expect to lose $16.67 from this game.