You play a gambling game with a friend in which you roll a die. If a 1 or 2 comes up, you win $8. Otherwise you lose $2. What is your expected value for this game?

Question

I have been trying to figure this out on my own. Can anyone help me?

Answer 1

If you roll the die once only, you have ⅓ probability of winning $8 and ⅔ probability of losing $2. So the odds of winning are 2:1 against. You can expect to lose!

If you roll the die twice you have ⅟₉ probability of winning $16, a probability of ⁴⁄₉ of winning $6, and a ⁴⁄₉ probability of losing $6. The odds of winning something (rather than losing) are 5:4 in favour.

For 3 rolls:

P=⅟₂₇ $24

P=²⁄₉ $14

P=⁴⁄₉ $4

P=⁸⁄₂₇ -$6 Odds of winning something 19:8 in favour. Expect to win!

Probability P=⅓ for rolling 1 or 2; P=⅔ for rolling 3-6.

Average all outcomes: ⅓(8+(-2)+(-2))=⅓(8-4)=4/3. So the average (expected value) is $4/3=$1.33 approx.