Rules for various games are given. Use expected value to help analyze the game. Notice that a "fair game" means all players' expected values are equal, or that the game's expected value is 0.

Roll 2d6 (2 dice with 6 sides).

- If sum is odd, player A gets 1 point.
- If sum is even, player B gets 2 points.

Is this game fair? If fair, explain why. If not, provide a scoring scheme to make the game fair.

No. Player A has expected value of 18/36 while player B has expected value of 36/36

Roll 2d6.

- If product is even, player A scores 1 point.
- If product is odd, player B scores 3 points.

Is this game fair? If fair, explain why. If not, provide a scoring scheme to make the game fair.

Yes, the expected value of both is 27/36.

Roll 1d6.

- Get a six and win $4.
- Get anything else and lose $1.

If 100 people play this game, how much money will the game makers (not the players) expect to make?

EV(each game) = \( (-$4)\frac{1}{6} + ($1)\frac{5}{6} = $\frac{1}{6} \); 100 games implies \( 100 ($\frac{1}{6}) ~= $16.67\)

Pay $1. Roll 1d6.

- Get a six and win $4.
- Get anything else and gain nothing.

If 100 people play this game, how much money will the game makers (not the players) expect to make?

EV(each game) = \( (-$4)\frac{1}{6} + ($1)\frac{6}{6} = $\frac{2}{6} \); 100 games implies \( 100 ($\frac{2}{6}) ~= $33.33\)

alt. EV(each game) = \( \frac{1}{6} (-$3) + \frac{5}{6} ($1) = $\frac{2}{6} \); 100 games implies above answer

There are three possible attackers at the end of this round. One die will be rolled and the damage of each attacker is listed below the dice faces.

We can block one of the attackers using the green dice at the bottom. Which attacker should we block and why?

One interpretion would be:

\( EV(Intercepter X) = (-1) \cdot \frac{5/6} = -\frac{5}{6} \)

\( EV(Assault Cruiser) = (-2) \cdot \frac{2/6} = -\frac{4}{6} \)

\( EV(Raiders) = (-2) \cdot \frac{2/6} = -\frac{4}{6} \),

which indicates the Intercepter X is the greatest threat. What would happen if we decide Raiders should be lower value because they ignore shields?

Pick a number and roll 2d6.

- Picked number comes up twice, win $4. \(EV(\text{twice}) = $4 \cdot \frac{1}{36} \)
- Picked number comes up once, win $1. \(EV(\text{once}) = $1 \cdot \frac{10}{36} \)
- Picked number doesn't appear, lose $2. \(EV(\text{none}) = -$2 \cdot \frac{25}{36} \)

What are the expected earnings for the player of this game? (hint: it doesn't matter which number picked)

The expected value of the game is -$1, so on average you will lose $1 per play.

Pick a number and roll 3d6.

- Picked number comes up thrice, win $9. \(EV(\text{thrice}) = $9 \cdot \frac{1}{216} \)
- Picked number comes up twice, win $4. \(EV(\text{twice}) = $4 \cdot \frac{15}{216} \)
- Picked number comes up once, win $1. \(EV(\text{once}) = $1 \cdot \frac{75}{216} \)
- Picked number doesn't appear, lose $2. \(EV(\text{none}) = -$2 \cdot \frac{125}{216} \)

How many outcomes are there? (hint: more than 18) 216

What are the expected earnings for the player of this game?

(hint: a weighted tree diagram is recommended because your number either appears, or doesn't)

The expected value of the game is -106/216, so on average you will lose 50 cents per play.

Review the odds for Powerball (https://wilottery.com/games/powerball). A ticket costs $2. Assume the jackpot is $250,000,000.

- Recall that odds represent different events, not total outcomes. For example, 1:39 odds means there is 1 chance to win and 39 chances to lose. Thus the probably of winning is 1/40. What are all the probabilities of winning money in the PowerBall?
- Based on your previous work, what are the expected earnings per ticket?

\( \$250{,}000{,}000 \frac{1}{292,201,339} + \$1{,}000{,}000 \frac{1}{11,688,055} + \$50{,}000 \frac{1}{913,131} + \$100 \frac{1}{36,527} + \$100 \frac{1}{14,496} + \$7 \frac{1}{581} + \$7 \frac{1}{703} + \$4 \frac{1}{93} + \$4 \frac{1}{40} - \$2 ~= -0.829 \)