### Instruction: In the Chicken Game involving two cars racing towards each other, if both drivers swerve away from each other, neither gets hurt and they both earn 2 points. However, if one driver swerves and the other doesn't, the swerving driver loses and gets 0 points while the non-swerving driver earns 3 points. If both drivers continue straight towards each other, they both lose and earn 1 point. If the probability of each driver swerving is 0.6, what is the expected outcome of the game in terms of points earned by each driver?### Response: Let's denote the probability of swerving as S and the probability of not swerving as N. We are given that the probability of swerving (S) is 0.6, so the probability of not swerving (N) is 1 - 0.6 = 0.4.

There are four possible outcomes in the Chicken Game:

1. Both drivers swerve (SS): They both earn 2 points.
2. Both drivers don't swerve (NN): They both earn 1 point.
3. Driver 1 swerves, and Driver 2 doesn't swerve (SN): Driver 1 earns 0 points, and Driver 2 earns 3 points.
4. Driver 1 doesn't swerve, and Driver 2 swerves (NS): Driver 1 earns 3 points, and Driver 2 earns 0 points.

Now, let's calculate the probabilities of each outcome:

1. Probability of SS: P(SS) = P(S) * P(S) = 0.6 * 0.6 = 0.36
2. Probability of NN: P(NN) = P(N) * P(N) = 0.4 * 0.4 = 0.16
3. Probability of SN: P(SN) = P(S) * P(N) = 0.6 * 0.4 = 0.24
4. Probability of NS: P(NS) = P(N) * P(S) = 0.4 * 0.6 = 0.24

Now, let's calculate the expected points for each driver:

Expected points for Driver 1:
E(D1) = P(SS) * 2 + P(NN) * 1 + P(SN) * 0 + P(NS) * 3
E(D1) = 0.36 * 2 + 0.16 * 1 + 0.24 * 0 + 0.24 * 3
E(D1) = 0.72 + 0.16 + 0 + 0.72
E(D1) = 1.6

Expected points for Driver 2:
E(D2) = P(SS) * 2 + P(NN) * 1 + P(SN) * 3 + P(NS) * 0
E(D2) = 0.36 * 2 + 0.16 * 1 + 0.24 * 3 + 0.24 * 0
E(D2) = 0.72 + 0.16 + 0.72 + 0
E(D2) = 1.6

So, the expected outcome of the game in terms of points earned by each driver is 1.6 points for Driver 1 and 1.6 points for Driver 2.