Let's Go Out to the MoviesProbability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Gretchen and Henry were sent to their rooms for fighting in the house. They each separately voiced their protest to their father, insisting that the fight was nothing more than healthy sibling competition, and they each wanted to go out that afternoon to see a movie. He was moved by their stories, but wouldn't simply set them free. Instead, he devised a system. He went to each child's room with a penny, and told them that they would have to show up in the den in 10 minutes, and could choose to bring the penny with them or leave it in their respective rooms. Dad would then flip the one or two pennies brought to the den, and if the pennies he flipped came up heads, the kids could go to the movies. If neither brought a penny, or if he flipped at least one tail, they would stay in their rooms until supper time.
The problem facing Gretchen and Henry was that neither knew what the other would do. It would be easy if they could collude -- one would bring a penny, and the other would not, giving them a 50% chance of going free -- but they did not have this luxury.
If they both acted optimally, what is the probability that they will be free in time to see the movie?
HintIf they both bring a penny 100% of the time, they will never be sent back without a flip, but they will only have a 25% chance of going free. Can they do better? Acting optimally means they will have to pursue some course of action some percentage of the time.
AnswerThey should each bring the penny with probability 2/3, and they will go free 1/3 of the time.
There are a couple of ways of tackling this. The first is to say that what is optimal for Henry must also be optimal for Gretchen, so whatever probability one has of bringing the penny should equal the probability that the other brings the penny. If we set that probability equal to 'p', then the probability that they go free ('f') is:
p^2*(1/4) + p*(1-p)*(1/2) + (1-p)*p*(1/2) + (1-p)*(1-p)*0
Since the last term goes to 0, this leads to an equation of:
p - (3/4)*(p^2) = f
From here, you could either plug in values of p from 0 to 1, finding that f is at its greatest (1/3) when p is equal to 2/3, or you could use calculus.
Warning: CALCULUS FOLLOWS!
Since this is a quadratic equation, the value will be at its maximum or minimum when its first derivative equals 0. The first derivative of f with respect to p is:
df/dp = (-3/2)p + 1
Setting this equal to 0, we get p = 2/3.
To determine if this is a maximum or a minimum, we need to take the second derivative:
d2f/dp2 = (-3/2)
Since the second derivative is negative, the value we found by setting the first derivative equal to 0 is a maximum, and we have our answer.
CALCULUS IS OVER
Another way we can solve this is to say that each needs to make a decision that will equalize their chance of survival REGARDLESS of the action taken by the other.
This point, where you get the same result regardless of the other person's actions, is known as the "Nash Equilibrium Point". The brilliant mathematician John Nash (who was made famous in the book and movie "A Beautiful Mind") showed that this point of equilibrium will always give the maximum overall result, both in a cooperative game like this one and in a competitive game such as poker, as long as the correct strategy is a mixed strategy like this.
So if Henry assumes that Gretchen will bring her penny with her none of the time, all of the time, or somewhere in between, he wants to pick a probable course of action (in game theory, a "mixed strategy") that will maximize his chances of going free. Let's just look at the two extremes (i.e., Gretchen will either bring her penny 100% of the time, or 0% of the time).
If Gretchen brings the penny 100% of the time, and Henry brings his penny with probability p, they will go free with probability:
p*1*(1/4) + (1-p)*1*(1/2)
If we assume Gretchen never brings her penny, then they will go free with probability:
Since Henry wants to be indifferent to Gretchen's actions, we can set these two probabilities of going free equal to each other. We then get:
(p/4) + (1-p)/2 = p/2 [setting equations equal]
(p/2) + (1-p) = p [multiplying through by 2]
1-p = (p/2) [subtracting p/2 from each side]
1 = (3/2)p [adding p to each side]
p = 2/3 [solving for p]
Gretchen will determine the same probability of bringing a penny by using the same logic.
Thus, they will each bring a penny 2/3 of the time, maximizing their joint chances of going free at 1/3.
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