"Extra" solutions

From HW 4 Suppose I want to go into the casino business with a simple gambling game..  In this game, you roll a fair, six-sided die.   You will win 1 dollar for each pip on the die if there are an odd number of pips.  You will lose one dollar for each pip if there is an even number of pips.   I'm going to charge some amount of money to play the game.  How much would I need to charge in order to make the game fair?  (A fair game is one in which neither side has an advantage -- so the expected value should be 0.)

Solution:

x	f(x)
1	1/6
-2	1/6
3	1/6
-4	1/6
5	1/6
-6	1/6

Let X represent the amount you win after rolling the die.  E(X) = 1(1/6) - 2(1/6) + 3(1/6) .... -6(1/6) = (1/6) *(9-12) = -.50

So you already start out at a disadvantage.  To make the game fair, *I* should pay *you* 50 cents.  Then your total winnings would be
X + .50  so E(X + .50) = -.50 + .50 = 0

I didn't intend this to be tricky.  It was a typo.  Should have been "you will lose one dollar for each odd pip, and win one for each even."


From HW 5 (the others are already in the on-line solutions)
 Suppose we toss a fair-coin 100 times.  Let X represent the number of heads.
    a) Use the normal approximation to find the probability that you will get more than 60 heads.
    b) Use the normal approximation to find the probability you'll get between 45 and 55 heads.  Between 40 and 60. Between 35 and 65.
    c) Compare your answers in (b) with the exact probabilities obtained from the binomial distribution.

Solution
a) X follows a binomial distribution with E(X) = 100*.5 = 50 and SD(X) = sqrt(100*.5*.5) = 5.
P(X > 60) = P(Z > (60-50)/5) = P(Z > 2) = 1-.977 = .023

b) P(45 < X < 55) = P(-1 < Z < 1) = .683
P(40 < X < 60) = P(-2 < Z < 2) = .954
P(35 < X < 65) = P( - 3 < Z < 3) = .997

c) P(45 < X < 55) = P(X = 45) + P(X = 46) + ... + P(X =54) = (100 choose 45) * (1/2)^2 100 + ... etc.
=.719
P(40 < X < 60) = 94.3
P(35 < X < 65) = .996

NOTE: you'll get slightly different probabilities if you use "less than or equal instead of equal")

From Homework 7
According to an urban legend, the Washington Redskins, a DC-based football team, are able to predict presidential elections.  The theory goes like this:  if the Redskins lose their last homegame before the election, the incumbant party loses.  If they win the last homegame, the incumbant party wins.  This method has been right 15 out of the last 16 times.   The only time it failed was this election.  On October 31, the Redskins lost to the GreenBay Packers and therefore the incumbant should have lost, according to this predictor.  However, he won.  Assume that the Redskins have no predictive ability whatsoever.  (In which case the string of successes is a coincidence.)  What's the probability of getting 15 or more correct predictions in 16 elections?

Solution
This is a pretty open-ended problem the way it is stated. But the point of this exercise is to see how tools you've learned can help you attack it.  First step is to recognize what we're trying to observe.

Teh Redskins play a game (one game in particular every four years) and we note whether or not this game makes a successful prediction.  So let X represent the number of successful predictions they make in n attempts.

The purpose of this problem is to compare a theoretical outcome with an observed outcome. We observed x = 15 out of 16.  What sort of theoretical model should we compare this to?

Well, if there is no real predictive ability, then the prediction should be just as likely to be right as to be wrong.  So the probability of a success should be p = .5.

You should see now where this is heading. X will be a binomial random variable. We have a fixed number of trials (n = 16), the outcome of each trial is a success (correct prediction) or failure (incorrect prediction), trials are independent (no way of knowing, but it seems safe to assume that whether or not the team makes a successful prediction in one election would not affect it in the future), and the probability of a success, p = .5, we will assume to be the same at each trial.

Now we're asked to find P(X >= 15) = P(X = 15) + P(X = 16) = (16 choose 15) * (1/2)^15 * (1/2) + (16 choose 16) * (1/2) ^ 16 = .0002

In other words, it is possible that the outcome we saw (15 correct predictions out of 16) happened by chance, but it happens very rarely.

What does this mean? I have no idea.  It's really hard to believe that this team is able to predict the outcome of the elections.  there is a subtle flaw in our logic, which I hope we get a chance to explore more later.  The flaw is that we're assuming that this "experiment" could be repeated infinitely many times.  So we play 16 Redskin games, observe the outcome of the election, and repeat that infinitely many times.  Obviously this isn't possible, and so it isn't really right to apply a probability model in this context.