Quiz #4 solutions
1. (Comic from xkcd.com)
See the comic to the side. The machine is able to
correctly detect whether or not the sun has gone
nova. It then rolls two six-sided dice, and if both
come up 6, it lies and if not then it tells the truth.
Suppose you initially believe that there is a
probability of .00001 that the sun will have gone
nova at the moment when you ask the machine.
Then when asked, the machine says that yes, the
sun has gone nova.
a. So, use Bayes’ Rule to find the probability
P(N|Y), the probability the sun has in fact gone
nova (N), given that the machine says that yes it
has (Y).
|
|
.
.
.
.
…
b. Now, suppose you then quickly ask the machine a second time if the sun has gone nova, and it
replies “yes” a second time. Using the answer for (a) as the new “prior probability”, use Bayes’ Rule
again to find the updated probability the sun has gone nova.
Replace .00001 and .99999 with .00034988 and .99965012 respectively to get .01210187.
c. How many times would the machine have to answer “yes” consecutively for you to conclude
there is at least a 50% probability that the sun has gone nova?
If you want to use R efficiently, you could do something like this:
> p = .00001
> p = (35*p/36)/(35*p/36 + (1-p)/36)
>p
[1] 0.000349881
> p = (35*p/36)/(35*p/36 + (1-p)/36)
>p
[1] 0.01210187
> p = (35*p/36)/(35*p/36 + (1-p)/36)
>p
[1] 0.3000896
> p = (35*p/36)/(35*p/36 + (1-p)/36)
>p
[1] 0.937525
After about three consecutive “yeses”, I’d start getting concerned, and if the machine
answers “yes” four times in a row, you’ve got about 9 minutes to make your peace with
whatever you need to make peace with.
2. If you follow sports, it can be extremely difficult to differentiate skill/ability from random
chance.
NFL teams play a 16-game regular season. Over the last five years the New England
Patriots, coached by Emperor Palpatine (Bill Belichick) have won approximately 63.5% of all
the games they have played. So, assume that next year, the Patriots would win any given
game that they play with probability .635, and each game is independent.
a. Find the probability the Patriots would win 8 or fewer of their 16 games.
> pbinom(8, size=16, prob = .635)
[1] 0.1928697
b. Find the probability the Patriots would win 12 or more games.
Either
> pbinom(11, size=16, prob=.635, lower.tail=FALSE)
[1] 0.2480011
or
> 1-pbinom(11, size=16, prob=.635)
[1] 0.2480011
gives the solution.
c. Find the probability the Patriots would win all 16 of their games.
> dbinom(16, size=16, prob=.635)
[1] 0.000698843
d. Imagine replaying the upcoming NFL season 100 times. Use R to simulate the number of
games won by the Patriots in each of those 100 seasons. In how many of the seasons in
your simulation did the Patriots win 8 games or fewer? Show the output and mark those
seasons clearly in some way.
Answers will of course vary here, but use
> rbinom(100, size=16, prob=.635)
to generate the data.
[1] 7 12 8 12 8 11 9 9 11 10 9 9 10 11 10 14 12 13 12 9 12 10 9 11 11
[26] 9 9 14 10 10 12 10 10 10 11 10 10 9 9 13 12 13 9 9 8 11 7 14 9 10
[51] 10 9 8 9 13 7 10 11 11 11 11 11 9 9 7 8 8 6 13 11 8 11 11 13 9
[76] 14 9 11 10 12 14 11 10 11 9 9 11 5 7 10 11 9 12 14 10
In this particular sample, I count 14 seasons where the Pats won eight games or
fewer.
3. Powerball is a lottery which can be played in Georgia among many other states. On
5/18/2013, the jackpot for top prize reached an amazing $591 million. That week,
approximately 232,000,000 people played (i.e. that many tickets were bought). Since the
probability of winning the jackpot is 1/175,223,510, and it costs $2 to play, it is tempting to
think that playing the game would be a good deal ($591M times 1/175,223,510 = $3.37,
which is higher than the cost of a $2 ticket). However, as we’ll see, this reasoning is not
quite correct.
Here’s the short version of the solution.
Generate the probabilities of having 0,1,2,3,4,5 winning tickets.
> part3 = dbinom(0:5, size=232000000, prob=1/175223510)
> part3
[1] 0.266062710 0.352273214 0.233208962 0.102924695 0.034068672 0.009021543
Add to that data the probability of having 6 or more winning tickets.
> part3 = c(part3, 1-sum(part3))
> part3
[1] 0.266062710 0.352273214 0.233208962 0.102924695 0.034068672 0.009021543
[7] 0.002440205
Find the average value of a winning ticket by taking the payoffs and weighting them
with these probabilities.
> payoffs = c(328333333, 164166666, 109444444, 82083333, 65666666, 54722222,
46904762)
> weighted.mean(payoffs, part3)
[1] 182005915
Expected value of a $2 ticket:
> 182005915/175223510
[1] 1.038707
Thus, even when the jackpot is announced at $591M, the odds for playing the game
are still not truly favorable, as the expected value of a $2 ticket is only $1.04.
BONUS: Assuming 232 million people play as in this problem, how much does the
announced jackpot have to be to make a $2 Powerball ticket “worthwhile”?