Day 02, MAT115

As you do this homework, remember our "lessons for life":

1. Keep an open mind.
2. Just do it. Jump in. Make mistakes and fail, but never give up.
3. Often when we've done it once, we do it again. Follow up one good deed with another!
4. Understand simple things deeply.
5. Break a difficult problem into easier ones.
6. Look for patterns and similarities.
7. Explore the consequences of new ideas. Generalize.
8. Examine issues from several points of view.
   Don't be a turkey -- be a dog! (this one may require a little explanation....)

Above all, please write things up yourself -- do your own work -- even if you're working with someone else. After all, you'll be writing the solutions up yourself on the exams....

We might start by taking a look at that first reading, on Probability. Let's see what lessons the author wants us to learn:

• Idea #31, a (a little history)
• Idea #31, b (ground rules)

• What's the probability that a die comes up 5? 6? 5 or 6? (This is the focus of your first reading.)

• Is there such a thing as "being due"? If my probability of making a freethrow in basketball is 1/2, and I miss this shot, will I (of necessity) make the next? If I miss three in a row, am I due?

• Big Ideas:
  1. Generally, if the situation is "fair" (meaning that all events have the same chance of happening), then
     

     The set of "OutcomesPossible" is called the universe, or sample space. We often draw a picture to help us do our calculations, showing the universe, as well as those events that are "desired".

  2. Probabilities are numbers between 0 and 1.
  3. The probability of a certain event is 1.
  4. The probability of an impossible event is 0.
  5. The sum of the probabilities of each separate outcome possible is 1.
  6. When occurrences cannot coincide, their probabilities can be added.

     This leads to the idea of complementary events: if two events are complementary, then one or the other must happen, but not both.

     So the sum of the probabilities of the complementary events must be 1:

     

• Assuming a fair six-sided die, let's calculate the probability that a die comes up 5, 6, and (either 5 or 6) on a given roll.
  • Draw the universe.
  • In each case, find those events that are desired.
  • Calculate the probability of the desired outcome.
  • Calculate the probability of the complement of the desired outcome, and describe it.

• Let's look at another problem: what's the probability of tossing a fair coin three times and getting exactly two heads?

  Let's use a tree to create the universe. [We'll find trees very useful throughout the course, and we'll actually be studying them later as graphs.]

• It's not always easy to figure out the theoretical probabilities (as we did in the last two examples).

  What's the probability that an 81-year-old female in Wood County, Ohio will survive another year?

  That's the kind of question an insurance adjuster is concerned about, as am I: that's my mom!

  An approximation can be obtained by looking at records for Wood County for the last year:

  

• Your reading told you about the game of craps. We were able to use a table to compute the probability of a 7 or 11:

  1 2 3 4 5 6 1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) 2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) 3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) 4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) 5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) 6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

  The sample space or universe for a roll of two dice.

  Let's do a few more calculations:

  1. 
  2. 
  Now that we've calculated those, here are some easier ones:
  1. 
  2. 
  These events are called complementary events: one or the other has to happen, but not both. We can compute the probability of a complementary event by doing a simple subtraction from 1:

1. Remember our "Question of the day" from last time:
   What's the probability that two people in this room have the same birthday?

   How much would you have bet against me? Would you have made a smart bet?

   "Odds" is an equivalent way of thinking about this probability: how much would you bet against my bet of $10 that there will be two people in this room with the same birthday?

   Even money (odds)? If so, then you would be saying that the probability of there being two people with the same birthday is 1/2.

   Essentially odds comes up when you look at an event and its complement. So in a horse race, for example, odds of 40 to 1 mean that in 41 races, your horse should win 1 time and other horses 40 times. So to get to odds from probability is pretty easy, actually: if the probabilities of the event is and hence the probabilities of the complement of the event is then the odds are 1 to 2 (just look at the numerators, as long as both probabilities share a common denominator).

2. What about leap years? Forget 'em! (Mathematicians often simplify a problem to make it "tractable" (that is, easier to deal with).)
3. Let's suppose there are 35 people in the room. How many total ways could we have birthdays? Since each person makes their "choice" independently, we multiply.
4. An event is either a success (two people share a birthday) or a failure (no common birthdays). These are complementary events. One or the other must happen, but not both. It's actually a lot easier to compute the number of ways that 35 people can have different birthdays (what I've called a "failure").
5. My birthday probability calculator gives me about 0.81 -- how do we get that?
   

6. So how much should one have bet against my $10 (so that the game would have been fair)?
7. I would have been right about 81 times out of 100. So if I bet $81 dollars, and Bobby would have bet $19, then over the long term (say 100 games) I'd pay to Bobby, on average, $81*19 whereas Bobby would pay to me $19*81 The same amount of money! That's a fair game. So I should have bet 81/19 as much as he bet me, or $4.26 to his $1. Whoops! I made a stupid bet (and Bobby made a very smart one!). Bobby would have been making a smart bet win or lose (and he was quite likely to lose, by the way -- he loses, on average 4.26 times to every time he wins).

8. General rule: only bet if the odds are in your favor (except to illustrate a concept in math class!;), and that Lotteries are a tax on people who are bad at math.

9. Notice how the author of the Birthday article (Tony Crilly) generalizes the problem: having solved one problem, he immediately begins thinking about generalizations of the game.
   Explore the consequences of new ideas. Generalize.

• The rules:
  1. There are three doors, behind which are two donkeys and a car.
  2. You choose one: Monty Hall then shows you a door behind which appears a donkey, then offers you the chance to either "stick" with the door you chose at first, or "switch" for the third door.

• The question: Do you stick or switch? How can we decide?

  Marilyn vos Savant said "Switch!" -- and she's the smartest person on the planet, right?

• The analysis: We should be able to do this theoretically:

  • What's the sample space?

  • What's your probability of success given that you stick? What's your probability of success given that you switch?

    Sticking and switching are complementary strategies. One strategy will win, and one strategy will lose. Hence,

    P(win by sticking)+P(win by switching)=1

    from which we conclude

    P(win by sticking)=1-P(win by switching)

  • The easy one to calculate is the probability of winning by sticking: 1/3. Hence, switching is better by twice, 2/3!

• Let's check our answer: we might simulate the game.

  Let's break into groups to play, and see if our theoretical answer is correct:

  • Three cards represent the three doors; choose one card to be the "car", and the other two cards to be donkeys.
  • Try 10 (or more) plays where one player always sticks and the other always switches.
  • Tally up the successes of each strategy (realizing that if sticking wins, switching would have lost, and vice versa -- the events are complementary).
  • Is one strategy better? If so, by how much? "What are the odds?"
  • Then we'll combine the class's data, to "see what the data says".
  • Now: Can you justify/explain your answer?

• Is it possible that by restricting the game, or by extending the game, we might learn more?

  To give you more confidence in the answer (and perhaps to learn more!), we may consider extending the game: what if we have ten doors, one car, and nine donkeys. After you pick your door, Monty shows you eight doors with donkeys behind them. Would you switch for the remaining door? How much better is one strategy over the other?

You rub an old oil lamp and a genie appears. You ask for a wish - since you let him out of the lamp, after all. Reluctantly the genie agrees. You tell him you want the Royal Black Diamond - a famous gem worth millions of dollars. At your feet appear 27 identical - looking gems. You ask Mr Genie what is going on and he replies that you may take any one of the stones with you. They are identical in appearance, but the genie explains that the Royal Black is slightly heavier than the 26 fakes. At this you say, " Great I wish I had a way to weigh them!" The genie grants your wish and a magic balance scale appears at your feet. The genie explains that upon being used three times the scale will disappear. Can you find a way to choose the right gem from the 27?

With 8 stones we could put half on one side and half on the other and the heavier side will contain the RBD. Repeating this twice will leave us with the heaviest stone after 3 uses of the scale. Now how do we adapt this idea to 27 stones?

We could put some, say 10 gems on one side, 10 gems on the other side, and leave 7 off the scale. If one side of the scale is heavier, then it must contain the RBD. If neither side is heavier, then the RBD is among the 7 left off the scale. So we can learn about gems not on the scale! But we need to divide the gems up more carefully....