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....)

Please don't just go out on-line and download someone else's solutions! Don't copy your friend's solutions: write things up yourself, even if you're working with someone else. After all, you'll be writing the solutions up yourself on the exams....

So I lost $20 to Kyle's $1, with my money betting on two people in here having the same birthday. How much would you have bet against me? Would you have made a smart bet? Kyle did!

"Odds" is an equivalent way of thinking about this probability: how much would you bet against my bet of $20 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.

• 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?

• 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".

• 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.
• Let's look at another problem: what's the probability of tossing a fair coin three times and getting exactly two heads?
• Now, it's not always easy to figure out the theoretical probabilities (as we did in the last two examples).

  What's the probability that a 78-year-old white 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:

  

• Now 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 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. What about leap years? Forget 'em! (Mathematicians often simplify a problem to make it "tractable" (that is, easier to deal with).)
2. Let's suppose there are 37 people in the room. How many total ways could we have birthdays? Since each person makes their "choice" independently, we multiply.
3. 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 37 people can have different birthdays (what I've called a "failure").
4. Birthday probability calculator gives me about 0.85.
5. So how much should one have bet against my $20 (so that the game would have been fair)?
6. I would have been right about 85 times out of 100. So if I bet $85 dollars, and Kyle would have bet $15, then over the long term (say 100 games) I'd pay to Kyle, on average, $85*15 whereas Kyle would pay to me $15*85 The same amount of money! That's a fair game. So I should have bet 85/15 as much as he bet me, or $5.67 to his $1. Whoops! I made a stupid bet (and Kyle made a very smart one!). Kyle would have been making a smart bet even had he lost (which he was quite likely to do, by the way -- he loses, on average 5.67 times to every time he wins).

7. General rule: only bet if the odds are in your favor (except to illustrate a concept in math class!;).

8. Lotteries are a tax in people who are bad at math.

• 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.
  3. Do you stick or switch? How can we decide?
  4. Marilyn vos Savant said "Switch!" -- was the smartest person on the planet right?
  • 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)=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!

  • To check our answer, we will 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 to be the donkeys.
    • Try 10 (or more) plays using just the sticking strategy.
    • 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?

• This problem moves us away from probability, and into the realm of number representation, which is our next topic.

  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?

• Now: one of our strategies is to look at things from different perspectives. Another is to break difficult problems into simpler ones. We could do this problem if... If there were only three stones! How?

  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....