Day 03, MAT115

• You have an assignment due Thursday -- see the assignment page, and make sure to keep up!
• I've also assigned some more readings. Thursday I'll assign a few more problems to be turned in next week, Thursday.

Then we'll wrap up the Birthday problem.

• 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 (maybe a tree) to help us do our calculations, showing the universe, as well as those events that give the "desired" result.

• Properties of probabilities:
  • Probabilities are numbers between 0 and 1.
  • The probability of a certain event is 1.
  • The probability of an impossible event is 0.
  • The sum of the probabilities of each separate outcome possible is 1.
  • The probability of a succession of independent events is the product of the probabilities.
  • When occurrences cannot coincide, their probabilities can be added.

    This last property 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:

    

• You roll one fair six-sided die and toss one fair coin [what's the universe look like?]. What's the probability of
  1. tails and a 3?
  2. an even number on the die and heads?

• You pull one card from a standard deck of 52 cards [what's the universe look like?]. What's the probability of
  1. a heart?
  2. a five?
  3. a heart or a five? (be careful!)

1. Last time we computed the probability of "no common birthdays":
   

   This was computed by using the independence of events.

   The complementary event, "at least two people have the same birthday", will have probability about 1-.19=.81.

   So if we played this bet 100 times, then I would have been right about 81 times out of 100. So if I bet $81 dollars, and Bobby had bet $19, then over the long term (say 100 games) I'd pay to Bobby, on average,

   $81*19 whereas Bobby would pay 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).

   In terms of odds, we say that the odds were 81 to 19 that I was going to win the bet (because the probabilities were

   and

2. Now we'll show how one might use trees, property of probability such as the multiplication of probabilities of independent events or adding up non-coincident events to get the same result. It's always nice in math to examine issues from several points of view.

   I begin with a coin-tossing example like we studied last time.

   We continue with a simpler example of the birthday problem: Suppose four people have birthdays in the first week of February. What's the probability that two (or more) of them have the same birthday? We'll use trees.

   Finally we consider the birthday problem itself.

3. 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 (e.g. "What's the probability of having three people with the same birthday?").
   Explore the consequences of new ideas. Generalize.

There are at least three good suggestions that I know of:

• Tallies (A couple of your assigned readings talk about tallies).

• One-to-one correspondence
  • using body parts -- "one hand" of sheep, say -- meaning five
  • cairns

• And then an unusual method of "counting by partitions" that Patricia Baggett and Andrzej Ehrenfeucht proposed at the 2011 National Math Meetings.
  • Let's begin by counting some students their way....

  • They proposed that primitive societies may have counted this way. Let's suppose you need to let the King know how many sheep you have:
    1. divide your sheep equally ("one for you, one for me") into two pens: either there is one left over, or not. You make a note of whether there is one left over or not.
    2. Send all the sheep in pen two (and any "left over") out to pasture, and then
    3. You divide the sheep in pen one into pens one and two: i.e., just do it again! And again, and again, and.... until you get down to a pen one with just sheep in it.
    4. Now let's see how we might record the results to send to the King.

       The easiest way to illustrate the counting method is via a tree -- which we've seen before, when doing probabilities. Finding the sample space of four coin tosses was best done using trees. But it's a little different, because not all the branches occur.

       Let's see how we might use a tree to represent the solution to the 22 counting problem: in the linked example, we would get 10110 by writing the remainders from left-to-right starting from the bottom of the tree. (The result should always start with a 1 if done correctly, since we always end with one sheep!)

       

       The answer will be written as 1, 0, 1, 1, 0

       That is, from the bottom up, left to right. This is important! We have to have a consistent scheme for writing.

       So how do we write

       • 9 sheep
       • 31 sheep
       • 54 sheep

    5. Can you go backwards? How many sheep is meant by
       1. 1,1,0,1,1,0,0
       2. 1,1,0,1,1,0,0
       3. 1,0,0,1,0,1,1,0,1

       Try making a tree with these remainders.

    6. and now let's count some coins:
       1. We'll be counting pennies - in small groups.
       2. Divide your penny rolls up (no need to do it evenly!)
       3. Get your answer, then swap with another nearby group.
       4. Compare (and record) your answers