Day 03, MAT115

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....

1. The set of all possible outcomes is called the universe of the experiment, or sample space.
   
2. Probabilities are numbers between 0 and 1.
3. The probability of an event that is certain 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 events cannot coincide, their probabilities can be added.
7. One special situation in which "occurrences cannot coincide" is where we break the universe into two pieces which we might call event E and event "not E". These are called complementary events, because they are non-overlapping, and every outcome falls into one event or the other, and we learned that
   

So the probability of two or more in a room having a common birthday is the complementary probability, or

Think about that -- it's pretty amazing! It turns out that if you have 23 strangers in a room, it's more likely than not that two of them will share a birthday.

So if you walk into a party with 23 or more people, people you don't know, and you make an even money bet that two people share a birthday, then over your lifetime you'll win more money than you lose.

That's inside information. Only gamble if you have inside information that gives you the edge.

Let's say Thad is going to play against me. How much should Thad bet against my $10 (so that the bet is "fair" in the sense that if you play over and over, on average you break even)?

Let's say we play this game 100 times (in 100 classes of 35 people each).

Based on the probabilities, I would have been right about 81 times out of the 100. So if I bet $81 dollars, and Thad bet $19, then over the long term I'd pay to Thad, on average,

If I bet $10, then Thad would have bet 19/81 of that: $2.35

• The odds of my winning are the ratio of the probabilities, mine to Thad's, and the ratio of the bets should be the same as the odds:
  

• only bet if the payoff is in your favor (except to illustrate a concept in math class!;), and that Lotteries are a tax on people who are bad at math.

• The rules:
  1. There are three doors.
  2. Behind two doors there are donkeys; behind the other door there's a car.
  3. You choose one door.
  4. Monty Hall then shows you that behind one of the doors which you didn't choose there is a donkey.
  5. Monty 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?

• 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? (solution)

  • Time for a contest....

• 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?