Day 03, MAT115

• Read Idea 01, Zero, pp. 4-7
• Read Idea 09, Primes, pp. 36-39

I've still got $10 that says that there are two people in here who have the same birthday. How much will you bet against me?

Let's take a poll, and see where folks sit:

$0
between $0 and $5
between $5 and $10
$10
between $10 and $15
between $15 and $20
greater than $20

Given equally likely outcome events, we first determine the number of outcomes possible in the sample space (e.g. toss of a single fair coin means two, roll of an ordinary fair die means six, craps -- roll of a red and blue die -- has a sample space of 36).

We then count how many of the outcomes consitute a "success"

1. What about leap years? Forget 'em!
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). It's actually easier to compute the number of failures.
4. Birthday probability calculator:

5. So how much should one have bet?

• 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. It turns out that there is an "optimal strategy". Your mission: determine what strategy is best.
  • 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?

• Why don't we 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 the sticking strategy.
  • Tally up the successes of each strategy (realizing that if sticking wins, switching would have lost, and vice versa).
  • 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, and, 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?