SchecterAdvancedMath

I am thinking of a number. It is either 1, 10 or 100. If I asked you to guess the number, you would have one out of three chances of getting it right, or simply 1/3. That is all there is to basic probability. You count how many possibilities there can be (3 in this case); you count how many of those make you successful (1 in this case); and you divide the number of successful outcomes by the total number of outcomes. If I said I was thinking of two numbers from the set 1, 10, 100 and you had to guess either one to win, then your chance to win would be 2/3.

This is a very simple idea but it can become very difficult as the things you are working with become harder to count. Here are some simple examples to make sure you get the idea:

What is the probability of throwing heads when you flip a coin? There are two outcomes (don't be a smartypants and suggest that the coin can land on it side) and you win with one of them. So the chances are 1/2.

What is the chance of throwing a 2 or a 5 when you roll one die? There are six outcomes, and you win with two of them, so the chances are 2/6 = 1/3.

Now for a harder example. You may recognize this as the start of a game of Craps.

What is the chance of throwing a 7 or 11 when you roll two dice? Let's first do it wrong, just to so you can see how it easy it is to make mistakes in probability. There are 11 possible sums 2, 3, 5, 6, 7, 8, 9, 10, 11, 12. We win with two of this so the probability is 2/11... which is completely wrong! The reason it is wrong is subtle. When you count the number if possible outcomes, you must be sure you are counting the right thing. The possible outcomes need to be things that equally likely to occur. We counted 2 and 7 each once in the list of outcomes, but there are many ways to get a 7 and only one way to get a 2. Really we should have counted 7 six times: 1-6, 2-5, 3-4, 4-3, 5-2, 6-1.

Here is the correct way to solve the problem. There are really 36 outcomes, as you can see in the table below:

One die is listed vertically and the other horizontally. Their sums appear in the table. Out of these 36 there are 8 that add up to either 7 or 11, so the chance of rolling a 7 or 11 is 8/36 = 2/9. You might have noticed that the chance of not rolling a 7 or 11 is 7/9, or simply 1 - 2/9.

This is general principle I in probability - the complement principle: The chance of not having something is equal to 1 minus the chance that it does happen.

Applications - Casinos:

Why do we care about probability? Because it helps up make predictions about the future. For a gambler, meteorologist, and investor, probability is their bread and butter. For example, let's say I want to bet on a game where I roll two dice and I win if I get a 7 or 11. We already calculated that the chance I win this game is 2/9. Does that mean that every nine times I play, I will win twice and lose seven times? Absolutely not! I might get lucky and win all nine times. It does mean, however, that in the long run my winning percentage is expected to be 2/9. So if I play 9 million times, I expect to win around 2 million times. It is much more unlikely to win 9 million times in a row, than it is to win 9 times in a row. This is called the law of large numbers, which is quite hard to rigorously state but much easier to understand inuitively. Let's get back to our game.

Say the casino offers to pay me 3:1 if I win. That is, if I bet $100 and I win, then they give me my bet back plus a gain of $300. What will happen if I play this game a lot? In the long run I will win 2/9 and lose 7/9. Assuming I bet $100 every time (big spender) then after nine times I gain $300 twice, and lose $100 seven times. This is a net loss of $100 for every $900 I spend on the game. I might win in the short run by luck, but in the long run I am playing a game that is hugely in favor of the casino. That is how casinos make money. It is a no-brainer business. Just sit back let probability and the law of large numbers do its work.

In real life, casinos make their games more favorable to make it more fun and exciting for the player. For example, in the real game of Craps the chance of you winning is about 493/1000, and the odds are 1:1. Hence if you spend $1000 on Craps then you expect to win around $493 and lose around $507 for a net loss of only about $14. For lots of people, this might be a fair price just for the entertainment value of sitting around a craps table and making $1000 worth of bets. Of course, you might get lucky and win money, or get unlucky and lose all your money, but in the long run, you will lose $14 per every $1000 you bet playing Craps. Of course, the casino lives for the long run, they don't care who gets lucky and unlucky as long as they take in lots of players and they get their consistent $14 payoff. If Foxwoods has 100 Craps tables, and each one takes in $1,000,000 worth of bets a day. Then Foxwoods expects to earn $1,400,000 a day from its Craps tables alone. This leaves a lot of profit even after you pay off the 500 or so employees needed to run the tables for the day.

In class we will analyze just why the odds of winning a game of Craps is about 493/1000.

Applications - Basketball and Freethrows

Here is a hard one because it is not clear what to count. Let's say I am a 60% freethrow shooter. That means that the number of freethrows I make divided by the number I attempt equals .60 or 3/5. What is the chance that I will make two freethrows in a row? Here too we can make a table where the first freethrow attempt is shown horizontally and the second vertically. An X marks that I made the freethrow, and an O means I missed. I put three X's and two O's randomly for each of the two throws. Inside the table, I put in a two letter combination showing what happened on the two throws.

	X	O	X	X	O
O	X, O	O, O	X, O	X, O	O, O
X	X, X	O, X	X, X	X, X	O, X
O	X, O	O, O	X, O	X, O	O, O
X	X, X	O, X	X, X	X, X	O, X
X	X, X	O, X	X, X	X, X	O, X

After two throws there are 9 times when both shots are made, 4 times when neither is made, and 12 times when one out of two is made. So the chance you make both freethrows is 9/25. You might have noticed that 9/25 is just 3/5 x 3/5.

This is general principle II of probability - the multiplication principle: The chance of two independent events both happening is the product of their individual probabilities.

Review Problem: You are sitting at home watching the NBA finals. Shaq is on the freethrow line with no time left on the clock. His freethrow shooting percentage is a dismal 68%. He needs to make both shots to win the game. Your Uncle says that he bets you $20 they don't win the game, and he will give you 2:1 odds. Do you take his bet?

Solution: The chance Shaq will make both shots is .68 x .68 = .4624. If you made this $20 bet 10,000 times then you would expect to win 4624 times, and lose 5376 times. You win $40 4624 times, and you lose $20 5376. This is a net gain of $77,440. If you make the bet once, you have an expected gain of about $7.74. Take the bet!

Putting all the Principles to Use

I can throw a crumpled piece of paper across the class into a garbage can 1 in 5 times. If I try five times, what's the chance I get it in at least once?

Let's do it wrong. If I have 1/5 chance and I do it five times, then I have 5 x 1/5 = 100% chance to get at least one. That means if I throw it five times, I will certainly get at least one in. Nonsense! Just watch me try - I miss five in a row fairly often. Another way to see how silly this thinking is, is to extend my attempts to ten throws. Is my chance to get at least one 10 x 1/5 = 200% Double nonsense. Probabilities only exist between 0% and 100% inclusive.

This leads to a non-principle of probability - the addition non-principle. It is not true that: The chance of at least one of two or more events occuring equals the sum of the chances that each event occurs. Please never use this hideous illogical but tempting non-principle.

Now let's do it right. This hard problem requires some careful thinking, a good plan, and the use of our two main principles. The chance of missing a throw is 4/5 (general principle I). The chance of missing all five throws is 4/5 x 4/5 x 4/5 x 4/5 x 4/5 = .32768 (general principle II). The chance of not missing all five, or equivalently the chance of making at least one, is 1 - .32768 = .67232 or about 67%.

Review Problem: Assuming that I am a 60% freethrow shooter, what is the chance I get at least one freethrow out of 5 shots?

Solution: The chance you miss a freethrow is 40% or 2/5 (general principle I). The chance you miss all 5 freethrows is 2/5 x 2/5 x 2/5 x 2/5 x 2/5 = .03125 (general principle II). The chance you don't miss all 5 freethrows is equal to 1 - .03125 = .96875 or about 97% (general principle I). Getting at least one is the same as not missing all five, so the answer is about 97%.

Review Problem: I play the lottery 10,000 times, and my chance of winning each time is 1 in 1,000,000, what's the chance I win at least once?

Solution: The chance you don't win each time is 999,999/1,000,000. The chance you don't win 10,000 times is (999,999/1,000,000)^10,000. This is what calculators are for! It comes to about .990498. The chance of winning at least once is .009502, less than 1 in a 100.