Notes for February 25 and 27

The dependent probabilities in a 52 card deck

One of the simplest mathematical models of dependency is sampling without replacement, which is the way most card games or lotteries or the game of Bingo works. You have a set of outcomes which get effectively randomized and a trial is performed, meaning a card is taken from the deck or a ping pong ball is removed from the hopper or a bingo marker is removed from the spinner. Once removed, the number of possible outcomes has been reduced by one and probabilities for success and failure of certain outcomes change.

Looking for an ace: There are 52 cards is a standard deck and 4 of them are aces. If I draw a card from a randomized deck, the chances are 4/52 = 1/13 ~= 7.7% that the card will be an ace. What are the chances the second card is an ace?

That depends on the first card.

Probability that the second card is an ace, given the first card is an ace is 3/51 = 1/17 ~= 5.9%.

Probability that the second card is an ace, given the first card is not an ace is 4/51 ~= 7.8%.

Unlike the mathematical model of free throw shooting where we re-calculate the probabilities by adding the most recent make or miss into the percentage, which means a miss brings the odds down and a make brings the odds up, not getting an ace makes the odds a little better next time, and getting an ace makes the odds worse.


This is the formula for the dependent probability model of sampling without replacement is given at the left. The two numbers in parentheses are a binomial coefficient, the numbers you get when you use nCr on your calculator, which I pronounce "n choose r" in class. The pairs of numbers that look like a base and an exponent, except that the exponent is underlined, is the convention developed by Donald Knuth at Stanford for writing the numbers that you get on your calculator using the nPr function, which I pronounce "n fall r", referring to the name "the falling factorial". If we think about a deck of cards, the lowercase letters refer to the size of the hand n, where r is the number of successful trials (r for right) and w is the number of unsuccessful trials (w for wrong), and r+w=n. The uppercase letters refer to the size of the deck, where T is the size of the deck, G is the number of cards we consider success if we draw them and B is the number of cards we consider a failed trial if we draw them. The letter T stands for Total, G for Good and B for Bad. Again, we have an equation, G+B=T.

Example: If we want consider drawing a heart a success and anything else a failure, what is the probability of drawing three hearts and two non hearts in a five card hand from a well-shuffled 52 card deck.

Here are the six numbers we need.
n = 5
r = 3
w = 2
T = 52
G = 13
B = 39

On a TI-30XIIs, here are the keys you would press.

5[prb][right]3×13[prb]3×39[prb]2÷52[prb]5[enter]

The calculator will read as follows.

5 nCr 3*13 nPr 3*39 nPr 2/52 nPr 5
0.081542617

This means the probability of exactly three hearts and two cards of some other suit is about 8.15%.

Let's say instead the deck had 10,000 cards and 2,500 hearts. our numbers would change.
n = 5
r = 3
w = 2
T = 10000
G = 2500
B = 7500

On a TI-30XIIs, here are the keys you would press.

5[prb][right]3×2500[prb]3×7500[prb]2÷10000[prb]5[enter]

The calculator will read as follows.

5 nCr 3*2500 nPr 3*7500 nPr 2/10000 nPr 5
0.0878613102

The difference is small, but the second number is much closer to the odds of 3 out of 5 when the probability of success is .25 every time


5 nCr 3*.25^3*.75^2
0.087890625

The point of this is that as the size of the deck gets larger, dependent and independent probabilities get closer together.

When we use categorical data, the most important statistic we try to predict is the proportion of a value in the population, which we call p, which we will estimate using the proportion from the sample, known as p-hat.

Again we will create a confidence interval, and the formula for standard deviation is very different.

sp-hat = sqrt(p-hat * q-hat/n)

The confidence level multipliers for xx% are taken from the z-score table (Table A-2) instead of the t-score table, and this is because the standard deviation for the sample and the standard deviation for the population are expected to be relatively close to one another. The values for the CLMxx% are given on the first page of your yellow sheets in the lower left hand corner.

CLM90% = 1.645
CLM95% = 1.96
CLM99% = 2.575

Example: Consider data sets #1 and #2, and the proportion of males. Let's find the 95% confidence interval for the underlying population, which we will limit to students at Laney who take statistics.

Data set #1:
n = 38
f(males) = 18
p-hat(males) = 18/38 ~= .474
q-hat(males) = 1 - p-hat(males) ~= .526

.474 - 1.96*sqrt(.474*.526/38) < p < .474 + 1.96*sqrt(.474*.526/38)
.315 < p < .633

Given this sample of 38 students, we are 95% confident the percentage of male students taking statistics at Laney is between 31.5% and 63.3%.


Data set #2: n = 42
f(males) = 12
p-hat(males) = 12/42 ~= .286
q-hat(males) = 1 - p-hat(males) ~= .714

.286 - 1.96*sqrt(.286*.714/42) < p < .286 + 1.96*sqrt(.286*.714/42)
.149 < p < .423

Given this sample of 42 students, we are 95% confident the percentage of male students taking statistics at Laney is between 14.9% and 42.3%.

Data sets #1 and #2 combined: n = 80
f(males) = 30
p-hat(males) = 30/80 = .375
q-hat(males) = 1 - p-hat(males) = .625

.375 - 1.96*sqrt(.375*.625/80) < p < .375 + 1.96*sqrt(.375*.625/38)
.269 < p < .481


Given this sample of 40 students, we are 95% confident the percentage of male students taking statistics at Laney is between 26.9% and 48.1%.

Notice how much our intervals disagree with one another. This is because our best point estimates from the three sets are .474, .286 and .375. Also notice that the width of the 95% confidence interval tends to be smaller as n gets bigger. When the sample size is 38, the width of the confidence interval is .318. At n = 42, it is .274 wide. At n = 80, the width is .212. The most common way to make a confidence interval narrower is to increase the size of the sample.

There are two other ways to change the width. If you ask for a higher confidence level, the interval will get wider. If p and q are close to 50%, the confidence interval will be wider than if the are both far away from 50%.

Class Notes for 4/27 and 4/29

Dependence and Independence
The first line of Leo Tolstoy's Anna Karenina is "All happy families are alike; each unhappy family is unhappy in its own way." In statistics, all independent trials are alike, in that the probability of a particular outcome of one trial does not effect the outcome of later trials, nor was it effected by earlier trials. With dependent probability, the outcome of one trial is effected by the outcome of previous trials, but how that effects things is not always the same.

For example, if we talk about a 70% free throw shooter taking two shots, does missing the first shot effect the probability of missing the second shot? Let's look at this simple problem three different ways.

Predicting the future mathematically by carefully studying the past: Let's say I called this person a 70% free throw shooter because so far in the season, she has made 7 of 10 shots from the line. If she misses, she has now made 7 of 11 shots from the line, and she is now a 63.6% shooter. Should we factor that in to be more precise? Some mathematical models would say yes.

Using this method, missing a shot would make her percentage worse and making one would make it better, no matter how many shots she had taken. But if she had made 70 of 100 so far in the season, one miss would make her 70 of 101, which would lower her percentage to 69.3%, a much smaller effect than it has if she had made 7 of 10. If instead we were looking at her entire career instead a single season, perhaps she has made 700 of 1000, and missing one would mean she was 700 of 1001, which changes her percentage to 69.9%. When discussing free throw percentage, announcers on TV usually round to the nearest percent, so the first example of 7 of 10 to 7 of 11 would be a drop from 70% to 64%, the second example would be a drop from 70% to 69%, and in the third example, 69.9% would round to 70% and the change would be too small to notice.

Factors effecting success and failure: The free throw shooter goes to the line for two shots and misses the first. Is there a physical reason? We might treat free throw shooting as we would picking a random number from 1 through 100, where we count any number from 1 through 70 as a success and 71 through 100 as failures, but shooting free throws includes the human factor. Maybe she missed because she is nervous or distracted. Maybe it's late in the game and she is tired or injured, changing her technique. If any of these are the case, it might make more sense for us to downgrade the probability of making the next shot, though exactly how much it should be downgraded is no longer some simple formula like turning a fraction into a percentage.

Compensating for failure: Again, let's add in the human factor in this problem. She misses the first free throw, and her coach notices her technique looks inconsistent. "Elbow up!" the coach shouts from the sidelines, and the shooter hears the coach and readjusts her technique to match the way she shoots in practice. Should this change bring her back to being a 70% shooter, or even upgrade her chance of success? That is uncertain, but the failure on the first shot and the diagnosis of at least one reason for the failure could effect the odds, and that effect means the second shot should not be considered independent of the first.

The dependent probabilities in a 52 card deck

One of the simplest mathematical models of dependency is sampling without replacement, which is the way most card games or lotteries or the game of Bingo works. You have a set of outcomes which get effectively randomized and a trial is performed, meaning a card is taken from the deck or a ping pong ball is removed from the hopper or a bingo marker is removed from the spinner. Once removed, the number of possible outcomes has been reduced by one and probabilities for success and failure of certain outcomes change.

Looking for an ace: There are 52 cards is a standard deck and 4 of them are aces. If I draw a card from a randomized deck, the chances are 4/52 = 1/13 ~= 7.7% that the card will be an ace. What are the chances the second card is an ace?

That depends on the first card.

Probability that the second card is an ace, given the first card is an ace is 3/51 = 1/17 ~= 5.9%.

Probability that the second card is an ace, given the first card is not an ace is 4/51 ~= 7.8%.

Unlike the mathematical model of free throw shooting where we re-calculate the probabilities by adding the most recent make or miss into the precentage, which means a miss brings the odds down and a make brings the odds up, not getting an ace makes the odds a little better next time, and getting an ace makes the odds worse.


This is the formula for the dependent probability model of sampling without replacement is given at the left. The two numbers in parentheses are a binomial coefficient, the numbers you get when you use nCr on your calculator, which I pronounce "n choose r" in class. The pairs of numbers that look like a base and an exponent, except that the exponent is underlined, is the convention developed by Donald Knuth at Stanford for writing the numbers that you get on your calculator using the nPr function, which I pronounce "n fall r", referring to the name "the falling factorial". If we think about a deck of cards, the lowercase letters refer to the size of the hand n, where r is the number of successful trials (r for right) and w is the number of unsuccessful trials (w for wrong), and r+w=n. The uppercase letters refer to the size of the deck, where T is the size of the deck, G is the number of cards we consider success if we draw them and B is the number of cards we consider a failed trial if we draw them. The letter T stands for Total, G for Good and B for Bad. Again, we have an equation, G+B=T.

Example: If we want consider drawing a heart a success and anything else a failure, what is the probability of drawing three hearts and two non hearts in a five card hand from a well-shuffled 52 card deck.

Here are the six numbers we need.
n = 5
r = 3
w = 2
T = 52
G = 13
B = 39

On a TI-30XIIs, here are the keys you would press.

5[prb][right]3×13[prb]3×39[prb]2÷52[prb]5[enter]

The calculator will read as follows.

5 nCr 3*13 nPr 3*39 nPr 2/52 nPr 5
0.081542617

This means the probability of exactly three hearts and two cards of some other suit is about 8.15%.

The Expected Value (EV) of a two outcome game

Let us assume we have a game that has only two outcomes, winning and losing. Let us further assume that two players have decided to wager on this game, both putting money into a pooled amount and the winner taking all at the end.


If we look at the game from the point of view of one of the players, we need to know the probability of winning p, how much that player put in, which we call Risk and how much the opponent put in, called Profit. The expected value EV equals the probability of victory p times the sum of Profit and Risk divided by Risk.

Different books use different formulas for this. Some do not divide by Risk. By dividing, the number we get is a percentage of return, and a game of flipping coins for $1 a game is equivalent to a game of flipping coins for $100 a game. Some subtract 1 from this formula. This just changes the most important number in identifying results from 1 to 0.

If EV = 1, we consider this a "fair game". For every $1 risked on this game, the expected value is that you will have that dollar returned to you, breaking even. Notice if we are flipping coins, that event never happens on any single play. Either the player makes a dollar profit or a dollar loss, but expected value is about the long run.

If EV > 1, the game is advantageous to the player. If EV < 1, the game is disadvantageous to the player.

In the game of roulette, there are 38 slots where the ball can land, and for simplicity's sake we will assume each has an equal chance of showing up, so p = 1/38. For every $1 you risk, you can make a profit of $35 if you correctly guess in the exact slot where the ball will land. To find the expected value using the TI-30xIIs, you should type in this.

1÷38×(35+1)÷1[enter]

The calculator will read as follows.

1/38*(35+1)/1
0.947368421

What this number means is that for every dollar risked on the spin of a roulette wheel, you should expect about 94.7 cents returned to you in change. In other words, about 5.3 cents is lost from every dollar you bet on every spin of the wheel.

Another way to play the game is to bet red or black. Of the 38 compartments, 18 are red and 18 are black and 2 are green. The probability of victory on betting one of the two major colors is 18/38 = 9/19 ~= 0.473684211. The profit and risk are now both $1. Here's what to type on the TI-30xIIs.


16÷38×(1+1)÷1[enter]

The calculator will read as follows.

16/38*(1+1)/1
0.947368421

The game has changed, both in probability and amount of profit compared to risk, but from the player's point of view, the expected value is precisely the same and still in favor of the casino.

No matter what the levels of profit and risk are, we can find a probability p that will make the expected value equal to 1, and that is p = Risk/(Profit+Risk). If the probability is increased with the profit and risk remaining unchanged, the game becomes advantageous. If is is decreased, the game becomes disadvantageous.

Modern and Classic Parimutuel odds

Profit and risk are listed either in classic form like 3-1 or 2-7 (or sometimes with colons 3:1 or 2:7), where profit is the first number and risk is the second.

In online betting sites, the numbers are given as numbers with absolute value greater than 100, with either a + or - in front of them. +250 means 250 is the profit and 100 is the risk, while -250 means 10o is the profit and 250 is the risk. The fourth page of the yellow sheet explains this in greater detail and shows how to switch back and forth between the two systems.

Practice problems

1. With a well-shuffled 52 card deck, find the probability of getting exactly r hearts in a five card hand when
a) r = 0
b) r = 1
c) r = 2
d) r = 3 (already solved above)
e) r = 4
f) r = 5

2. Find the break-even p when Profit and Risk are as given. Round to three places after the decimal point.

a) Modern parimutuel = +150
b) Modern parimutuel = -110
c) Classic parimutuel = 5:3
d) Classic parimutuel = 5:11

Answers in the comments.

Class Notes for 4/20

We dealt with probability in a single instance earlier in the class when we had the relative frequencies of the values of categorical variables. Relative frequencies, listed in a population as p and in a sample as p-hat, are numbers between 0 and 1. If we take all the relative probabilities of all the values of a variable, they will add up to 1, or something very close to 1 depending on rounding error.

We will now talk about probability in multiple event experiments, like flipping ten coins or rolling five dice or drawing a hand of four cards from a 52 card deck. The first important split in the types of multiple event experiments is between independent and dependent events.

Events are independent if the probability of a later event does not change based on the result of an earlier event. For example, if I flip a coin that I can assume is fair, there is a 50% chance of heads and a 50% chance of tails every time I flip it. If by chance, the coin comes up heads ten times in a row, even though earlier testing had shown it to be a 50%-50% chance each time, the eleventh flip is still 50%-50%. Unusually long runs of all heads or all tails are rare, but they are not impossible. Flipping coins and rolling dice are typical examples of independent random events.

Events are dependent if the probability of a later event changes based on the result of an earlier event. The typical example of this is drawing cards from a shuffled deck. If the deck has 52 cards and 4 aces, the probability of drawing an ace from the deck is 4/52 = 1/13 ~= .0769...

If the card has been drawn, what is the probability of the second card being an ace? That depends on what the first card is. If the first card is an ace, there are only 3 left in the deck, which now has 51 cards, so the probability is 3/51 = 1/17 ~= .0588..., which is a lower probability than getting an ace the first card.

If the first card wasn't an ace, the odds are 4/51, ~= .0784..., a slightly higher probability than drawing an ace the first time.

If I say someone is a 70% free throw shooter, is every free throw attempt independent of what happened before? Often, we set up such an experiment assuming independence just to make our work simpler, but the human factor is involved, so in reality it's very likely to be dependent. Some people get frustrated after a few misses and will do worse. Others will learn from the mistakes of a few misses and figure out what they are doing wrong and make improvements. A player might be having a bad day for some reason, or might instead have excellent concentration or just really good luck that day. But again, these kinds of experiments are often set up as though each free throw trial is independent of what came before.

Let's look at flipping coins. A list of all possible events is called the event space. Here are some examples of event spaces.

Event space for flipping one coin
Heads (H)
Tails (T)

ways to get one head = 1
ways to get no heads = 1

Event space for flipping two coins
HH
HT
TH
TT

ways to get two heads = 1
ways to get one head = 2
ways to get no heads = 1

Event space for flipping three coins
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT

ways to get three heads = 1
ways to get two heads = 3
ways to get one head = 3
ways to get no heads = 1


The list of numbers of ways to get r successes in n trials is often written in the pattern of the picture shown here, and this pattern is called Pascal's Triangle, at least in most of the world. The Italians call it Tartaglia's Triangle and the Chinese call it Yanghui's Triangle. None of these people actually invented it or claimed to have invented it. It's been around since before the time of Christ, and it has been studied all around the world.

While it is very common to see it presented in the form here as an equilateral triangle, it can also be presented where the first numbers in each row are lined up straight as follows

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
... etc.

It is standard to start counting the top row as row 0, and the left most column as column 0. For example, the 6 we see in the middle of the last row I typed in is row 4, column 2. Instead of having a copy of Pascal's Triangle around, our calculators have these numbers available. On Texas Instruments calculators, the function is under the probability menu. On the TI-30XIIs, the way to get that 6 is to type

4 [prb][right arrow]2[enter]

The calculator will read

4 nCr 2
6

All scientific calculators should have this function available, but all of them are slightly different. The TI-89 writes it as nCr(4,2) and Casio calculators write it as 4 C 2. I will pronounce it "4 choose 2", and when I type on the blog, I will type C(4,2). When I write it on the board or on tests, I will put a 4 on top of a 2 and surround both numbers with a large parentheses. These numbers are called the binomial coefficients.


The formula for finding the probability for exactly r successes in n independent trials where the probability of success on any single trial is p is shown here. In some books, they don't use the letter q, instead replacing it with (1-p). Likewise, sometimes w is replaced with (n-r). I use the extra letters and include the relationships between them. The letters r and w stand for right and wrong. The letter p and q are standard in probability texts for the probability of a success or a failure.

Let's do an example. You are given a four question multiple choice test, each question having five possible answers. The test is given in a language you do not read, so all you can do is guess. Each question is independent from the others, meaning that if C is the right answer to the first question, it's also possibly the answer to the second. The probability p of a correct guess is 1 chance in 5, or .2, The probability of failure q is 1-.2 = .8, and of course p + q = 1.

Probability of no correct answers = C(4,0)*.2^0*.8^4 = .4096
Probability of exactly one correct answer = C(4,1)*.2^1*.8^3 = .4096
Probability of exactly two correct answers = C(4,2)*.2^2*.8^2 = .1536
Probability of exactly three correct answers = C(4,3)*.2^3*.8^1 = .0256
Probability of four correct answers = C(4,4)*.2^4*.8^0 = .0016

The expected value of correct answers is n*p, so in this case it's 4*.2 = .8, which isn't possible. You can't get a fraction of correct answers on a multiple choice test. The expected value in this case says that over the long run, a test like this should average .8 right answers out of four. As we can see, the most likely thing to happen is actually a tie for first, where getting either no answers right or one answer right both have a probability of about 41%. If you need to get three answers right to pass the test, the odds are less than 3% to get either three or four right, and the odds of getting everything right by chance is a very slim 16 chances in 10,000.

If you have a TI-83 or TI-84, there is a function under the distribution menu called binompdf(n,p,r). All you have to is enter the function, then the three values in the order given, separated by commas.

The function for three right in four trials with probability .2 at each trial is binompdf(4, .2, 3), which as we see above is .0256.


Practice problem.
The test is changed. There are now five multiple choice questions and four choices for each, but it is still given in a language you do not read.

Round the probabilities to four places after the decimal.

1. What is the expected value?

2. What is the probability of no correct answers?

3. What is the probability of exactly one correct answer?

4. What is the probability of exactly two correct answers?

5. What is the probability of exactly three correct answers?

6. What is the probability of exactly four correct answers?

7. What is the probability of five correct answers?

Answers in the comments.