Probability

I am offered two lotto cards:

– Card 1: has numbers

– Card 2: has numbers

Which card should I take so that I have the greatest chance of winning lotto?

Lotto

In the casino I wait at the roulette wheel until I see a run of at least five reds in a row. I then bet heavily on a black.I am now more likely to win.

Roulette

Coin TossingI am about to toss a coin 20 times. What do you expect to happen?

Suppose that the first four tosses have been heads and there are no tails so far. What do you expect will have happened by the end of the 20 tosses ?

Coin Tossing• Option A

– Still expect to get 10 heads and 10 tails. Since there are already 4 heads, now expect to get 6 heads from the remaining 16 tosses. In the next few tosses, expect to get more tails than heads.

• Option B– There are 16 tosses to go. For these 16 tosses I

expect 8 heads and 8 tails. Now expect to get 12 heads and 8 tails for the 20 throws.

• In a TV game show, a car will be given away.– 3 keys are put on the table, with only one of them

being the right key. The 3 finalists are given a chance to choose one key and the one who chooses the right key will take the car.

– If you were one of the finalists, would you prefer to be the 1st, 2nd or last to choose a key?

TV Game Show

Let’s Make a Deal Game Show• You pick one of three doors

– two have booby prizes behind them– one has lots of money behind it

• The game show host then shows you a booby prize behind one of the other doors

• Then he asks you “Do you want to change doors?”– Should you??! (Does it matter??!)

• See the following website:• http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

Game Show Dilemma

Suppose you choose door A. In which case Monty Hall will show you either door B or C depending upon what is behind each.

No Switch Strategy ~ here is what happens

Result A B C

Win Car Goat Goat

Lose Goat Car Goat

Lose Goat Goat Car

P(WIN) = 1/3

Game Show Dilemma

Suppose you choose door A, but ultimately switch. Again Monty Hall will show you either door B or C depending upon what is behind each.

Switch Strategy ~ here is what happens

Result A B C

Lose Car Goat Goat

Win Goat Car Goat

Win Goat Goat Car

Monty will show either B or C.

You switch to the one not shown

and lose.

Monty will show door C, you switch to B and win.Monty will show door B, you switch to C and win.

P(WIN) = 2/3 !!!!

Matching Birthdays• In a room with 23 people what is the

probability that at least two of them will have the same birthday?

• Answer: .5073 or 50.73% chance!!!!!

• How about 30? • .7063 or 71% chance!• How about 40? • .8912 or 89% chance!• How about 50? • .9704 or 97% chance!

ProbabilityWhat is Chapter 6 trying to do? – Introduce us to basic ideas about probabilities:

• what they are and where they come from• simple probability models (genetics)• conditional probabilities• independent events• Baye’s Rule Teach us how to calculate probabilities:• tables of counts and using properties of

probabilities such as independence.

ProbabilityI toss a fair coin (where fair means ‘equally likely outcomes’)

What are the possible outcomes? Head and tail ~ This is called a “dichotomous experiment”

because it has only two possible outcomes. S = {H,T}. What is the probability it will turn up heads? 1/2

I choose a patient at random and observe whether they are successfully treated.

What are the possible outcomes?

“Success” and “Failure” What is the probability of successful treatment?

?????

What factors influence this probability? ?????

What are Probabilities?• A probability is a number between 0 & 1

that quantifies uncertainty.

• A probability of 0 identifies impossibility

• A probability of 1 identifies certainty

Where do probabilities come from? • Probabilities from models:

The probability of getting a four when a fair dice is rolled is 1/6 (0.1667 or 16.7% chance)

• Probabilities from data or Empirical probabilities

What is the probability that a randomly selected patient is successfully treated?– In a clinical trial n = 67 patients are “randomly”

selected.– 40 of these patients are successfully treated.– The estimated probability that a randomly chosen

patient will have a successful outcome is 40/67 (0.597 or 59.7% chance)

Where do probabilities come from?

• Subjective Probabilities– The probability that there will be another

outbreak of ebola in Africa within the next year is 0.1.

– The probability of rain in the next 24 hours is very high. Perhaps the weather forecaster might say a there is a 70% chance of rain.

– A doctor may state your chance of successful treatment.

Where do probabilities come from?

For equally likely outcomes, and a given event A:

Simple Probability Models

“The probability that an event A occurs” is written in shorthand as P(A).

P(A) =Number of outcomes in A

Total number of outcomes

1. Heart DiseaseIn 1996, 6631 Minnesotans died from coronary heart disease. The numbers of deaths classified by age and gender are:

SexAge Male Female Total< 45 79 13 92

45 - 64 772 216 98865 - 74 1081 499 1580> 74 1795 2176 3971

Total 3727 2904 6631

Let A be the event of being under 45B be the event of being maleC be the event of being over 64

1. Heart Disease

SexAge Male Female Total< 45 79 13 92

45 - 64 772 216 98865 - 74 1081 499 1580> 74 1795 2176 3971

Total 3727 2904 6631

Find the probability that a randomly chosen member of this population at the time of death was:

a) under 45 P(A) = 92/6631 = 0.0139

1. Heart Disease

SexAge Male Female Total< 45 79 13 92

45 - 64 772 216 98865 - 74 1081 499 1580> 74 1795 2176 3971

Total 3727 2904 6631

Conditional Probability• We wish to find the probability of an event

occuring given information about occurrence of another event. For example, what is probability of developing lung cancer given that we know the person smoked a pack of cigarettes a day for the past 30 years.

• Key words that indicate conditional probability are:“given that”, “of those”, “if …”,

“assuming that”

“The probability of event A occurring given that event B has already occurred” is written in shorthand as P(A|B)

Conditional Probability

P(A|B) =__________ , P(B) > 0

Conditional Probability and Independence

P(A and B) P(B)

Two events A and B are said to be independent if

P(A|B) = P(A) and P(B|A) = P(B)i.e. knowing the occurrence of one of the events tells you nothing about the occurrence of the other.

1. Heart Disease

SexAge Male Female Total< 45 79 13 92

45 - 64 772 216 98865 - 74 1081 499 1580> 74 1795 2176 3971

Total 3727 2904 6631

Find the probability that a randomly chosen member of this population at the time of death was:

b) male assuming that the person was younger than 45.

SexAge Male Female Total< 45 79 13 92

45 - 64 772 216 98865 - 74 1081 499 1580> 74 1795 2176 3971

Total 3727 2904 6631

Find the probability that a randomly chosen member of this population at the time of death was:

b) male given that the person was younger than 45. P(B|A) = 79/92 = 0.8587

2. Heart Disease

P(B|A) = P(A and B)/P(A) = (79/6631)/(92/6631) = 79/92

SexAge Male Female Total< 45 79 13 92

45 - 64 772 216 98865 - 74 1081 499 1580> 74 1795 2176 3971

Total 3727 2904 6631

Find the probability that a randomly chosen member of this population at the time of death was:

c) male and was over 64.P(B and C) = (1081 + 1795)/6631= 2876/6631=.434

1. Heart Disease

SexAge Male Female Total< 45 79 13 92

45 - 64 772 216 98865 - 74 1081 499 1580> 74 1795 2176 3971

Total 3727 2904 6631

Find the probability that a randomly chosen member of this population at the time of death was:

d) over 64 given they were female (not B).

1. Heart Disease

SexAge Male Female Total< 45 79 13 92

45 - 64 772 216 98865 - 74 1081 499 1580> 74 1795 2176 3971

Total 3727 2904 6631

P(C|not B) = (499+2176)/2904 = .9211

1. Heart DiseaseFind the probability that a randomly chosen member of this population at the time of death was:

d) over 64 given they were female (not B).

2. Hodgkin’s Disease

Type None Partial

Positive

RowTotals

LD 44 10 18 72LP 12 18 74 104MC 58 54 154 266NS 12 16 68 96ColumnTotals

126 98 314 n = 538

Response to Treatment

2. Hodgkin’s Disease

2. Hodgkin’s Disease

TypeNone Partial Positiv

e

RowTotals

LD 44 10 18 72LP 12 18 74 104MC 58 54 154 266NS 12 16 68 96ColumnTotals

126 98 314 n = 538

Response to Treatment

a) Had positive response to treatment

P(pos) = 314/538 = .584 or 58.4% chance

2. Hodgkin’s Disease

TypeNone Partial Positive

RowTotals

LD 44 10 18 72

LP 12 18 74 104

MC 58 54 154 266

NS 12 16 68 96

ColumnTotals

126 98 314 n = 538

Response to Treatment

b) Had at least some response to treatmentP(par or pos) = (98 + 314)/538 = 412/538 = .766 or 76.6% chance

2. Hodgkin’s DiseaseType

None Partial PositiveRowTotals

LD 44 10 18 72

LP 12 18 74 104

MC 58 54 154 266

NS 12 16 68 96

ColumnTotals

126 98 314 n = 538

Response to Treatment

c) Had LP and positive response to treatmentP(LP and pos) = 74/538 = .138 or 13.8%

2. Hodgkin’s Disease

TypeNone Partial Positiv

e

RowTotals

LD 44 10 18 72LP 12 18 74 104MC 58 54 154 266NS 12 16 68 96ColumnTotals

126 98 314 n = 538

Response to Treatment

d) Had LP or NS as there histological type.

P(LP or NS) = (104 + 96)/538 = .372 or 37.2% chance

2. Hodgkin’s Disease

TypeNone Partial Positiv

e

RowTotals

LD 44 10 18 72LP 12 18 74 104MC 58 54 154 266NS 12 16 68 96ColumnTotals

126 98 314 n = 538

Response to Treatment

What conditional probabilities would be of interest?

EXAMPLES IN NOTES

3. Right Heart Catheterization and 30-day Mortality (Conners, et al. 1996)

Catheter?

YES NORow Totals

RHC 830 1354 2184

No RHC 1088 2463 3551ColumnTotals 1918 3817 5735

RHC = patient had catheter put inNo RHC = patient did not have catheterYES = Died within 30 daysNO = Survived 30 days P(YES) = 1918 / 5735 = .3344 or

33.44%

What is the probability that a heart patient in this study died?

Died within 30 days?

3. Right Heart Catheterization and 30-day Mortality (Conners, et al. 1996)

Catheter?

YES NORow Totals

RHC 830 1354 2184

No RHC 1088 2463 3551ColumnTotals 1918 3817 5735

RHC = patient had catheter put inNo RHC = patient did not have catheterYES = Died within 30 daysNO = Survived 30 days

P(RHC) = 2184 / 5735 = .3808 or 38.08%

What is the probability that a heart patient had a right heart catheter put in during treatment?

Died within 30 days?

3. Right Heart Catheterization and 30-day Mortality (Conners, et al. 1996)

Catheter?

YES NORow Totals

RHC 830 1354 2184

No RHC 1088 2463 3551ColumnTotals 1918 3817 5735

RHC = patient had catheter put inNo RHC = patient did not have catheterYES = Died within 30 daysNO = Survived 30 days

P(YES | RHC) = 830 / 2184 = .3800 or 38.00%

What is the probability that a patient would die within 30 days given that they had a right heart catheter put in?

Died within 30 days?

3. Right Heart Catheterization and 30-day Mortality (Conners, et al. 1996)

Catheter?

YES NORow Totals

RHC 830 1354 2184

No RHC 1088 2463 3551ColumnTotals 1918 3817 5735

RHC = patient had catheter put inNo RHC = patient did not have catheterYES = Died within 30 daysNO = Survived 30 days

P(YES | No RHC) = 1088 / 3551 = .3064 or 30.64%

What is the probability that a patient would die within 30 days given that they did not have a right heart catheter put in?

Died within 30 days?

3. Right Heart Catheterization and 30-day Mortality (Conners, et al. 1996)

How many times more likely is a patient who had a right heart catheter put in to die within 30 days than patient who did not have a Swan-Ganz line put in?P(YES | RHC) = .3800P(YES | No RHC) = .3064

.3800/ .3064 = 1.24 times more likely. This is called the relative risk or risk ratio (denoted RR).

Risk of death is 24% greater for those that had a Swan-Ganz line put in.

3. Right Heart Catheterization and 30-day Mortality (Conners, et al. 1996)

The shading for 30-day mortality is 1.24 times higher for the RHC group than for the No RHC group (recall RR = 1.24).

Patients having a Swan-Ganz line put in have 1.24 times higher risk of death within 30-days of initial treatment.

Building a Contingency Table from a Story

4. HIV ExampleA European study on the transmission of the HIV virus involved 470 heterosexual couples. Originally only one of the partners in each couple was infected with the virus. There were 293 couples that always used condoms. From this group, 3 of the non-infected partners became infected with the virus. Of the 177 couples who did not always use a condom, 20 of the non-infected partners became infected with the virus.

Let C be the event that the couple always used condoms. (NC be the complement)

Let I be the event that the non-infected partner became infected. (NI be the complement)

C NC

NI I

4. HIV Example

Total

Total

Condom UsageInfectio

n Status

A European study on the transmission of the HIV virus involved 470 heterosexual couples. Originally only one of the partners in each couple was infected with the virus. There were 293 couples that always used condoms. From this group, 3 of the non-infected partners became infected with the virus.

C NC

NI I

4. HIV Example

Total

Total

Condom UsageInfectio

n Status

470293

3

Of the 177 couples who did not always use a condom, 20 of the non-infected partners became infected with the virus.

C NC

NI I

4. HIV Example

Total

Total

Condom UsageInfectio

n Status

470293

3 20

177290 157

23447

a) What proportion of the couples in this study always used condoms?

C NC

NI I

Total

Total

Condom UsageInfection

Status

470293

3 20

177290 157

23447

4. HIV Example

P(C )

a) What proportion of the couples in this study always used condoms?

C NC

NI I

Total

Total

Condom UsageInfection

Status

470293

3 20

177290 157

23447

4. HIV Example

P(C ) = 293/470 (= 0.623)

b) If a non-infected partner became infected, what is the probability that he/she was one of a couple that always used condoms?

4. HIV Example

C NC

NI I

Total

Total

Condom UsageInfection

Status

470293

3 20

177290 157

23447

P(C|I ) = 3/23 = 0.130

4. HIV Examplec) In what percentage of couples did the non-

HIV partner become infected amongst those that did not use condoms?

P(I|NC) = 20/177 = .113 or 11.3%• Amongst those that did where condoms? P(I|C) = 3/293 = .0102 or 1.02%• What is relative risk of infection associated

with not wearing a condom?RR = P(I|NC) / P(I|C) = 11.08 times

more likely to become infected.

4. HIV Example

The percentage of couples where the non-HIV partner became infected in the non-condom user group is 11 times higher than that for condom group.

Relative Risk (RR) and Odds Ratio (OR) Example: Age at First Pregnancy and Cervical

Cancer A case-control study was conducted to determine whether there was increased risk of cervical cancer amongst women who had their first child before age 25. A sample of 49 women with cervical cancer was taken of which 42 had their first child before the age of 25. From a sample of 317 “similar” women without cervical cancer it was found that 203 of them had their first child before age 25.

Q: Do these data suggest that having a child at or before age 25 increases risk of cervical cancer?

Relative Risk (RR) and Odds Ratio (OR)The ODDS for an event A are defined asOdds for A = _______P(A)

1 – P(A)For example suppose we roll a single die the odds for a 3 are:

Odds for 3 = P(3)/(1 – P(3)) = = (1/6)/(1 – (1/6)) = 1/5 1 three for every 5 rolls that don’t result in a six.(Odds for a 3 are 1:5 and odds against are 5:1)

Relative Risk (RR) and Odds Ratio (OR)The Odds Ratio (OR) for a disease associated with a risk

factor is ratio of the odds for disease for those with risk factor and the odds for disease for those without the risk factor

OR = _________________________

P(Disease|Risk Factor)1 – P(Disease|Risk

Factor)

_____________________

P(Disease|No Risk Factor)1 – P(Disease|No Risk

Factor)

_______________________

The Odds Ratio gives us the multiplicative increase in odds associated with having the “risk factor”.

Odds for disease amongst those with risk factor present

Odds for disease amongst those without the risk factor.

Relative Risk (RR) and Odds Ratio (OR)Age at 1st Pregnancy Case Contro

l

Row Totals

Age < 25

42 203 245

Age > 25

7 114 121

ColumnTotals 49 317 n = 366

Cervical Cancer

a) Why can’t we calculate P(Cervical Cancer | Age < 25)?Because the number of women with disease was fixed in advance and therefore NOT RANDOM !

Relative Risk (RR) and Odds Ratio (OR)Age at 1st Pregnancy Case Contro

l

Row Totals

Age < 25

42 203 245

Age > 25

7 114 121

ColumnTotals 49 317 n = 366

Cervical Cancer

b) What is P(risk factor|disease status) for each group?P(Age < 25|Case) = 42/49 = .857 or 85.7%P(Age < 25|Control) = 203/317 = .640 or 64.0%

Relative Risk (RR) and Odds Ratio (OR)Age at 1st Pregnancy Cas

eContr

ol

Row Totals

Age < 25

42 203 245

Age > 25

7 114 121

ColumnTotals 49 317 n = 366

Cervical Cancer

c) What are the odds for the risk factor amongst the cases? Amongst the controls?Odds for risk factor cases = .857/(1-.857) = 5.99Odds for risk factor controls = .64/(1- .64) = 1.78

Relative Risk (RR) and Odds Ratio (OR)Age at 1st Pregnancy Case Contro

l

Row Totals

Age < 25

42 203 245

Age > 25

7 114 121

ColumnTotals 49 317 n = 366

Cervical Cancer

d) What is the odds ratio for the risk factor associated with being a case?

Odds Ratio (OR) = 5.99/1.78 = 3.37, the odds for having 1st child on or before age 25 are 3.37 times higher for women who currently have cervical cancer versus those that do not have cervical cancer.

Relative Risk (RR) and Odds Ratio (OR)

Odds RatioThe ratio of dark to light shading is 3.37 times larger for the cervical cancer group than it is for the control group.

e) Even though it is inappropriate to do so calculate P(disease|risk status).

P(case|Age<25) = 42/245 = .171 or 17.1%

P(case|Age>25) = 7/121 = .058 or 5.8% Now calculate the odds for disease

given the risk factor statusOdds for Disease for 1st Preg. Age < 25 = .171/(1 - .171) = .207Odds for Disease for 1st Preg. Age > 25 = .058/(1 - .058) = .061

Relative Risk (RR) and Odds Ratio (OR)

f) Finally calculate the odds ratio for disease associated with 1st pregnancy age < 25 years of age. Odds Ratio = .207/.061 = 3.37

This is exactly the same as the odds ratio for having the risk factor (Age < 25) associated with being in the cervical cancer group!!!!

Relative Risk (RR) and Odds Ratio (OR)

Final Conclusion: Women who have their first child at or before age 25 have 3.37 times the odds of developing cervical cancer when compared to women who had their first child after the age of 25.

Relative Risk (RR) and Odds Ratio (OR)

Risk FactorStatus

Case Control

Risk Factor Present a bRisk Factor Absent c d

Disease Status

OR = _____a X db X c

Much easier computational formula!!!

Relative Risk (RR) and Odd’s Ratio (OR)When the disease is fairly rare, i.e. P(disease)

< .10 or 10%, then one can show that the odds ratio and relative risk are similar.

OR is approximately equal to RR when P(disease) < .10 or 10% chance.

In these cases we can use the phrase:“… times more likely” when interpreting the OR.

Relative Risk (RR) and Odds Ratio (OR)Age at 1st Pregnancy Case Contro

l

Row Totals

Age < 25

a42

b203 245

Age > 25

c7

d114 121

ColumnTotals 49 317 n = 366

OR = (42 X 114)/(7 X 203) = 3.37 Because less than 10% of the population of women develop cervical cancer we can say women who have their first child at or before age 25 are 3.37 times more likely to develop cervical cancer than women who have their first child after age 25.

More About RR and OR• The most commonly cited advantage of the RR over the OR

is that the former is the more natural interpretation. The relative risk comes closer to what most people think of when they compare the relative likelihood of events.

e.g. suppose there are two groups, one with a 25% chance of mortality and the other with a 50% chance of mortality. Most people would say that the latter group has it twice as bad. But the odds ratio is 3, which seems too big.

RR = .50/.25 = 2.00 OR = P(death|high mortality)/P(survive|high mortality) P(death|low mortality)/P(survive|low mortality)

= .50/(1 - .50) = 3.00 .25/(1 - .25)

More About RR and OREven more extreme examples are possible. A

change from 25% to 75% mortality represents a relative risk of 3, but an odds ratio of 9. A change from 10% to 90% mortality represents a relative risk of 9 but an odds ratio of 81.

More About RR and OR• OR’s arise as part of logistic regression

which we will study later in the course.

• Despite their pitfalls OR’s are really the only option when case-control studies are used.

• Any study of risk needs to adjust for potential confounding factors which is typically done using logistic regression.