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Philosophy 200Bayes Theorem
Common fallacies of probability: The Gambler’s Fallacy
◦ Is assuming that the odds of a single truly random event are affected in any way by previous iterations of the same or other truly random event.
Common fallacies of probability: The Gambler’s Fallacy
◦ Is assuming that the odds of a single truly random event are affected in any way by previous iterations of the same or other truly random event.
Ignoring the Law of Large Numbers◦ Is assuming there must be other explanations for very
improbable events.
a priori probability: ◦ The sort of probability achieved by dividing the
number of desired outcomes vs. the total number of outcomes.
◦ Applies to random events. Statistical probability:
◦ The frequency at which a given event is observed to occur.
◦ Applies to events that are not truly random.
Two Kinds of Probability
What is the a priori probability (expressed as a percent) of a batter in baseball getting a hit in one at-bat?
An Example
What is the a priori probability (expressed as a percent) of a batter in baseball getting a hit in one at-bat?◦ 50%
An Example
What is the a priori probability (expressed as a percent) of a batter in baseball getting a hit in one at-bat?◦ 50%
What is the statistical probability (expressed as a percent) of a major league batter getting a hit in one at-bat?
An Example
What is the a priori probability (expressed as a percent) of a batter in baseball getting a hit in one at-bat?◦ 50%
What is the statistical probability (expressed as a percent) of a major league batter getting a hit in one at-bat?◦ 25.4%
An Example
• Wendy has tested positive for colon cancer. • Colon cancer occurs in .3% of the population
(.003 statistical probability)• If a person has colon cancer, there is a 90%
chance that they will test positive (.9 statistical probability of a true positive)
• If a person does not have colon cancer, then there is a 3% chance that they will test positive (3% statistical probability of a false positive)
• Given that Wendy has tested positive, what is the statistical probability that she has colon cancer?
A Case Study in probability:
The correct answer is 8.3%
Answer:
The correct answer is 8.3% Most people (including many doctors)
assume that the chances are much better than they really are that Wendy has colon cancer. The reason for this is that people tend to forget that a test must be absurdly specific to give a high probability of having a rare condition.
Answer:
h = the hypothesise = the evidence for hPr(h) = the statistical probability of hPr(e|h) = the true positive rate of e as
evidence for hPr(e|~h) = the false positive rate of e as
evidence for h
Formal Statement of Bayes’s Theorem:
Pr(h|e) = Pr(h) * Pr(e|h)
[Pr(h) * Pr(e|h)] + [Pr(~h) * Pr(e|~h)]
h ~h Total
e True Positives
False Positives
Pr(e)*Pop.
~e False Negatives
True Negatives
Pr(~e)*Pop.
Total Pr(h)*Pop. Pr(~h)*Pop.
Pop. = 10^n
The Table Method:
n = sum of decimal places in two most specific probabilities.
h ~h Total
e = Pr(e|h) * [Pr(h)*Pop.]
= Pr(e|~h) * [Pr(~h)*Pop.]
Total of this row
~e = below - above
= below - above
Total of this row
Total Pr(h)*Pop. Pr(~h)*Pop.
Pop.
The Table Method:
h ~h Total
e = Pr(e|h) * [Pr(h)*Pop.]
= Pr(e|~h) * [Pr(~h)*Pop.]
Total of this row
~e = below - above
= below - above
Total of this row
Total Pr(h)*Pop. Pr(~h)*Pop.
Pop.
The Table Method for Wendy:
has CC ~ have CC
Total
e = Pr(e|h) * [Pr(h)*Pop.]
= Pr(e|~h) * [Pr(~h)*Pop.]
Total of this row
~e = below - above
= below - above
Total of this row
Total Pr(h)*Pop. Pr(~h)*Pop.
Pop.
The Table Method for Wendy:
has CC ~ have CC
Total
tests positive
= Pr(e|h) * [Pr(h)*Pop.]
= Pr(e|~h) * [Pr(~h)*Pop.]
Total of this row
~ test positive
= below - above
= below - above
Total of this row
Total Pr(h)*Pop. Pr(~h)*Pop.
Pop.
The Table Method for Wendy:
has CC ~ have CC
Total
tests positive
= Pr(e|h) * [Pr(h)*Pop.]
= Pr(e|~h) * [Pr(~h)*Pop.]
Total of this row
~ test positive
= below - above
= below - above
Total of this row
Total .003*Pop. .997*Pop. 100,000
The Table Method for Wendy:
has CC ~ have CC
Total
tests positive
= Pr(e|h) * [Pr(h)*Pop.]
= Pr(e|~h) * [Pr(~h)*Pop.]
Total of this row
~ test positive
= below - above
= below - above
Total of this row
Total 300 99,700 100,000
The Table Method for Wendy:
has CC ~ have CC
Total
tests positive
= True Positive Rate (.9) * 300
= Pr(e|~h) * [Pr(~h)*Pop.]
Total of this row
~ test positive
= below - above
= below - above
Total of this row
Total 300 99,700 100,000
The Table Method for Wendy:
has CC ~ have CC
Total
tests positive
270 = Pr(e|~h) * [Pr(~h)*Pop.]
Total of this row
~ test positive
= below - above
= below - above
Total of this row
Total 300 99,700 100,000
The Table Method for Wendy:
has CC ~ have CC
Total
tests positive
270 = Pr(e|~h) * [Pr(~h)*Pop.]
Total of this row
~ test positive
30 = below - above
Total of this row
Total 300 99,700 100,000
The Table Method for Wendy:
has CC ~ have CC
Total
tests positive
270 = False positive rate (.03) * 99,700
Total of this row
~ test positive
30 = below - above
Total of this row
Total 300 99,700 100,000
The Table Method for Wendy:
has CC ~ have CC
Total
tests positive
270 2,991 Total of this row
~ test positive
30 = below - above
Total of this row
Total 300 99,700 100,000
The Table Method for Wendy:
has CC ~ have CC
Total
tests positive
270 2,991 Total of this row
~ test positive
30 96,709 Total of this row
Total 300 99,700 100,000
The Table Method for Wendy:
has CC ~ have CC
Total
tests positive
270 2,991 3,261
~ test positive
30 96,709 96,739
Total 300 99,700 100,000
The Table Method for Wendy:
has CC ~ have CC
Total
tests positive
270 (true positive)
2,991 (false positive)
3,261
~ test positive
30 (false negative)
96,709 (true negative)
96,739
Total 300 99,700 100,000
The Table Method for Wendy:
has CC ~ have CC Totaltests positive
270 (true positive)
2,991 (false positive)
3,261
What are Wendy’s chances?
•Wendy’s Chances given that she tests positive are the true positives divided by the number of total tests. That is, 270/3261, which is .083 (8.3%).•Those who misestimate that probability forget that colon cancer is rarer than a false positive on a test.
Note that testing positive (given the test accuracy specified) raises one’s chances of having the condition from .003(the base rate) to .083.
If we use .083 as the new base rate, those who again test positive then have a 73.1% chance of having the condition.
A third positive test (with .731 as the new base rate) raises the chance of having the condition to 98.8%
How about a second test?
![Gambler’s Ruin Bandit Problem · A. Gambler’s Ruin Problem If action F is removed from the GRBP, it becomes the Gambler’s Ruin Problem. In the model of Hunter et al. [10] of](https://img.pdfslide.us/doc/110x75/5f0c18f57e708231d433ba74/gambleras-ruin-bandit-problem-a-gambleras-ruin-problem-if-action-f-is-removed.jpg)
