Basic Probability—The Goat Problem (AKA The Monty Hall Problem)
You are presented with three curtains. Behind two of those curtains are goats, and behind the remaining door is a red convertible (or whatever type of expensive car that strikes your fancy).
1 |
2 |
3 |
You pick curtain 3, and the game show host is kind enough to reveal to you that behind curtain 1, there is a goat. Do you switch? Or stay? What is the probability now that the car is in fact behind curtain 3? ½ you say? Are you sure?
1 GOAT |
2 |
3 YOUR PICK |
What most people fail to recognize is that revealing one of the goats does not affect the probability of the car being behind any of the three curtains. At the beginning of the game, before you pick your curtain, and before a goat is revealed, there is a 1/3 chance of the car being behind any of the curtains.
1 CAR |
2 |
3 |
1 |
2 CAR |
3 |
1 |
2 |
3 CAR |
You picked curtain 3.
1 CAR |
2 |
3 YOUR PICK |
1 |
2 CAR |
3 YOUR PICK |
1 |
2 |
3 CAR YOUR PICK |
The game show host shows you a goat behind one of the curtains that is not curtain 3.
1 CAR |
2 GOAT |
3 YOUR PICK |
1 GOAT |
2 CAR |
3 YOUR PICK |
1 GOAT |
2 |
3 CAR YOUR PICK |
In two of these situations, you will win if you switch. In the remaining, you will win if you stay. So you have a 2/3 chance of winning if you switch, and a 1/3 chance of winning if you stay. This is not the ½ you assumed at the beginning. But don’t feel bad if you still don’t get it, sometimes even the brightest of mathematicians can’t seem to wrap their heads around it. This only goes to show that we do not reason about probability very well, and many of our intuitions about religion and evolution probabilities are flawed.
Basic Probability—Bayes Theorem
A French male would like to obtain a green card to enter the United States. He must first test negative for HIV according to government standards. The French male has no history of risky behavior that would make him more likely to carry the virus. However, he tests positive. Knowing the following probabilities:
Probability of being HIV+ given French male with no risk factors
P(HIV+|French male w/no risk factors) = 0.01%
Therefore, probability of being HIV- given French male with no risk factors
P(HIV-|French male w/no risk factors) = 99.99%
Probability of testing positive given HIV+
P(test+|HIV+) = 99.9%
Therefore, probability of testing negative given HIV+
P(test-|HIV+) = 0.1%
Probability of testing negative given HIV-
P(test-|HIV-) = 99.99%
Therefore, probability of testing positive given HIV-
P(test+|HIV-) = 0.01%
What is the likelihood that this man does indeed carry the virus given that he is a French male with no risk factors and he tested positive? A probability tree might make the solution more obvious.
10,000 French men w/no risk factors
↓ ↓
1 HIV+ 9999 HIV-
↓ ↓ ↓ ↓
1 test+ 0 test – 1 test+ 9998 test –
We know that this man tested positive, so he can fall into 1 of two options—testing positive when HIV+ and testing positive when HIV-. Given that these options are on the same level of the probability tree, they have equal weight. This man therefore has a ½ chance of actually carrying the virus.
Another way to approach this problem is to use Bayes theorem. Bayes theorem evaluates the probability of a hypothesis (H) being true given data (D). It states,
So for this situation, the likelihood of the man carrying the virus (our hypothesis) given he is a French male with no risk factors and tested positive (our data) would be calculated:
P(HIV|French with no risk factors and tests+) =
Given that this semester focuses on Chance, a basic understanding of probability is key. There are essentially two schools of thought in the realm of probability—the Bayesians, and the Frequentists.
E. Sober’s Four Approaches to Probability
Kolomogorov’s three axioms of probability state that:
P() is a probability measure precisely when the following conditions are satisfied for any propositions A and B:
0≤ P(A) ≤ 1.
P(A) = 1, if A must be true.
If A and B are incompatible, then P(A or B) = P(A) + P(B).
1. Frequentists-objective probability
The frequentist approach to probability takes the actual frequency of events to be the probability of occurrence. For example, say you toss a coin 100 times and, in those 100 tosses, 62 land on heads. From a frequentist perspective, the probability of getting heads would be P(H) = 0.62. In the example of the French male above, a frequentist would disagree about the probability of being HIV+. The probability of being HIV+ could only be 1 (if he has it), or 0 (if he does not).
2. Bayesians-subjective probability
The Bayesian approach claims that we can have a reasonable degree of belief to the outcome of an event. This confidence is based on previous evidence or observations. A Bayesian says that we can make inferences on the probability of landing heads in a coin toss. Based on previous observations, we know that the coin will either land heads (1) or tails (0), so the probability of landing heads should be somewhere between 0 and 1. Also, with the HIV problem above, using observed frequencies, we were able to calculate a probability of French male being HIV+ that is not 0 or 1. Not coincidentally, this probability was calculated using Bayes Theorem.
3. Hypothetical relative frequency-somewhere in between
Most of us fall into the category of “What is the probability of landing heads in a coin toss?”—“Why, ½ of course.” Where does this come from? Well we assume that hypothetically, if you toss a coin an infinite number of times, the frequency of heads in a fair coin toss should converge to 0.5.
Just as the probability of landing heads converges to ½, the probability of any specific sequence of heads and tails converges to 0 as n increases. The problem is, however, that one of these sequences DOES happen…so we cannot equate a probability of 0 with impossibility. The Law of Large Numbers states that
if and only if
So the frequency of landing heads does not have to converge to 0.5 only approach it, there is wiggle room—namely e. Notice that a probability appears on both sides of the if and only if statement. Unfortunately, our “somewhere in between” interpretation of probability is circular…it defines probability in terms of probability.
4. Propensity interpretation of probability-“The Imaginary Invalid”
Propensities refer to dispositional properties can be described with if/then statements. For example:
If a compound is soluble, then it will dissolve in water.
But, dissolving in water is the definition of solubility. Sober uses the analogy of a part from the play “The Imaginary Invalid.” A quack doctor claims that he discovers why opium puts people to sleep. He gives opium the property of a “dormative virtue.” This simply means that opium has the ability to put people to sleep. Propensity is a redundant way to say probability.
We like to talk about what will probably happen because (due to chaos described below), we will probably never have a way of knowing with absolute certainty.
Chaos
Chaos theory deals with phenomena that appear chaotic, but are in truth deterministic. These occurrences are unpredictable, the weather being a perfect example. The most well known description of chaos theory is the butterfly effect which states that a butterfly flapping its wings in Taiwan will bring about a chain of events that ultimately cause a hurricane in Brazil. Most people believe that, even if the universe is entirely deterministic, we can never know enough to make absolute predictions.
Summary provided by L.B.
Admin apologizes for the weirdly formatted equations. If they really bother you, let me know in the comments, and I will come back to pretty them up a bit.
I still feel the French HIV example would have been more effective against frequentism had different numbers been used… for despite the technical difference between “1 or 0” and “50/50” probabilities, both the frequentist and the Bayesian end up giving the Frenchman the same answer: “either you have it or you don’t; care to flip a coin?”
Plugging in slightly different numbers (.02/99.98%, .5/99.5%, and .03/99.97%, say) returns more “nuanced” probabilities that, at least speaking for myself, would have made a more effective demonstration of the superiority of the Bayesian approach.