1. INTRODUCTORY REMARKS
2. QUESTIONS
3. ANSWERS
4. MATHEMATICAL FORMULAE
5. OPINIONS ON SOME WELL-KNOWN PROBLEMS AND ISSUES
6. PHILOSOPHIZING ABOUT PROBABILITY

I. INTRODUCTORY REMARKS

Many people see a sharp distinction between puzzles or problems they do for "fun" and math problems, which can be laborious, boring and/or tedious. Of course, mathematicians often find complex problems to be fun whereas there are many non-mathematicians who find any sort of puzzle to be reprehensible labor. These are subjective responses — "in the eye of the beholder". The French mathematician/physicist/theologian Blaise Pascal (the co-founder of probability theory who was "notorious" for Pascal's Wager) gave up on mathematics for a period because he thought it was a form of sexual indulgence. Once, when I was calculating my nightly earnings in the taxi office, I heard another taxi-driver doing his timesheet say that mathematics is better than sex.

I have tried to construct twenty questions, most of which I believe can be solved and understood "intuitively" without knowledge of formulas from probability theory. The last four questions require a knowledge of the mathematical formulae for permutations and combinations — so it might be a good idea to read Section IV before attempting these problems it the reader is not already familiar with the formualae.

I am anxious to see if these problems are laborious ("too mathematical") or are "fun".

(Be sure to notice my pedantic emphasis on the distinction between combinations and permutations. People often use the phrase "combinations and permutations" because they don't know the difference. A permutation is an arrangement or an ordering — there are 6 ways to order [permute] 2 out of 3 objects: ABC :=> AB, BA, AC, CA, BC, CB. A combination pays no attention to ordering so there are only 3  ways to combine 2 out of 3 objects: ABC :=> AB, AC & BC.)

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II. QUESTIONS

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III. ANSWERS

     ABCDE (women)   FGHIJ (men)
        
        ABC             FG
        ABD             FH
        ABE             FI
        ACD             FJ
        ACE             GH
        ADE             GI
        BCD             GJ
        BCE             HI
        BDE             HJ
        CDE             IJ

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IV. MATHEMATICAL FORMULAE

or odds of 1 to 2 that a number greater than 4 will be rolled (the same as odds of 2 to 1 against a number greater than 4 being rolled). When the division of the numerator by the denominator is carried-out, the resulting fraction is the odds ratio

Using the P(event) notation is helpful for describing two important types of events in probability theory: mutually exclusive events and independent events. If a single die is rolled (the singular of dice is die, not douse as with mouse) the event P(one) = 1/6 and the event P(five) = 1/6 are mutually exclusive events because it is not possible to roll both a one and a five on a single roll of a die, ie, P(one AND five) = 0. The probability of the union of two mutually exclusive events — P(one OR five) — is the sum of their probabilites, ie, P(one OR five) = 1/6 + 1/6   = 2/6 = 1/3.

By contrast, if a single die is rolled twice, the results of the first roll has no influence on the results of the second roll — the two events are independent. It is quite possible to roll a one on the first roll and a five on the second roll — and the probability of this occurring is the product of the two probabilities, ie, P(one 1st AND five 2nd)  = 1/6 x 1/6 = 1/36. Similarly, the probability of NOT rolling a one on either roll is P(not one 1st) x P(not one 2nd)  = 5/6 x 5/6 = 25/36 as in question (6). The complement of rolling a one is not rolling a one, ie,
5/6 = P(not one)  = P(complement of one) = 1  - P(one) = 1 - 1/6.

In question (13) the probability of drawing a heart, P(heart) is the same as the probability of drawing a diamond, P(diamond), that is, P(heart) = P(diamond)= ½ and these are mutually exclusive — like flipping a balanced coin and getting a head or a tail, P(head) = P(tail)= ½. The probability of getting an even card and the probability of getting a heart or a diamond are independent events, so the product of the probability of an even heart or an even diamond is ½ x 4/9 = 2/9 or ½ x 2/5 = 1/5, respectively. Thus the total probability of getting an even card is the sum of the probabilities of the mutually exclusive events of drawing a card from either deck: 2/9 + 1/5.

Days of the week are mutually exclusive events, like numbers on a die, except that the probability of a particular day is 1/7 rather than 1/6. Question (7) asks for the probability of an event (rain) for 3 days where the probability is one-in-a-million for any particular day. Thus, P(rain) for 3  days in a week is

       1              1              1             3
   ------   +  ------   +  ------  =  ------
   million      million       million      million

The probability of rain and the probability of playing golf are independent events for the dedicated golfer in this problem, therefore P(rain AND golf) for a given week is the product of their independent probabilities:

 2             3
---   X  -------
 7         million

The law of averages states that as the number of independent trials of an event increases, the observed fraction of occurences of that event increasingly approaches the probability of the event in one trial. Thus, if a die is rolled 60 times the number of times a one (1) is rolled should be close to 10 and if a die is rolled 600 times, the number of times a one (1) is rolled should be even closer (relatively speaking) to 100.

Independent events cannot be mutually exclusive and mutually exclusive events cannot be independent. However, events that are not independent may be dependent rather than mutually exclusive. Two rolls of a balanced die (independent events) can be thought of as selecting with replacement. Selecting without replacement results in dependent events.

Thus, in question (1) the probability of selecting an apple on the second selection is dependent upon the first selection. If an apple is selected first, then the probability of getting an apple second, P(apple 2nd), is 5/9, but if an orange is selected first, then P(apple 2nd) is 6/9. Formally, P(apple 2nd GIVEN apple 1st)  = 5/9 and P(apple 2nd GIVEN orange 1st)  = 6/9. The probability of both the dependent events is the product of their probabilities, ie, P(apple 1st) x P(apple 2nd GIVEN apple 1st) = 6/10 x 5/9 =  P(apple 2nd AND apple 1st) — as in question (1).

The equivalence of P(apple 1st) x P(apple 2nd GIVEN apple 1st) and P(apple 2nd AND apple 1st) is relevant to the last question, which asked for P(diseased GIVEN positive) on the basis of the information P(diseased), P(positive GIVEN diseased) and P(positive GIVEN not diseased).

The last two events are mutually exclusive, therefore they can be added:

P(positive GIVEN diseased)  +  P(positive GIVEN not diseased) = P(positive).

P(diseased) and P(positive GIVEN diseased) are dependent rather than independent events, therefore the product

P(diseased)  x P(positive GIVEN diseased) = P(positive AND diseased).

Thus,

                                                                       P(diseased)  x P(positive GIVEN diseased)
  P(diseased GIVEN positive)  =  --------------------------------------------------------------------
                                                     P(positive GIVEN diseased) +  P(positive GIVEN not diseased)

                                                      P(positive AND diseased)
                                                  =  ---------------------------
                                                               P(positive)

This is the idea behind Bayes' Theorem (or Bayes's Theorem).

Note that:

P(A GIVEN B) x  P(B) =  P(A AND B) =  P(B AND A) =  P(B GIVEN A) x  P(A)

In selecting with replacement the apple of question (1) would have been returned to the bowl after the first selection before making the second selection — resulting in a

  6         6        36        9
---   X  ---  =  ---   =  ---
10        10      100      25

chance of selecting an apple both times. Conversely, selecting without replacement resulted in the problem as originally stated in question (1). Similarly, if an ace of diamonds has been removed from a deck of 52 cards, the chance of drawing a king of diamonds has become 1/51 -- selection without replacement.

Selection without replacement leads to dependent events whose total probability is the product of those events. This fact leads to the formula for permutations, because the number of permutations of people standing in a line can be modeled as choosing a first person for the line, then selecting among those remaining (selecting without replacement) the second person for the line, then the third person, etc.

Imagine two people (Ben & Charlie) standing side-by-side. The number of permutations possible is 2: BC & CB. (Note that there is only one combination of B & C, since order does not matter when combining.)

Now imagine 3 people, Alice, Ben and Charlie. Again, there is only one way to combine 3 people, but there are 6 ways to permute them: ABC, ACB, BAC, BCA, CAB, CBA. Notice that in the first case A stands to the left while B & C display their 2 permutations. In the second case, it is B that stands to the left and in the third case C stands to the left. Adding a third person has provided 3 new possibilities for permuting 2 people. So the formula is 3x2=6 permutations. Similarly, when Danièle joins the group, there are now 4 new possibilities for permuting 3 people, resulting in 4x3x2=24 permutations.

Understanding the above logic should make it clear that 7 people can arrange themselves in 7x6x5x4x3x2=5,040 ways, even though it would be painfully tedious to enumerate all the permutations. The mathematical notation for 7x6x5x4x3x2x1 is 7!, where the exclamation mark has the name factorial — representing the product of the integer with all the smaller positive integers.

If Alice & Ben join hands to form a circle and start dancing around the center, there is only one permutation (one arrangement), since neither Alice nor Ben can be said to be on the "end". If Alice is taken as a point of reference, Ben has only one permutation relative to Alice — he is both to the right of her and to the left of her. If Charlie and Danièle join the circle, Danièle becomes a point of reference (like an endpoint) and ABC can permute themselves in 3! ways relative to Danièle — A, B or C can be to her right or left.

When pairing up to ride the motorcycle, a person has one chance in 4 of being the driver, and the remaining people have one chance in 3 of being the rider, so each driver-rider permutation has 1/4 X 1/3 = 1/12 chance of happening — implying that there are 4x3=12 possible permuations. Similarly, a 1-in-6 chance of winning first prize, a 1-in-5 chance of winning second prize and a 1-in-4 chance of winning third prize results in 1/6 x 1/5 x 1/4 = 1/120 chances of any particular 1st-2nd-3rd prize permutation of happening (6 x 5 x 4). With 4 contestants and 2 prizes this can be more easily visualized:

    AAAA  BBBB  CCCC  DDDD  EEEE    1st prize

    BCDE  ACDE  ABDE  ABCE  ABCD    2nd prize

Notice that for each of the five choices of the first prize there are four remaining choices for the second prize — given all possible arrangements of 2 objects taken from 5 objects — ie, selection without replacement.

As was mentioned, the general formula for the number of permutations of n objects is n! (n factorial) — ie,
n x (n - 1) x (n - 2) ...  x 3 x 2 x 1.
The general formula for the number of permutations of r objects selected from n objects can be understood by thinking in terms of selecting r objects from n objects without replacement. If there are n objects from which to make selection 1, there will be n - 1 objects from which to make selection 2, n - 2 objects from which to make selection 3, and n - [r + 1] objects from which to make selection r.

Thus, the number of permutations of r objects selected from n objects is
n x (n - 1) x (n - 2) ...  x (n - [r + 1])

A more concise permutation formula results from multiplying the numerator and dividing the denominator by (n - r)!

n x (n - 1) x (n - 2) ...  x (n - [r + 1])  x (n - r)!             n!
---------------------------------------------------------    =    ---------
                              (n - r)!                                         (n - r)!

Thus, for question (10), selecting a 1st, 2nd and 3rd prize winner (r = 3 selections) from n = 6 contestants results in

             6!             6!
         --------   =    -----    =    6 x 5 x 4 permutations
         (6 - 3)!         3!

It is a convenient mathematical "fact" that both 1! = 1 and 0! = 0. This means that if we are looking for the number of permutations of 4 objects (n = 4 and r = 4) we get

             4!             4!
         --------   =    -----    =    4!
         (4 - 4)!         0!

What if we simply wanted to know the number of ways we could 3 people from 6 contestants without assigning them to be 1st, 2nd or 3rd prize winners? In other words, how many combinations of 3 people can be selected from 6 people without regard to their arrangement? Because the arrangement of the r = 3 people does not matter, the general formula for combinations can be derived from the general formula for permutations by dividing-out the unwanted r! permuations:

               n!
         -----------
         r!(n - r)!

Question (14) is partly a combination problem that can be solved by the combinations formulas. Thus, to select 2 men from 5 men there are

               5!              120
         -----------   =    -----    =    10 combinations
         2!(5 - 2)!         12

and to select 3 men from 5 men there are

               5!              120
         -----------   =    -----    =    10 combinations
         3!(5 - 3)!         12

Selecting the combinations of men and selecting the combinations of women are independent events — and each combination has a 1/10 probability, so the total probability of any one committee is 1/10 x 1/10 = 1/100 — which is another way of saying that there are 100 possible committees.

             Now you can test yourself:              WORKING WITH EVENTS .

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V. OPINIONS ON SOME WELL-KNOWN PROBLEMS AND ISSUES

The St. Petersburg paradox was proposed in the 18th century by the Swiss mathematician Nicholas Bernoulli. Bernoulli proposed a game based on payoffs associated with doubling payoffs and flipping a coin until a tail appears. If a tail appears on the first flip, the player receives $2. If a tail appears on the second flip, the player receives $4. Tail on third flip: $8. Tail on fourth flip: $16. Head on any flip simply means waiting for the next flip and knowing that the payoff will be doubled if a tail appears. Thus, a player would have a 50-50 chance of winning $2, a 1/4 chance of winning $4, a 1/16 chance of winning $16, a 1/64 chance of winning $64, and in general a 1/2^k chance of winning $2^k.

Bernoulli asked: "How much would you pay to play this game?"

How much would you pay?

Reputedly, most people would not pay as much as $25 to pay the game. Yet, the expected value (payoff times probability of winning) of the playing the game is infinite:

 = (1/2)($2.00) + (1/4)($4.00) + (1/8)($8.00) + (1/16)($16.00) + ...

 = $1.00 + $1.00 + $1.00 + $1.00...

although the probability of winning as much as $25 is less that 1/16.

Many people believe that Nicholas's cousin Daniel Bernoulli solved the paradox by invoking the marginal utility of money — that the subjective utility of money decreases as the amount of money increases. But $25 is far below the margin of utility for money for most people. Nicholas Bernoulli suggested that most people greatly discount improbable events, which seems like a more plausible explanation. But then why do so many people seem motivated to buy a lottery ticket that has a high payoff, despite the fact that the probability of winning is so low (although may not be known)?

The marginal utility of money explanation makes more sense for a coin flip in which two people of equal net worth risk half their net worth on the flip. The expected value of playing the game is the same as the expected value of not playing the game, yet it likely that for both players, the benefit of gaining half-again their net worth would not compensate for the risk of losing half-again their net worth. This is "common sense psychology"...based on the marginal utility of money (value of extra money compared to value of existing money).

The assymetry between the benefit associated with a gain and the harm associated with a loss of equal value has formed the basis of prospect theory. Most people will choose a 1 in 1,000 chance of gaining $5,000 over a gain of $5, but prefer a sure loss of $5 to a 1 in 1,000 chance of losing $5,000 — even though the expected values of the gains and losses are equivalent. A $5 gain may seem negligible and a $5,000 loss could be devestating (depending on wealth). One prospect theory study showed that women were more likely to do breast self-examination if warned of the dangers of cancer without self-examination than if told of greater protection against cancer with self-examination. Prospect theory also predicts that most people would pay more to remove the only bullet from a gun in Russian roulette than they would pay to remove one of four bullets ("the certainty effect"), even though the probability of being shot is reduced by the same amount in both cases.

In his book FOOLED BY RANDOMNESS, Nassim Taleb cites an example of a quiz given to medical doctors to demonstrate the failure of most people to grasp Bayesian probability:

A test of a disease presents a rate of 5% false positives. The disease strikes 1 out of 1,000 of the population. People are tested at random, regardless of whether they are suspected of having the disease. A patient's test is positive. What is the probability that the patient has the disease?

Reportedly, the majority of doctors answered 95% and less than a fifth of the doctors got the right answer, which indicates a psychological tendency to forget or ignore the context. The answer becomes more apparent if you remember that there is a 99.9% chance that any patient came from the portion of the population that does not have the disease, although a positive test does significantly increase the chance that the patient came from the 0.1% group that has the disease. About 50 out of 999 people who do not have the disease will be expected to test positive, and the 1 out of 1,000 who has the disease is expected to test positive (no percentage of false negatives were given in the problem). Thus, 1/(50+1) gives a roughly 2% chance that a person tested at random who tests positive will have the disease.

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VI. PHILOSOPHIZING ABOUT PROBABILITY

Rolling dice, flipping coins, drawing colored balls, etc., are reasonably treated as quantifiable probablistic random processes because the outcomes are so definite and uncontrollable. The decay rate of radioactive isotopes can become predictable on the basis of past events, while the underlying causes remain unknown. It is too easy see quantifiable patterns in phenomena with unknown causes, and extrapolating to the future. Adding straws to the back of a camel will not break the camel's back until the last straw does so. Events for which underlying causes are unknown are described as "luck" (good or bad) when the outcome is deemed favorable or unfavorable.

Events can be quantitatively described as probable or improbable when compared to the total number of possible outcomes. But when the total number of outcomes cannot be specified, there are no grounds for calculation. Intuitively, it seems like a very improbable event that I would unexpectedly meet a Canadian acquaintance in a British train station. But how can I quantify this probability? Do I include all of the other people I have ever known and the total number of people in British train stations? Improbable events occur — coincidences — even though they occur with low probability. Given a high number of possible improbable events, it is highly probable that some of them will occur. We only notice the "coincidences" that do occur, not the ones which do not occur. The occurance of improbable events do not necessarily require paranormal or supernatural explanations.

Additionally, some coincidences are more probable than we might expect due to our lack of appreciation of actual probabilities. The probability of two people not having the same birthday is 364/365 (multilpying the 364 days remaining for the second person times the 1/365 probability of the birthday of the first person). 1 - 364/365 => 0.275% chance of having the same birthday. But for 23 people there is a greater than 50% chance that at least two of them will have the same birthday because

P(same birthday) = 1 - (364/365)(363/365)...(343/365) > 0.5

Although flipping a coin or rolling dice are treated as random processes they are not. Whether a coin comes up heads or tails is determined by the trajectory of the coin, the speed of rotation, the angle of rotation, air resistance, material characteristics of the surface on which the coin is thrown, the force of gravity in the location, etc. The same can be said for a roll of dice. There are so many variables, the variables are so hard to measure and the interaction of the variables is so complex that the flipping of a coin or rolling of dice are practically speaking "random". Said another way, the processes can be treated epistemologically as being random although metaphysically they are not — they are deterministic.

In 1961 a research meteorologist at MIT named Edward Lorenz was using a set of equations to model weather on a computer when he discovered that rounding his initial numbers to three decimal places produced dramatically different results from those obtained by using six decimal places. Systems so sensitive to small variations in initial conditions have been called "chaotic", but they are more accurately described as pseudo-random — just as so-called random numbers generated by computer are called pseudo-random. Again, the phenomena are metaphysically deterministic, but their unpredictability renders them epistemologically random — no different from rolling dices or flipping coins.

In arguing against the Copenhagen Interpretation of Quantum Mechanics Albert Einstein made the infamous remark, "God does not play dice with the universe." I call the remark infamous because it is usually quoted to display how irrational & disreputable Einstein's beliefs were. Neils Bohr, Werner Heisenberg and others in the Copenhagen School proposed that randomness is a metaphysical condition of subatomic particles, whereas Einstein argued that randomness as a phenomenon is an artifact of our ignorance of underlying deterministic processes & forces — limitations on our knowledge. His careless attempt at emotional/theological appeal by a reference to God & dice probably cause many to discredit the more rational arguments Einstein made.

If 50 red marbles are placed in a glass bowl on top of 50 green marbles and the bowl is shaken, the green and red marbles will mix. The probability that the marbles will separate again into all green in one section and all red in another section is very low. This fact about mixing underlies the second law of thermodynamics, which is sometimes described as stating that the disorder of the universe must inevitably increase. If the bowl is dropped causing the glass to shatter and the marbles to disperse, the disorder is increased much further — it is easier to destroy than to create.

Evolution seems to disprove the second law. More generally, the existence of forces might drive organization rather than disorganization. But the development of an organism or species can only occur by the dissipation of energy — meaning that the organization of the organism has increased only at the cost of greater disorder in the universe at large. Those who argue for the prospect of physical immortality face an uphill battle against the second law of thermodynamics.

Probability bridges the gap between descriptive statistics (average, standard deviation, histograms, etc.) and inferential statistics (decision-making statistics). The fact that a sample average tends to have a normal probability distribution means that highly improbable sample averages suggest that the population average is different from the assumed value — grounds for hypothesis tesing and decisions about hypotheses concerning a population known only through sampling. (Philosopher David Hume denied that any correlation — probable or improbable — would be grounds for deducing causation.)

Decision-making is based on the probability of an event occuring times the payoff of the event — a cost/benefit decision. More formally:

Expected value = probability X payoff

It would seem advantageous to wager $1 on the chance of winning $10 by rolling a snake-eye (one) with a single die because the expected value is

probability X payoff = (1/6) X $10 = $1.67

which is greater than the $1 cost. But there is a 5/6 chance of losing the $1, which could be critical if you need the money to buy bus-fare. Non-monetary factors are often important in cost/benefit calculations — with benefits more often being more difficult to quantify and calculate than costs.

Pascal's Wager is based on an expected value calculation. Pascal set the probability for God's existence at ½ and the payoff for believing in God's existence as infinite reward. The Wager fails to distinguish between the Gods of Catholicism, Mormonism or Islam — or the costs of making the wrong choice (which could be infinite punishment). The Wager also fails to consider whether it matters to God that a person has chosen to believe in Him based on a mercenary expected value calculation.

The decision on whether to make cryonics arrangements can be compared to Pascal's Wager. The payoff of successful cryonic reanimation to a condition of enduring youth is extremely high (if not infinite) to some people, even if the probability of success is low. Because probability of success is based on expectations about future science, there can be no objective means of determining the probability — the determination is subjective/intuitive.

(For more on the subject of subjective probability, see Some Philosophizing about Subjective Risk.)

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