- INTRODUCTORY REMARKS
- QUESTIONS
- ANSWERS
- MATHEMATICAL FORMULAE
- OPINIONS ON SOME WELL-KNOWN PROBLEMS AND ISSUES
- PHILOSOPHIZING ABOUT PROBABILITY

Puzzles & games are forms of mental work that many do for "fun". There is an emotional/mental gratification that can be achieved by solving a problem — or by understanding the solution when the effort to solve has failed. In competititve games, the problem solving has a win/lose pride element usually resulting in greater gratification for the "winner". Puzzles are more like solitaire where the adrenalin of competition is missing and the gratification is more purely mental — lacking even the sense of accomplishment that accompanies solving problems that improve the conditions of life.

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

There are six chances in ten that the first choice is an apple. With 5 apples and 4 oranges left in the bowl, there are five chances in nine that the second choice is an apple. The product of the two choices is:

6 5
30
1

--- X --- = ---
= --- (one chance in three)

10 9
90 3

If the chance that CB will win is 1 then the chance that SS will win is 2 and for GH it is 4. If GH has 4 out of a total of 4 + 2 + 1 = 7 chances of winning, that means 3 out of 7 = 3/7 chances of not winning.

There are six possible permutations of any three different numbers. Only one of those permutations will be strictly increasing. Therefore, there is only one chance in six (1/6) that any sequential selection of three cards will be strictly increasing. Note that the size of the hand makes no difference to the answer as long as the hand contains three or more cards. Any three distinct numbers have 6 distinct arrangements.

There are 6 x 6 = 36 possible outcomes, but only the following 6 outcomes produces totals to seven:

1,6 2,5 3,4

6,1 5,2 4,3

That means that there are 36 - 6 = 30 ways
in which the two dice can ** not** roll a seven. Therefore,
the odds against rolling a seven are 30-to-6 or (reducing) 5-to-1

One third of the people are men and half have gray hair

1 1
5

--- + --- = ---

3 2
6

But this will double-count men with gray hair (who are in both categories). Three of the eighteen (1/6) are men with gray hair, so

5
1 2

--- - --- = ---

6
6 3

is the chance that a person chosen at random with be either a man or a person with gray hair or both.

A simpler solution is to exclude women without grey hair (6/18 excluded, so 12/18 = 2/3 are to be counted.)

If one die is thrown, the chances of not rolling a one (1) are

5

--- = 0.8333

6

If two dice are thrown, the chances of not rolling a one (1) are

5 5

--- X -- = 0.6944

6 6

If three dice are thrown, the chances of not rolling a one (1) are

5 5
5

--- X -- X --
= 0.5787

6 6
6

If four dice are thrown, the chances of not rolling a one (1) are

5 5
5 5

--- X -- X --
X -- = 0.4823

6 6
6 6

So four dice give better than even chance (over 50% chance) of rolling a one (1).

To view the problem in another way, note that the probability of rolling
a one (1) is 1/6, ie, P(** one**) = 1/6. If
you roll a die ("one dice") 60 times you will, on average,
get 10 ones (1s). But that does not mean that if you roll a die
6 times that you will have to wait until the 6th roll to get the first
one (1). Because 3/6 = 1/2 it might seem that on average
the first one (1) would appear halfway between the first and
the sixth roll:

1st 2nd 3rd 4th 5th 6th

but the middle is **between** the 3rd and the 4th!

If it rains three times weekly and the probability of a golfer getting hit by lightning on a rainy day is one-in-a-million, then the probability of getting hit by lightning would be 3-in-a-million for a golfer who played every day. For a golfer who only plays 2 days per week, it is necessary to multiply by 2/7:

2
3
1

--- X -------- =
------------

7
million
1,166,666

There are 24 arrangements (24 **permutations**)
of 4 people on 4 different sides of a square table:

ABCD BACD CABD DABC

ABDC BADC CADB DACB

ACBD BCAD CBAD DBAC

ACDB BCDA CBDA DBCA

ADBC BDAC CDAB DCAB

ADCB BDCA CDBA DCBA

For a circular table there are an infinite number of arrangements if exact position in space matters — e.g., that Danièle can sit on the North side ot the table, the East side or at any of the infinite number of positions in between. If the only thing that matters is the relative position of each person (who is sitting next to whom), then the position of Danièle is arbitrary and there are 6 arrangement (permutations) of the other 3 people in relation to Danièle:

ABC

ACB

BAC

BCA

CAB

CBA

In this analysis there is no difference between Danièle sitting on the SouthWest side, the NorthEast side or the North by NorthWest side — all that matters is the relative positions of the others. If Ben sits to Danièle's right, then if Alice sits to Ben's right, the arrangement will be BAC — with Charlie to the left of Danièle.

Assuming that Danièle has extra pink motorcycle helmets that she can lend to other drivers & riders, there are 12 possible permutations (the person on the left being the driver and the one on the right being the rider):

AB AC AD BC BD CD

BA CA DA CB DB DC

Or, more simply, after one of 4 possible drivers are selected, one of 3 possible riders can be selected: 4 x 3 = 12

Using the same logic as in the previous problem, after one of six contestants is selected for 1st prize and one of five contestants is selected for 2nd prize, then one of four contestants is left for 3rd prize: 6 x 5 x 4 = 120

Probability of having the disease and testing positive: 9/10

Probability of NOT having the disease and testing positive: 2/10

Probability of having the disease: 1/100

Probability of NOT having the disease: 99/100

Probability of having the disease x Probability of having the disease and testing positive = 1/100 x 9/10 = 9/1,000

Probability of NOT having the disease x Probability of NOT having the disease and testing positive = 99/100 x 2/10 = 198/1,000

Probability of testing positive = 9/1,000 + 198/1,000 = 207/1,000

Therefore, probability of having the disease if test positive:

9 / 1,000
9
1

---------------- =
-------- = -----

207 / 1,000
207
23

Another way to view this problem is to imagine:

Of 1,000 people, 10 have the disease and 990 do NOT have the disease

9 of the 10 test positive ------------ and 198 of the 990 test positive

Therefore 9 of the 9+198=207 people who test positive have the disease.

This problem is similar to problem (6). The chance of one of of the dice matching any arbitrary number is the same as the chance that it will match some particular number, such as the one (1) in problem (6). The probability of matching is the complement of the probability of not matching on three rolls:

5
5 5

1 - -- X -- X
-- = 1 - 0.5787 = 0.4213

6
6 6

There are four even hearts {2,4,6,8} and two even diamonds {2,4} so the chances of getting an even heart is

1 4
2

--- X --- = ---

2 9
9

and the chances of getting an even diamond is

1 2
1

--- X --- = ---

2 5
5

The total chances of getting an even card is

2
1
10
9 19

--- + --- =
--- +
--- = ---

9
5
45
45 45

Therefore, the chance that an even card is a heart will be the chance of getting an even heart (2/9) divided by the chance of getting an even card (19/45), which equals 10/19.

There are 6 x 6 x 6 = 216 possible outcomes, but only the following 20 outcomes produces the desired result:

1,2,3 1,3,5 2,3,4 2,5,6

1,2,4 1,3,6 2,3,5 3,4,5

1,2,5 1,4,5 2,3,6 3,4,6

1,2,6 1,4,6 2,4,5 3,5,6

1,3,4 1,5,6 2,4,6 4,5,6

20 5

---- = ---- =
0.09259

216 54

There are 6 x 6 x 6 = 216 possible outcomes, but only the following 27 outcomes produces the desired result:

1,3,6 2,3,5 3,2,5 4,1,5 5,1,4 6,2,2

1,4,5 2,4,4 3,3,4 4,2,4 5,2,3 6,3,1

1,5,4 2,5,3 3,4,3 4,3,3 5,3,2

1,6,3 2,6,2 3,5,2 4,4,2 5,4,1

2,2,6 3,1,6 3,6,1 4,5,1 6,1,3

27 1

---- = ----

216 8

1
1

------ = ----

8 - 1 7

The odds are 7-to-1 against rolling a ten.

There are ten ways (ten **combinations**) to choose
three women from five and there are ten ways to choose two men
from five, therefore there are 10 X 10 = 100 different
committees that could be formed when any of the groups of three
women can be paired with any of the groups of two men.

Visually:

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

Because the order of the cards does not matter, this question is asking for the number of combinations (not permutations) of five different objects selected from 52 objects. Using the mathematical formula for combinations:

**
52!
52!
-------------
= -------
= 2,598,960 ** combinations

5!(52 - 5)! 5! 47!

The total number of hands possible is the number of combinations
of **5** cards selected from **13** cards, ie,

**
13!
13!
-------------
= -------
= 1,287 ** combinations

5!(13 - 5)! 5! 8!

There is only one way of selecting 3 face cards from 13, but there are 45 ways of selecting 2 different cards from the remaining 10:

**
10!
10!
-------------
= -------
= 45 ** combinations

5!(10 - 2)! 2! 8!

45
5

------- = ---- =
0.034965

1,287 143

There are 6 ways to select 2 aces from the 4 aces
in the deck (ie, 6 possible ** combinations**) and there
are 13 different kinds of cards, so the total number of combinations
possible of 2 cards is 6 x 13 = 78. There are
4 ways to choose 3 Queens from 4 Queens, but because
the 3-of-a-kind suit must be different from the 2-of-a-kind suit, the
possible combinations of the 3-of-a-kind must be taken from the
12 remaining cards, ie, 4 x 12 = 48. Thus, the
total number of combinations of 2-of-a-kind and 3-of-a-kind are
48 x 78 = 3,744. To find the probability of a
full house one must divide the number of combinations resulting in a
full house by the total number of combinations — determined in
question (17):

3,744
6

------------- = ------- =
0.00144

2,598,560 4,165

As in question (19) there are 6 ways to select 2 aces from
the 4 aces in the deck (ie, 6 possible ** combinations**)
and there are 13 different kinds of cards, so the total number of
combinations possible of 2 cards is 6 x 13 = 78.
There are 48 possible choices for the 3rd card, 44 possible
choices for
the 4th card and 40 possible choices for the 5th card,
but the these last 3 cards can be chosen in any order, so it is
necessary to divide by the number of permutations possible for
3 cards — ie, divide by 3! = 6:

78x48x44x40

------------------ = 1,098,240 combinations
resulting in pairs

3!

To find the probability of a pair one must divide the number of combinations resulting in a pair by the total number of combinations — determined in question (17):

**1,098,240 352
------------- = ------- =
0.42257
2,598,560 833
**

Mathematical proofs can be not only boring & tedious, but unconvincing. A mathematical formula can be a handy tool for solving a problem when the problem is an obvious candidate. Proving the truth of the formula is a matter of clear understanding. The puzzles I provided were intended to be simple enough to solve by enumeration, but also solvable by formula. In this section I will attempt to provide "ordinary language" explanations and rationales for formulae.

The probability of an event in probability theory is denoted by the
notation P(** event**). The probability of any event is the
number of times that event can occur divided by the total number of
possible outcomes. Probabilities can be expressed as fractions or
decimals. The total probability of all possible outcomes always
sums to one (1). Thus, the probability of getting a head on the
flip of a balanced coin,
P(

Again, probability is usually a number between
**0** and **1**.
If the probability of an event is **0**, then the event is
** impossible** — such as rolling a seven with a single die
("one dice"). If the probability of an event is

Number of outcomes resulting in the event

P(** Event**)

Total number of possible outcomes

Thus, the probability of rolling a number greater than 4 with a single die will be:

P(5,6)
2
1

---------------- = ---
= --- =
0.33333....

P(1,2,3,4,5,6) 6
3

**Odds** are similar to probability, but relate the chance of an
event occuring to the chance of the event ** not** occuring
rather than to total possible outcomes:

Number of outcomes resulting in the event

Odds(** Event**)

Number of outcomes NOT resulting in the event

Thus, the odds of rolling a number greater than 4 with a single die will be:

P(5,6)
2
1

----------- = ---
= ---

P(1,2,3,4) 4
2

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:

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

5/6 = P(

In question (13) the probability of drawing a heart,
P(** heart**) is the same as the probability of drawing a
diamond, P(

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

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

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(

The equivalence of
P(** apple 1st**) x P(

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

P(** positive** GIVEN

P(** diseased**) and
P(

P(** diseased**)

Thus,

P(** diseased**)

P(

P(

P(** positive** AND

P(

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

Note that:

P(** A** GIVEN

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

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 .

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.

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**)

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

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

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.)