by Ben Best

### I. INTRODUCTORY REMARKS

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 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.)

### (1) A fruit bowl has 6 apples and 4 oranges. If one piece of fruit is selected at random and then a second piece of fruit is selected at random, what are the chances that both pieces of fruit will be apples?

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

### (2) Golden Hoof (GH) is twice as likely to win a horse race than Silver Saddle (SS) and Silver Saddle is twice as likely to win as Chomping-at-the-Bit (CB). If only the 3 horses are in the race, what is the probability that Golden Hoof will not win?

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.

### (3) Given a hand containing the nine numbered hearts from a deck of cards [2 to 9, with Ace counting as one (1)] — what is the probability of drawing three cards in sequence from the hand such that each selection results in a number larger than the previous one?

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.

### (4) What are the odds against rolling a seven with a single throw of a pair of dice [2 dice]?

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

### (5) There are six men and twelve women. Half of the men have gray hair and so do half of the women. What are the chances that a person chosen at random will be a man, a person with gray hair or both?

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.)

### (6) How many dice must be thrown to give a better than 50% chance that at least one of the dice will show a one (1) (a single "snake eye")?

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!

### (7) If the chances of a golfer being struck by lightning on a rainy day are one-in-a-million, if it rains an average of three days per week and if you play golf every Saturday & Sunday — rain or shine — what are your chances of being struck by lightning in a given week? (Assume lightning only strikes on rainy days.)

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

### (8) How many ways can Alice, Ben, Charles and Danièle arrange themselves around a square table? How many ways can they arrange themselves around a circular table?

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

ABCD  BACD  CABD  DABC
ACDB  BCDA  CBDA  DBCA

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.

### (9) How many arrangements of rider & driver are there if Alice, Ben, Charlie and Danièle pair up on Danièle's motorcycle — one person driving and one person riding behind the driver while holding on?

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

### (10) If there are 6 contestants in a contest, in how many potential ways can 1st, 2nd and 3rd prize be awarded?

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

### (11) One person per hundred people has infectious foot-in-mouth disease. The probability of a person with infectious foot-in-mouth disease testing positive is 9/10 and the probability of a person who does not have infectious foot-in-mouth disease testing postive is 2/10. What is the probability that a person who tests positive has the disease?

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.

### (12) If two dice are rolled three times, what is the probability that the two dice will match (ie, display the same number) on one of the three rolls?

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

### (13) There are two hands of cards, one consisting of nine hearts numbered 1 to 9 (ace is 1) and another hand consisting of five diamonds numbered 1 to 5. If there is an equal chance of choosing a card from either hand, what is the probability that an even card will be a heart?

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.

### (14) If a single die (one "dice") is rolled three times, what is the probability that each roll will result in a number larger than the previous roll?

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

### (15) If three dice are thrown (rolled), what are the odds against rolling a total of 10?

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.

### (16) Out of five men and five women, how many ways are there to form a committee consisting of three women and two men ?

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
BCD             GJ
BCE             HI
BDE             HJ
CDE             IJ
```

### (17) How many different poker hands (5 cards) are possible with a standard 52-card deck?

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!

### (18) If 5 cards are drawn from a deck consisting of only the 13 hearts from a 52-card deck, what is the probability of getting all of the face cards (King, Queen and Jack)?

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

### (19) What is the probability of getting a "full house" in five cards drawn in a poker game from a standard 52-card deck? [A full house consists of 3 cards of the same kind (eg, 3 Queens) and 2 cards of another kind (eg, 2 Aces).]

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

### (20) What is the probability of getting a "pair" in five cards drawn in a poker game from a standard 52-card deck? [A pair consists of 2 cards of the same kind (eg, 2 Queens) and 3 cards that are different from the kind of the pair (eg, different from Queens) and that are all different from each other.]

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

### IV. MATHEMATICAL FORMULAE

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(head) = ½ = 0.5 because 2 outcomes (heads or tails) are equally possible when a balanced coin is flipped. If there is an equal probability of Alice, Ben, Charlie or Danièle being the driver of Danièle's motorcycle, then the probability of Ben being the driver can be denoted P(Ben) = ¼ = 0.25.

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 1, then the event is certain — such as the probability of getting a head or tail with a coin that will not stand on its edge. The probability of an event is calculated as:

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

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

Thus,

P(diseasedx 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 .

### 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 1/2k chance of winning \$2k.

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= 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.) HOME PAGE