FIBONACCI NUMBERS, PROBABILITY, AND ...

Proceedings, ICMET’ 13 (17-20 Dec 2013), 13-21, KERALA, INDIA 2015

FIBONACCI NUMBERS, PROBABILITY, AND GAMBLING ANDREAS N. PHILIPPOU University of Patras, Greece One-Day Seminar at All Saints College, Trivandrum - Kerala, India, December 16, 2013Invited Speaker at ICMET’ 13, Trivandrum - Kerala, India, December 17-20, 2013

Abstract. The Fibonacci and Lucas numbers are briefly introduced and their relationship to the golden mean and the geometric distribution of order k is presented. Three gambling systems are also touched upon, as well as the odds in some odd-even games.

1. Introduction Leonardo Fibonacci (c.1175-c.1240) was born in Pisa, Italy, but had a Muslim teacher in sea side Bugia of what is to day Algeria when his father was a customs officer there. He studied and travelled extensively in the Mediterranean, becoming one of the best mathematicians of the Middle Ages [Eves (1990)]. In 1202 he introduced the Hindu-Arabic numerals to Europe with his book Liber Abaci (Book of Calculation), which also includes his rabbit problem. The numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …, formally defined by the recurrence relation (1.1)

𝐹1 = 1, 𝐹2 = 1, 𝐹𝑛+2 = 𝐹𝑛+1 + 𝐹𝑛 ,

𝑛 ≥ 1,

have been named Fibonacci numbers by nineteenth century French mathematician Edouard Lucas to honor Fibonacci. We now know, however, that the Fibonacci numbers have been known before Fibonacci by Indian scholars who had been interested in rhythmic patterns formed from one-beat and two-beat notes or syllables. The astronomer Johann Kepler rediscovered them in 1611, and since then several renowned mathematicians, including J. Binet and E. Catalan, have dealt with them. Lucas studied Fibonacci numbers extensively, and the sequence 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, …, formally defined by the recurrence (1.2)

𝐿1 = 1, 𝐿2 = 3, 𝐿𝑛+2 = 𝐿𝑛+1 + 𝐿𝑛 , 𝑛 ≥ 1,

bears his name.

1

During the twentieth century, interest in the Fibonacci numbers and their applications rose rapidly. In 1961 Vorobyov published his Fibonacci Numbers, and Hoggatt, Jr., followed in 1969 with Fibonacci and Lucas Numbers. Meanwhile, in 1963, Hoggatt, Jr. and his associates founded The Fibonacci Association and started publishing The Fibonacci Quarterly. Finally, in 1984, the First International Conference on Fibonacci Numbers and Their Applications was held in Patras, Greece, and the Proceedings were published by Reidel. The Second was held in Santa Clara, California, in 1986, the Third in Pisa, Italy, in 1988 (one every two years in Europe and the USA, respectively), and so on. The Proceedings have been published by Kluwer until 2004 and by a new publisher thereafter. See, for example, Philippou, Bergum and Horadam (1986, 1988). The Fibonacci numbers appear in sunflowers, pineapples, and phyllotaxis. They appear in geometry and architecture, in computer science and probability theory, in gambling. The scrambled version 13, 3, 2, 21, 1, 1, 8, 5 of the first eight of them appears in The Da Vinci Code, a novel by D. Brown (2003) which is also a well known Hollywood film. 2. The Rabbit Problem The rabbit problem which appears in Liber Abaci is trivial and may be stated as follows. On January 1, there is a pair of adult rabbits in an enclosure. This pair produces one pair of baby rabbits on February 1st, and one pair of baby rabbits on the first day of each month thereafter. Each baby pair grows to be an adult pair in one month, and produces a baby pair on the first day of the third month of their life as well as on the first day of each month thereafter. Find the number of pairs of rabbits in the enclosure a year later on January 1st after the day’s births. Counting them, or otherwise, the number of pairs of adult rabbits is A 13 = 233 = F13, the number of pairs of baby rabbits is B13 = 144 = F12, and the total number of pairs of Rabbits is T13 = A13 + B13 = 377 = F14 = F12 + F13. 3. The Golden Mean (or Golden Ratio, Golden Section, Divine Proportion) According to Euclid's Elements (Στοιχεῖα in Greek) "A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less." Suppose we want to divide a line segment AB into two line segments AS and SB so that AB AS = . AS SB Setting 𝑥 = AS/SB, we get

2

𝑥=

𝐴𝑆 + 𝑆𝐵 1 1 = 1+ = 1 + 𝑖𝑓𝑓 𝑥 2 − 𝑥 − 1 = 0. 𝐴𝑆 𝐴𝑆 𝑥 𝑆𝐵

The equation 𝑥 2 − 𝑥 − 1 = 0, has the roots 1 + √5 𝛼= ( ) = 1.61803 … , 2

1 − √5 𝛽= ( ) = −0.61803 … . 2

The first, denoted here by α, even though it is usually denoted by φ in honor of the Greek sculptor Phidias (Φειδίας in Greek), has been called golden section, golden ratio, golden mean, or divine proportion, since it is associated with our perception of beauty. The numbers α and β are irrational. A geometric construction of S given the line segment AB can be done as follows. a. Construct a perpendicular BC at point B, with BC half the length of AB. Draw the hypotenuse AC. b. Draw an arc with center C and radius CB. This arc intersects the hypotenuse AC at point D. c. Draw an arc with center A and radius AD. This arc intersects the original line segment AB at point S. Point S divides the original segment AB into line segments AS and SB with lengths in the golden ratio. Prove it. 4. Two Results Relating the Fibonacci Numbers and the Golden Mean Proposition 4.1. The ratios of consecutive Fibonacci numbers converge to the golden mean or divine proportion 𝐹𝑛+1 1 + √5 𝑙𝑖𝑚 =𝛼=( ) = 1.61803 … . 𝑛→∞ 𝐹𝑛 2 Proof. Let 𝑥𝑛 = 𝐹𝑛+1 ⁄𝐹𝑛 , 𝑛 ≥ 1. Then, 𝑥1 = 1,

𝑥3 = 3⁄2 ,

𝑥2 = 2,

𝑥4 = 5⁄3 ,

𝑥5 = 8⁄5,

𝑥6 = 13⁄8 , …

which indicates (and it may be proven so) that 𝑥𝑛 converges to a positive number, say x. Consequently, 𝑥 = lim 𝑥𝑛 = lim ( 𝑛→∞

𝑛→∞

= lim (1 + ( 𝑛→∞

𝐹𝑛 + 𝐹𝑛−1 ), 𝐹𝑛

𝑏𝑦 (1.1),

1

1 )) = 1 + ( ) 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥 2 − 𝑥 − 1 = 0, 𝑥𝑛−1 𝑥

which establishes the proposition, since the positive root of 𝑥 2 − 𝑥 − 1 = 0 is α. 3

The Binet formulas to which we turn now provide the following closed expressions for the Fibonacci and Lucas numbers Proposition 4.2 (Binet Formulas). Let α be the golden mean and β = 1 – α. Then 𝐹𝑛 =

𝛼 𝑛 − 𝛽𝑛 , 𝛼−𝛽

𝑛 ≥ 1,

𝐿𝑛 = 𝛼 𝑛 + 𝛽 𝑛 ,

𝑛 ≥ 1.

Proof. Since α and β are roots of 𝑥 2 − 𝑥 − 1 = 0, it follows that (4.1)

𝛼 2 = 𝛼 + 1,

𝛽 2 = 𝛽 + 1.

Multiplying each side of the two equations by 𝛼 𝑛 , 𝛽 𝑛 , respectively, we get (4.2) which imply (4.3)

𝛼 𝑛+2 = 𝛼 𝑛+1 + 𝛼 𝑛 ,

𝛽 𝑛+2 = 𝛽 𝑛+1 + 𝛽 𝑛 ,

𝛼 𝑛+2 − 𝛽 𝑛+2 𝛼 𝑛+1 − 𝛽 𝑛+1 𝛼 𝑛 − 𝛽 𝑛 = + , 𝛼−𝛽 𝛼−𝛽 𝛼−𝛽

𝑛 ≥ 1,

𝛼 𝑛+2 + 𝛽 𝑛+2 = 𝛼 𝑛+1 + 𝛽 𝑛+1 + 𝛼 𝑛 +𝛽 𝑛 ,

𝑛 ≥ 1.

and (4.4)

We show now the first Binet formula. Setting 𝑈𝑛 = (𝛼 𝑛 − 𝛽 𝑛 )⁄(𝛼 − 𝛽) , 𝑛 ≥ 1, it suffices to show that 𝑈𝑛 satisfies the defining relationship of Fn. It does, since 𝑈1 = (𝛼 − 𝛽)/(𝛼 − 𝛽) = 1, 𝑈2 = 𝛼 + 𝛽 = 1 and 𝑈𝑛+2 = 𝑈𝑛+1 + 𝑈𝑛 , 𝑛 ≥ 1, by (4.3). We next show the second. Setting 𝑉𝑛 = 𝛼 𝑛 + 𝛽 𝑛 , 𝑛 ≥ 1, it suffices to show that 𝑉𝑛 satisfies the defining relationship of Ln. It does, since V1 = α + β = 1, 𝑉2 = 𝛼 2 + 𝛽 2 = 𝛼 + 𝛽 + 2 = 3 𝑏𝑦 (4.1), and 𝑉𝑛+2 = 𝑉𝑛+1 + 𝑉𝑛 , 𝑛 ≥ 1, by (4.4).

5. A Few Fibonacci Identities The following identities may be established by induction on n. 𝑛

(5.1)

∑ 𝐹𝑖 = 𝐹𝑛+2 − 1,

𝑛≥1

𝑖=1 𝑛

(5.2)

∑ 𝐹𝑖2 = 𝐹𝑛 𝐹𝑛+1 ,

𝑛≥1

𝑖=1

(5.3)

𝐹𝑛−1 𝐹𝑛+1 − 𝐹𝑛2 = (−1)𝑛 ,

𝑛≥1

4

[𝑛/2]

(5.4)

𝐹𝑛+1 = ∑ ( 𝑖=0

𝑛−𝑖 ), 𝑖

(5.5) 𝐹𝑛 𝐿𝑛 = 𝐹2𝑛 , We give an even easier proof of the first.

𝑛≥1

𝑛≥1

Proof of (5.1). By means of the definition of the Fibonacci numbers, we have for 𝑛 ≥ 1 𝑛

𝑛

∑ 𝐹𝑖 = ∑(𝐹𝑖+2 −𝐹𝑖+1 ) = (𝐹3 −𝐹2 ) + (𝐹4 −𝐹3 ) + ⋯ + (𝐹𝑛+2 −𝐹𝑛+1 ) = 𝐹𝑛+2 − 1 . 𝑖=1

𝑖=1

6. Infinite Series and Fibonacci Numbers ∞

(6.1)

∑ 𝑖=2 ∞

(6.2)

∑ 𝑖=0

(−1)𝑖 =𝛼−1 𝐹𝑖 𝐹𝑖−1 1 7 − √5 = 𝐹2𝑖 2

Proof. The second is due to Millin (1974), a high school student at that time (see also Good (1974)). We prove the first. In fact, we have by (5.3) and Proposition 4.1 as 𝑛 → ∞ 𝑛

𝑛

𝑛

𝑖=2

𝑖=2

𝑖=2

(−1)𝑖 𝐹𝑖−1 𝐹𝑖+1 − 𝐹𝑖 2 𝐹𝑖+1 𝐹𝑖 𝐹𝑛+1 𝑆𝑛 = ∑ =∑ = ∑( − )= − 1 → 𝑎 − 1. 𝐹𝑖 𝐹𝑖−1 𝐹𝑖 𝐹𝑖−1 𝐹𝑖 𝐹𝑖−1 𝐹𝑛

7. Probability and Fibonacci Numbers – The Geometric Distribution of Order k Problem 7.1. Toss a fair coin until a head (H) appears two consecutive times. Denote by E n the event that this will happen at the n-th tossing. What is the probability of En, say P(En)? What if the coin is not fair? What if the unfair coin is tossed until H appear k consecutive times? Solution. For n = 2, E2 = {HH}. Therefore 1 1 1 𝐹1 𝑃(𝐸2 ) = 𝑃({𝐻𝐻}) = ( ) ( ) = 2 = 2 . 2 2 2 2 For n = 3, E3 = {THH}. Therefore 1 1 1 1 𝐹2 𝑃(𝐸3 ) = 𝑃({𝑇𝐻𝐻}) = ( ) ( ) ( ) = 3 = 3 2 2 2 2 2 For n = 4, E4 = {HTHH, TTHH }. Therefore 1 1 𝐹3 𝑃(𝐸3 ) = 𝑃({𝐻𝑇𝐻𝐻}, {TTHH}) = 4 + 4 = 4 2 2 2 In general, it can be proven that

5

𝐹𝑛−1 , 2𝑛 In fact, much more can be shown, the following. 𝑃(𝐸𝑛 ) ==

𝑛≥1

Theorem 7.1 [Philippou and Muwafi (1982)]. For any positive integer k, denote by Tk the number of independent trials with success probability p (0 < p < 1) until the occurrence of the kth consecutive success, and set q = 1 – p. Philippou and Muwafi (1982) found that, for n = k, k + 1, … , 𝑛1 + ⋯ + 𝑛𝑘 𝑞 𝑛1 +⋯+𝑛𝑘 𝑛 𝑓𝑘 (𝑛) = 𝑃(𝑇𝑘 = 𝑛) = 𝑝 ∑ ( 𝑛 , … , 𝑛 ) ( ) 𝑝 1 𝑘 and 0 otherwise, where the summation is taken over all k-tuples of non-negative integers n1, n2, …, nk such that n1 +2n2 + …+ knk = n – k. Proof. It is based on the observation that a typical element of the event (𝑇𝑘 = 𝑛) is an arrangement 𝐴 = 𝑥1 𝑥2 … 𝑥𝑛1 +𝑛2 +⋯+𝑛𝑘 𝑆𝑆 … 𝑆 (𝑘 𝑆 ′ 𝑠), such that n1 of the x’s are E1 = F, n2 of the x’s are E2 = SF, …, nk of the x’s are Ek = SS…SF (k-1 S’s), and n1 + 2n2 + …+ knk = n – k. Fix n1, …, nk. Then the number of the A’s is 𝑛 + ⋯+ 𝑛 ( 1𝑛 , … , 𝑛 𝑘 ) 1 𝑘 and each one has probability 𝑃(𝐴) = [𝑃(𝐸1 )]𝑛1 [𝑃(𝐸2 )]𝑛2 … [𝑃(𝐸𝑘 )]𝑛𝑘 𝑃(𝑆𝑆 … 𝑆) (𝑘 𝑆 ′ 𝑠) 𝑞 𝑛1 +⋯+𝑛𝑘 = 𝑞 𝑛1 (𝑝𝑞)𝑛2 … (𝑝𝑘−1 𝑞)𝑛𝑘 𝑝𝑘 = 𝑝𝑛 ( ) . 𝑝 Therefore, 𝑞 𝑛1 +⋯+𝑛𝑘 𝑛 +⋯+𝑛 𝑃(𝑎𝑙𝑙 𝐴′ 𝑠: 𝑛𝑖 ≥ 0 𝑎𝑛𝑑 𝑓𝑖𝑥𝑒𝑑, 0 ≤ 𝑖 ≤ 𝑛) = ( 1𝑛 , … , 𝑛 𝑘 ) 𝑝𝑛 ( ) . 𝑝 1 𝑘 But the non-negative integers 𝑛1 , 𝑛2 , … , 𝑛𝑘 may vary subject to the condition n1 + 2n2 + …+ knk = n – k, and this completes the proof of the theorem. To Theorem 7.1, we have the following corollary. Corollary 7.1 [Philippou and Muwafi (1982)]. Let 𝑇𝑘 be as in Theorem 7.1 and assume that 𝑝 = 1/2. Then (𝑘) 𝐹 𝑃(𝑇𝑘 = 𝑛) = 𝑛−𝑘+1 , 𝑛 ≥ 𝑘, 2𝑛 and 6

𝑃(𝑇2 = 𝑛) =

𝐹𝑛−1 , 2𝑛

𝑛 ≥ 2,

(𝑘)

where 𝐹𝑛 is the nth Fibonacci number, and 𝐹𝑛 is the nth Fibonacci number of order k. (2) Proof. It follows from Theorem 7.1 for 𝑝 = 1/2, since 𝐹𝑛 = 𝐹𝑛 , and 𝑛 + ⋯+ 𝑛 (𝑘) F𝑛+1 = ∑ ( 1𝑛 , … , 𝑛 𝑘 ) , 1 𝑘

𝑛 ≥ 0,

where the summation is taken over all k-tuples of non-negative integers n1, n2, …, nk such that n1 +2n2 + …+ knk = n [Philippou and Muwafi (1982), Philippou (1983)]. The above (𝑘)

formula for F𝑛+1 generalizes (5.4). Is 𝑓𝑘 (𝑛) a proper probability mass function? The answer is yes, by means of the transformation 𝑘

𝑛𝑖 = 𝑚𝑖 (1 ≤ 𝑖 ≤ 𝑘),

𝑛 =𝑚+∑

(𝑖 − 1)𝑚𝑖

𝑖=1

and the multinomial theorem. Theorem 7.2 [Philippou, Georghiou and Philippou (1983)]. Let 𝑓𝑘 (𝑛) be as in Theorem 7.1. Then ∞

∞

∑ 𝑛=𝑘

𝑓𝑘 (𝑛) = ∑

𝑃(𝑇𝑘 = 𝑛) = 1. 𝑛=𝑘

The probability generating function (pgf) of 𝑇𝑘 , say 𝑔𝑘 (𝑠), and hence its mean 𝜇𝑘 and variance 𝜎𝑘2 are ∞

𝑔𝑘 (𝑠) = ∑ 𝜇𝑘 =

𝑛=𝑘

𝑠 𝑛 𝑓𝑘 (𝑛) =

1 − 𝑝𝑘 , 𝑞𝑝𝑘

𝜎𝑘2 =

(1 − 𝑝𝑠)𝑝𝑘 𝑠 𝑘 , 1 − 𝑠 + 𝑞𝑝𝑘 𝑠 𝑘+1

|𝑠| ≤ 1,

1 − (2𝑘 + 1)𝑞𝑝𝑘 − 𝑝2𝑘+1 . 𝑞 2 𝑝2𝑘

They named the distribution of 𝑇𝑘 𝑡ℎ𝑒 geometric distribution of order k with parameter p, since for k = 1 it reduces to the geometric distribution with pmf 𝑓1 (𝑛) = 𝑞 𝑛−1 𝑝,

𝑛 ≥ 1.

A different derivation of gk(s) was first given by Feller (1968). Alternative simpler formulas for calculating 𝑓𝑘 (𝑛) have been found. The following recurrence, for example, is very efficient.

7

Theorem 7.3 [Philippou and Makri (1985)]. Let 𝑓𝑘 (𝑛) be the probability mass function of 𝑇𝑘 . Then 𝑓𝑘 (𝑘) = 𝑝𝑘 ,

𝑓𝑘 (𝑛) = 𝑞𝑝𝑘 ,

𝑘 + 1 ≤ 𝑛 ≤ 2𝑘,

𝑓𝑘 (𝑛) = 𝑓𝑘 (𝑛 − 1) − 𝑞𝑝𝑘 𝑓𝑘 (𝑛 − 𝑘 − 1),

𝑛 ≥ 2𝑘 + 1.

8. Probability, Fibonacci, and Gambling There are two well publicized cases of individuals who beat the casinos and won themselves quite a lot of money. There are also many who won once or a few times. Millions, however, lose on the average every day. That is why casinos exist and thrive. The following three systems of gambling enjoy some popularity among gamblers. (a) Double your bet system. In any game of chance you start by betting (say on red in American roulette) a certain amount of money, say A dollars. If you win, you stop. If you lose, you double your bet by wagering 2A dollars. If you win, you stop. If you lose, you bet 4A dollars, and so on. It appears that the system is unbeatable earning you A dollars when you win and stop, but it is not. Pretty soon you may end up not being able to double your bet for lack of money, or you may reach the upper limit posed by the casino. Either way you are a loser. (b) The Fibonacci system. It resembles the double your bet one, but it is less aggressive. In any game of chance you start by betting (say on red in the American roulette) a certain amount of money, say A dollars. If you win, you stop. If you lose, you bet A dollars again. If you gain, you are at the beginning. If you lose, you bet 2A dollars. If you win, you are at the beginning. If you lose, you bet 3A dollars, and so on, moving one step to the right through the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …, when losing and two steps to the left when winning. As in the double your bet system, you may end up pretty soon without money. Moreover, one win is not sufficient to win you money, unless this happens at the first time. (c) The d’Alembert system. According to this system, in any game of chance (say the roulette) you increase or decrease the amount of your bet in a round, depending on whether you lost or won the previous round. The “logic” is that if you lose one round, you are more likely to win on the next, and you should increase the amount of your bet. But this is completely erroneous. In a game of chance, say the roulette, the outcome of any spin is independent of the outcomes of previous spins. The odds at the roulette are always the same. 9. The Odds in Some Odd-Even Games [Schuster and Philippou (1975)] Problem 9.1. In tossing an unbiased six-sided die until a six appears, is the best bet “odd” = 1, 3, 5, … or “even” = 2, 4, 6, …? What if the die is biased?

8

The answer is given in the following Theorem 9.1 (Bernoulli Odd-Even Game). In an odd-even game of counting independent Bernoulli trials with constant positive success probability p (q = 1 – p) until the occurrence of success, the best bet is “odd”, since 𝑃("even") =

𝑞 , 𝑞+1

𝑃("even") − 𝑃("odd") =

𝑞−1 . 𝑞+1

Problem 9.2. In counting the number of phone calls arriving at the switch board of Mohandas College of Engineering and Technology (or Kerala University or All Saints College) on December 16, 2013, is the best bet “odd” = 1, 3, 5, … or “even” = 0, 2, 4, ... ? The answer is given in Theorem 9.2 (Poisson Odd-Even Game). In an odd-even game of counting the number of occurrences of an event in positive time t following the Poisson distribution with mean rate λt, the best bet is “even”, since 𝑃("𝑒𝑣𝑒𝑛") = (

1 + 𝑒 −2𝜆𝑡 ), 2

𝑃("𝑒𝑣𝑒𝑛") − 𝑃("𝑜𝑑𝑑") = 𝑒 −2𝜆𝑡 .

Lest one think that life is always so simple, consider the following generalization of Problem 9.1, which is a sucker bet. Problem 9.3. In tossing an unbiased six-sided die until a “six” appears r times, is the best bet “odd” = 1, 3, 5, … or “even”= 2, 4, 6, …? What if the die is biased? The answer is based on Theorem 9.3 (Negative Binomial Odd-Even Game). In an odd-even game of counting the number of independent Bernoulli trials with positive success probability p until the occurrence of the r-th success, the best bet is “odd” or “even” providing r is odd or even, respectively, since 𝑞−1 𝑟 1 + (𝑞 + 1) 𝑃("𝑒𝑣𝑒𝑛") = ( ), 2

𝑞−1 𝑟 𝑃("𝑒𝑣𝑒𝑛") − 𝑃("𝑜𝑑𝑑") = ( ) . 𝑞+1

Additional sucker bets can be based on Theorem 9.4 (Binomial Odd-Even Game). In an odd-even game of counting the number of successes in n independent Bernoulli trials with constant positive success probability p, the best bet is “even” except when n is odd and 𝑝 ≥ 1/2, since

9

1 + (𝑞 − 𝑝)𝑛 𝑃("𝑒𝑣𝑒𝑛") = ( ), 2

𝑃("𝑒𝑣𝑒𝑛") − 𝑃("𝑜𝑑𝑑") = (𝑞 − 𝑝)𝑛 .

REFERENCES 1. Eves, H.W. (1990). An Introduction to the History of Mathematics, Saunders College Publications, 6-th edn., Philadelphia. 2. Feller, W. (1968). An introduction to probability theory and its applications, Vol I, 3rd edn., Wiley, New York. 3. Good, I.J. (1974). A Reciprocal Series of Fibonacci Numbers. The Fibonacci Quarterly, 12, 346. 4. Millin, D.A. (1974). Problem H-237. The Fibonacci Quarterly, 12, 309. 5. Philippou, A.N. (1983). A note on the Fibonacci sequence of order k and multinomial coefficients. Fibonacci Quarterly, 21, 82-86. 6. Philippou, A.N., Bergum, G.E. and Horadam, A.F., eds. (!986). Fibonacci numbers and their applications. Mathematics and Its Applications, 28. D. Reidel, Dordrecht. 7. Philippou, A.N., Bergum, G.E. and Horadam, A.F., eds. (1988). Applications of Fibonacci numbers. Kluwer Academic Publishers, Dordrecht. 8. Philippou, A.N., Georghiou, C. and Philippou, G.N. (1983). A generalized geometric distribution and some of its properties. Statistics and Probability Letters, 1, 171175. 9. Philippou, A.N. and Makri, F.S. (1985). Longest success runs and Fibonacci-type polynomials. The Fibonacci Quarterly, 23, 338-346. 10. Philippou, A.N. and Makri. F.S. (1986). Successes, runs and longest runs. Statistics and Probability Letters, 4, 211-215. 11. Philippou, A.N. and Muwafi, A.A. (1982). Waiting for the kth consecutive success and the Fibonacci sequence of order k. The Fibonacci Quarterly, 20, 28-32. 12. Schuster, E.F. and Philippou, A.N. (1975). The odds in some odd-even games. American Mathematical Monthly, 82, 646-648.

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