323

BALANCED INCOMPLETE BLOCK DESIGNS USED

IN THE METHOD OF SIMULATION

BY A. W. JOSEPH, M.A., B.Sc., F.I.A.

The method of simulation THE method of simulation, alternatively known as the Monte Carlo method, to which Sidney Benjamin drew our attention(1) is a lazy way of solving problems that could be solved accurately but for the large number of different logical situations that have to be considered. The method traces out what actually happens if the conditions of a problem are observed and if certain unknown events are presumed to occur at random. The solution is taken to be the average of the results of all the experiments. Inevitably, however, this method gives rise to random error. Now certain actuarial problems depend on making assumptions on such things as the mortality to be expected to be experienced by a life or set of lives. It is of little consequence if random error is imposed on top of such assumptions. On the other hand the shape of the probability distributions that the ran- dom variables in the problem are to follow may be known exactly. For such a problem the introduction of random error is a blemish that should be made as small as possible. In this note one such problem will be considered. The argument will introduce the subject of balanced incomplete block designs, which, as far as the writer knows, have only once been touched on by an actuary and that over one hundred years ago. The problem is the chance that a Jackpot may be opened at the game of Poker, discussed by Redish and ROSS.(2)

The chance of a Jackpot’s being opened at poker In poker all possible five-card hands are ranked in order of precedence.

The Jackpot is considered opened if at least one player is dealt a pair of knaves or any higher ranking hand. Redish and Ross used the method of simulation to determine the chance that a Jackpot could be opened for two, three, four or five players. Each set of five hands of five cards was obtained by a separate pseudo-random process. The process was carried out 150,000 times.

Now the population out of which these 150,000 samples were taken, viz. the totality of all possible different sets of five hands of five cards each, has characteristics only approximately fulfilled (although very nearly so) by the 150,000 samples themselves. For example in the full population each card will appear the same number of times as any other, each different pair of cards will appear the same number of times as any other pair and so on. It seemed possible that a kind of stratified sampling where exactly the same number of each card appeared in the set of samples, might, by eliminating one kind of random error, give better results than completely independent samples, and in a note (3) it was shown that in fact this

324 Balanced Incomplete Block Designs used in the

hypothesis seemed to be true. It was conjectured that if the sampling secured that each different pair of cards appeared an equal number of times the results might be even better and the present note investigates this conjecture. It was first necessary to consider balanced incomplete block designs or b. i. b. d.s for short.

Balanced incomplete block designs A set of samples where each pair of objects appears an equal number of

times is called balanced. The objects themselves (v in number) are called the varieties, and each sample consisting of the same number (say k) of varieties is called a block, the number of blocks being denoted by b. No variety appears more than once in a block and no two blocks are identical. All varieties appear in the same number (say r) of blocks. From these definitions it follows that

bk=vr (1)

The number of times that each variety appears with any other (i.e. the number of blocks in which each pair of varieties appears) is denoted by . It is easy to see that

r (k–1) = 1 (v-l) (2)

The set of all possible blocks (C in number) is called complete and is, of course, balanced. Any lesser set of blocks that is balanced is called a balanced incomplete block design. S. Vajda has recently written a useful monograph(4) on the subject containing an excellent bibliography.

One of the earliest papers to appear anywhere on balanced incomplete block designs was written by W. S. B. Woolhouse over one hundred years ago.(5) A dozen years earlier Kirkman had propounded and solved the problem now well known under the title ‘Kirkman’s fifteen school-girl problem’ which, as its title suggests, concerns fifteen school-girls, who were marched each day for a week in five sets of three girls. The problem was to show how this could be done so that each girl was accompanied exactly once by every other schoolgirl. Woolhouse analysed this problem in considerable detail. The full set of 35 blocks of three girls is clearly a b.i.b.d. where v = 15, k = 3, b = 35, r = 7, = 1.

Furthermore a b.i.b.d. in which its blocks can be grouped into sets so that the blocks of each set contain between them all the varieties once is said to be resolvable. Since this is what happens each day, Kirkman’s schoolgirl design is resolvable. The Kirkman design is a particular case of what is now called a Steiner triple system.

The greater part of the quite considerable and elegant work on b.i.b.d.s has been concerned with finding designs where b is small. This is because b.i.b.d.s only came into prominence about thirty years ago when they were studied so as to be applied to statistical experiments in agricultural field trials, and clearly to be of practical use the number of blocks had to be

Method of Simulation 325

small. For the Jackpot problem a b.i.b.d. was required where v = 52 and k = 5 and this could not be found in Fisher and Yates Statistical Tables,(6) the natural reference book to turn to. But for computer work larger values

of b can be tolerated. Could a general solution be found for b =

i.e. one block for each different pair of varieties? It was easy to see, how- ever. that this was not possible if v was even and k odd because when

b = equation(1) shows that which is not integral.

Attempts were made, therefore, to find a general solution for b = v(v–1) when v was even and k odd and for b = ½v(v–1) for other values of v and k. Only partial success was achieved. The solutions obtained are given in Appendix I and are listed in the following table. p denotes an odd prime.

Values of k for which Balanced Incomplete Block Designs are given in Appendix I

v= P 2P 4P P+1 b = ½v(v–1) All values 4 4 b = v(v–1) 3,5 3,4,5 3

The failure to find a solution for v = 4p, k = 4, b = ½v(v–1) was disappointing since Hanani (7) has shown that such a solution exists. It may be possible to extricate a construction out of Hanani’s paper but unfortunately theorems of existence are very much intermingled with constructions in that paper. The case v = 4p, k = 5, b = v(v–1) was solved by using a computer to give a display of a large number of possible patterns from which it was possible to recognize a few patterns that led to success.

For the Jackpot problem it would have been desirable to obtain a b.i.b.d. for v = 52, k = 25 which could be split up into five k = 5, b.i.b.d.s. In default of such a solution five b.i.b.d.s were found for v = 52, k = 5, b = 2,652 such that each of the 13,260 blocks were different, and which could be strung together into 2,652 blocks of 25 objects, each of the 25 objects being different. The blocks of 25 objects did not themselves form a b.i.b.d. This solution is given in Appendix II.

The 2,652 blocks are really formed from 51 basic designs each one giving rise to 52 designs by cyclic permutations. This has a storage advantage for computer work, and in fact on the small IBM 1401 4K machine to which I have access it was necessary to store the 51 blocks on punched cards. Pseudo-random numbers were calculated by the formula un+1 = 371,283 un (mod 524,287).* The first such number was divided by 52 and the remainder, a pseudo-random number in the range 0–51 was used to pick out a number from a starting list of the numbers l-52 thus reducing the list to 51 numbers.

* 524,287 = Mersenne prime 219 – 1; 371,283 = 175 (mod 524,287); 17 is a primitive root of 524,287.

326 Balanced Incomplete Block Designs used in the

The second pseudo-random number was divided by 51, and the remainder was used to pick out a second number from the list, and so on. In this way a pseudo-random permutation of the 52 numbers was obtained. This permutation was applied to the blocks of Appendix II to obtain an iso- morphic representation of Appendix II having, of course, the same properties as Appendix II. The last stage was to allocate to each of the elements of the blocks a card from an ordinary pack of playing cards and then to assess the rank of each of the five sets of poker hands so produced in order to determine whether it would be possible to open a Jackpot for one player (the first hand), two players (first two hands), three players (first three hands), four players (first four hands), five players (the five hands). The number of such cases was tabulated. The process was repeated 29 times more by calculating new pseudo-random permutations of 52 numbers.

The results of the 79,560 experiments are set out in the following table to which is added the results of the experiments by Redish and Ross (2) and by myself (3).

Probability of opening a Jackpot

Number of players 1 2 3 4 5

R. and R. 50,000 deals (1) ·3728 ·5019 ·6108 ·6929 R. and R. 50,000 deals (2) ·3716 ·5046 ·6098 ·6958 R. and R. 50,000 deals (3) ·3763 ·5009 ·6146 ·6940 R. and R. 150,000 deals -3736 -5025 ·6118 ·6943

J (1968) 26,000 deals (1) ·2060 ·3707 ·5012 ·6078 ·6916 J (1968) 26,000 deals (2) ·2060 ·3722 ·5036 ·6085 ·6917 J (1968) 26,000 deals (3) ·2022 ·3696 ·5056 ·6100 ·6951 J (1968) 78,000 deals ·2047 ·3708 ·5034 ·6088 ·6928

J (now) 26,520 deals (1) ·2069 ·3712 ·5042 ·6125 ·6966 J (now) 26,520 deals (2) ·2056 ·3710 ·5049 ·6090 ·6925 J (now) 26,520 deals (3) ·2040 ·3687 ·4994 ·6026 ·6877 J (now) 79,560 deals ·2055 ·3703 ·5029 ·6080 ·6923

Exact (where known) ·2063 ·3718

The results are inconclusive. As mentioned in my earlier note there is indication of less dispersion from the mean and better accuracy to known exact results when each card appears an equal number of times in the blocks. The modification to make each pair of cards appear an equal number of times gives no evidence of extra accuracy.

My thanks are due to the Wesleyan and General Assurance Society for the use of their computer.

Method of Simulation 327

REFERENCES

(1) BENJAMIN, S. Putting computers on to actuarial work. J.I.A. 1966, 92, 134. (2) REDISH, K. A. AND Ross, A. S. C. The chance of the Jackpot being opened at poker.

J. R.S.S. Series A (General), 1967, 130,423. (3) JOSEPH, A. W. A Criticism of the Monte Carlo method as applied to mathematical

computations. J.R.S.S. Series A (General), 1968, 131,226. (4) VAJDA, S. The Mathematics of Experimental Design: Incomplete Block Designs and

Latin Squares. Charles Griffin & Co. (5) WOOLHOUSE, W. S. B. On triadic combinations of fifteen symbols. J.I.A. 1862, 10,

275. (6) FISHER, R. A. AND YATES, F. Statistical Tables for Biological, Agricultural and Medical

Research, Oliver & Boyd, 1963. (7) HANANI, HAIM. The Existence and Construction of Balanced Incomplete Block

Designs. Ann. Math. Statist. 1961,32, 361.

328 Balanced Incomplete Block Designs used in the

APPENDIX I

BALANCED INCOMPLETE BLOCK DESIGNS

General It is assumed that the varieties are represented by the numbers 0, 1, . . .

(v-l). A new block may be formed from a given block by adding 1 to each

member and replacing the member v, if it appears, by 0. Up to v different blocks are obtained by repeating the process. This is called cycling.

A block of k members has k(k– 1) mutual differences, half being positive and half negative. If v is added to the negative differences the k(k– 1) differences are brought (mod v) within the range 1,(v– 1). Since they appear in pairs whose sum is v we can restrict ourselves to the ½k(k– 1) differences in the range 1, ½ v. It is clear that all cycled blocks have the same differences.

Consider odd v. If we can find a set of blocks such that (1) Cycling each block produces v different blocks and (2) The differences of the set of blocks comprise all the integers 1 to ½ (v– 1), repeated an equal number of times ,then the set of blocks and their cycles form a b.i.b.d. with E.g. for v = 15 the blocks (0,1,2); (0,1,6); (0,2,5); (0,2,7); (0,3,7); (0,3,9); (0,4,11) produce differences 1,1,2; 1,5,6; 2,3,5; 2,5,7; 3,4,7; 3,6,6; 4,4,7, and, therefore, when cycled produce a b.i.b.d. with v = 15, k = 3, b = 105, r = 21, = 3.

Consider even v. If we can find a set of blocks such that (1) Cycling each block produces v different blocks and (2) The differences of the set of blocks comprise all the integers 1 to ½ (v – 2), repeated an equal number of times 2 ( and the integer ½ v, repeated times then the set of blocks and their cycles form a b.i.b.d. with = 2 . E.g. for v = 14 the blocks

(0,1,2,7); (0,1,3,10); (0,1,4,6); (0,1,5,6); (0,1,7,11); (0,1,7,12); (0,1,8,11); (0,1,8,12); (0,1,9,11); (0,1,10,12); (0,2,4,8); (0,2,5,11); (0,2,6,11) produce the required differences and, therefore, when cycled produce a b.i.b.d. with v = 14, k = 4, b = 182, r = 52, = 12.

There are, of course, other ways of producing b.i.b.d.s. We now proceed to particular cases.

v = p (an odd prime), k any, b = ½ v(v– 1) If A is a block, denote by CA the set of p blocks obtained by cycling A

and denote by 2A, 3A etc. the blocks obtained by multiplying (mod p) the members of A by 2, 3 etc. If CA, C2A, C3A, --- C(p–1)A are all distinct, A is called perfect. If CrA = C(p–r)A and the resulting ½(p–1) sets of cycled blocks are all distinct, A is called semi-perfect. Otherwise A is imperfect. Examples:p = 13, k = 4 Perfect (0,1,2,4), (0,1,2,5), (0,1,2,6)

Method of Simulation

Semi-perfect (0,1,2,3), (0,1,3,4) Imperfect (0,1,3,9), (0,1,5,6)

329

The full set of cycled blocks from a semiperfect A or one-half the set of cycled blocks from a perfect A gives a b.i.b.d. for v = p, k = any, b = ½p(p–1), r = ½ k(p–1), = ½ k(k–1) E.g. p = 13, k = 4 If A = Semi-perfect (0,1,2,3) Blocks (0,1,2,3); (0,2,4,6); (0,3,6,9); (0,4,8,12); (0,2,5,10); (0,5,6,12) all cycled give a b.i.b.d. for p = 13, k = 4, b = 78, r = 24, = 6 If A = Perfect (0,1,2,4) Blocks (0,1,2,4); (0,2,4,8); (0,3,6,12); (0,3,4,8); (0,5,7,10); (0,6,11,12) all cycled also give a b.i.b.d. for p = 13, k = 4, b = 78, r = 24, = 6

The existence of a b.i.b.d. for v = p, any k and b = ½ v(v–1) is shown by verifying that the block (0,1,2, --- k–1) is semi-perfect.

v = 2p, k = 3, b = v(v–1) The term ‘bicycling’ will be used to describe the same process as cycling

except that 2 is added to each member of a block instead of 1. For p > 7 the solution is

(0,1,p) (0,2,p+2) (0,3,P) (0,4,P+4) (0,5,P) (0,6,p+6)

plus 3A, 5A, --- (p–2)A; (p+2)A --- (2p–1)A 3B, 5B, --- (p–2)B; (p+2)B --- (2p–1)B 3C, 5C, --- (p–2)C; (p+2)C --- (2p–1)C 3D, 5D, --- (p–2)D; (p+2)D --- (2p–1)D

where A = (0,2,6) B = (0,1,p+6) C = (0,3,p+4) D = (0,5,p+2)

the whole set, 4(p – 2) + 6 = 4p – 2 blocks in number being bicycled. The proof lies in noting that the blocks A, 3A . . . (p – 2)A; (p + 2)A . . .

(2p–1)A etc. give the right differences apart from differences p, and that the blocks (0,1,p) etc. provide the missing differences p together with the same differences as blocks A, B, C, D. A little care has to be taken in bicycling instead of cycling.

For completeness a solution for v = 14 is (0,1,3) (0,1,5) (0,1,7) (0,1,9)

(0,1,11) (0,3,13) (0,5,13) (0,7,13) (0,9,13) (0,11,13) (0,9,11) (0,7,9) (0,5,7) (0,3,5) (0,3,7) (0,3,9) (0,5,11) (0,7,11) (0,1,2) (0,3,6) (0,4,9) (0,2,4) (0,2,6) (0,2,8) (0,2,10) (0,4,8), the whole set being bicycled.

It is not worth while giving a solution for v = 10, b = 90 since a simpler b.i.b.d. with b = 30 is given in Fisher and Yates (1), which can be expressed

330 Balanced Incomplete Block Designs used in the

as (0,3,6) (0,1,2) (0,1,5) (1,3,5) (0,3,4) (0,2,7), the whole set being bicycled.

v = 2P, k = 4, b = ½ v(v–1) The solution is

(0,2,3,p) (0,2,p,p+1)

plus 3A, 5A, --- (p–2)A 3B, 5B, --- (p–2)B

where A = (0,1,2,p–1)

B = (0,2,3,p+1) the whole set being cycled

plus (0,1,p,p+1) the latter being bicycled.

v = 2p, k = 5, b = v(v–1) The solution is

(0,3,9,12,p) (0,9,12,p,p+3) (0,6,9,p+3,p+6) (0,3,6,p+3,p+9) (0,1,2,3,p) (0,3,4,P,P+1) (0,1,3,p+1,p+2) (0,2,3,p+1,p+3) (0,2,6,p,p+6)

plus 5A, 7A --- (p–2)A 5B, 7B --- (p–2)B 5C, 7C --- (p–2)C 5D, 7D --- (p–2)D

where A = (0,1,4,p+2,p+3) B = (0,1,2,3,p+4) C = (0,1,2,4,p+3) D = (0,1,3,p+2,p+4)

the whole set being cycled.

v = 4p, k = 3, b = v(v–1) The solution is

(0,–3,p) if p = 4m+1 or (0,3,p) if p = 4m–1 (0, – 2,p) if p = 4m+1 or (0,2,p) if p = 4m–1

(0,1,p) (0,–1,p) (0,1,2p) (0,2,2p)

(0,p,2p) plus 3A, 5A --- (p–2)A

Method of Simulation

B, 3B --- (p–2)B 3C, 5C --- (p–2)C; (p+2)C --- (2p–1)C 3D, 5D --- (p–2)D;(p+2)D --- (2p–1)D 3E, 5E --- (p–2)E; (p+2)E --- (2p–1)E

where A = (0,1,2p–1) B = (0,–1,2p+1) C = (0,2,p+1) D = (0,1,3) E = (0,1,p,+3) if p = 4m+1 or (0,1,p–2) if p = 4m–1

the whole set being cycled.

v = 4p, k = 4, b = v(v–1) The solution is

(0,1,4,p+2) if p = 4m+1 or (0,3,4,p+2) if p = 4m–1 (0,–1,p,2p) if p = 4m+1 or (0,1,p,2p) if p = 4m–1

(0,–2,p,2p) (0,2,p,2p) (0,1,p,3p) (0,3,p,3p)

(0,4,p,3p)

plus 3A, 5A --- (p–2)A; (p+2)A --- (2p–1)A 3B, 5B ---(p–2)B; (p+2)B --- (2p–1)B 3C, 5C --- (p–2)C; (p+2)C --- (2p–1)C 3D, 5D --- (p–2)D; (p+2)D --- (2p–1)D

where if p = 4m+1 or if p = 4m–1 A = (0,1,4,2p+2) A = (0,1,4,2p+2) B = (0,1,p+2,p+4) B = (0,1,p–1,p–3) C = (0,3,p–1,p+1) C = (0,3,p+2,p+4) D = (0,4,p+1,p+2) D = (0,4,p+2,p+3)

the whole set being cycled.

v = 4p, k = 5, b = v(v–1) The solution is

(0,3,p,p+3,2p+6) (0,6,p,p+6,2p+9) (0,12,p,p+9,2p+12) (0,9,p+9,2p2p+3) (0,4,p+4,p+2,2p) (0,1,p,2p+1,2p+2) (0,6,p,2p,3p) (0,2,p–1,p+2,3p+2) (0,1,3,p,2p) (0,1,2p+3,p,3p) (0,3,2p+1,p,3p)

plus 5A, 7A --- (p–2)A; (p+2) A --- (2p–1)A

K

331

332 Balanced Incomplete Block Designs used in the

5B, 7B --- (p–2)B; (p+2) B --- (2p–1)B 5C, 7C --- (p–2)C; (p+2) C --- (2p–1)C 5D, 7D --- (p–2)D; (p+2) D --- (2p–1)D

where A = (0,1,2,p – 2,2p – 1) B = (0,4,p + 3,2p + 1,2p + 2) C = (0,2,3,p + 1,p + 4) D = (0,2,p–1,2p+1,3p+3)

the whole set being cycled.

v = p + 1, k = 3, b = v (v–1) The solution is

2A, 3A, . . . . . . . . (p–1)A cycled w.r.t. p plus (0,α) cycled with p added

(0,β) cycled with p added (0,β – α) cycled with p added

where A ( = (0,α,β) ) is perfect e.g. A = (0, 1, 3)

v = p+1, k = 4, b = ½ v (v–l) The solution is

2A, 3A, . . . . . . . . ½ (p–1)A cycled w.r.t. p plus (0,α,β,) cycled with p added

(0,α,γ) cycled with p added where A( = (0,α,β,γ) ) is semi-perfect e.g. A = (0,1,2,3) or (0,1,3,4)

Method of Simulation 333

334 Block Designs used in the Method of Simulation