The Annals of Human Genetics has an archive of material originally published in print format by the Annals of Eugenics ( ).

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1 The Annals of Human Genetics has an archive of material originally published in print format by the Annals of Eugenics ( ). This material is available in specialised libraries and archives. We believe there is a clear academic interest in making this historical material more widely available to a scholarly audience online. These articles have been made available online, by the Annals of Human Genetics, UCL and Blackwell Publishing Ltd strictly for historical and academic reasons. The work of eugenicists was often pervaded by prejudice against racial, ethnic and disabled groups. Publication of this material online is for scholarly research purposes is not an endorsement or promotion of the views expressed in any of these articles or eugenics in general. All articles are published in full, except where necessary to protect individual privacy. We welcome your comments about this archive and its online publication.

2 SOLUTION TO A GEOMETRICAL PROBLEM IN PROBABILITY BY W. L. STEVENS The Galton Laboratory, Rothamsted Experimental Station, Harpenden, Herts. 1 STATEMENT OF PROBLEM THE following problem in geometrical probability is one which has been put to the author a number of times, most recently by H. Jeffreys. So far as we know, no discussion has yet appeared in the literature, and the method of solution gives it a special interest. On the circumference of a circle of unit length, n arcs each of length x are marked off at random. What is the probability that every point of the circle is included in at least one of the arcs! 2 REPRESENTATION AND TERMINOLOGY The location of each arc is represented on the diagram by its first point, i.e. the point at the clockwise end. The first arc is labelled 1, the remaining arcs are labelled from 2 to n, in the order in which they occur anti-clockwise round the circle. There is said to be a gap after the rth arc, if for a distance greater than x there is no other arc. This implies that after the rth arc there is a portion of the circle uncovered by any arc. For example, in Diagram 1 there are gaps after 2 and 4. 3 PROBABILITY OF GAPS AFTER SPECIFIED ARCS Let us find the probability f (1) that there is a gap after the rth arc. We shall show that the restriction that there shall be a gap after r is equivalent to the restriction that no arc shall occur in the portion of the circumference of length x terminating at 1. For any configuration obeying the former restriction may be converted to one obeying the latter restriction by moving the arcs r + I, r + 2,..., n clockwise through a distance x. Conversely, the latter may be converted to the former by moving arcs r + 1, r + 2,..., n anti-clockwise through a distance x. The equivalence is made clearer by Diagram 2. The probability required is therefore the probability that the n- 1 arcs, excluding the first, shall fall in a portion of the circumference of length 1 - x, i.e. f(1) = (1-xy-,......(3.1) Notice that the possible configurations contributing to this probability include those which contain gaps elsewhere than in the specified place. In the same way, we find the simultaneous probabilityf(2) of gaps after r, and rz, where r2>rl. Here again we see that this restriction is equivalent to the restriction that no arc EUGENICS IX, IV 21

3 I X S 4 Diagram 1. Gaps after 2 and 4 when x = a. 4 Diagram 2. Gap after 3, when 2 = &. Arcs 4, 5, 6, 7 can be moved back through one right angle to guarantee that the last quarter, shown shaded, contains no arcs. Gap after 4 remains. Diagram 3. Gaps after 2 and 5. Hence 3, 4, 5 are moved through one right angle and 6, 7 through two right angles to clear last half of circle, shown shaded on the right.

4 W. L. STEVENS 317 shall occur in the portion of the circumference of length 2x terminating at 1. For a configuration of the former kind can be converted to one of the latter by moving arcs rz+ 1,... n clockwise through a distance 2x, and arcs r1+ 1,... rz clockwise through a distance x, or conversely, as is shown by Diagram 3. Hence the required probability is f(2) = (1-2xp (3.2) As before, configurations which have gaps in the specified places and elsewhere are included in the above probability. The generalization now follows. If there are gaps in i specified places, the distances between each pair of arcs enclosing a gap may be reduced by x, to produce a configuration with the portion of length ix terminating at 1 containing no arcs; and conversely. Hence where IC is the greatest integer less than l/x. f(i) = (1 - ix) -l i < k, =o i>k,... (3.3) 4 A GENERAL THEOREM 4.1. To proceed with the solution, we shall develop a probability theorem, which may be used in this and similar situations. Let us suppose that n events can each happen in two ways, denoted by H and T. The events are not independent, but the probability that any h specified events are H, whatever the remainder, is known to be f(h) h=0,1,2...,n. We wish to develop formulae giving the probability f(h,t) h,t=o,1,2... n, h+t<n, that any h specified events are H, any other t specified events are T, and the remaining events either H or T. In particular we shall findf(h, n-h), the probability that h specified events, and only these, are H First evaluate the probabilityf(h, 1) of h specified H s and one specified T. The admissible configurations are all those with the h specified H s, excluding those which have an H at the event which is to be a T, i.e. excluding configurations with h+ 1 specified H s. Hence f(h 1) =f@) -,f(h+ 1) = -Af(h)....(4.21) We use, here and later, the notation of the Calculus of Finite Differences, but our proof is not dependent on the properties of these operators. If we write down in vertical column the probabilities f(h), where h = n, n - 1,... 0, of h specified H s, whatever the other events, then the first differences are the probabilities f (h, 1) of h specified H s and one specified T, whatever the others. Continuing this argument, we find the probability f(h, 2). The admissible configurations are those with the h specified H s and one of the two specified T s, excluding those in which 21-2

5 318 GEOMETRICAL PROBLEM IN PROBABILITY the other event destined to be a T is an H instead; i.e. those with h specified H s and one specified T, excluding those with h+ 1 specified H s and one specified T. Hence f(h, 2) =f(k 1)-f(h+ 1, 1) = -Af(h, 1) = Ll Zf (h) =f(h)-2f(h+ l)-f(h+2).... (4.22) Therefore the second differences are the required probabilities f (h, 2) of h specified H s and two specified T s The generalization is now obvious. We May write f(h3 t) = (-d)tf(h)... (4.31) = (1 -E)if(h) t(t - 1) =f(h)-tf(h+l)+ 2 f (h + 2) -... ( - ) f(h + t).... (4.32) Assuming the above formula to be true for all h and a particular t, we may find the probability f(h, t + l), of h specified H s, (t + 1) specified T s, and the remaining events either H or T. Select any t out of the (t+ 1) events to be T. Then the admissible configurations are those which have the h specified H s, and these t specified T s, excluding those in which the other event, which should be a T, is an H instead; i.e. those with h specified H s and t specified T s, excluding those with h + 1 specified H s and t specified T s. Hence f(h, t+ 1) =f(h, t)-f(h+ 1, t) = -Llf(h, t) = (-Ll)l+lf(h)....( 4.33) Avoiding the use of operators, and substituting directly from equation (4-32), we would find t(t - 1) f(h, t + 1) =f(k) - tf(h + 1) + 2 f(h+2)... (-)lf(h+t) -f (h + 1) + tf (h + 2)... ( - ) t tf (h + t) ( - ) t+1 f (h, t + 1) (t+ 1) t =f(h)-(t+ l)f(h+l)+,--f(h+2)... (-)l+lf(a, tfl).... (4.34) Hence, with or without using operators, the general formula (4.33 or 4-34) is proved by induction In particular the probability that h specified events are H, and the remainder all T, is (n-h) (n-h- 1) f(h, n - h) = f(h) - (n-h)f(h+l)+- -f(h+2)... 2 (-)fi-? f(n)....(4.41)

6 ~ jf(h) (1 ~.(4-61) ~. (k ~ ~~- W. L. STEVENS 319 These quantities are the leading differences of the function f (h), h = n, n - 1, f (n) f(n-1,1) f(n-1) f (n- 2, 2) f(n- 2, 1).f(n- 3, 3) f(n-2) f(n-3, 2)... f(n- 3, 1) f (n- 4, 3) f(n-3) f (n- 3, 2) --. f(0, n) Conventional formulation. It is worth noting that we may write f(h, t) =p (l-p)l,...(4.51) where, after expansion, we make the substitution Pr =f (r)....(4.52) In other words, we can use formulae which would be appropriate to the case when the events are independent, and each have a probability p of being H and (1-p) of being T, provided that we make the above substitution after the expansion Probability of JL unspecijied H s. The probability of exactly 1~ H s and n - h T s is obtained directly by noting that there are n! ways... h! (n-h)! of selecting h events from n events. The probability of exactly h H s is therefore n! (n-h)(n-jl- 1) - (n-~i)f(h + 1)+ - f(h + 2)... (-)7t-hf(n))...(4.62) h! (n-h!)\ 2 5 FREQUENCY DISTRIBUTION OF NUMBER OF GAPS Considering the existence or non-existence of a gap after any specified arc to be the alternatives H and T respectively, we may immediately write down, from equations (3.3) and (4.62), the frequency distribution of the number of gaps, n(n n(n- 1) ) f(o) (n-k+ 1) = 1 -n(l -x) -l+ (1- hqfi-1, ~ * ~~ 2 k! (n- 1) (n-2)... (n-k+ 1) f(1) =n{(l-x)rl-1-(n-1) (1-2x)m ~ - 1)! n(n- 1)... f(k) = -~ (n- +1) (1- kx)n-l. k!... (5.1)

7 320 GEOMETRICAL PROBLEM IN PROBABILITY 6 EXAMPLE If x = 1/5, how many arcs must be taken to make the probability of no point being uncovered less than 50 %? Substituting in the above formula for f(0) we find Hence 17 arcs would be required. n P yo o/o

The Annals of Human Genetics has an archive of material originally published in print format by the Annals of Eugenics ( ).

The Annals of Human Genetics has an archive of material originally published in print format by the Annals of Eugenics (1925-1954). This material is available in specialised libraries and archives. We