Pillow-Problems: Problem #45

Problem #44 | Problem #47
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

Problem:

If an infinite number of rods be broken: find the chance that one at least is broken in the middle.

Answer:

.6321207 &c.

Solution:

Divide each rod into (n + 1) parts, where n is assumed to be odd, and the n points of division are assumed to be the only points where the rod will break, and to be equally frangible.

The chance of one failure is

(n − 1)/

(n)

;

∴ " " n failures is (

(n − 1)/

(n)

)ⁿ

= (1 −

(1)/

(n)

)ⁿ

Now, if m =

(1)/

(n)

; then, when n =

(1)/

(0)

, m = 0;

∴ the chance that no rod is broke in the middle = (1 − m)

(1)/

(m)

, when m = 0;

i. e. it approaches the limit (1 − 0)

(1)/

(0)

.

And Ans. = 1 − (1 − 0)

(1)/

(0)

.

Now (1 − 0)

(1)/

(0)

= e. Hence if, in the series for e, we call the sum of the odd terms 'a', and, of the even terms 'b'; then e = a + b; and (1 − 0)

(1)/

(0)

= a − b = 2a − e.

Q.E.F.

[N.B. What follows here was not thought out.]

Now a = 1 +

(1)/

(2!)

+

(1)/

(4!)

+ &c.