Lewis Carroll wrote "Pillow Problems", a collection of 72 logic and probability puzzles, while lying in bed at night. Many had clever but flawed solutions due to Carroll's limited understanding of modern probability concepts. For example, in one problem about breaking rods, Carroll incorrectly assumed the probability of breaking at the middle was nonzero. Overall, "Pillow Problems" reflects the nascent state of English probability in Carroll's time and his personal difficulties with more rigorous concepts like continuous probabilities.

1. Lewis Carroll’s ”Pillow Problems”:
On the 1993 Centenary
Eugene Seneta
Statistical Science, Vol. 8, No. 2, 180-186
Collegio Carlo Alberto
Bayesian Statistics
April 21, 2015

2. Overview
• Eugene Seneta: Professor Emeritus, School of Mathematics
and Statistics, University of Sydney. He is known for the
variance gamma model in financial mathematics (the
Madan-Seneta process).
• Lewis Carroll: pen name of Charles Lutwidge Dodgson (1832
1898), was an English writer, mathematician, logician and
photographer. His most famous writings are Alice’s
Adventures in Wonderland and its sequel Through the
Looking-Glass → wordplays, logic, and fantasy.

3. Pillow Problems
• 72 problems: formulated and worked out at night while in
bed, with the answer written down afterwards
some are difficult, interesting and with correct solutions
some are badly posed, with imaginative but incorrect
solutions
• Lennon (1945): ”His range was that of a freshman today in a
good technical school, though the freshman would have
clearer ideas about elementary things than CLD.”
• Weaver (1954): ”He lagged behind the best knowledge of his
time.”

4. English Probability
• As a probabilist LC may not be not important
→ ”..., but his work reflects the nature, standing and
understanding of probability within the wider English
mathematical community of the time”.
• De Morgan (1086 - 1871): ”inverse probability” (obsolete
term for the probability distribution of an unobserved variable)
→ urn models
• Venn (1834 - 1923): approach to probability through logic
→ improvement of ”Venn diagrams”
• Little or nothing of Laplace (1749 - 1827) and Legendre (1752
- 1833)

5. No. 45 (1)
• If an infinite number of rods be broken: find he chance that
one at least is broken at the middle.
divide each rod into n + 1 parts, where n is odd.
assume the n points of division are the only (equally likely)
points where a rod can break
⇒ P(rod not breaking in the middle) = 1 − n−1
assume the same number of rods as breakpoints (!)
⇒ P(no rod breaking in the middle) = (1 − n−1)n
⇒ answer = limn→∞ 1 − (1 − n−1)n = 1 − e−1 = 0.6321207

6. No. 45 (2)
• Density over rod length to the breakpoint position (e.g.
uniform)
⇒ answer = 0
• Countability and uncountability literature at its infancy,
Cantor (1845 - 1918)
• In ignorance of ”continuous probabilities” through probability
densities, Laplace

7. A Controversy
• Rev. Simmons: a random point being taken on a given line,
what is the chance of it coinciding with a previously assigned
point?
using the uniform distribution: if the point is k, probability
of taking a point to its left is k and to its right is 1 − k
⇒ answer = 0
LC: if the probability of a specific point is zero, the the
probability of any point is zero, yet some point is chosen
⇒ ”... when an event is possible, its chance of happening is
not zero”

8. No. 50
• There are two bags, H and K, each containing 2 counters:
and it is known that each counter is either black or white. A
white counter is added to bag H, the bag is shaken up, and
one counter is transferred (without looking at it) to bag K,
where the process is repeated, a counter being transferred to
bag H. What is now the chance of drawing a white counter
from bag H?
3 counters in bag H where white can be (0, 1, 2, 3) with
prior distribution (0, 1/4, 1/2, 1/4)
2 counters in bag K where white can be (0, 1, 2, 3) with
prior distribution (1/4, 1/2, 1/4, 0)
a counter is transferred in turn between bags
⇒ correct solution = 17/27 → urn mixing model (Ehrenfest,
Markov chain models)

9. No. 72 (1)
• a bag contains 2 counters, as to which nothing is known
except that each is either black or white. Ascertain their
colours without taking them out of the bag

10. No. 72 (2)
• a bag contains 2 counters, as to which nothing is known
except that each is either black or white. Ascertain their
colours without taking them out of the bag
assume a binomial prior distribution: (1/4, 1/2, 1/4) for
(BB, BW, WW)
suppose to add a black counter to the ball
⇒ change of drawing a black counter = 2/3
⇒ there must be two blacks and a white
⇒ before the black was added there must be one black and
one white

11. No. 72 (3)
• a bag contains 3 counters, as to which nothing is known
except that each is either black or white. Ascertain their
colours without taking them out of the bag
assume a binomial prior distribution: (1/8, 3/8, 3/8, 1/8)
for (BBB, BBW, BWW, WWW)
suppose to add a black counter to the ball
⇒ change of drawing a black counter is
1
8 ∗ 1 + 3
8 ∗ 3
4 + 3
8 ∗ 2
4 + 1
8 ∗ 1
4 = 5
8
5
8 does not coincide with the probability of drawing a black
from any one number of the partition

12. Thank you for your attention