Seneta 1993
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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.
Seneta 1993
- 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