This is an instance of the so-called Three Prisoners Problem. See, e.g., Martin Gardner, Mathematical Games: Problems Involving Questions of Probability and Ambiguity, Sci. Am., Oct. 1959, at 180–82; Martin Gardner, Mathematical Games: How Three Modern Mathematicians Disproved a Celebrated Conjecture of Leonhard Euler, Sci. Am., Nov. 1959, at 188; George Casella & Roger L. Berger, Statistical Inference 21–23 (2d ed. 2002). It is similar to the well-known Monty Hall Problem. See, e.g., Steve Selvin, A Problem in Probability, 29 Am. Statistician 67 (1975); Steve Selvin, On the Monte Hall Problem, 29 Am. Statistician 134 (1975); Marilyn vos Savant, Ask Marilyn, Parade Mag., Sept. 9, 1990, at 15.
Larry is correct. Regardless of which son will be removed as a beneficiary, it must be the case that Bill or Chris will remain as a beneficiary. Therefore, Larry's answer to Al that Bill will remain has no bearing on Al's chances of being removed; they continue to be 1/3.
Before Larry answers, there are four possible cases.
|Case||Son to Be Removed||Larry'sAnswer||PriorProbability|
Larry's answer eliminates cases II and III, leaving cases I and IV as the only possibilities.
|Case||Son to Be Removed||Larry'sAnswer||PriorProbability||Posterior Probability|
The chances that it is case IV (1/3) are twice the chances that it is case I (1/6), so the chances that Chris will be removed are two to one, or 2/3, leaving Al's removal chances at 1/3.
Although, strictly speaking, Larry is correct, there is a sense in which Dawn is correct. Larry's answer gives Al information about whether his brothers will be removed as beneficiaries. Specifically, it informs Al that Bill's and Chris's removal chances are zero and 2/3, respectively. Insofar as it is foreseeable that Al will recount his conversation with Larry to his brothers, Larry knows that his answer to Al will give Bill and Chris information about their removal chances.