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Then considering the case with p = a and q = b, the last vote counted is either for the first candidate with probability a/(a + b), or for the second with probability b/(a + b). So the probability of the first being ahead throughout the count to the penultimate vote counted (and also after the final vote) is:
Pages in category "Probability problems" The following 31 pages are in this category, out of 31 total. This list may not reflect recent changes. B. Balls into bins ...
The problem is considered a paradox because two seemingly logical analyses yield conflicting answers regarding which choice maximizes the player's payout. Considering the expected utility when the probability of the predictor being right is certain or near-certain, the player should choose box B.
The a needle lies across a line, while the b needle does not. In probability theory, Buffon's needle problem is a question first posed in the 18th century by Georges-Louis Leclerc, Comte de Buffon: [1] Suppose we have a floor made of parallel strips of wood, each the same width, and we drop a needle onto the floor.
Sleeping Beauty problem: A probability problem that can be correctly answered as one half or one third depending on how the question is approached. Three Prisoners problem , also known as the Three Prisoners paradox: [ 3 ] A variation of the Monty Hall problem .
As the warden says B will be executed, it is either because C will be pardoned ( 1 / 3 chance), or A will be pardoned ( 1 / 3 chance) and the coin to decide whether to name B or C the warden flipped came up B ( 1 / 2 chance; for an overall 1 / 2 × 1 / 3 = 1 / 6 chance B was named because A will be ...
The goal is to compute P(B), the probability that at least two people in the room have the same birthday. However, it is simpler to calculate P(A′), the probability that no two people in the room have the same birthday. Then, because B and A′ are the only two possibilities and are also mutually exclusive, P(B) = 1 − P(A′).
As already remarked, most sources in the topic of probability, including many introductory probability textbooks, solve the problem by showing the conditional probabilities that the car is behind door 1 and door 2 are 1 / 3 and 2 / 3 (not 1 / 2 and 1 / 2 ) given that the contestant initially picks door 1 and the ...