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The answer to the first question is 2 / 3 , as is shown correctly by the "simple" solutions. But the answer to the second question is now different: the conditional probability the car is behind door 1 or door 2 given the host has opened door 3 (the door on the right) is 1 / 2 .
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.
Berkson's paradox arises because the conditional probability of given within the three-cell subset equals the conditional probability in the overall population, but the unconditional probability within the subset is inflated relative to the unconditional probability in the overall population, hence, within the subset, the presence of decreases ...
Banach's match problem is a classic problem in probability attributed to Stefan Banach. Feller [ 1 ] says that the problem was inspired by a humorous reference to Banach's smoking habit in a speech honouring him by Hugo Steinhaus , but that it was not Banach who set the problem or provided an answer.
Although the conditional and unconditional probabilities are both 2/3 for the problem statement with all details completely specified - in particular a completely random choice by the host of which door to open when he has a choice - the conditional probability may differ from the overall probability and the latter is not determined without a ...
Each scenario has a 1 / 6 probability. The original three prisoners problem can be seen in this light: The warden in that problem still has these six cases, each with a 1 / 6 probability of occurring. However, the warden in the original case cannot reveal the fate of a pardoned prisoner.
Given two events A and B from the sigma-field of a probability space, with the unconditional probability of B being greater than zero (i.e., P(B) > 0), the conditional probability of A given B (()) is the probability of A occurring if B has or is assumed to have happened. [5]
The probability of drawing another gold coin from the same box is 0 in (a), and 1 in (b) and (c). Thus, the overall probability of drawing a gold coin in the second draw is 0 / 3 + 1 / 3 + 1 / 3 = 2 / 3 . The problem can be reframed by describing the boxes as each having one drawer on each of two sides. Each ...
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