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As in the Monty Hall problem, the intuitive answer is 1 / 2 , but the probability is actually 2 / 3 . The Three Prisoners problem, published in Martin Gardner's Mathematical Games column in Scientific American in 1959 [7] [55] is equivalent to the Monty Hall problem. This problem involves three condemned prisoners, a random one of ...
Text from another version, posted under the title Monte Hall Problem. Charles Matthews 20:57, 10 Jun 2004 (UTC) The “Monte Hall Problem” (Applied Probability) On the old game show “Let’s Make a Deal”, the final segment involved two contestants who had won the most money that day. There were 3 closed doors with prizes hidden behind ...
If the player (in 2/3) however picked one of the two goats, then the host had to show the second goat, and in that 2/3 switching will win the car (Carlton). Switching doubles the chance to win from 1/3 to 2/3 (Carlton). Without having better evidence, probability to win by switching (Pws) will be 2/3, and so the guest always should switch.
Start with an ace and two jokers. You're the dealer. Shuffle the 3 cards. Deal them face down left to right. The leftmost card corresponds to "door 1", the middle one "door 2" and
1 Problem description. 2 Solution. 3 Vos Savant's solution. 4 Aids to understanding. 5 Simulation. 6 Increasing the number of doors. 7 Sources of confusion. 8 Other ...
I believe in constant refinement. 'Monty Hall problem', Wikipedia, There are 3!=6 sequences for arrangement of 3 prizes behind 3 doors. The prizes are {goat1, goat2, car}. For any 1 sequence there are 4 possible games (series of choices) by the player 'p' and host 'h' including an option for the player to change their choice. For {g1 g2 car}:
The difference between these problems is confusing to many (per Morgan et al. - "the distinction between the conditional and unconditional situations here seem to confound many") and after "seeing" the simple solution as correct switching back to focusing on the conditional problem is also extremely difficult (per Morgan et al. - "F1 [which ...
If the host reveals that the player initially picked a goat (2/3 probability), the player still has a 50-50 chance of picking the car behind one of the two remaining doors. (His total odds of winning over the course of this game are 2/3: 1/3 initial win plus 50% chance of winning in the 2/3s of cases where he switches: 1/3 + 50% * 2/3.)