Monty Hall strikes again
March 10th, 2020 at 2:22:54 PM
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In the Monty Hall show there are 10 cups, and there is only one cup hiding a prize. The cups are in random order. A contestant selects two cups and marks them. Then the host of the show, who knows what is under each of the 10 cups, removes six of the unmarked cups not hiding the prize. There are now four cups left, the prize is under one, and two of them are the contestant's original choice. Here is the offer: the contestant may retain the original two marked cups or the contestant may exchange these two marked cups for only one of the two other cups left. What to do in order to maximize the probability of getting the prize?
March 10th, 2020 at 2:41:36 PM
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Ah, this sounds very similar to the original Monty Hall problem, but with a twist. And 9 goats instead of just 2! ;-)
still exchange my two cups for one of the remaining cups. Odds of winning with my original picks: 2 out of 10 (20%). Odds of one of the remaining cups being a winner after 6 were removed: 1 out of 4 (25%).
"Dealer has 'rock'... Pay 'paper!'"
