Klopp
Klopp
Joined: Mar 10, 2020
  • Threads: 7
  • Posts: 16
March 10th, 2020 at 2:22:54 PM permalink
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?
Joeman
Joeman
Joined: Feb 21, 2014
  • Threads: 35
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Romes
March 10th, 2020 at 2:41:36 PM permalink
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!'"
drebbin37
drebbin37
Joined: Feb 7, 2012
  • Threads: 0
  • Posts: 8
March 10th, 2020 at 3:19:27 PM permalink
I agree with Joeman, except...

I think the chance of winning after a switch is 40% (half of the 80% chance of being wrong with your initial guess). Fun twist!

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