Monty Hall Strikes Again
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Description: Dr Math answers a reader question and explains why the puzzle works the way it does.
Subject: Probability Question from OZ! Date: Wed, 2 Nov 1994 17:13:05 +1100 (EST) From: "Sean Pryor" Hi... Basically the problem goes like this. There are three cups, one of which is covering a coin. I know the whereabouts of the coin, but you don't. You pick a cup, and I take one of the remaining cups, one which DOESN'T contain a coin. Both you and I know the cup I pick doesn't contain a coin. You then have the option to swap your cup with the third, remaining cup, or keep your first choice. What is the probability of the coin being in the cup if you keep your first choice, or if you decide to swap them? To summarise: Three cups with a coin under one. You pick one, I pick one that DOESN'T have the coin. You then either stay with your choice, or swap it with the remaining cup. What is the probability of getting the coin, either way? BTW the answer I get is 50% either way - though it has been suggested that you have a 2/3 chance if you swap cups.... I disagree with this, but I don't think my maths is capable of giving a definitive answer (which is why you're reading this!) (This problem has placed a cool Australian $50 on the line here in a bet - so I need an answer proving me right! :-) My mate would never let me forget it if he was right...) Thanxx Sean Pryor
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| Page title: | Math Forum - Ask Dr. Math |
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| Description: | There are three cups, one of which is covering a coin. I know the whereabouts of the coin, but you don't. You pick a cup, and I take one of the remaining cups, one which DOESN'T contain a coin. Both you and I know the cup I pick doesn't contain a coin. You then have the option to swap your cup with the third, remaining cup, or keep your first choice. What is the probability of the coin being in the cup if you keep your first choice, or if you decide to swap them? |
| IP-address: | 144.118.94.19 |