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anyone fancy a gamble

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jimmer | 18:42 Thu 02nd Dec 2004 | Science
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ok, so i'm on a quiz show and to win the star prize of say 20 quid i have to guess which door its behind. there are three doors and i have two chances. my first guess, door number one. The quiz show host opens up door number three which is empty and gives me the chance to stick with door number 1 or change my mind and select door number two. what should i do?

 

jim

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Stick with your first instinct - it's there for a reason...of course, going with the "second respose," the letter 'B' on a multiple choice test, always felt safe. Whether or not is was actually right is a different story...
Just because it's not behind door number 3 doesn't effect the likeliness of which of the other doors it is behind, it's not going to move around as the quiz master shows you empty doors.
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Change your mind - because as a pure statistical comparison only, you have a 1 in 2 chance if you change your mind, rather than a 1 in 3 chance
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Yes, but there are not 52 doors. The host also stands a 1 in 2 chance of getting the right door after you pick the first choice out of the three. So actually, if you try to "get his side of the deal" you are choosing a response that only stood a 50% chance of being correct (1 in 2). So really it's not any safer to select from the host's chances, because he only stood half a chance to be right.

The magic here is that the quiz show host opens a door that he knows is empty, and thereby has given you a massive clue as to where the prize really lies. In the original example you chose door 1 so, as said, you have a 1 in 3 chance of the prize. Then the host opens door 3 (empty), thus showing that of the two doors available to them they did not choose door 2.

 

They had a 2 in 3 chance of getting an empty door (even though they knew where the prize lay) and using their 2 in 3 chance the did not open door 2, so switching to door 2 gives you that same and much better odds of winning.

 

Message to hippy - Thanks for that. I have known about this for some time, and never understood the logic. Thanks to your well-worded answer, I have now grasped it. Well done, however I still don't agree, I just understand the logic better!

If anybody wants this explained even more clearly (no offense ansteyg or Hippy) you should read the brilliant "The curious incident of the dog in the night-time" by Mark Haddon, where this problem is also mentioned. In fact you should read the book anyway!
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thanks for your answers. its all sounds a bit nutty but it does actually improve your chances to switch on the second chance. the way i understand it is through a change in the probability: by sticking to my first guess i have a one in three chance of winning by switching doors my chances improve to 50/50 but i must actually switch doors to take advantage of these odds.

 

jim

Yes - change every time, no question.
By the way I agree with AlFas - the Mark Haddon book contains an excellent analysis of this conundrum, both by mathematical formula and by tree diagram (very illustrative) but I also agree that you should just read the book anyway.
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I understand the shift from 1 in 3 odds to 1 in 2 odds, so theoretically it's a better situation the second time. But, what would keep you from choosing door 1 from the 3 doors at first (1 in 3), and then also choose door 1 from the 2 doors the second time (1 in 2). Inevitably, you will face a situation with 1 in 2 odds (ie the "second round").

It doesn't matter what you do the first time. It only mattters what you do the second time. Choose door 1 both times, and your fine (1 in 2 chance). Choose door 1 the first time and door 2 the second time, and your still facing the exact same odds (1 in 2). It doesn't matter.
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