You are a contestant on a gameshow. The gameshow host shows you three doors, and tells you that behind one of them is the star prize, while behind the other two is nothing - not even your bus fare home. The game works like this, every time. You pick one of the doors. The gameshow host then opens one of the remaining two doors, to show you that the star prize is not behind that one. Since he knows where the star prize is and two of the doors have nothing behind them, he is always able to do this, and he always does. The host then offers you a choice. Do you want to stay with your original door, and win what's behind it? Or do you want to swap for the other door that remains unopened? Does it affect your chance of winning, if you do?
kankoo Wed 03/09/08 17:47
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Your first choice was a 3 to 1 shot. Once one of the losing doors is revealed, you're faced with a second choice; now a 2 to 1 shot, whether you consider it a choice between 2 doors or whether or not to stay with your original choice. Short answer: no, it makes no difference to your chance of winning
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It's the famous Monty Hall problem. Based n probabilities you should always switch as it will take you from a 1/3 chance to a 2/3 chance.
The door you choose first has a 1/3 chance of being right. The door the host opens is a 0/3 chance of being right. Sice sum of probabbilities must equal 1, the remaining door has a 2/3 probability of leading to the star prize.
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Alot of people tend to think that as there are only 2 doors left, it must be a 50-50 chance and swapping dooors will make no difference.
Factor30 has already stated that this is the classic Monty Hall problem and you have a 2 in 3 chance of winning if you swap doors.
Check out the following link for a full explanation:
http://en.wikipedia.org/wiki/Monty_Hall_proble m
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I'd never heard of this puzzle and I'm glad to read here http://en.wikipedia.org/wiki/Monty_Hall_proble m that lots of cleverclogs people don't get it.
I could hide behind the fact that kankoo doesn't seem to have stated the problem correctly - but in truth I'll have to admit that I would still have got it wrong... I'm not sure I understand it even now!
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Sorry but don't get the logic behind the explanation given.
Once the first door is opened, there are only 2 doors left. You now have 2 choices - stay with your original choice or change it. Statistics may say you if you switch you have a 2 in 3 chance of getting it right - but one of the doors is now completely out of the equation. So there's then a 50/50 chance that you'll select the right door - either by staying with your original choice or choosing to switch.
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Hello firetracie. Others can explain it better than I can but i'll have a go at demonstrating it another way.
Imagine the problem were the same except that there were 100 doors (1 prize, 99 blanks). You choose a door (very unlikely to be the prizewinning door). The host, who knows where the prize is then opens 98 doors which have no prize behind them and reminds you that the prize lies either behind your original unopeneddoor or the remaining unopened door. Would you feel the chances are still 50/50 for each door?
No shame in getting this wrong. Many experts got this Monty Hall problem wrong when it was first published, and although I'm a maths teacher my instinctive initial answer was wrong when someone asked me this one a few years ago.
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I must admit I don't see it either. If you have 2 doors, the odds are surely 50/50 as the other door is no longer in play?
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I have a horrible feeling you either get this or you don't. Some of the correspondence that went on in the States over this question was really vicious with mathematicians writing in and calling other people liars and cretins. Anyhow, here's my way of rationalising it.
Your original choice only had a 1/3 probability of being right so regardless of anything else that happens you can't improve on those odds by sticking. Your odds of making a better decision by switching weren't any better to start with because with two other doors to choose from you had only a 50:50 chance of picking the right door out of the remaining 2/3 probability, i.e. 1/3 again. The game show host has just removed that problem for you - he's told you which is the incorrect door of the remaining pair so your odds of picking the correct one (assuming one of them is correct) just went up to 2/3.
None of this alters the fact that 1/3 of the time you had already picked the correct door and will lose by switching, it's just that 2 out of every 3 times you didn't get it right first time, but now you can.
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Question Author
Thanks very much everyone. I think that I now understand! Probability theory never was my strong point.
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