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Topic: $1,000,000 box (Read 13879 times) |
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mingus33
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$1,000,000 box
« on: Jan 15th, 2007, 8:50pm » |
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you are asked by the game show host to select one box of 3. One of the three contains $1,000,000 and the other two contain nothing.
After you have selected your box, the host removes one of the two remaining boxes and tells you that the box he removed does NOT contain the $1,000,000.
Now he asks you if you would like to keep your box or switch.
what should you do and why?
thanks
tony
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Whiskey Tango Foxtrot
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Re: $1,000,000 box
« Reply #1 on: Jan 15th, 2007, 9:32pm » |
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I think a similar problem has made its way through the forum before. I can't come up with a link but I'm pretty sure I've seen it around.
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Ghost Sniper
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Re: $1,000,000 box
« Reply #3 on: Jan 16th, 2007, 11:01am » |
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Did you get this idea out of "Deal or no deal"?
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denis
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Re: $1,000,000 box
« Reply #4 on: Jan 16th, 2007, 12:17pm » |
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Actually this puzzle existed many years ago before Deal or No Deal was created. It is more of an offshoot of the Monty Hall Show.
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rmsgrey
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Re: $1,000,000 box
« Reply #5 on: Jan 16th, 2007, 5:02pm » |
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The puzzle pre-dates Monty Hall by some time. Joseph Bertrand's Calcul des probabilites in 1889 contains it (according to The Magical Maze by Ian Stewart)
It's one of the classic examples of the counter-intuitive nature of probability, and has, in various forms, kept cropping up over the years - and has tripped up a number of qualified mathematicians who should have remembered better...
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Locke64
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Re: $1,000,000 box
« Reply #6 on: Jan 23rd, 2007, 4:01pm » |
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I first saw this one on Numb3rs. 
You switch I don't have time to explain; I gtg.
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Icarus
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Re: $1,000,000 box
« Reply #7 on: Jan 23rd, 2007, 5:10pm » |
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on Jan 16th, 2007, 12:17pm, denis wrote:| It is more of an offshoot of the Monty Hall Show. |
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Which was called "Let's Make a Deal". Monty wasn't actually required to open a gag door and offer a switch, and often just went with the original pick. That changes things considerably.
But if you know your host follows a policy of opening a gag door, then offering a switch to the remaining door, then 2/3 of the time switching is to your advantage.
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Locke64
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Re: $1,000,000 box
« Reply #8 on: Jan 23rd, 2007, 6:18pm » |
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Here's the visual way.
A=$1,000,000, B and C=nothing
|A| |B| |C|
1: You pick A, B/C is eliminated, you pick C/B, you lose.
2: You pick B, C is eliminated, you pick A, you win.
3: You pick C, B is eliminated, you pick A, you win.
You win 2 out of the three times. If you don't switch, however, you only have a one-in-three chance of winning, which is obvious.
I forgot how to do the math version. :(
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"Not really, but-" ~ANTISNAKERZ_RULE
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CowsRUs
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Re: $1,000,000 box
« Reply #9 on: Feb 7th, 2007, 8:15pm » |
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I say, Nuke it along with the Statue of Liberty and biobomb the host 
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JiNbOtAk
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Re: $1,000,000 box
« Reply #10 on: Feb 16th, 2007, 3:22am » |
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What if the host would only offer to reveal one door, on the condition that you chose the right one ?
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towr
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Re: $1,000,000 box
« Reply #11 on: Feb 16th, 2007, 3:53am » |
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on Feb 16th, 2007, 3:22am, JiNbOtAk wrote:| What if the host would only offer to reveal one door, on the condition that you chose the right one ? |
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?
Then if he opens a door you know you have the right one?! That sort of gives it away, doesn't it?
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rmsgrey
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Re: $1,000,000 box
« Reply #12 on: Feb 16th, 2007, 10:47am » |
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on Feb 16th, 2007, 3:53am, towr wrote:
?
Then if he opens a door you know you have the right one?! That sort of gives it away, doesn't it? |
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It does rather depend on whether you know that's what he's doing. If you know that's his strategy, then (assuming you always get the chance to switch) you can still win 2/3
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Random Lack of Squiggily Lines
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Re: $1,000,000 box
« Reply #13 on: Mar 15th, 2007, 11:24am » |
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dump the first part, part 2 you have 2 boxes, 1 in 2 chances to get the $1kk( $1,000,000)
50%
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You can only believe i what you can prove, and since you have nothing proven to cmpare to, you can believe in nothing.
I have ~50 posts to hack a "R" into a "D". Which one?
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rmsgrey
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Re: $1,000,000 box
« Reply #14 on: Mar 15th, 2007, 12:49pm » |
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on Mar 15th, 2007, 11:24am, tiber13 wrote:dump the first part, part 2 you have 2 boxes, 1 in 2 chances to get the $1kk( $1,000,000)
50% |
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A lot of people think that, which is why the question is a classic. However, the correct answer is that you win twice as often when you switch.
Consider a variant where, instead of opening one of the other boxes, the Host instead always gives you the chance to switch your box for both the others. Better off switching then, right? How about if, when you agree to take the other two boxes instead, you have to pick which one (of those two) you actually want after seeing them both opened? Still a 2/3 chance of winning when you switch, right? How about if the Host doesn't open any boxes, but tells you: "if the prize is in either of these two boxes, it's in this one" after you pick the pair of boxes.
Another variation is to take a pack of playing cards and try and pick the Ace of Spades by the following procedure:
Pick one card and set it aside.
Get someone else to look through the rest of the cards and, if the Ace of Spades is in that pile, set it aside; otherwise, set a random card aside.
You then have two cards, one of which is the Ace of Spades. If you're right, then it doesn't matter where those two cards came from and you have a 50% chance of having the Ace if you don't switch; if I'm right, then the card you picked is only the Ace if you picked it (just under 2% of the time).
If you like, you can try it as an experiment - if you go through the process 100 times, if I'm right, you'll get the Ace roughly twice; if you are it'll be around 50 times.
Of course, there is a variation where you do have the Ace 50% of the time - instead of getting someone to look through the cards to pick which ones to reveal, reveal them randomly until there's only one and the one you picked left - that way about 2% of the time you'll have the Ace; 2% the other card will be the Ace, and 96% you'll have revealed it earlier and have to start over. If you repeat that process until you end up with 2 cards without seeing the Ace, it is then 50% each way of either being the Ace - because you learned something about your card every time you didn't have to start over after revealing a card...
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srn437
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Re: $1,000,000 box
« Reply #16 on: Aug 29th, 2007, 6:59pm » |
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It's a trick, they want you to think that. It's 50-50. Besides, you could have started with the one he gives you the option to switch with and had the same situation.
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Aryabhatta
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Re: $1,000,000 box
« Reply #17 on: Aug 30th, 2007, 12:59am » |
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on Aug 29th, 2007, 6:59pm, srn347 wrote:| It's a trick, they want you to think that. It's 50-50. Besides, you could have started with the one he gives you the option to switch with and had the same situation. |
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Wrong again, the 1/0th time. Wrong again, the 1/0th time. Wrong again, the 1/0th time. Wrong again, the 1/0th time. Wrong again, the 1/0th time. Wrong again, the 1/0th time. Wrong again, the 1/0th time. Wrong again, the 1/0th time. Wrong again, the 1/0th time. Wrong again, the 1/0th time. Wrong again, the 1/0th time. Wrong again, the 1/0th time. Wrong again, the 1/0th time. Wrong again, the 1/0th time.
I feel the strange urge to respond to srn347's posts... I think i should stop now.
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| « Last Edit: Aug 30th, 2007, 12:59am by Aryabhatta » |
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towr
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Re: $1,000,000 box
« Reply #18 on: Aug 30th, 2007, 1:00am » |
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on Aug 29th, 2007, 6:59pm, srn347 wrote:| It's a trick, they want you to think that. It's 50-50. Besides, you could have started with the one he gives you the option to switch with and had the same situation. |
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I can't imagine why I'm still bothering to enlighten you; but the situations would not be the same.
Imagine a pack of cards, your goal is to get the ace of spades. You select one card, X, without seeing at; from the other 51 cards I remove 50 cards that are not the ace of spades. Do you stick to your choice, or switch to the other card, Y?
If you had by chance picked Y at the start, how likely do you think it that at the end, the other card would be A?
For the Monty Hall problem the only difference is that it's equivalent to a three card game, and not a full deck one. But the end situation would still not be the same if you had picked the other; and switching is still better on mathematical grounds.
However based on prior experience I have little hope you'll actually consider any argument contrary to the answer you set your mind on.
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JiNbOtAk
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Re: $1,000,000 box
« Reply #19 on: Aug 30th, 2007, 4:25am » |
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on Aug 30th, 2007, 12:59am, Aryabhatta wrote:| I feel the strange urge to respond to srn347's posts... I think i should stop now. |
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I know the feeling, it's like when you have an itch between your shoulderblades; you could never reach it to scratch it properly, but it doesn't mean you'll stop trying.
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mikedagr8
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Re: $1,000,000 box
« Reply #20 on: Aug 30th, 2007, 4:29am » |
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Oh yes, now that's a metaphor I have to use.
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JiNbOtAk
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Re: $1,000,000 box
« Reply #21 on: Aug 30th, 2007, 4:47am » |
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on Aug 30th, 2007, 4:29am, mikedagr8 wrote:| Oh yes, now that's a metaphor I have to use. |
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I think the word you're looking for is simile. But I could be wrong. 
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mikedagr8
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Re: $1,000,000 box
« Reply #22 on: Aug 30th, 2007, 4:53am » |
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Wouldn't surprise me if your correct. I generally can't remember all the terms like alliteration and all that crap. All I know is that I'll be using it.
Guess what, you are.
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| « Last Edit: Aug 30th, 2007, 4:56am by mikedagr8 » |
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SMQ
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Re: $1,000,000 box
« Reply #23 on: Aug 30th, 2007, 4:55am » |
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on Aug 29th, 2007, 6:59pm, srn347 wrote:| It's a trick, they want you to think that. It's 50-50. |
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If it's really 50-50, why would "they" try to trick you, since it wouldn't matter if you switched or not?
--SMQ
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srn437
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Re: $1,000,000 box
« Reply #24 on: Aug 30th, 2007, 10:08am » |
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So your answer to the riddle would be wrong. And perhaps it's an idiom.
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| « Last Edit: Aug 30th, 2007, 10:10am by srn437 » |
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