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Interesting Math Riddle...
Well, not really a riddle, but a rather interesting and not so easy to understand fact....
Let's say there's a show on TV where the participant has to chose 1 out of 3 closed boxes. In one of them there are $10k, and the other 2 are empty. He makes a choice, and before opening the box the host of the show opens one of the two other boxes which is empty (there must be at least one, because a total of 2 boxes out of the 3 are empty). now, should the participant change his choice to the other last box that the host didn't open? or should he stay with his choice? most would think it doesn't matter at all... but the correct answer is that he should switch... why? at the beginning the participant was to chose one of 3 boxes... one of the three contained the cash, therefore he had a 33.333...% of chosing the right one... now, after the participant has opened one of the empty boxes, if he changes his choice... then he has to chose 1 out of 2 boxes, and has a 50% chance to get it right... but if he stays with his old choice, he still has a 33.333% chance of course... |
that hurt my head
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What if he chooses the same box, he still has the 50%
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This was posted on Oprano a few weeks ago.
Regards, Lee |
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Regards, Alex |
you got me..... the percentage is higher but i dont know if i'd switch, because its only higher now there are 2 options left
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Switching doesn't increase his odds, removing one box did.
I wouldn't switch. I always go with my gut instinct. |
After one of the boxes is eliminated you are left with 2 boxes, so your chances of getting it right are 50% NO MATTER which box you choose.
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There is no reason to switch it's now 50 50 chance either way...
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He could elect to change his box, then chose the same one and he'd be guessing at 50%. |
It's a brain teaser to make you think the odds of it being his original choice is lower than the odds of it being the other box.
it isn't. |
1 of 3 boxes...
picks one. one's opened that is empty leaving two left. the one he picked and the one that's left... either of which contain the money. what difference does it make if he switches he's still picking one of two boxes left... sounds like 50/50 proposition to me? =D |
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but it's not very accurate... i mean, you already chose it at the beginning, and had a 33.333% chance of winning, and you stay with the same choice.. nothing changed...you still have 33.33% how can it be different? |
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you had 33.33% when there was 3 boxes! now that he opened 1....there are 2...no matter which one you pick you have 50/50! (unless of course you switched to the one he just opened...then you are stupid) |
it's 50% before and after :2 cents:
This is a 100% useless... jk :winkwink: |
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I was indeed wrong... It's not 50%, It's 66.66%! read here: http://256.com/gray/teasers/#deal your intuition doesn't always work... this is pure statistics.. |
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It's actually a 66.66666...% chance of the other box winning.
Why? Because the information changed. Essentially, you gave the participant a choice of choosing ONE box (which would be the choice if he stuck with the original box) or choosing TWO boxes, one of which can't win and that one would be revealed as a loser (at which point the participant would switch their choice from the original to the other one). Run the stats, and switching will always reveal a greater chance of a win. What would you pick... one box with a 33.3333...% chance of Winning, or TWO boxes, each with a combined chance of 66.66666...% chance of winning? |
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YOu silly monkey. :2 cents: |
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now I'm sorry for the repost :1orglaugh |
say the boxes are:
A B C say i choose A, and box C is removed ... why should i move to B? I could have easily selected B in the begining and C was removed... Then I'd go back to A using your theory. My point? same odds. |
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http://256.com/gray/teasers/lets_deal.html |
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Short answer? In your example, would you rather choose just A, or BOTH B and C, knowing that one of the losing choices (between B and C) would be revealed? In revealing a losing choice, you are gaining information. A still has a 33.333...% chance of winning. I'd go with the other 66.66...% switch, personally. |
This 'riddle' is older them Methusalem himself and is nothing more then statistics. Get back to school :321GFY
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((A+B)/C + ((n)^2 / C-B+A) + 1) * (n)^2 - C - (A + B^2) = A. You're right, he should switch.
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It makes as much sense as the fact that it makes no sense and makes sense at the same time....if that makes any sense.
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and what kind of education do you have? |
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<---- i swear i didnt see other answers before i posted. honest. really.
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that could explain why you are wrong |
Here's another way of looking at it.
There are a million doors. You choose one of them. Monty opens 999,998 empty doors leaving the one you chose and another one. Do you really think you picked the car on your first try and shouldn't switch. |
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I have the second highest level of highschool education you can get over here in Holland and did college, computer science, which involved a lot of math. And? |
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