12-11-2001, 01:47 AM | #141 |
Jack Burton
Join Date: November 10, 2001
Location: Bathurst & Orange, in constant flux
Age: 37
Posts: 5,452
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Simulator-er.... rubish.
This is the mathematical way of figuring it out: Useful Choices : Possible Choices. IN the first problem, there are 3 doors, 1 is useful. So the chances of getting the useful option are 1:3 or 33.33...%. Then One useless choice is removed, thus affecting the entire problem. NO matter what we said before we now have a right and a wrong door. 2 choices, one is useful. therefore the chances of getting the right one are 1:2, or 50%. Therefore it doesnt matter what happens, we have equal chances of being right or wrong. When all else fails, use maths, its pretty!! |
12-11-2001, 02:45 AM | #142 |
Jack Burton
Join Date: November 10, 2001
Location: Bathurst & Orange, in constant flux
Age: 37
Posts: 5,452
|
Ok..
Ive sent an email out to everyone who i know, they are going to run it as a survey. This contains the puzzle and the 2 possible answers. Ill have their results back in a few days, well say that the majority of people win, ok?? (as ib the same set up as an election). This will include people at the forum, so cast ur vote soon if u havent already done so! BTW- The results from the emails will be fair, and unaltered. I will give the exact figures from the email, so ill say something like CHANGE: 5 50-50:5. I think that this should end this battle once and for all. |
12-11-2001, 09:22 AM | #143 |
Jack Burton
Join Date: March 31, 2001
Location: The zephyr lands beneath the brine.
Age: 40
Posts: 5,459
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Being not so very good with pascal myself I am still trying to create a simulator myself. We'll talk later [img]smile.gif[/img]
But seriously, from what I see of the way your simulation works, you let it pick a certain door every time, and check to see if the prize is behind one of the other two doors. Did you think about removing one completely, or are the possible answers Door 1, Door 1, not (2 or 3) Door 2, Door (2 or 3), not 1 Door 3, Door (2 or 3), not 1 Obviously, if that is the case you count door 2 and 3 double. Then there's the random number generator, which sometimes does not go from 1-6 but from 0-5, or the other way around, and could be used wrong, and so on. There's a good chance it contains a flaw. On the other hand, there's a good chance it's right as well Back to my own... |
12-11-2001, 09:51 PM | #144 |
Iron Throne Cult
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quote: Um, you are joking right? You can't solve a maths puzzle by majority rules [img]graemlins/uhoh1.gif[/img] What if the majority of people are wrong? What about when Gallileo (?) said the sun was the centre of the universe? Were the people who had him killed right because they were in the majority?
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12-12-2001, 01:22 AM | #145 |
Jack Burton
Join Date: November 10, 2001
Location: Bathurst & Orange, in constant flux
Age: 37
Posts: 5,452
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quote: Yes, ofcourse im joking- I know im right, so id tamper with the results to reflect that. In the mean time, im trying to figure out how to word the correct answre correctly. |
12-12-2001, 04:24 AM | #146 |
Jack Burton
Join Date: November 10, 2001
Location: Bathurst & Orange, in constant flux
Age: 37
Posts: 5,452
|
OK. Ill repeat the puzzle here for clarity...
U are at a game show. U choosec one door- possibly the one of the 3 that has a prize behind. Then 1 wrong door is eliminated. U can now switch doors. Should u, and y? The correct answer is it doesnt matter. NO matter what happens you are left to choose between what must be a right door and a wrong door. This means u have 2 choices- right or wrong. 1 of these is useful, one is not. Therefore refering to every good probablility book i know of- Probabliity= Useful Choices/ Possible Choices. this ends up being 1/2, or 50%. |
12-12-2001, 07:12 AM | #147 |
Symbol of Cyric
Join Date: March 1, 2001
Location: Outside my place
Age: 42
Posts: 1,283
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Maths and statistics are sooooo annoying sometimes arent they ? Look at it this way:
Lets say you pick the right door the first time (33% chance of doing so). Then, after you change your opinion, you ALWAYS fail, so you have 33% chance of failing when you change. The other choice is to pick a wrong door the first time (66% chance of doing so). Then, after you change your pick, you ALWAYS win, as the other wrong door has been opened by the quiz master, so you win the 66% of the time when you change. Clear enough ?
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Previously known as Ertai_OHF |
12-12-2001, 07:20 AM | #148 |
Symbol of Cyric
Join Date: March 1, 2001
Location: Outside my place
Age: 42
Posts: 1,283
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BTW this has been the first time i think that i read 6 pages of a post. Riddles are sooo cool [img]smile.gif[/img] .
About the Cleopatra riddle, i think it lacks info, as you can suggest LOTS of causes that would have made them die (each of them sillier than the others [img]smile.gif[/img] ), all of them more or less correct, but youll say NO till you hear the one you wanna hear [img]smile.gif[/img] .
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Previously known as Ertai_OHF |
12-12-2001, 04:22 PM | #149 |
Jack Burton
Join Date: November 10, 2001
Location: Bathurst & Orange, in constant flux
Age: 37
Posts: 5,452
|
NO matter what happens, ur left with 2 doors- one right, one wrong. FLip a coin- Heads or Tails?? its 50-50, same situation different wording. The chances will have to change in the second choice because the situation that ur in will change.
I belive that the mistake Ertai is making is the same one that was made by whoever made this up. They try to recreate it, but dont try to prove what their saying with another example. Heres What Will Happen: Say u pick the right door. ur chances are 1 in 3. When the quiz master eliminates one door, and u change or stay-u have 2 choices so ur chances are 50-50. If you pick a wrong door, the other wrong door is eliminated so again you may stay or change. 2 choices- 50-50 |
12-12-2001, 05:27 PM | #150 |
Dracolisk
Join Date: September 16, 2001
Location: Bellingham, WA, USA
Age: 47
Posts: 6,901
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Simulators, mathematicians, and smart people in general agree that the odds of winning are 66.6666% if you change from your original guess.
Let's say there were a million doors. The odds of your picking the right one on the first try are one in a million. Then, the host opens all the other doors but one, and you are left with a choice: Do you want to stick with the door that has a 0.0001% chance of holding the prize, or do you want to change to the ONE door that, for SOME reason, was left closed out of all the 999,999 others? I'll run a simulator with those odds, too, if you want.
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