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Post by jpg on Sept 6, 2014 3:20:14 GMT -5
Well you are switching from a 1/3 to a different 2/3 probability divided by 2 (if I understand what Baba is saying) so it isn't 1/2 really because you are 'loosing' a 1/3 chance of being right to gain a 1/2 chance of being right?
What I really want to know is how this will help me become a better trader (my aim in life...).
4balance: if baba makes us look to foolish we just have to put him on ignore.. Oh wait. Wrong board...
JPG
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Post by liane on Sept 6, 2014 5:22:18 GMT -5
Wow ! What a conundrum! I was in the camp that your chances do not change with switching. But for anyone really bored on a Saturday, here's some food for thought. I think I'm convinced enough to switch: Monty Hall problem
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Post by brentie on Sept 6, 2014 7:27:55 GMT -5
I'm a simple man, I say it doesn't matter. You started with a 1 out of 3 chance but when one door was revealed your chances went to 50/50. Whether you stay with the original or switch doors, you still have that 50/50 shot.
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Post by spiro on Sept 6, 2014 8:22:42 GMT -5
You guys are killing me with all this door talk. For months my wife has been trying to get me to replace the front door. I keep telling her, I really don't want to switch. In this case there is a 90% chance that I will have to switch doors. Hopefully, I can delay the switch for several years. It's hard to imagine how someone can get tired of a perfectly good front door. But then again, I never seem to get tired of listening to those damn conferences.
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Post by daduke38 on Sept 6, 2014 8:49:47 GMT -5
18. UK and EU Approval (in Phase III now?) I think it is just UK. EMA hasn't even been filed to my knowledge.
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Post by Deleted on Sept 6, 2014 8:53:24 GMT -5
Here is a related one: We have three cards. One has an "X" printed on both sides, another has an "O" on both sides and the last has an "x" on one side and an "O" on the other. Cards unseen, we draw one and place it on the table. An "X" is seen. Question: What are the odds that the other side is also an "X" ?
To save time, this is the WRONG answer: We see an "X" and thus know we didn't draw the "O-O" card. There is a 50-50 chance we drew either the "X-X" card or the "X-O" card so the odds are even money that an "X" is on the other side. As I said, this is WRONG'.
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Post by BD on Sept 6, 2014 9:26:48 GMT -5
I thought 4balance and brentie's analylsis is correct. We're left with 50/50 odds, so the good prize is just as likely to be behind the door already picked as behind the other one. There's no benefit, probability-wise, of switching. Just to be extra sure, I wrote and ran a C++ simulation just now that confirms our thesis However, this result is at odds with liane's article, so that's a conundrum. I'm attaching the program in case anyone's a software nerd and wants to try it...maybe someone can tell me what I did wrong, if in fact the simulation is not really simulating the circumstances correctly... Attachment Deleted
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Post by BD on Sept 6, 2014 9:48:49 GMT -5
Ha, I just realized that even the result my simulation gives (correct 50% of the time) doesn't agree with the "correct" vos Savant predicted result of a 66% chance of being right by switching.
When I eliminate the "switch", I get the expected result of being correct 33% of the time.
The logic of the article makes sense to me, so there must be some sort of violation of the assumptions (mentioned in the article) in my simulation model for the case where the contestant switches.
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Post by babaoriley on Sept 6, 2014 10:19:15 GMT -5
When I first heard vos Savant's riddle a million years ago, I too, thought it wouldn't matter whether you switch or not, 50/50 all the way, and the person who told it to me, didn't know how to explain the answer. And supposedly, PhD's argued with vos Savant, some telling her in nasty terms she was wrong. It was all very fascinating. I don't need a computer program to prove it to you all, but imagine this, suppose there were a million doors, and only one good prize, you will pick one, Monty will then show you every other door (all 999,998 of them) and just two doors remain. What do you do? Well, you'd have to feel that first pick of yours was 1 in a million, or you'd switch in a heartbeat.
Another way of looking at it is this, you will surely be wrong to switch 1/3 of the time, but you'll be right 2/3 of the time, cuz, when you think about it, the rules of the game actually make Monty tell you the correct curtain every time you've picked wrong initially (2/3 of the time).
I enjoy such things, and that one is the single most counter-intuitive one I've ever come across.
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Post by BD on Sept 6, 2014 10:29:29 GMT -5
Nice, baba, good simple explanations. The fact the contestant is correct only 33% of the time when they don't switch pretty clearly implies that they'd be right 66% of the time when they DO switch... thus there's clearly a bug in my simulation in that situation.
The article offers another simple way to look at it: because we know Monty (or whoever, lol) is going to reveal a goat door, the contestant would be just as well off switching their guess before Monty shows what's behind any door...because switching essentially means trading one door (the original guess) for both of the other two. And you clearly have a better chance with two doors than with just one.
Fun stuff.
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Post by rockyp on Sept 6, 2014 11:12:23 GMT -5
If the contestant can improve his/her chances of winning by switching once, it follows logically that switching a second time would further improve his/her chances of winning. By mathematical induction, it can be shown that as the number of times they switch their guess the probability of winning approaches 100%. The moral of this story is that high frequency trading is the only way to make money. --- Alternatively, you may wish to consider the case where there are two contestants. Contestant A selects Door #1. Contestant B selects Door #2. Monty opens Door #3 to reveal a goat. Right now, there are two unknown doors, Door #1 and Door #2. One has a car, one has a goat. If Contestant A can improve her chances of winning the car by switching to Door #2, can Contestant B also improve his chances of winning the car by switching to Door #1? --- Finally, what happens to the odds if one of the contestant already has a nice car but really, really wants a pet goat?
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Post by thekindaguyiyam on Sept 6, 2014 11:16:30 GMT -5
You guys are killing me with all this door talk. For months my wife has been trying to get me to replace the front door. I keep telling her, I really don't want to switch. In this case there is a 90% chance that I will have to switch doors. Hopefully, I can delay the switch for several years. It's hard to imagine how someone can get tired of a perfectly good front door. But then again, I never seem to get tired of listening to those damn conferences. For your anniversary or her birthday take her to lowes or home depot. Tell her; sweetheart at home we have door #1. What you see before you are 40 other doors and it's your gift from me to select a door #2 and a door #3. Then take her through the routine of what has been presented. When she rolls her eyes... that's when you get your butt quickly to the airport to go to Siberia. As an ancient philosopher once said; when one door closes another door opens. BTW.. the gift IS the Option to pick a door; not to necessarily buy one.
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Post by BlueCat on Sept 6, 2014 12:29:13 GMT -5
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Post by jpg on Sept 6, 2014 13:15:43 GMT -5
Baba,
Does that mean I get a goat or get to learn how this makes us become better traders?
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Post by paradigm on Sept 6, 2014 14:51:07 GMT -5
I thought 4balance and brentie's analylsis is correct. We're left with 50/50 odds, so the good prize is just as likely to be behind the door already picked as behind the other one. There's no benefit, probability-wise, of switching. Just to be extra sure, I wrote and ran a C++ simulation just now that confirms our thesis However, this result is at odds with liane's article, so that's a conundrum. I'm attaching the program in case anyone's a software nerd and wants to try it...maybe someone can tell me what I did wrong, if in fact the simulation is not really simulating the circumstances correctly... BD, I believe the bug in your program is when you assign which (incorrect) door to show. You are randomly picking between the two incorrect doors, which makes it possible you are showing the door the person originally chose (when they choose an incorrect door).
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