LOL - Math Edition
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Category: News And Views
Forum Name: Thoughts and Opinions
Forum Description: Got something you need to say?
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Topic: LOL - Math Edition
Posted By: Darur
Subject: LOL - Math Edition
Date Posted: 08 December 2007 at 1:22am
So tonight my family wanted to watch "Deal or No Deal", for those not familiar its another mindless game show where everything is almost always pure luck. Tonight's show was interesting though with a classic Monty Hall problem.
For those that didn't see it, the contestant went through all the cases down to hers and the last on the board. The two possible values were $200,000 and $1,000. The host offered her a chance to switch cases which she declined. Had she accepted, she would have had a 96% chance of selecting the $200,000 case with her original odds being 1/26 of picking the $200,000 case first and 25/26 odds with it being in the last case opened.
Proof of how valuable math can be 
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Replies:
Posted By: carl_the_sniper
Date Posted: 08 December 2007 at 1:37am
So how do you figure 96%?
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Posted By: pntbl freak
Date Posted: 08 December 2007 at 1:40am
Since you have two cases left, wouldnt it be a 50% chance?
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Posted By: Darur
Date Posted: 08 December 2007 at 1:47am
pntbl freak wrote:
Since you have two cases left, wouldnt it be a 50% chance?
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Nope.
Initially, the odds of selecting the $200,000 case is 1/26. That means there is a 25/26 chance of the $200,000 case being in the cases to be opened. As the player eliminates cases, the odds remain 1/26 and 25/26 for the player and the unopened cases respectively regardless of how many cases are open. In the end the case the player selected still has a 1/26 chance of holding $200,000; meanwhile, the last unopened case has a 25/26 chance of holding $200,000 because it keeps the initial odds of all the unopened cases.
25/26 = 96.15% 1/26 = 3.85%
http://en.wikipedia.org/wiki/Monty_Hall_problem - Wikipedia for more info
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Posted By: carl_the_sniper
Date Posted: 08 December 2007 at 1:52am
Aww crap I knew that sounded familiar.
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Posted By: RoboCop
Date Posted: 08 December 2007 at 6:42am
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So did she not choose correctly or did she take the deal?
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Posted By: Skillet42565
Date Posted: 08 December 2007 at 9:44am
Posted By: .357 Magnum
Date Posted: 08 December 2007 at 9:49am
Darur wrote:
game show where everything is almost always pure luck.
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It is pure luck.
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Posted By: Darur
Date Posted: 08 December 2007 at 6:47pm
RoboCop, she decided to go with what was in her case, which was $1,000, hoping for the 4% chance she had the money. A bad move.
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Posted By: oreomann33
Date Posted: 08 December 2007 at 6:50pm
.357 Magnum wrote:
Darur wrote:
game show where everything is almost always pure luck.
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It is pure luck.
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No it's not, math comes into play like Darur said. DUH
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Posted By: pntbl freak
Date Posted: 08 December 2007 at 7:17pm
Maybe Im still missing something.
She selected the $1,000 dollar case to begin with. Once she got down to the final two cases she had a chance to switch giving her a 96% chance that the case contains 200,000. However she doesnt know what she picked.
Lets say she picked the 200,000 case to begin with. She gets down to the final two cases again and the 1,000 case is still out there. She would then have a 96% that the case contained 1,000. She would have no clue if its a good move to switch or not.
Since there are two cases left I still think she would have a 50% chance.
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Posted By: FROG MAN
Date Posted: 08 December 2007 at 7:30pm
pntbl freak wrote:
Maybe Im still missing something.
She selected the $1,000 dollar case to begin with. Once she got down to the final two cases she had a chance to switch giving her a 96% chance that the case contains 200,000. However she doesnt know what she picked.
Lets say she picked the 200,000 case to begin with. She gets down to the final two cases again and the 1,000 case is still out there. She would then have a 96% that the case contained 1,000. She would have no clue if its a good move to switch or not.
Since there are two cases left I still think she would have a 50% chance.
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its how you look at it.
if I roll a dice there is a 1/6 chance i will get 1, if I roll it now 10 times, and 1 has yet to come up, the odds of rolling it have increased. tho if you ignore those previous 10 rolls, it is still 1/6 chance.
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Posted By: pntbl freak
Date Posted: 08 December 2007 at 7:33pm
FROG MAN wrote:
pntbl freak wrote:
Maybe Im still missing something.
She selected the $1,000 dollar case to begin with. Once she got down to the final two cases she had a chance to switch giving her a 96% chance that the case contains 200,000. However she doesnt know what she picked.
Lets say she picked the 200,000 case to begin with. She gets down to the final two cases again and the 1,000 case is still out there. She would then have a 96% that the case contained 1,000. She would have no clue if its a good move to switch or not.
Since there are two cases left I still think she would have a 50% chance.
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its how you look at it.
if I roll a dice there is a 1/6 chance i will get 1, if I roll it now 10 times, and 1 has yet to come up, the odds of rolling it have increased. tho if you ignore those previous 10 rolls, it is still 1/6 chance.
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Maybe I just feel like arguing right now and Im just looking for an error, but I still dont see how this changes anything.
Just because a 1 has yet to be rolled doesnt change the probability that you will roll at 1 on your next roll. The dice doesnt care what you rolled previously and you will always have a 1/6 chance.
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Posted By: FROG MAN
Date Posted: 08 December 2007 at 8:27pm
pntbl freak wrote:
FROG MAN wrote:
pntbl freak wrote:
Maybe Im still missing something.
She selected the $1,000 dollar case to begin with. Once she got down to the final two cases she had a chance to switch giving her a 96% chance that the case contains 200,000. However she doesnt know what she picked.
Lets say she picked the 200,000 case to begin with. She gets down to the final two cases again and the 1,000 case is still out there. She would then have a 96% that the case contained 1,000. She would have no clue if its a good move to switch or not.
Since there are two cases left I still think she would have a 50% chance.
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its how you look at it.
if I roll a dice there is a 1/6 chance i will get 1, if I roll it now 10 times, and 1 has yet to come up, the odds of rolling it have increased. tho if you ignore those previous 10 rolls, it is still 1/6 chance.
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Maybe I just feel like arguing right now and Im just looking for an error, but I still dont see how this changes anything.
Just because a 1 has yet to be rolled doesnt change the probability that you will roll at 1 on your next roll. The dice doesnt care what you rolled previously and you will always have a 1/6 chance.
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the probability of throwing a dice 6 times and not getting a 1 is .335, the probability of getting at least one 1, is .665
with a mutiple independent out comes, you can still find a single probability that is more likley. Though if you treat each throw as an independent, it stays at 1/6 chance to get a 1.
ps. im agreeing with you.
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Posted By: Darur
Date Posted: 08 December 2007 at 9:46pm
pntbl freak wrote:
Maybe Im still missing something.
She selected the $1,000 dollar case to begin with. Once she got down to the final two cases she had a chance to switch giving her a 96% chance that the case contains 200,000. However she doesnt know what she picked.
Lets say she picked the 200,000 case to begin with. She gets down to the final two cases again and the 1,000 case is still out there. She would then have a 96% that the case contained 1,000. She would have no clue if its a good move to switch or not.
Since there are two cases left I still think she would have a 50% chance.
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Your trying to look at the odds from all angles, which doesn't work. We're looking for the specific odds of the contestant having picked $200,000. In the second to last round, there was $10, $1000 and $200,000 in play. Because she was shooting for $200,000, we consider the odds of her selecting that case, which is 1/26. We lump together all the other cases as n/26 (n for the number of cases below $200,000, I don't remember how many this is). The odds of her selecting one of these lower cases however is much much higher.
Regardless, when we look at her situation at the last round, the odds of her having the $200,000 case remain at 1/26, and the odds the last case to open is $200,000 remains 25/26.
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Posted By: Bolt3
Date Posted: 08 December 2007 at 10:45pm
It's called conditional probability and its a real thing.
It's looking at the situation from a mathematical standpoint.
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Posted By: Dye Playa
Date Posted: 09 December 2007 at 12:13am
My brain hurts.
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Posted By: RoboCop
Date Posted: 09 December 2007 at 12:17am
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It's an easy concept of the math invovled. I'm surprised teachers haven't taught you these things before.
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Posted By: Dye Playa
Date Posted: 09 December 2007 at 12:22am
RoboCop wrote:
It's an easy concept of the math invovled. I'm surprised teachers haven't taught you these things before. |
They probably did (no pun intended), but whether or not I was paying attention at the time is a different story.
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Posted By: carl_the_sniper
Date Posted: 09 December 2007 at 12:24am
I understand it but I could never explain it.
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Posted By: Glassjaw
Date Posted: 09 December 2007 at 12:30am
I don't understand it, but could probably explain it.
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Posted By: folder504
Date Posted: 09 December 2007 at 12:33am
There's a 1/26 chance that the initial guess was correct. Then, it's like "Here, these 24 are wrong." Since you likely didn't guess correctly initially, chances are good (25/26) that switching will win.
http://en.wikipedia.org/wiki/Monty_Hall_problem - http://en.wikipedia.org/wiki/Monty_Hall_problem
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Posted By: Susan Storm
Date Posted: 09 December 2007 at 11:02am
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I am not sure that this is a correct application of Monty Hall...
Because in Monty Hall, two of the choices are interchangeable, and one is different. One of the interchangeable choices is eliminated intentionally, leading to the favored odds for switching.
Here, the cases are all "interchangeable", because they are eliminated randomly rather than intentionally.
If Monty Hall chose a door randomly to open, rather than always opening an empty door, then the probabilities would not apply, and the odds would be 50/50 on the switch.
I am not terribly familiar with Deal/No Deal, but based on what I have seen, Monty Hall does not apply because of the randomness of case elimination.
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Posted By: Snake6
Date Posted: 09 December 2007 at 11:42am
I'm Lost?
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Posted By: The Guy
Date Posted: 09 December 2007 at 12:25pm
Because the contestant is the one that chooses which cases to eliminate, Monty Hall does not apply here.
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Posted By: Darur
Date Posted: 09 December 2007 at 3:17pm
Susan Storm wrote:
I am not sure that this is a correct application of Monty Hall...
Because in Monty Hall, two of the choices are interchangeable, and one is different. One of the interchangeable choices is eliminated intentionally, leading to the favored odds for switching.
Here, the cases are all "interchangeable", because they are eliminated randomly rather than intentionally.
If Monty Hall chose a door randomly to open, rather than always opening an empty door, then the probabilities would not apply, and the odds would be 50/50 on the switch.
I am not terribly familiar with Deal/No Deal, but based on what I have seen, Monty Hall does not apply because of the randomness of case elimination. |
For the most part in the game show, Monty Hall does not apply, so you would be correct. However, at the third or even fourth to last picking in this particular case, there are 3 values which we can consider interchangeable because they are all below $200,000 and $200,000 itself is not interchangeable.
By the last round, dumb luck had left the $200,000 in play with the rest of the interchangeable amounts below $200,000 eliminated. At this point in time, the odds of the $200,000 case being selected is 1/26, and the odds of having selected an interchangeable value below $200,000 is n/26 where n > 1 (I believe its 20, but don't quote me on that).
Its an unorthodox example, but the probabilities still apply at the last case selection, which is all I'm talking about it.
The Guy wrote:
Because the contestant is the one that chooses which cases to eliminate, Monty Hall does not apply here.
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Not quite. Monty Hall only requires that an interchangeable value and a prize remain, one of which selected by the contestant initially with the option for the contestant to switch. For the sake of ensuring this is true every time, we allow the host to remove all options except for one unopened case/door and one in the contestant's hand.
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Posted By: Susan Storm
Date Posted: 09 December 2007 at 4:33pm
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Darur wrote:
Monty Hall only requires that an interchangeable value and a prize remain, one of which selected by the contestant initially with the option for the contestant to switch.
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Not true. Monty Hall requires that the eliminated option is always a null (not the prize in question). This requires that the elimination be non-random.
In the classic example, Monty Hall opens a door; not the contestant. And Monty Hall ALWAYS opens an empty door - that's the way the game is set up.
And that is required for the math to work. If alternate choices are randomly eliminated, then Monty Hall does not apply.
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Posted By: RoboCop
Date Posted: 09 December 2007 at 4:48pm
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The math behind Monty Hall is basically the same.
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Posted By: Susan Storm
Date Posted: 09 December 2007 at 6:41pm
Not the same. An essential element (non-random elimination) is missing.
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Posted By: Darur
Date Posted: 09 December 2007 at 9:28pm
Susan Storm wrote:
Not true. Monty Hall requires that the eliminated option is always a null (not the prize in question). This requires that the elimination be non-random. |
The example applies because the eliminated options are still null. They are values which are less then $200,000, which is the prize.
The only reason non-random elimination is needed is so the eliminated options are not the prize, which is precisely what you said. The systematic elimination has no bearing on the final odds so long as an interchangeable value and the prize remains. This is true in this particular event on the show, so Monty Hall is relevant.
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Posted By: Jack Carver
Date Posted: 09 December 2007 at 9:36pm
Ok I looked at the wiki and I think it does apply.
The game begins with the contestant picking a case which he or she
believes will have the highest value. During the rest of the game, the
contestant selects the rest of the cases one at a time for rejection,
the value of each case being revealed after it is selected.
So she picked one, then picked 24 more to open, none of them were the 200K case, then she had her decision to make and she was a noob. Right?
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Posted By: Man Bites Dog
Date Posted: 09 December 2007 at 9:58pm
I can honestly say I don't understand a damn thing in this thread.
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Posted By: RoboCop
Date Posted: 09 December 2007 at 10:03pm
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Susan Storm wrote:
Not the same. An essential element (non-random elimination) is missing. | Does it really matter? The other cases are being eliminated either way.
The same concept of switching cases or doors will give you a higher percentage rate of getting the better deal.
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Posted By: Susan Storm
Date Posted: 09 December 2007 at 10:05pm
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Darur wrote:
The only reason non-random elimination is needed is so the eliminated options are not the prize, which is precisely what you said. The systematic elimination has no bearing on the final odds so long as an interchangeable value and the prize remains. This is true in this particular event on the show, so Monty Hall is relevant.
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No. You are wrong. If the elimination is random, then the prize will not always remain. If the prize does not always remain, then the entire paradigm is off. Monty Hall does not apply.
But don't take my word for it. Should be fairly simple to program a Deal/No Deal simulator. And unlike a Monty Hall simulator, with a 26-1 odds prediction, it shouldn't take a very large n to see significant results.
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Posted By: Darur
Date Posted: 09 December 2007 at 10:12pm
Susan Storm wrote:
No. You are wrong. If the elimination is random, then the prize will not always remain. If the prize does not always remain, then the entire paradigm is off. Monty Hall does not apply. Yet the prize does remain here. We've seen the end results. We know that the contestant will not eliminate the prize, its kept in play. Essentially luck plays the role of Monty Hall. In the grand scheme of things it does not apply. Your nitpicking pointless details.
But don't take my word for it. Should be fairly simple to program a Deal/No Deal simulator. And unlike a Monty Hall simulator, with a 26-1 odds prediction, it shouldn't take a very large n to see significant results. Your completely missing the point here. The game itself is NOT a Monty Hall problem. This particular event is a Monty Hall problem. I would love to write up a simulation but I have no experience in programing.
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Please, prove to me mathematically what effect randomness has here? We've seen the end results, we KNOW the contestant will be left with a choice between an interchangeable value and a prize, one of which was selected from 26 other possibilities and the other was narrowed down to as the only alternative.
We are simply skipping the part where Monty Hall eliminates the cases and doing it ourselves. The odds remain.
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Posted By: Susan Storm
Date Posted: 09 December 2007 at 10:18pm
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I will go ahead and write up a simulator when I have a moment.
In the meantime, consider this: How is randomly eliminating 24 of the 25 cases different from simply selecting a case at random right off the bat?
(There is no difference)
Would you feel that MH applied if instead of eliminating cases, we just skipped to the end by picking a case and asking if you wanted to switch with that one?
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Posted By: Jack Carver
Date Posted: 09 December 2007 at 10:19pm
It's different because we saw that in all 24 eliminated, the prizes were NOT the big one. The contents are revealed after chosen to be eliminated.
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Posted By: Darur
Date Posted: 09 December 2007 at 10:35pm
Jack Carver wrote:
It's different because we saw that in all 24 eliminated, the prizes were NOT the big one. The contents are revealed after chosen to be eliminated.
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Jack hit the nail on the head.
This is one of those "Ends justify the means" problems. The values which were not the prize were eliminated one by one, leaving the precise end results needed for Monty Hall to apply. I don't think the contestant thinking "this case doesn't hold $200,000" would be any different then thinking "I HOPE this case doesn't hold $200,000" while she selected them, so long as she eliminated the cases which were interchangeable.
We are not just skipping to the end here though, we are selecting, eliminating all but one and then leaving the chance to switch. The contestant chose a case when the odds were 1/26 to get $200,000 and wound up with a case with the original 1/26 odds and the chance to switch to a case with 25/26 odds of holding the $200,000. She eliminated all the interchangeable values by luck, but that is not relevant to any simulation.
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Posted By: Susan Storm
Date Posted: 09 December 2007 at 11:40pm
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My simulator below. Please feel free to check the code. Javascript - just load it into your browser.
It randomly assigns values to cases, then asks you to pick a case. It then randomly eliminates 24 cases. It then displays the value of the case you selected, and the value of the remaining case.
Go nuts.
<html> <body>
<script type="text/javascript">
//This establishes the values for the cases
var vals = new Array(); vals[0] = 1; vals[1] = 5; vals[2] = 10; vals[3] = 20; vals[4] = 30; vals[5] = 40; vals[6] = 50; vals[7] = 75; vals[8] = 100; vals[9] = 250; vals[10] = 500; vals[11] = 750; vals[12] = 1000; vals[13] = 2500; vals[14] = 5000; vals[15] = 7500; vals[16] = 10000; vals[17] = 25000; vals[18] = 40000; vals[19] = 50000; vals[20] = 75000; vals[21] = 100000; vals[22] = 250000; vals[23] = 500000; vals[24] = 750000; vals[25] = 1000000;
//This randomly assigns values to cases var cases = new Array(); for (i=0;i<26;i++) { cases=vals[Math.round(25*Math.random())]; }
//This asks the user to select a case
var selection = prompt("Please selected a case (1-26)",0); document.write("You selected Case #"+selection + "<br />");
//Now we start eliminating cases //24 cases will be eliminated
var j=1; for (i=0;i<cases.length-2;i++) { j=1; while (j=1) { r = Math.round(25*Math.random()); if (cases[r]>0 && r!=selection-1) { break; } } cases[r]=0; }
//Now identify the remaining case
for (i=0;i<cases.length;i++) { if (cases>0 && i!=selection-1) { var finalcase=i; break; } }
//Now lets print out the results
document.write("<br />"); document.write("<br />"); document.write("Your case contained $"+cases[selection-1]+"<br />"); document.write("The remaining case contained $"+cases[finalcase]+"<br />"); </script>
</body> </html>
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Posted By: Susan Storm
Date Posted: 09 December 2007 at 11:45pm
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Jack Carver wrote:
So she picked one, then picked 24 more to open, none of them were the 200K case, then she had her decision to make and she was a noob. Right?
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No. Because she didn't know up front that the 200k case would survive to the end.
In Monty Hall, the prize is NEVER eliminated. It is ALWAYS either the remaining option or the original selection. This is an essential element.
You guys are all wrong. But don't take my word for it - try my simulator. Paste into browser, and keep hitting reload. If Darur is right, then the remaining case should be greater in value than the selected case 25 times out of 26. If I am right, then it'll be about 50/50.
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Posted By: Darur
Date Posted: 10 December 2007 at 12:12am
Susan, your script is flawed. Its assuming the $200,000 case is no longer in play in the end. In order for it to work properly, it needs to mimic the exact circumstances. I've said time and time again, this situation alone, what happened with the lady is a Monty Hall problem. The game show itself is NOT a Monty Hall problem.
Your script would have to have the contestant select from n + 1 number of cases, where n is the number of cases below $200,000 and the 1 is for the $200,000 case. It would then have to keep the $200,000 case in play until the end along with one of the cases below $200,000 because thats precisely what happened.
I'm sorry, but your completely wrong here.
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Posted By: Susan Storm
Date Posted: 10 December 2007 at 12:20am
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Sure - but your scenario isn't Deal or No Deal, because there is no guarantee that the 200k case will stay in play.
If there were such a guarantee, then MH would apply, because ELIMINATION WOULD NOT BE RANDOM.
But the actual eliminations in the game are random, unlike the scenario you are describing.
EDIT - I can easily tweak the script to never eliminate the 200k case, in which case Monty will apply. But that script will bear no relation to Deal or No Deal. Which reinforces my point: Non-random elimination is required for Monty to apply.
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Posted By: Gatyr
Date Posted: 10 December 2007 at 12:32am
Susan Storm wrote:
Sure - but your scenario isn't Deal or No Deal, because there is no guarantee that the 200k case will stay in play.
If there were such a guarantee, then MH would apply, because ELIMINATION WOULD NOT BE RANDOM.
But the actual eliminations in the game are random, unlike the scenario you are describing. |
I think you are looking at it as the 200K case is specifically what the
woman wanted from the beginning, which isn't the case (lolpun?). the
case she is going for at the end could be any case, but in this one
specific instance, it is the 200k.
Starting over and fast forwarding to the end where there are two cases left, one with a
high dollar amount, one with a low dollar amount, the MH problem comes
into play because there is a 25 in 26 chance she did not pick the high
dollar amount case, with the 1 in 26 odds that she did.
The randomness of the case-picking does nothing to eliminate the higher
dollar amount case that she goes for in the end, it just eliminates
specific choices and narrows down what she will be hoping for during
the final part of the show.
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Posted By: Susan Storm
Date Posted: 10 December 2007 at 12:37am
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You are correctly describing exactly what my script does, Gatyr. Random elimination, leaving two cases - one higher value than the other.
And it shows that Monty does not apply. The only way we get the 25-1 results is to change to non-random elimination.
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Posted By: Darur
Date Posted: 10 December 2007 at 12:52am
Susan Storm wrote:
Sure - but your scenario isn't Deal or No Deal, because there is no guarantee that the 200k case will stay in play.
If there were such a guarantee, then MH would apply, because ELIMINATION WOULD NOT BE RANDOM.
But the actual eliminations in the game are random, unlike the scenario you are describing.
EDIT - I can easily tweak the script to never eliminate the 200k case, in which case Monty will apply. But that script will bear no relation to Deal or No Deal. Which reinforces my point: Non-random elimination is required for Monty to apply. |
But the eliminations aren't random in this case!
We know the $200,000 case and $1,000 case are in play at the end. We are just looking at the odds of switching cases yielding a better result at the end. In the game itself, yes, the cases are eliminated randomly. In this particular case, everything except for the $200,000 case and the $1,000 case were eliminated by chance, I'll give you that. However, the end result is identical to if the cases were eliminated by the host selecting them.
All I'm talking about is the odds of the final outcome in this particular event, how we get there doesn't matter so long as the $200,000 case and a case of lesser value remain in play. We know that this will be the final outcome, because thats what happened on the show. The question is simply would it have been better for her to switch or not.
------------- Real Men play Tuba
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PH33R TEH 1337 Dwarf!
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Posted By: Gatyr
Date Posted: 10 December 2007 at 12:59am
I re-read the thread and I'm still not seeing how the odds aren't what Darur is saying they are, or how randomness matters after 24 of the 26 cases are opened.
At the end of the segment, when there are two cases unopened (the one the contestant picked and the last case remaining of the cases she didn't pick or open yet), is there not a 1 in 26 chance that the woman picked the higher of the two cases, and a 25 in 26 chance that she picked something other than the higher dollar amount?
Not worrying about a specific dollar amount, only the value relationship between the two, how can the odds of her picking a case with a dollar amount different than the greater of the two amounts not be 25 - 1?
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Posted By: Susan Storm
Date Posted: 10 December 2007 at 12:59am
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Seriously, Darur, you are making no sense.
Elimination is either random or non-random.
Whether we only consider the cases where some particular case happens to be the remaining case is irrelevant. If the elimination process, a priori, was random, then Monty does not apply. If we knew ahead of time that case X would remain, then elimination was not random.
I can also tweak the script (a little more complicated) to restart the elimination process if the final two cases do not include 200k. Thus we will only "consider" cases where 200k is randomly not eliminated. But the result will be the same: 50/50
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: Susan Storm
Date Posted: 10 December 2007 at 1:02am
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Gatyr wrote:
At the end of the segment, when there are two cases unopened (the one the contestant picked and the last case remaining of the cases she didn't pick or open yet), is there not a 1 in 26 chance that the woman picked the higher of the two cases, and a 25 in 26 chance that she picked something other than the higher dollar amount?
|
Because random elimination is the same as just picking a case right away.
So look at it this way: You pick a case, and I pick a case. What are the odds that your case has a higher value than mine? That's right: 50/50
Eliminating the other cases one by one (RANDOMLY) is just for show.
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: Darur
Date Posted: 10 December 2007 at 1:47am
Alright, I'm going to try and explain everything again.
The contestant came on the show, selected her case and began to eliminate cases. Before long the highest value on the board was $200,000 with a number of lower values including $10 and $1,000. By sheer luck, she managed to eliminate all the cases except for $200,000 and $1,000.
At this point in time, the host offered her a chance to exchange her case with the last remaining case. She declined and won $1,000, the $200,000 case happened to be the other case.
The odds of the $200,000 being in her case were 1/26. The odds of a smaller value being in her case were n/26 where n is the number of cases with a lower value then $200,000.
If you like, we can simplify the odds even more and redefine the odds:
n/26 chance of her case holding < $200,000, where n is the number of cases with less then $200,000 m/26 chance of her case holding > $200,000 where m is the number of cases with more then $200,000 n > m
Initially the odds of her selecting a case greater then $200,000 is m/26 and the odds of her selecting a case less then $200,000 is n/26.
As the game progresses, we eliminate all cases of value > $200,000 and most cases of value < $200,000 and the $200,000.
In the final round, the odds she picked a case with value > $200,000 is still m/26, and (m/26) < (n/26). Therefore at this point in time, the odds of her having the $1,000 case on her are much higher then the odds of her having the $200,000. At this point, it is to her advantage to switch.
The reason random elimination doesn't matter is we are simply trying to find the odds at this particular moment. The cases were eliminated one by one, leaving only the $200,000 and $1,000 in play. For our purposes we can pretend they were eliminated purposely to leave one value below $200,000 and $200,000. The odds of her originally selecting a case with a higher value is significantly lower then the odds of her selecting a case with a lower value.
A simulator for this event has to produce this final outcome. The $200,000 case is in play and a case of lesser value is in play. There are more cases below $200,000 then the lone $200,000 case. What are the odds the $200,000 is in her case and the odds $1,000 is in her case.
Honestly, I cannot spell it out clearer then that. It doesn't matter if the cases were eliminated randomly or not because the end result is a value of $200,000 and a value less then $200,000. The initial odds of selecting a value less then $200,000 were greater then selecting $200,000 or any larger value.
------------- Real Men play Tuba
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Posted By: Susan Storm
Date Posted: 10 December 2007 at 2:13am
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You are wrong, Darur. Dead wrong. Non-random selection is required for MH. You are describing random selection.
Just because we non-randomly focus on a particular subset of situations (where 200k happened to survive) doesn't change the fact the selection was random. And random selection means no Monty.
Would you like me to tweak my script to focus on that subset? If that doesn't do it, please describe what you would like the simulator to do.
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: Darur
Date Posted: 10 December 2007 at 2:18am
Yes, I know non-random selection is needed, but the point remains. We are left with two specific cases, so the selection can be considered non-random because every event requires the same out come.
I'll try to explain this tomorrow better.
------------- Real Men play Tuba
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PH33R TEH 1337 Dwarf!
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Posted By: Susan Storm
Date Posted: 10 December 2007 at 2:30am
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Except that you aren't changing the selection process - you are simply cherrypicking the trials that yielded the results you liked. That isn't non-random selection.
Non-randomness has to PRECEDE the selection.
Here is what you are doing: roll a die a bunch of times. Write down whenever you roll a six, and ignore all the other rolls. After a while, you declare the dice loaded, because of all the rolls were sixes.
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: Ticalxx421
Date Posted: 10 December 2007 at 2:40am
...WoW
...simply mind boggling
------------- [IMG]http://i14.tinypic.com/73e0l8j.jpg">
Represent!
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Posted By: pb125
Date Posted: 10 December 2007 at 2:52am
Darur, think about it. Don't think hard. Barely think.
It's random.
End of story.
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Posted By: High Voltage
Date Posted: 10 December 2007 at 3:02am
pb125 wrote:
Darur, think about it. Don't think hard. Barely think.
It's random.
End of story.
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Posted By: Susan Storm
Date Posted: 10 December 2007 at 3:21am
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Ok - I think I figured out Darur's perspective.
Take the following scenario.
1. I randomly pick a case.
2. I randomly designate a "target" - say, 200k
3. We start randomly eliminating cases.
4. If 200k gets eliminated, we start over (or turn off the TV until the next episode)
5. Therefore, we are only concerned with cases where 200k remains one of the final two cases.
And in this scenario, YES - there is only a 1/26 chance that I hold the 200k case. Absolutely.
BUT - and here is my point - this is irrelevant, because you are merely defining yourself into truth. You are doing my deal with the dice.
This works equally well with a target of 1, or 25k, or whatever. Once you eliminate all the situations where you are wrong, you are always right.
You aren't changing the process - you are merely choosing to select your data after the fact.
In the more general scenario of "there are two cases left, should I switch", the answer is 50/50.
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: scotchyscotch
Date Posted: 10 December 2007 at 5:12am
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Its either $200,000 or it's not 50/50. Whats the diffference between the two boxes besides there value? same odds don't argue just accept it.
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Posted By: carl_the_sniper
Date Posted: 10 December 2007 at 11:23am
scotchyscotch wrote:
Its either $200,000 or it's not 50/50. Whats the
diffference between the two boxes besides there value? same odds don't
argue just accept it. |
No
If you had started from that point with 2 cases than you would be correct.
But with 24 cases, the odds change when you get down to the last two
because of the potential for an odd to be fulfilled is increased.
I really can't explain it well.
------------- <just say no to unnecessarily sexualized sigs>
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Posted By: Susan Storm
Date Posted: 10 December 2007 at 11:45am
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You are still wrong, Carl.
You (and Darur) are committing a sampling bias. If you take the universe of D/ND endgames (2 cases left), and select out only the ones where 200k is one of the remaining cases, then yes - only 1/26 chance that you are holding the 200k.
BUT consider this: Take the same universe of D/ND endgames, and this time select out only the ones where $1 is one of the remaining cases, then there is also only a 1/26 chance that you are holding $1. EVEN IF THE OTHER NUMBER IS 200K.
In other words, in the situations where the final two cases are $1 and $200k, the probabilities that you hold one or the other - in the very same actual situations - will vary depending solely upon what you were selecting for. The $1/200k subset will always be included, whether you are sampling for $1 or 200k, yet the probabilities will shift.
It is all about what you are selecting for. By selecting out only the sample that meets your critieria, you have a sample that 100% meets your critieria. That's fine, but you cannot then conclude that ALL cases meet your criteria.
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: procarbinefreak
Date Posted: 10 December 2007 at 12:00pm
guys... i think we're forgetting the most important part of said episode...
all the girls holding the suitcases were in bikinis.
also, they need more asian girls.
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Posted By: Panda Man
Date Posted: 10 December 2007 at 12:09pm
counting Cards anyone?
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Posted By: Gatyr
Date Posted: 10 December 2007 at 2:35pm
Susan Storm wrote:
4. If 200k gets eliminated, we start over (or turn off the TV until the next episode)
5. Therefore, we are only concerned with cases where 200k remains one of the final two cases. |
You're still being too selective with the data, and I think this is where the disagreement is stemming from. For all intensive purposes regarding Monty Hall type situations, we are only worried about the relationship between cases, not the dollar amount, and equate the greater of the two amounts to the door with the prize behind it, and the lesser amount to the door with nothing behind it.
All that matters is that at the end, one of the cases has a greater amount of money in it, and that there is a 1 in 26 chance the person picked that case, with a 25 in 26 chance that they didn't (and through the random elimination, we know that there is a 25 in 26 chance they picked something lower because all of the higher amounts have been eliminated). Specifics don't matter.
I don't really understand what we are missing in eachother's arguments...at all. I still think you are hanging up on the fact that you believe we must know the case we will be shooting for at the end at the begining of the show, which isn't the case.
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Posted By: Jack Carver
Date Posted: 10 December 2007 at 2:44pm
Wait, so what if we REALLY wanted that case with $1000? Are the odds that we have the 1k case 1/26 and the odds that we have the 200k is 25/26 just because that's the one we don't want?
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Posted By: Gatyr
Date Posted: 10 December 2007 at 3:10pm
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Jack Carver wrote:
Wait, so what if we REALLY wanted that case with $1000? Are the odds that we have the 1k case 1/26 and the odds that we have the 200k is 25/26 just because that's the one we don't want? |
EDIT: scratch that, I think you just said what Susan has been trying to say and it made more sense.
I'll think about it after I'm done studying/taking my final.
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Posted By: Susan Storm
Date Posted: 10 December 2007 at 3:20pm
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JC did in fact just restate my point, using far fewer words.
Well done. :)
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: Funky
Date Posted: 10 December 2007 at 3:51pm
Hah, way to be Jack.
And yes, Susan is right. You don't need to have taken a few university statistics classes to see it (but they help).
MH does not apply and wow, Susan, nice on the javascript.
-------------
"Don't you hate pants?"
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Posted By: Darur
Date Posted: 10 December 2007 at 9:44pm
Susan, I spoke with my AP Calc teacher about the problem and he agrees that MH does indeed apply.
Your trying to apply the particular odds to multiple events when in truth I'm just looking at the lone odds of this one event.
I'll see if I can work out a proof for you.
------------- Real Men play Tuba
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PH33R TEH 1337 Dwarf!
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Posted By: Funky
Date Posted: 10 December 2007 at 10:27pm
http://en.wikibooks.org/wiki/Transwiki:Deal_or_No_Deal_game_strategy - http://en.wikibooks.org/wiki/Transwiki:Deal_or_No_Deal_game_ strategy
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"Don't you hate pants?"
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Posted By: carl_the_sniper
Date Posted: 10 December 2007 at 10:52pm
Susan Storm wrote:
You are still wrong, Carl.
You (and Darur) are committing a sampling bias. If you take the universe of D/ND endgames (2 cases left), and select out only the ones where 200k is one of the remaining cases, then yes - only 1/26 chance that you are holding the 200k.
BUT consider this: Take the same universe of D/ND endgames, and this time select out only the ones where $1 is one of the remaining cases, then there is also only a 1/26 chance that you are holding $1. EVEN IF THE OTHER NUMBER IS 200K.
In other words, in the situations where the final two cases are $1 and $200k, the probabilities that you hold one or the other - in the very same actual situations - will vary depending solely upon what you were selecting for. The $1/200k subset will always be included, whether you are sampling for $1 or 200k, yet the probabilities will shift.
It is all about what you are selecting for. By selecting out only the sample that meets your critieria, you have a sample that 100% meets your critieria. That's fine, but you cannot then conclude that ALL cases meet your criteria. |
Still wrong?
------------- <just say no to unnecessarily sexualized sigs>
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Posted By: pntbl freak
Date Posted: 10 December 2007 at 10:55pm
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When only three cases remain, Deal or No Deal might seem like a version of the http://en.wikipedia.org/wiki/Monty_Hall_problem" class="extiw" title="w:Monty_Hall_problem - Monty Hall problem . Consider a Deal or No Deal
game with three cases (similar to the three doors in the Monty Hall
problem). The contestant has one case. Then, one of the two other cases
is opened. Finally, the contestant is given the option to trade his or
her case for the one unopened case remaining.
The Monty Hall problem gives the contestant a 2/3 chance of winning
with a switch and a 1/3 chance of winning by keeping his or her case.
However, there is a critical difference between http://en.wikibooks.org/w/index.php?title=Let%27s_Make_a_Deal&action=edit" class="new" title="Let's Make a Deal - Let's Make a Deal and Deal or No Deal.
In the Monty Hall problem, the host has used his secret knowledge of
what lies behind each of the three doors to cause a bad choice to
always be revealed. This non-random selection of a bad choice is what
causes the difference in odds of winning between switching and not
switching on Let's Make a Deal.
In Deal or No Deal the odds behave exactly as you would
instinctively expect them to: 2 boxes with a fifty-fifty chance of the
top prize being in either of them.
So I was right to begin with. At least I still think so.
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Posted By: High Voltage
Date Posted: 11 December 2007 at 1:21am
So can we /thread now plox?
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Posted By: Kayback
Date Posted: 11 December 2007 at 6:49am
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I dunno about the rest of you, but i see a 1/2 as better odds than the original 1/26 chance, so would probably swap...
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Posted By: High Voltage
Date Posted: 11 December 2007 at 7:33am
Kayback wrote:
I dunno about the rest of you, but i see a 1/2 as better odds than the original 1/26 chance, so would probably swap... |
Did you even bother to read the thread?
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Posted By: Susan Storm
Date Posted: 11 December 2007 at 10:45am
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carl_the_sniper wrote:
Still wrong? |
Based on your previous post, yes: you are still wrong.
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: carl_the_sniper
Date Posted: 11 December 2007 at 11:15am
Susan Storm wrote:
carl_the_sniper wrote:
Still wrong? |
Based on your previous post, yes: you are still wrong. | Explain please.
------------- <just say no to unnecessarily sexualized sigs>
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Posted By: Susan Storm
Date Posted: 11 December 2007 at 11:18am
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Monty Hall still does not apply. The odds are 50/50, not 25-1. If you disagree, then you are wrong.
/explanation
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: carl_the_sniper
Date Posted: 11 December 2007 at 11:22am
Susan Storm wrote:
Monty Hall still does not apply. The odds are 50/50, not 25-1. If you disagree, then you are wrong.
/explanation |
Not what I asked.
I know that what I posted was wrong.
How at that post was I still wrong?
Also, I knew there was something fishy, when I tried to explain it. I tried to explain exactly why it wasen't 50/50, but I couldn't, it didn't work.
------------- <just say no to unnecessarily sexualized sigs>
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Posted By: Susan Storm
Date Posted: 11 December 2007 at 11:32am
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carl_the_sniper wrote:
If you had started from that point with 2 cases than you would be correct.
But with 24 cases, the odds change when you get down to the last two because of the potential for an odd to be fulfilled is increased.
|
In this post, Carl, you basically repeated what Darur has been saying - Monty Hall.
And you are still wrong, because the eliminations have been random. There is no distinction between "my case" and "the other cases".
What D/ND comes down to is randomly selecting two cases. First you select a case, then you spend the rest of the show selecting another case. But all you have done is select two cases at random.
And if you select two cases at random from a pile, why would you think that the case you selected last is somehow better than the case you selected second?
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: jerseypaint
Date Posted: 11 December 2007 at 12:21pm
Wait. For Monty Hall, doesn't the host needs to know where the duds are and reveal one of the duds to the person?
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Posted By: Susan Storm
Date Posted: 11 December 2007 at 12:26pm
|
jerseypaint wrote:
Wait. For Monty Hall, doesn't the host needs to know where the duds are and reveal one of the duds to the person? |
Exactly.
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: Kayback
Date Posted: 11 December 2007 at 12:59pm
High Voltage wrote:
Did you even bother to read the thread? | of course i did. So far there has been a long argument about if this fits the maths of the old Monty Hall problem or not.
Either way the chances of making a good choice are better at the last two cases than at the start.
I'm sure we'd all like to believe we are lucky enough to choose the right case off the bat, but thats most likely not the case. Neither side of the argument has proven it's a bad idea to swap.
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Posted By: Susan Storm
Date Posted: 11 December 2007 at 1:25pm
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It's not a "bad" idea to swap, it is just a waste of time. It's 50/50.
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: carl_the_sniper
Date Posted: 11 December 2007 at 3:55pm
Susan Storm wrote:
carl_the_sniper wrote:
If you had started from that point with 2 cases than you would be correct. But with 24 cases, the odds change when you get down to the last two because of the potential for an odd to be fulfilled is increased. |
In this post, Carl, you basically repeated what Darur has been saying - Monty Hall.
And you are still wrong, because the eliminations have been random. There is no distinction between "my case" and "the other cases".
What D/ND comes down to is randomly selecting two cases. First you select a case, then you spend the rest of the show selecting another case. But all you have done is select two cases at random.
And if you select two cases at random from a pile, why would you think that the case you selected last is somehow better than the case you selected second? |
So?
I said that I was never quite sure if it was actually monty hall.
You still haven't answered how I was "still wrong".
------------- <just say no to unnecessarily sexualized sigs>
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Posted By: Susan Storm
Date Posted: 11 December 2007 at 4:07pm
I'm not sure what you mean here, Carl... The post I quoted was what I was referring to. The statement you made in that post is incorrect.
------------- "No society can surely be flourishing and happy, of which the far greater part of the members are poor and miserable."
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Posted By: carl_the_sniper
Date Posted: 11 December 2007 at 5:10pm
Susan Storm wrote:
I'm not sure what you mean here, Carl... The post I quoted was what I was referring to. The statement you made in that post is incorrect. |
You said I was "still wrong"
When you should have just said that I was wrong.
Why did you say still?
------------- <just say no to unnecessarily sexualized sigs>
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Posted By: Gatyr
Date Posted: 11 December 2007 at 5:40pm
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Goodness Carl, you sure are annoying as hell.
He said "still" because there was more than one point in time where you were wrong, both in this thread and in others. Maybe not more than one post in this thread, but you were wrong at on point, posted at another point, and you were still wrong.
That that (you being wrong, caring, what Susan said, and what I said) for what its worth: nothing.
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Posted By: carl_the_sniper
Date Posted: 11 December 2007 at 9:43pm
Gatyr wrote:
Goodness Carl, you sure are annoying as hell.
He said "still" because there was more than one point in time where you were wrong, both in this thread and in others. Maybe not more than one post in this thread, but you were wrong at on point, posted at another point, and you were still wrong.
That that (you being wrong, caring, what Susan said, and what I said) for what its worth: nothing. |
Ok thank you.
------------- <just say no to unnecessarily sexualized sigs>
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Posted By: Kayback
Date Posted: 11 December 2007 at 9:55pm
Susan Storm wrote:
It's not a "bad" idea to swap, it is just a waste of time. It's 50/50.
|
Except it's not a waste of time. Like I said earlier, while we'd like to believe we have the skill, luck or 1337ness to select the high price case the first time, the odds were you didn't. So it's probably a good idea to change, even if you say it's a 1/2 chance. It's an improvement on the original 1/26.
You can choose to keep your case, because you said it's 1/2 chances it's the money bag, but its ALSO the original case you chose, which means you're betting a 1/2 chance on a 1/26 choice.
Smart?
KBK
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Posted By: jerseypaint
Date Posted: 11 December 2007 at 10:06pm
Kayback wrote:
Susan Storm wrote:
It's not a "bad" idea to swap, it is just a waste of time. It's 50/50.
|
Except it's not a waste of time. Like I said earlier, while we'd like to believe we have the skill, luck or 1337ness to select the high price case the first time, the odds were you didn't. So it's probably a good idea to change, even if you say it's a 1/2 chance. It's an improvement on the original 1/26.
You can choose to keep your case, because you said it's 1/2 chances it's the money bag, but its ALSO the original case you chose, which means you're betting a 1/2 chance on a 1/26 choice.
Smart?
KBK |
The 1/26 choice doesn't even matter. When with two cases left, whether you switch or not, the probability will always be 50/50
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Posted By: High Voltage
Date Posted: 12 December 2007 at 7:42am
Kayback wrote:
Susan Storm wrote:
It's not a "bad" idea to swap, it is just a waste of time. It's 50/50.
|
Except it's not a waste of time. Like I said earlier, while we'd like to believe we have the skill, luck or 1337ness to select the high price case the first time, the odds were you didn't. So it's probably a good idea to change, even if you say it's a 1/2 chance. It's an improvement on the original 1/26.
You can choose to keep your case, because you said it's 1/2 chances it's the money bag, but its ALSO the original case you chose, which means you're betting a 1/2 chance on a 1/26 choice.
Smart?
No.
KBK |
Ok, I must ask another question. Have you seen the show?
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Posted By: scotchyscotch
Date Posted: 12 December 2007 at 8:32am
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Smarter people than you developed this show.
I told you all to just accept it!
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Posted By: benttwig33
Date Posted: 12 December 2007 at 11:45pm
There is only one real way to solve this.
One of us has to go on the show, get down to the last cases, and switch and see how you do.
Go!
------------- Sig is WAY too big.
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Posted By: carl_the_sniper
Date Posted: 12 December 2007 at 11:59pm
benttwig33 wrote:
There is only one real way to solve this.
One of us has to go on the show, get down to the last cases, and switch and see how you do.
Go! |
I'll do it
Sign me up
------------- <just say no to unnecessarily sexualized sigs>
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