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LobowolfXXX
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On 2012-10-15 19:22, S2000magician wrote:
Quote:
On 2012-10-15 19:03, LobowolfXXX wrote:
Oh, I don't know...without the "which is why," it's just kind of cute trivia.

The interesting thing about all this is that it's pretty easy to verify empirically, yet still receives steadfast opposition.

Not unlike the two-children-one's-a-boy-what's-the-probability-that-the-other's-a-boy problem.


I heard that one as "A parent has two children who are identical twins. One is a boy. What is the probability that the other is a boy?". Stumped me for years.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley.

"...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us."
S2000magician
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Quote:
On 2012-10-16 12:20, LobowolfXXX wrote:
Quote:
On 2012-10-15 19:22, S2000magician wrote:
Quote:
On 2012-10-15 19:03, LobowolfXXX wrote:
Oh, I don't know...without the "which is why," it's just kind of cute trivia.

The interesting thing about all this is that it's pretty easy to verify empirically, yet still receives steadfast opposition.

Not unlike the two-children-one's-a-boy-what's-the-probability-that-the-other's-a-boy problem.

I heard that one as "A parent has two children who are identical twins. One is a boy. What is the probability that the other is a boy?". Stumped me for years.

Phone me about a tutoring session.
Adrian Fern
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The answer to that depends on how you arrive at the knowledge that one of the twins is a boy.

If there are two doors, and the twins each stand behind a door, and you open one door at random and find a boy, then the probability of the other door revealing a boy is 1/2. This is because there are only two possible outcomes on opening the second door: B or G.

If there are two doors, and the twins each stand behind a door, and each is given a button that will (when pressed either on its own or when the other is also pressed) light a single lamp dangling in front of you, and each is told to press the button only if they are a boy, and you see the lamp light up, then you know that one (at least) of the twins is a boy, and you also then know that the probability of the other being a boy (i.e. of both twins being boys) is 1/3. This is because there are three possible permutations that cause the lamp to light: B-B, B-G and G-B, and in only one of those three are both twins boys.
S2000magician
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On 2012-10-16 16:34, Adrian Fern wrote:
The answer to that depends on how you arrive at the knowledge that one of the twins is a boy.

If there are two doors, and the twins each stand behind a door, and you open one door at random and find a boy, then the probability of the other door revealing a boy is 1/2. This is because there are only two possible outcomes on opening the second door: B or G.

If there are two doors, and the twins each stand behind a door, and each is given a button that will (when pressed either on its own or when the other is also pressed) light a single lamp dangling in front of you, and each is told to press the button only if they are a boy, and you see the lamp light up, then you know that one (at least) of the twins is a boy, and you also then know that the probability of the other being a boy (i.e. of both twins being boys) is 1/3. This is because there are three possible permutations that cause the lamp to light: B-B, B-G and G-B, and in only one of those three are both twins boys.

Lobo's problem doesn't depend on any of that.

You missed a word.
Adrian Fern
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It's perfectly possible for a boy and a girl to look identical Smile
S2000magician
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Uh huh.

;)
Adrian Fern
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Monozygotic twins are another matter, though, for which we get an answer of 1 Smile
S2000magician
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I believe that that's the answer which Lobo sought.
msmagic1
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Really
landmark
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On 2013-02-02 13:10, msmagic1 wrote:
Really

Post of the year?
glowball
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Back to the Monte Hall Dilema. The simplest way to explain this is the following:

I have 3 cards and only one is the joker.

I place the 3 cards face down in front of a spectator and say "point to one card".

Then I say "which do you want? the card you pointed to or do you want the other two cards?".
I ask you which choice (one card or two cards) are the better odds of getting the joker?

This is what the Monty Hall issue is doing. The fact that he turns one of the wrong cards face up changes nothing.
Michael Daniels
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Good explanation!

Mike
harbour
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Interesting
Slim King
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Thanks!!!
THE MAN THE SKEPTICS REFUSE TO TEST FOR ONE MILLION DOLLARS.. The Worlds Foremost Authority on Houdini's Life after Death.....
landmark
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Just as a little addendum to glowball's excellent explanation--suppose you were able to keep your one card or switch to the choice of two facedown cards. Let's say you switch to the two cards. Are you disappointed if Monte now turns over the one he knows is not the joker and shows it to you? Of course not, you already knew that one of the cards would not be the joker--his turning it over gives you no new information, since he knew which card to turn over.
Michael Daniels
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The other variation discussed earlier in the thread is still worth considering:

Suppose that you make your choice of card and then one of the other two cards is turned over AT RANDOM (i.e., without knowing what the card is). On some occasions the joker is turned over, so you immediately lose. On other occasions an indifferent card is turned over.

On these occasions you may or may not have the joker. Is it better on these occasions to change to the other face-down card?

You could argue that, at the outset, there was a two-thirds probability that the joker was one of the non-selected cards. Turning the indifferent card face up hasn't changed that initial probability (as in the standard Monty Hall situation). So it is better to change your selection.

Or you could argue that turning the card over has changed the probabilties. There is now a 50:50 chance that each of the two remaining cards is the joker. So it makes no difference whether or not you switch your choice.

Which argument is correct and why? If turning a card over changes the probabilities in this situation, why doesn't it change the probabilities in the standard Monty Hall situation.

Also, suppose that you do not know whether the indifferent card was turned over at random, or because it was known to be indifferent. What difference does that make and why? How can your uncertainty about how the card was turned over make any difference?

Mike
LobowolfXXX
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If the card is turned over at random, it doesn't matter whether you switch or not (50-50). If you don't know whether the card was turned over at random, you should switch, which can't hurt (if it was random) and might help (if it was a deliberate choice to reveal a losing card).

I'll hold off on the "Why does it matter if it was random?" explanation until I have more time and/or those who think it doesn't matter weigh in.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley.

"...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us."
magicgerry06
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For those interested in this problem (like mI am), I have tried since a couple of years to present this plot as a "magic trick".
After 2 versions, the first one was good but too "gimmicky", the second one very good but too expensive, I have come up with a satisfactory solution. easy to do and costs nothing, it works in most languages. There is no maths or memorization or difficulty of any sort. It does require some one-time preparation, but just with some index cards (with lottery numbers on it), a single sheet of paper (for the list of prizes), and a normal pen, nothing else, no gimmick or funny business. I always keep it in my wallet. It can be adapted to your style, venue, weddings, birthday parties, Halloween, Valentine day, corporate events, etc. The list of booby prizes can also be very easily customized or changed for strolling, and be a good opportunity for comedy Magic! If you are interested in this plot (which I believe is a little gem that has not been exploited yet), pm me. It's also available now on lybrary.com. As it's my birthday month, I am offering a 50% discount on it up to the end of this month.
pm me and I will tell you how to get it).
Here is the effect:
Three business cards are put on the table, along with a (hidden) list of random lottery numbers and associated prizes. Two of the cards have a lottery number written on the back, and the third one has a car. The spectator chooses one of the cards, WITHOUT LOOKING AT IT. The performer looks at the three cards, and turns over one that he knows is a funny booby prize card of course. The spectator is now invited to either keep his original card, or to swap it against the other one. Whatever he does, he will always (100%) lose (and win a funny booby prize (-:.
(Note: You can also make the spectator win all the time, if you prefer. The performer can also predict whether the spectator would switch or not.) "
Gerard ZITTA
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"If Magic be the food of love, play on..."
dlhoyt
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This thread is testimony to the subtlety of the Monty Hall Problem. This recent web post is a summary of the history of the problem. It's well worth reading.
http://priceonomics.com/the-time-everyon......martest/
conrad3000
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Has any one looked at Gerard Zitta's work on this? Would love to hear any thoughts.
- Ryan
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