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Larry Davidson Inner circle Boynton Beach, FL 5270 Posts |
[quote]On 2003-06-01 11:37, The Original Countelmsley wrote:
For example, the probability of a coin being heads "or" tails is 1/2 + 1/2 = 1. The chance of a tail "and" a head would be based on 2 of 4 outcomes, giving you a probability of 1/2. The odds would therefore be 1 to 1. [quote] Actually, that isn't true, because the "head" side of most coins is heavier than the "tail" side. As a result, the tail side will land uppermost more often than the head side will. Re. cutting to four acers, the probability is directly proportional to the amount of time you practice and/or the availability of a reliable gimmick. |
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pinkbrain New user 10 Posts |
"If you change your original pick, your odds are increased from 1 out of 3 to 2 out of 3. Those who got it wrong were not interpreting Baye's Theorem on Conditional Probability correctly. So if Monte lets you pick another door, always do so."
I wrote a paper on the counterintuitive nature of probability a couple years back and studied this problem and several others like it. If you RANDOMLY choose one of the two remaining doors AFTER an empty door is revealed, your odds are 1/2. If you stay with your original choice, it is 1/3. And if you switch, it is 2/3. Oh, and to answer the original question of this thread, here's the other recent thread on this topic: http://www.themagiccafe.com/forums/viewt......2&10 |
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Larry Barnowsky Inner circle Cooperstown, NY where bats are made from 4770 Posts |
"Actually, that isn't true, because the "head" side of most coins is heavier than the "tail" side. As a result, the tail side will land uppermost more often than the head side will."
Larry, I should have said that this was a "fair" coin except if the effect is being performed in zero gravity. |
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prescott New user 33 Posts |
You know what you could do? Make the four aces short cards, and then cut directly to them. But as far as the odds of this happening naturally, I will leave that to a math guy.
Magic Happens
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Larry Barnowsky Inner circle Cooperstown, NY where bats are made from 4770 Posts |
"Larry, yes, I remember the answer and, if I am ever on 'Let's Make a Deal,' I'll be sure to do just that. I just did not, frankly, follow the reasoning very well. Eventually, I understood it....but frankly, after learning the answer, I tossed the reasoning right out of my brain to make room for other things like my phone number, social security number, street address, State Bar number, checking account number..."
John, the Monte Hall problem is solved by looking at outcomes. Let's assume you pick #1. If the jackpot is in 2 or 3, you lose by switching. If the jackpot is behind 2 and Monte shows you 3, you win by switching. If the jackpot is behind 3 and Monte shows you 2, you again win by switching. Therefore, once you select a door, there are only 3 outomes if you switch each time. With two of them you win and with one of them you lose. So by switching selections, your probability of winning is 2/3. Our intuition makes us think that it should be 1/3 whether we switch or not, but as you know as a magician, things are not always as they appear. |
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therntier Special user 681 Posts |
I think the easiest way to look at the Monte Hall problem is without mathematics. If we imagine that there are 100 doors, and Monte knows the winner, we will all agree that we have almost no chance that the door we picked will win. In other words, there is a good chance that the prize is behind one of the other 99 doors and Monte knows which one it is. If he goes and shows us 98 of those doors, I think we would all agree that switching would be the best thing to do.
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Larry Barnowsky Inner circle Cooperstown, NY where bats are made from 4770 Posts |
"If we imagine that there are 100 doors, and Monte knows the winner, we will all agree that we have almost no chance that the door we picked will win. In other words, there is a good chance that the prize is behind one of the other 99 doors and Monte knows which one it is. If he goes and shows us 98 of those doors, I think we would all agree that switching would be the best thing to do."
Great explanation. By taking the extreme situation, you made a very persuasive case for why switching is always the better choice. |
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cgscpa Elite user Ashton, MD 447 Posts |
Quote:
On 2003-06-01 11:10, The Original Countelmsley wrote: Thanks, Count. I meant for my formula to multiply each fraction not add. (1/52*1/51*1/50*1/49) But the fractions I was using would still have been incorrect as explained by Mr. Clarkson. Thanks for the intellectual challange, though. |
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Paul S New user Scotland UK 100 Posts |
lol!
And now that we all understand the Monte Hall problem, may I confuse everybody again by asking what happens if Monte only opens one of the remaining 99 doors. (kidding) PS |
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jacksorbetter Regular user Philadelphia 121 Posts |
Actually, you guys are assuming that when a deck is cut, all positions are equally likely to be cut to. While a reasonable assumption mathematically, in the real world I don't think it makes sense. I have never seen a person cut just one card, or 51 cards, and very rarely have I seen someone cut less than ten cards when asked to cut the deck.
the point that I am making goes beyond this. When going from an abstract set of prbabilities to a real world situation you must first verify that the assumptions of the computation apply. While most people assume that a coin tossed will land 50/50, this is clearly not always so. It can be loaded, or someone can be trained in flipping to get a desired outcome every time, or at least to deviate from "randomness." As someone pointed out earlier, using short aces will naturally increase the probability that they will be cut to, regardless of wether the person cutting realizes they are short and so on. There is an even more serious problem, namely that the word "random" does not completely define a situation, even theoretically. See Bertrand's paradox for more info. http://www.cut-the-knot.org/bertrand.shtml Anyway, the big point is that before you apply probabilities in the real world, you have tyo make sure they apply. |
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therntier Special user 681 Posts |
I think that all we are assuming is that, once a position of the deck is cut to, there are 52 possible cards that could occupy that position. It makes no difference where the deck is cut
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10cardsdown Special user Out There Somewhere 664 Posts |
Thanks pinkbrain! That's exactly the link to the thread I was looking for. Much appreciated!
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jacksorbetter Regular user Philadelphia 121 Posts |
therntier, that's saying the same thing.
Iam sure you got my point, but the assumption that after a shuffle any card can be at any position is not always valid. For example, say you open a new deck, and give it an overhand shuffle of say, 8 cutting actions, and then cut the cards. Unless the card that you cut at happens to be one of the cards that you cut at when you did the OH shuffle, the cards above and below it are the same cards as in the NDO. Even if you cut at such a card, one of the cards, either above or below will be a neighbor from the NDO. It is definitely not the case that the cards are uniformly ditributed in the sense that the probability of any card being in a given position is 1/52. Actually now that I think of it, this may be a useful idea for key-card effects. Another related example is one encountered occasinaly in magic, namely that riffle shuffles don't disturb the relative orders of the halves. If you have each half of the deck in order and separated by color, then after a riffle shuffle, the color groups will still be in the same order. Anyway, the point is a minor one, but worth keeping in mind I think. |
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pinkbrain New user 10 Posts |
Jacks or Better: I have had people just take one card off and have also seen them take all but one card when offered an oppurtunity to cut.
And your only talking about short term. Of course any probability cannot be held to be true in the short term. Its in the long run that the probabilities work out. Its not terribly odd to see 7 out of 10 heads on a coin flip but it is to see it 7000 out of 10000. Of course a simple shuffle (or 8 faros) will not make a fairly distributed deck, but over many shuffles it will. |
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Larry Barnowsky Inner circle Cooperstown, NY where bats are made from 4770 Posts |
Jacksorbetter,
To make the problem simpler, assume your spectator cuts always in the middle of the deck. If the deck were shuffled so that the mix is random then selecting a card from the middle should not bias the selection. I am assuming that certain cards don't have a predilection for seeking the center once shuffled. I am assuming the deck is thoroughly shuffled say my spreading the cards around as they do in some casinoes. I agree that normal shuffling creates patterns that could not be considered random. |
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therntier Special user 681 Posts |
If the probability of cutting to a certain card isn't one out of 52, then that must mean certain cards get cut to more than others. Which card would you bet on then?
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jacksorbetter Regular user Philadelphia 121 Posts |
"If the deck is shuffled so that the mix is random"
then everything follows. This however is an assumption, and not such an easy one to check, I think. I mean, one way to do it is to start the deck in some fixed arrangement, give it a shuffle and track the position of each card. Then reset the deck into the same original arrangement and reshuffle and so on. After say 1,000,000,000 trials you will have a distribution P(card #x=name of a card)=? but I have little doubt that for an accurate simulation of an OH shuffle it would not be 1/52 for every card. Nor would it be 1/52 for a riffle shuffle, or even putting the cards down on the table and "mixing" them up. |
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Joshua Quinn Inner circle with an outer triangle 2054 Posts |
Here's another question for the statistically-oriented: what are your chances of successfully guessing the color of each card in a deck without looking, as in OOTW? I assume that, since for any given card the chances are 1 in 2, then for a whole deck the chances would be 1 in 2^52, or about 1 in 4.5 quadrillion. (I realize the chances would be slightly better in an actual OOTW scenario since you wouldn't have to guess the leader cards, but I'm going for the general principle here.) Is this correct, or is there something more to it that I'm not accounting for?
Quinn
Every problem contains the seeds of its own solution. Unfortunately every problem also contains the seeds of an infinite number of non-solutions, so that first part really isn't super helpful.
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MacGyver Inner circle St. Louis, MO 1419 Posts |
Ok, everyone has made some really good points, and I'm going to try to set down some rules which would make it more accurate:
First we have to get a range of cards that the average person will cut from the center. I'd say about 15 cards in either direction, but someone should do trials with just telling the person to cut the deck, then recording what # they cut at, perhaps we could figure out what range normal people cut to, it should be a bell curve. Next, while i Know overhand shuffling and riffle shuffling isn't all that random, lets say that you take a face down deck and just keep shuffling it for a couple minutes, and with different kinds of shuffles, you should something ABOUT random.(Might I suggest throwing the cards onto the floor then mixing them up and putting them back in as random as you can without looking at the faces(or do it blindfolded). Then, we have to get the probability that 4 specific cards are in that range of normal cut to cards. THEN, we can start doing normal probability, but it might be more like 4/20, 3/19, 2/18, 1/17 depending both on the average range that people cut to and the probability that all 4 aces are in that range. Of course, that is assuming that the spectator has been told to cut the deck and not to cut to an ace... If you were TRYING to cut to an ace, then whoever was cutting would relize that all the aces are probably not in the center and therefore througout his 4 cuts he would try to hit as much of the deck as possible. As such, this problem is not math or probabilty, but human engineering that will really throw off the probabilities. |
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therntier Special user 681 Posts |
Probabilities are mathematical predictions of human engineering. Whenever you talk about a probability, you are admitting error, unless you are doing an infinite amount of trials. Without that, probably is just that, probable.
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