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Geek Culture / My Magic Trick - Idea

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A r e n a s
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Posted: 25th Oct 2009 17:12 Edited at: 25th Oct 2009 17:30
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Hey guys. I was thinking about that idea where you have three doors:

Behind 2 of them are goats, and behind the third is a car (or what ever you want). You pick one of the three doors (you have a 1/3 chance of winning right?).

(G) (G) (C)

The host then says behind this door is a goat (not the door you picked) and he opens it to show you. He then gives you the option to switch or stick. Heres the part which is difficult to grasp:

If you switch you have a 2/3 chance of winning and if you don't you have a 1/2 chance of winning (some people disagree and say its 1/2 and 1/2).

Here's my idea. Could you change that mathematical probability scenario into a card trick? Think about it. You have three cards; 2 kings and 1 Ace. You need to pick the ace. I'm still tying to figure out how it would work. If anyone has any ideas, just post

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Venge
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Posted: 25th Oct 2009 17:50 Edited at: 25th Oct 2009 17:51
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Watched the movie "21", didja?

http://answers.yahoo.com/question/index?qid=20090118005708AAVu7nC

I will live forever or die trying.
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xplosys
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Posted: 25th Oct 2009 18:25 Edited at: 25th Oct 2009 18:30
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Quote: "If you switch you have a 2/3 chance of winning"


I see what they are trying to do, but once you delete one of the possibilities, there are only two left. It's not the number of possible combinations of what is behind each door that determines whether you win or loose but the number of choices you have to pick from.

Let's just imagine that you picked a different one before the host removed one of them. Do you still have a 2/3 chance of winning by switching your pick - and why?

Brian.

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A r e n a s
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Posted: 25th Oct 2009 18:41 Edited at: 25th Oct 2009 18:44
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But this is where I got confused when I first heard it. I argued for the same cause that you are now proposing Xplosys. The thing is that you don't have 1/2 chance you have 2/3 chance because you must treat the two options as one choice. If you treat it as a 1/2 then the problem is that there is no point in any of it. You need to treat the two picks as one.

This is what was on the link posted by Venge:

Quote: "Funnily enough, a variation of this was discussed in Derren Brown's book. It's all about choice. If he didn't swap, he's only got one out of 3 chance of getting it, if he always swaps, it's 2/3 because he may choose:
Goat 1, then swaps it for a car
Goat 2, then swaps it for a car
A car, then swaps it for a goat"


Quote: "Watched the movie "21", didja?"


Nope. I heard it in maths. But I think I know what pert of 21 you mean... (I was playing a game on my laptop at the time I was watching it so I cant recall these things ) Also, I know it works i was just wondering if it would work in the form of a magic trick. You have a 2/3 chance of picking a red when you flip over the 'hosts revealing card' and so wouldn't that mean that using that method you could find a way of turning the odds from 1/3 to 2/3 of randomly choosing the correct card from 3 face down cards?

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xplosys
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Posted: 25th Oct 2009 18:47
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LOL.

I'm not sure what you just said, but let's take it to a simpler level. If you have two doors to pick from, one with a goat and one with a car, and you get to pick one, what are your chances of picking the car?

Now if you get to change your pick, do the odds of winning increase?

Brian.

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demons breath
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Posted: 25th Oct 2009 18:53
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I was never too comfortable with the whole scenario. I kind of get why it works, but I still would think of the second round as a new set with a one in 2 chance of having the car behind each door. I understand the logic behind it, but I'm a little superstitious with that and even if logic dictated that for some reason I only had a 1 in 60 chance of winning then I would quite possibly stick with my initial pick. Loyal to the end.

It is a bit easier to understand if you think of a higher number than 3 doors though, I always thought. If there were say 20 doors, then switching at the end would obviously be the best option. Nevertheless, like I say... faithfulness is a virtue

"A West Texas girl, just like me"
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Sven B
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Posted: 25th Oct 2009 18:57
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Quote: "I'm not sure what you just said, but let's take it to a simpler level. If you have two doors to pick from, one with a goat and one with a car, and you get to pick one, what are your chances of picking the car?"


1/2 of course.

But that's not the deal here.
(I argued with my teacher over this too).

The thing is, the host KNOWS where the car is, and won't open that door. So you can abuse the hosts knowledge by picking the door that he left untouched.

Sven B

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xplosys
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Posted: 25th Oct 2009 19:01
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Quote: "So you can abuse the hosts knowledge by picking the door that he left untouched."


He left two untouched. The one you picked, and the one he didn't pick. Unless you're saying that he picks the you did and shows you that it's goat. If that's what you mean, then I agree with you.

Brian.

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A r e n a s
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Posted: 25th Oct 2009 19:15 Edited at: 25th Oct 2009 19:29
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But the fact that the door opened by the host resets the whole field. If you take the entire thing in one probability tree its very easy to see how people come to the answer of 2/3 if you switch. Im gonna post a pic in a sec.

I have attached a pic to explain it as best as possible. (I think the size of the pic is within forum rules as my comp has a high resolution and the rules are for low res computers.)





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demons breath
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Posted: 25th Oct 2009 19:53
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Quote: "Unless you're saying that he picks the you did and shows you that it's goat."


Haha I think very few people would stick with the answer if they already knew 100% that they were going to lose.

Unless they really wanted a goat.

"A West Texas girl, just like me"
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Fallout
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Posted: 26th Oct 2009 10:49
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I found the best way to explain this one to people is to up the odds. -You have 100 boxes (1 car, 99 goats).
-You pick 1
-The host opens 98 others boxes with goats in, leaving 2 boxes (the one you originally picked, and one other)
-Now do you stick or swap?

In this example, you know the chance of you picking the car first time was extremely remote (1 in 100), so the car is far more likely to be in the swap box. That helps clarify it.

The next thing to consider is, you're not randomly labeling two boxes STICK and SWAP, and saying pick one. That would be 50-50. The boxes are labeled specifically according to previous events and information. You can visualize it this way ...

- I have two boxes, A and B
- I roll a dice with 100 sides.
- If it lands on 1-99, I put the car in box A. Otherwise I put it in box B.
- Now, with this information known to you, do you choose box A and B? You choose A, as it's a 99/100 chance.
- If this information WASN'T known to you, you have a 50-50 chance.

The reason for this is not knowing the information means you are guessing. If you do know the information, you're not guessing.

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A r e n a s
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Posted: 26th Oct 2009 10:54
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That is quite a good way of explaining it. But then could you turn it into a magic trick? Here some figures that i came up with that could help turn it into the trick. Bear in mind that you act as the host and do not turn over any of the cards.

2/3 of picking a goat on the first pick.
2/3 of picking a goat while you act as a host.
switching it then gives a 2/3 chance of winning.

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demons breath
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Posted: 26th Oct 2009 13:16
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How? There's too much chance and coincidence.

"OK I bet you a tenner that the next card I turn over has a slightly above average chance of being the Ace of Spades"

Just can't see it working.


Also the host doesn't have a 2/3 chance of picking a goat. He has to know EXACTLY where everything is or the whole exercise is obsolete.

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xplosys
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Posted: 26th Oct 2009 14:07 Edited at: 26th Oct 2009 14:10
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It doesn't work, of course. Each time a choice is removed, the odds of winning go up, but they can not be better than the number of choices. If there are two choices, one is a win and one is not, then the odds of winning are 1/2. It doesn't matter what happened a minute ago, which one you initially picked, or what the host knows.

In Arenas example, the final odds are based on 3 choices, but there are only two.

Brian.

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BatVink
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Posted: 26th Oct 2009 14:23
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Xplosys, look at in another way. You are picking 2 doors, and get to take the best prize from either of them.

If you pick 2 doors and the car is behind one of them, there is 100% chance that it will still be there when the host has finished.

So you have a 2 in 3 chance of picking the car by swapping - you picked 2 doors, and got to take the best prize from them.

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A r e n a s
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Posted: 26th Oct 2009 14:23
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Quote: "How? There's too much chance and coincidence."


It doesn't need to be definitely a win for you every time but a 66% chance of success is pretty good.

Quote: "Also the host doesn't have a 2/3 chance of picking a goat. He has to know EXACTLY where everything is or the whole exercise is obsolete."


I'm referring to the magic trick here where you dont know which is a goat and so you have a 2/3 chance of acting as a host and choosing the correct one as a goat.

Quote: "In Arenas example, the final odds are based on 3 choices, but there are only two."


Your treating it as a win loose situation the whole way throughout, where as in fact it isn't 50/50 its 1/3 to 2/3

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xplosys
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Posted: 26th Oct 2009 14:42
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I understand the theory, I just have a problem with the numbers. Does anyone know, is there a practical exercise done where the results can be viewed?

Brian.

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Fallout
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Posted: 26th Oct 2009 14:47 Edited at: 26th Oct 2009 14:49
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Here's a super extreme example, to hopefully clarify it once and for all ....

- Say I know last weeks lottery numbers, and you don't.
- I ask you to pick 6 numbers and guess last weeks lottery result
- You guess 6 numbers.
- I then tell you last weeks lottery numbers are either (STICK) the ones you guessed, or (SWAP) 6,17,22,24,28,32.

So, obviously to everyone in the world (except complete mathematical dunces), SWAP is going to be your best choice. There is no way you had a 50/50 chance of guessing the lottery results correctly the first time round.

THIS IS THE KEY BIT ...
Though you are being presented with 2 boxes to choose from, you are actually choosing between either "your original guess" or "every other box". In my example about, you're choosing between 1 lottery number guess, or every other number except your one. The fact that all the wrong ones, except 1, in that group have been revealed is irrelevant.

In the original example:
You made a choice at 1/3 odds. The remaining boxes therefore have a 2/3 odds chance. Forget the reveal step completely. Your choice is simply your original box, or the 2 remaining boxes. Which do you choose? Clearly the 2 remaining boxes.

It's the opening of one of the false boxes that confuses the issue and makes it an interesting logic problem, but that's all part of the fun.

That's all I can do to explain it now. I'm out of ideas.

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demons breath
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Posted: 26th Oct 2009 14:48 Edited at: 26th Oct 2009 14:52
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The reason the odds stay weighted is because of the host's knowledge. So for Fallout's example with the 100 doors, if you had the car there were 99 combinations he could have picked to open 98 doors with goats behind. If you had a goat, there was only one possible combination (as he has to leave the door with the car closed). That's why there are the different odds.

Quote: "I'm referring to the magic trick here where you dont know which is a goat and so you have a 2/3 chance of acting as a host and choosing the correct one as a goat."


Well that has nothing even remotely impressive about it.

"Here I have a bag with 3000 blue marbles in it and 1 red marble. I am going to put on this blindfold and select from this bag at random ANY MARBLE. I will bet anyone 10 pounds that it will be blue."

How many people do you think would go for that?


EDIT: Didn't see Fallout's new post.

And fallout, I like that example with the lottery. How about I bet you next time it's on something along those lines. Would be good. Bet 100 quid - if my numbers come up I win the lottery. If not I still get a consolatory 100 quid.

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Drew Cameron
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Posted: 26th Oct 2009 14:49
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The numbers are correct; when this was first published some of the worlds best mathematicians argued with the person who published it only to have to apologise when they realised they were wrong too.

I THINK its called "The Monty Hall Problem".

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Fallout
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Posted: 26th Oct 2009 14:51
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Colour blind people might fall for it!

Yeah, I too think it doesn't lend itself well to a trick.

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A r e n a s
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Posted: 26th Oct 2009 15:16 Edited at: 26th Oct 2009 15:17
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Quote: "[quote]I'm referring to the magic trick here where you dont know which is a goat and so you have a 2/3 chance of acting as a host and choosing the correct one as a goat."


Well that has nothing even remotely impressive about it.[/quote]

That was part of it. Using that idea you could choose the ace out of 2 kings and an ace and then that could possibly be adapted to picking an ace out of 3 kings and an ace and so if you could use the principle in theory you could change the odds of picking an ace from a pack of cards from 1/13 to 2/3. And successfully picking an ace out of a pack of cards has to be pretty impressive.

Quote: ""The Monty Hall Problem""


Ye, thats the name of the problem for sure.

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Fallout
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Posted: 26th Oct 2009 16:49
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So the magic trick would have to be similar to this ...

- The magician puts an Ace into a deck of cards
- A participant shuffles the deck
- They then hold out the fanned cards and the magicians picks one
- The participant then discards all the cards except the one the magician picked and one other
- The magician then picks the other one and shows everyone how he got the ace!
- Everyone shout obscenities from the crowd and demands their money back

I don't think this idea has any potential!

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A r e n a s
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Posted: 26th Oct 2009 16:57
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But the thing is that you dont need to have the participant throw away any cards. In theory you need to just pick the cards that would be thrown away yourself. After all there is a 51/52 chance that you will pick the same cards that would be thrown out anyway. So it would go like this:

- The magician puts an Ace into a deck of cards,
- A participant shuffles the deck,
- They then hold out the fanned cards and the magicians picks one having repeated the idea of the host showing the incorrect card in their mind,
- Everyone gasps in astonishment from the crowd and demands to see the tick again.

I tried it with just three cards and i tried it 10 times. I picked the correct card 7 times (which is brilliant considering that the odds are 1/3 and i should, in theory, have picked the cards about three to four times successfully). The problem was that after the seventh try i realised that i was falling into a pattern. I was starting with the right card, saying the left card was the hosts and saying my final choice was the middle card. So i start to get into a pattern...

But I do think that this trick has potential.

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Fallout
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Posted: 26th Oct 2009 18:15
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I'm not sure how you are imagining the trick would work. After the magician has selected a card from the deck, how is he discarding the 50 remaining wrong cards, without looking at them?

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A r e n a s
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Posted: 26th Oct 2009 18:21
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Using probability you could simulate the chances of acting as the dealer and simulating several of the wrong cards.

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demons breath
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Posted: 26th Oct 2009 19:51
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Nevertheless as a magician you should be aiming to get the trick right every time. Not more times than you don't.

"A West Texas girl, just like me"
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A r e n a s
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Posted: 26th Oct 2009 20:43
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Well with most times its a start. The problem is that i dont know how to increase my chances :S

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demons breath
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Posted: 26th Oct 2009 20:56
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This is the problem. Sorry I'm coming off as aggressively discouraging here aren't I... I don't mean to, I just think that as far as I know it's impossible to increase the odds.

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Posted: 26th Oct 2009 21:14
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Quote: "The problem is that i dont know how to increase my chances"

You have to cheat. That's sort of how magic works.

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xplosys
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Posted: 26th Oct 2009 23:38
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Quote: "You made a choice at 1/3 odds. The remaining boxes therefore have a 2/3 odds chance. Forget the reveal step completely. Your choice is simply your original box, or the 2 remaining boxes."


That's a very good explanation, and since I've now run the simulations.... I have to agree.

Brian.

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A r e n a s
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Ah Ha! I hav just managed to get it successfully working with a small trick. It works down to the last 10 or so cards (well it did for me 4 out of the 10 times i tried it ). I chose to pick the king of diamonds out of a pack of cards I split the pack into 3 piles, wrote a program on DB Pro so select one (so it was truely random) and then used the idea of choosing what the hoast would pick and choosing the other pile. I got the king of diamonds as the third card in the second try, the 7 card in the fith try, the fourth card in the eighth try and as the sixth card in the nineth try. What does it tell you?

Firstly, there may be some potential in my view of turning this into a full magic trick, im just gonna actually make some calculations to do with the matter.

And Secondlly, I truely have no life. (It was either IT coursework for a whole 3 hours straight or for a bit and then get distracted and start having a go with my idea)

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demons breath
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Posted: 26th Oct 2009 23:43
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Oh it's all about the coursework avoidance mate. I have so much coursework at the moment - one of the main reasons I'm back on these forums if I'm honest. Was avoiding work and popped in here, remembered how good it was...

One of my courseworks which I've been set at the moment is amazing though - literally as long as we write the article using Tex they don't care what it's on - needs a few equations and suchlike, but that's about it... Free rein pretty much. It's good. And not really what I expected from a maths course...

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Posted: 27th Oct 2009 00:18
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Quote: "That's a very good explanation, and since I've now run the simulations.... I have to agree."

No! It doesn't work - because the reveal is vital. If you remove the reveal, there is still a chance that the car is behind the one you chose, but if it's revealed not to be, you know it's not.

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A r e n a s
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Dam! I hit a road block. I have it so that once you have to view as the dealer you have a 1/2 chance of picking the none car. That put into the maths sets the end result up as 1/3 of winning which resets the whole thing back t just pick. I need to set it up as 2/3 and that keeps it in the realms of 2/3...

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demons breath
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Posted: 27th Oct 2009 00:39
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The dealer NEEDS to know where it comes from. The problem with this as a program is that the program would need to see what card was where. Makes the trick less impressive if you add in unnecessary steps to make the program less reliable, which is essentially what I think you're doing.

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A r e n a s
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Posted: 27th Oct 2009 01:05
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The program just generated random numbers. That is all it did.

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Mnemonix
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This is just the monty hall problem and I have already adapted it into a card based brainteaser to the annoyance and confusion of everybody I have tried it upon

Mnemonix
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A r e n a s
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Posted: 27th Oct 2009 09:50
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Okay then, fire away (what is it and how does it work?).

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Mnemonix
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Well I get a standard deck of cards that comes with 2 jokers and I use the ace of spades as the card that has to be "found".

I lay the 3 cards faced down in random order in front of my subject and I peek at the cards and I explain that I know where the ace is. I ask them to point to the card that they think is the ace and I flip up a joker(I cannot pick the card that they are pointing too, I explain this too). I then ask them if they want to switch or stick. I do this 2 or 3 times and I then give them the brainteaser of "Is it to your advantage to stick or to switch, and why?"

This leads to much confusion and their debate of probability when I reveal the answer that switching has a 2/3 chance of winning and sticking has a 1/3 chance of winning, when they confidently believe that neither option has a statistical advantage.

Mnemonix
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Fallout
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Posted: 27th Oct 2009 11:16 Edited at: 27th Oct 2009 11:18
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Quote: "No! It doesn't work - because the reveal is vital. If you remove the reveal, there is still a chance that the car is behind the one you chose, but if it's revealed not to be, you know it's not."


That's incorrect, unless we're thinking about something different. The reveal is always 100% fool proof. The host can always reveal all but one of the remaining boxes. He will either:
-Reveal all but one LOSE box (then you chose the winning box first)
-Or all remaining LOSE boxes and leave the winning box (then you chose a LOSE box first)

He will always unveil n-2 lose boxes (where n is the number of boxes in the game), leaving the winning box and one lose box. Therefore the unveil is irrelevant since you're choice is:
- Your box, or
- All the remaining boxes, since the host will take care of discounting all but the winning box (if it exists in that set of remaining box) for you.

In the example of 100 boxes, there is no way, if the winning box exists in the remaining 99 boxes, that the host can provide you with any other option than your first choice, or the winning box. Therefore the unveil itself means nothing.

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A r e n a s
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Posted: 27th Oct 2009 11:51
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Heres what is the difficult part of the magic trick. Say you have 10 cards and you start by spitting the deck into a half (with you trying to get a half with the card you have chosen; ace of spades). Are you more likely to have the win card if you just cut the deck in two, or if you were to lay the deck out face down and select one card (5 times) to be removed. Here are the probability calculations I have come up with. I disagree with the end result but cant find out why it doesn't work!

By splitting the 10 cards in half:

You have a 1/2 chance of the card you want to be remaining in your pack.

By taking one card out at a time:

9/10 X 8/9 X 7/8 X 6/7 X 5/6 = 1/2

You have a 1/2 chance of the card you want to be remaining in your pack.

Therefore it is equally likely to be remaining if you do it either way...

Then if you do that calculations for the whole sum:

By picking one card at random:

You have a 1/10 chance of the card you want being picked.

By taking one card out at a time:

9/10 X 8/9 X 7/8 X 6/7 X 5/6 X 4/5 X 3/4 X 2/3 X 1/2 = 1/10

You have a 1/10 chance of the card you want being picked.


So by these calculations there is no way to effect the probability of picking a card out at random...

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Posted: 27th Oct 2009 20:48
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Wow! I was just getting ready to write a post to argue the probability of the 3 door scenario. None of the explanations that I read here convinced me. I then read the Wiki page on the Monty Hall Problem and can see how it works now. Awesome stuff!
http://en.wikipedia.org/wiki/Monty_Hall_problem


a.k.a WOLF!
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A r e n a s
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Joined: 9th Jun 2008
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Posted: 28th Oct 2009 22:56
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Not a bad explanation. But then again if we'd wrote sommin that long no one would read it I just scanned it so im not sure about the explanation. I just looked at the pictures

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