Odds and probability

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Odds and probability

Postby Mikey.666 » Aug 16th, '07, 19:29



I didn't like maths in school. Now I regret it.

I was wondering if anybody could tell me how to figure out odds or the probabilty of something happening.

Example: The odds of a spec dealing ten face down cards in two piles red and black.

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Mike

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Postby Beardy » Aug 16th, '07, 19:48

depends if you switch the cards without them seeing ;)

(apologies for the pointless of the above post...I promise whole-heartedly, that a post as pointless as the above shall no longer be not seen on Talk Magic)

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Chris
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Postby moonbeam » Aug 16th, '07, 20:32

Just to clarify:

Are you after the odds of a spec placing 2 rows of 10 cards (20 in total) down on top of 2 marker cards, 1 black and 1 red, using a standard pack of 52 cards ??

It's been a while since I did maths, but I may be able to help - although I'm sure someone on here will prob beat me to it :? .

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Postby Mikey.666 » Aug 16th, '07, 20:45

Ok I'll try to be a bit more specific.

There is a red and black marker card. I tell them I'm going to take out ten cards, some red some black. I then say imagine 5 are black and 5 are red, the odds of you dealing them out correctly are...

That's it :)

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Postby lindz » Aug 16th, '07, 20:51

Read tricks of the mind there's a little bit on probality in there but not much but if you already have it it's worth a look. I think it say's along the lines of if you flip six coins and then it gives you the odds, there's also a little more in there aswell.

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Postby Neyak » Aug 16th, '07, 22:00

Do you mean that you have 10 randomly selected cards (so there may be 7 reds, 3 blacks or 5 and 5 or 1 and 9) and the spectator deals them out randomly left and right just how he feels like, somewhat like a typical OOTW?
That would be possible to work out, but it may turn out rather complicated as one has a factor of what combination of cards are chosen (how many red/how many black) and a factor of laying them out in the right piles. Not impossible though.

Or do you mean the spectator deals them out left-right-left-right and there is an equal number of reds and blacks? (that would be the easier exercise to calculate)

If it's the former, I'll try to work it out if I have a few minutes to spare tomorrow, or someone else may want to do so. I've just been told off (well, kind of) for being too mathematical in another thread. But I'll give it a go anyway.

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Postby Mikey.666 » Aug 16th, '07, 22:25

It's the first one.

However, it's all hyperthetical, I say " I won you to deal these cards face down in any order you like. If you think it's red put it by the red card, if black put it by the red card. Lets say I picked 5 reds and 5 blacks, the odds of you getting the whole thing correct are.....lets say I pick 7 red and 3 black the odds would be......"

That type of thing.

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Postby Beardy » Aug 16th, '07, 22:45

wouldnt the odds of him getting it all correct be the same either way? if you didnt tell him how many of what you had, so he had to guess himself? whther you had 10 reds and no blacks, or 50/50, either way he doesnt know that - so statistically he has even chances no matter how many of each you have. It is only the overall odds that play in effect here

Love

Chris
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"I hope to shake your hand before I die" - Derren Brown
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Postby Mikey.666 » Aug 16th, '07, 23:07

Well I just need some dogs b*ll*cks patter with lots of bigs numbers for the effect :P

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Postby Rob » Aug 17th, '07, 07:49

Anything odds, or maths-related, just ask Lawrence - he the man, in that field :wink:

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Postby connor o'connor » Aug 17th, '07, 08:20

I don't think you want an acurate reply, but if you have a random effect that can turn out either of two ways n=2 and you repeat this x number of times then you can't go far wrong with 2^x
A three way bet 3^x, etc
So for your little bet the odds could be said to be 2^10 or 1 in 1024.
This is not a statisticaly correct answer by any means. But it will be the one most likely given to you by a spec.
The problem with probabilitys is that very very few people understand them. So you don't really need a definative answer, just patter.
You could say "the chances of this happening is well over a thousand to one"
You could say "the chances of this happening are well over a hundred to one"
You could lie and say "it's just under a million to 1"
but if you are using a psycological effect like ootw. The lower the probability the more it lets their mind think that it could perhaps have been chance. It's that not being sure that lets the trick linger, especialy if your doing a 'test' rather than 'trick'.
Your patter can then conclude "now obviously just performing this experiment on one subjest is in no way scientific proof, but it's pretty darn convincing as I seem to have just over a 95% success rate"

Telling a spec it's a million to 1 chance and telling a spec it's a ten million to one chance will not make the effect better. In fact they may well switch off. The reason for this is that once you get into big numbers the general public have no experience of them. Hundreds or thousands are within their realm and so the can reconstruct the trick or think about the numbers more.

I do not speak as a magician on this, so it would be nice to get a magicians perspective on what I have just said. I speak as a statastician and the biggest problem the statistics have is the fact that people cannot comprehend there meaning without being trained in statistics. This is why I would suggest the lower numbers and false statistical theory.

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Postby Yorkshire Pudding » Aug 17th, '07, 08:23

Mathematicians (or mathemagician) correct me if I'm wrong please but I think the odds would be approximately 2 to the power of the number of cards you are dealing. So in this case it would be 2 to the power of 10 (or 2x2x2x2x2x2x2x2x2x2) which is 1024. So the odds would be approximately 1024:1 against. I say 'approximately' as when you deal one red card, for example, the odds of the next card also being red are slightly less than 50% as there are less red cards than black left in the deck... but you get the idea.

Yours tryingtoremembermymathsfromschoolingly,

Keith

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Postby Yorkshire Pudding » Aug 17th, '07, 08:27

DARN! That always happens to me, someone posts a much better answer just as I'm composing mine... Still, at least I got the numbers right <phew>

Yours tooslowattypingly,

Keith

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Postby hacker » Aug 17th, '07, 08:50

i guess you have a 50/50 chance on having the two piles of red and black cards but sometimes its just bad luck.

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Postby Neyak » Aug 17th, '07, 09:58

In principle it's right to say that you've got a 50/50 chance for each card to get it right. Except the first one, which doesn't matter where it goes (assuming you don't tell them to put the "reds left and the blacks right", but just tell them to put the cards into two piles). So for ten cards it's only a 1 in 2^9 chance, that's 1 in 512. But that's only a theoretical value and would be true if people decide by tossing a coin for example, if head the card goes left, otherwise right, or so. In reality people will not act completely random, so it is likely that they'll create two piles of about equal size, for instance. Hence the chances of getting it right are higher if there are about an equal number of reds and blacks, since it is impossible to get it right if you've only got 1 red card but the spectator deals two piles of 5. The psychological factor is hardly quantifiable, but if you want it theoretically, just stick to the 1 in 512 or more generally in the 1 in 2^(n-1) where n is the number of cards.

I must apologise for my earlier post when I announced it all to be more complicated but I was thinking about probabilities of getting certain combinations of reds and blacks out of a deck of 52 and so on, which, admittedly, is not exactly necessary to provide a theoretical value. Only if you know that the spectator will deal equal or about equal size piles or so it would matter, but as I said, for the number to include in your patter it doesn't matter.

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