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The Magic Cafe Forum Index » » All in the cards » » What are the odds of this (Simon Lovell "trick")? (0 Likes) Printer Friendly Version

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RogerTheShrubber
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In Simon Lovell's "How to Cheat at Everything," he describes something he uses as a trick over the radio: you take a shuffled deck, name two denominations (king and six, ten and eight, or whatever your two choices are) and start flipping the cards over to see if at any point those two denomiations will be found together in the deck in either order.

My question is this: what's the math on that? What is the probability that it will work? I actually wrote to Mr. Lovell a while back to ask him, and he said that by his experience it was about 75% but he didn't know the exact number. I often flip through the deck doing this when I'm bored and done practicing, and from keeping track over my first 500 trials I saw that it worked 409 times for me. However, my advanced math isn't up to figuring out the exact probability here, and my dabbling with search engines led to practically nothing (one person claimed that the odds were 56%, which seems awfully low, and she didn't show her math). I'm hoping that some of you are good enough to do the math and show me how it's done.

Cheers,

Roger

(Apologies if this is the wrong forum for my question - Mr. Lovell does pass it off as a trick, and it's hard to imagine a "trick" more "all in the cards" than this one.)
snm
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That's an 81.8% success rate. 409/500 = 0.818 and you move the decimal two places to the right for the 81.8%.
magicfish
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Tom Ransom provided data for this in Ted Lesley in his Paramiracles.
RogerTheShrubber
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Quote:
On Aug 27, 2015, snm wrote:
That's an 81.8% success rate. 409/500 = 0.818 and you move the decimal two places to the right for the 81.8%.


Thank you for the answer, but I meant overall - the exact odds of a 52-card deck. If were bad enough at math to need help figuring out the success rate of 500 trials, I'd be too embarrassed to post on the topic.
RogerTheShrubber
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Quote:
On Aug 27, 2015, magicfish wrote:
Tom Ransom provided data for this in Ted Lesley in his Paramiracles.


I believe that was for the odds of the same card showing up in the same place in two different shuffled decks. But if he also covered my question and you happen to have the figures and formula handy, I'd really appreciate it. I don't think this qualifies as exposure as it has nothing to do with a method, just the likelihood of something happening.
magicfish
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You're right Roger.
RogerTheShrubber
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On Aug 29, 2015, magicfish wrote:
You're right Roger.


That's what I thought but I wasn't completely sure. Thanks, magicfish.
FoggFactor
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I don't have any math to back this up but...the chances of two named values being next to each other OR separated by only one card in a shuffled deck is VERY high. I would be interested to know how many times out of your 500 that the cards were only one apart.

Maybe taking this question and emailing it to a math professor at a local university could give you a better answer. In my experience in doing similar things, the university is always happy to help!

- Frank
snm
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Given your data, 81.8% is the exact odds of a 52-card deck out of 500 trials

Quote:
On Aug 27, 2015, RogerTheShrubber wrote:
Quote:
On Aug 27, 2015, snm wrote:
That's an 81.8% success rate. 409/500 = 0.818 and you move the decimal two places to the right for the 81.8%.


Thank you for the answer, but I meant overall - the exact odds of a 52-card deck. If were bad enough at math to need help figuring out the success rate of 500 trials, I'd be too embarrassed to post on the topic.
WilburrUK
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My answer, which comes out significantly lower than Roger's experimental results.. 61.5%

If you want to check my work, here it is, there's every chance I made a mistake...

I'll assume the 2 values you choose are Ace and King, for the sake of argument.

There are 52! total orders that the deck can be in (I Won't list them all, it would take ages).

Of those, how many satisfy the requirement that there's an ace next to a king?

There are:
4x4 = 16 possible pairings that would satisfy our "Ace and King together" requirement
AS+KS, AS+KH, AS+KC, AS+KD
AH+KS, AH+KH, AH+KC, AH+KD
AC+KS, AC+KH, AC+KC, AC+KD
AD+KS, AD+KH, AD+KC, AD+KD

And for each of those there are 2 orders (A+K and K+A) making 32 ordered pairs.

For each of THOSE 32, there are 51 positions in the deck that are valid (that assumes that an ace on top and a king on the bottom DOESN'T count as together).

So that's 51x32 = 1632 ways of being right (eg KC at position 21 and AD at position 22).
But for each of THOSE 1632, there are 50! ways of arranging the rest of the deck.

Making 50! x 1632 arrangements that satisfy the Ace next to a King criteria.

To get the probability of it happening, just divide by the total number of possible deck arrangements (52!) and you get

1632 x 50!/51!
which comes out at = 0.615 or a 61.5% chance of hitting it.
WilburrUK
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Hmmm, I just convinced myself that the above post it nonsense and is under-estimating the probability! So please feel free to ignore it.
If you apply the approach to a 3-card deck (AKQ), it gives a provably wrong answer - back to the drawing board.
RogerTheShrubber
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WilburrUK, I really appreciate you trying. Math isn't my strong suit and the one time I tried reasoning the way you did I got a number around 48% myself, which is obviously too low. Thanks.
RogerTheShrubber
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Quote:
On Aug 30, 2015, FoggFactor wrote:
I don't have any math to back this up but...the chances of two named values being next to each other OR separated by only one card in a shuffled deck is VERY high. I would be interested to know how many times out of your 500 that the cards were only one apart.

Maybe taking this question and emailing it to a math professor at a local university could give you a better answer. In my experience in doing similar things, the university is always happy to help!

- Frank


Truth be told I would shuffle, and then look through the deck lightning fast to see if they were next to each other. During the times it failed, whenever of the two values got used up I stopped right there. I never looked for the "separated by one card" thing at all, it could have happened once, fifty times or never at all. I'll keep it in mind for future trials - I use this exercise to pass time often (in line, waiting for appointments, etc.).
Andy Moss
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I asked this same question in an old thread on the 'magical equations' forum (search for it under the title "Ninety-nine percent".)

According to TomasB (who painstakingly undertook a ten million simulation of the input data on his computer)the answer is as follows-

The probability of two cards being adjacent=48.6%
The probability of there being at most a single card between the two selections=73.6%

Hope this helps. Best wishes Andy.
RogerTheShrubber
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Quote:
On Sep 12, 2015, Andy Moss wrote:
I asked this same question in an old thread on the 'magical equations' forum (search for it under the title "Ninety-nine percent".)

According to TomasB (who painstakingly undertook a ten million simulation of the input data on his computer)the answer is as follows-

The probability of two cards being adjacent=48.6%
The probability of there being at most a single card between the two selections=73.6%

Hope this helps. Best wishes Andy.


Thanks for the reply, I appreciate it. The number 48.6 sounds awfully low to me given my own results and the fact that Simon Lovell uses this as a trick, but ten million tries are a lot harder to argue with than 500. Cheers.
landmark
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48.6% sound very low to me also, but in matters like this, it's always wise to follow Tomas's lead. Smile

But I'm still trying to figure out the flaw in WilburrUK's thinking. He had convinced me before he retracted his answer.
eralph357
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I think WilburrUK is using metric odds. To convert to the American Imperial system, I think you have to divide by 12 and multiply by 2.2, then subtract 40. :-P
MeetMagicMike
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Here's a "seat of the pants" way to estimate the odds:

Let's use Aces and Kings

Assume the Aces are evenly spread out in the pack. Each ace is touching two other cards. Each of those eight other cards has roughly a four out of 51 chance of being a King.

That's (8 X 4)/51 = .63

Not perfectly accurate because the first and last card dealt have lower odds AND because the aces might end up next to each other rather than spread out but those are small factors so it's a pretty good guestimate.

This bet is usually given as the two values next to each other OR one card apart as others have mentioned and that is where it gets counter intuitive. It's also sometimes given as one card between them or at the most two which would make the odds really high. I wonder if that is what the OP did in his experiment and just forgot.
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RogerTheShrubber
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No, I never even considered the idea of one (or two) cards between the cards of two named values. In fact, before this thread I never even heard of that being considered for any of this, so I couldn't have forgotten it if I wanted to. I do appreciate your answer, by the way, because any stab at the math is a better one than I've come up with.

In the Lovell book, on page 239, he says the odds are "heavily in my favor" (his words) that the two named values will appear together somewhere in the deck. I assumed that meant at least 70% and when he wrote back to answer my question he said it was about 75%, but didn't know the exact math. I'm such a dullard at math that I can't even come up with a reliable starting point, but after reading your 63% figure (which one could argue is enough to be called "heavily in my favor," I suppose, especially if math in the long run is being considered), I broke out a deck and ran it through ten trials. I got eight direct hits, to include two where the two named values were together TWICE in the deck. Now obviously an additional ten trials doesn't establish anything, and I certainly can't argue with any of the math that's been quoted anywhere in this thread because I'm not qualified to do so, but if figures like 63% and lower are close to the truth, my results are pretty bizarre. For all I know you're right, don't get me wrong, I just find myself utterly confused here in an area where I'm definitely out of my element to begin with.

Thanks again, by the way, to everyone who has chimed in so far. I really do appreciate this.
R2D2
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Yup I also got 0.486 from a Monte Carlo simulation with 10 million trials.

Working this out analytically is really tedious unless I'm missing some shortcut. You really have to watch out for double counting outcomes and then you end up looking a many many cases.
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