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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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On Oct 30, 2015, R2D2 wrote:
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On Oct 30, 2015, RogerTheShrubber wrote:
I'm certainly not qualified to refute the math, but Lovell says it's about 75% and my results have been even higher. Never before has such an informative series of answers to anything left me more confused. I can't help but think that Lovell would have noticed if the probability was a little less than half, but again, the math is above my head and the calculations here seem very solid. I've never suffered from headaches before but I'm starting to get one now Smile


In that case, we probably misunderstood the question. Let me know if I'm stating it correctly.

1. Pick two ranks (e.g. Kings and Queens).
2. Look through the deck to see if there is a K next to a Q (in either order).

You're asking the probability that this happens at least once.

Is that correct?


Yes. With no cards between them.
RogerTheShrubber
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On Oct 30, 2015, alecStephenson wrote:
Interesting.

To recap, TomasB and R2's simulation, R2's recursion methodology and my working, have shown that the chance is 48.6%.

As for the book, I have now found the part: it is under the TWO CARDS TOGETHER section.
And RogerTS is correct - the quote in my edition is

"Think of two card values. Not suits, just values. So you may think of K or 3 ot 5 and 9 for example. [...] I'll bet you even money that in that deck, right here and right now, two cards of the values you are thinking of will be together in that deck. The odds are heavily in my favor that they will be so."

Firstly, it's a great book, and a really great and easy read, and also communicates the basics of odds calculations. But the author has simply made a mistake here.

It is understandable: most odds things in the book are either fairly straightforward (to a maths/stats person) or they are complex but based on well known statistical problems (bithday problem, coupon matching etc). In this case, it is neither. It is a tricky problem that you are unlikely to find in any standard probability textbook.


I like the book too (although I could do without all the silly "Freddy" crap), but one thing I noticed is that he gives the odds on almost everything in the book except for this, leaving me of course no less confused than I was when I started reading it. I had heard of the "two cards" thing from my aunt years ago, back when the Monkees were regularly on the charts, and when I saw this section of the bookmy hopes shot through the roof because he was giving odds on almost everything else. No such luck.
RogerTheShrubber
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On Oct 30, 2015, alecStephenson wrote:
RogerTS: On your experiments, I can think of two possible explanations.

The second one is perhaps more likely.

1) You said that you run through the deck lightning fast. You may have identified cards as being together when they are not.

2) You are doing the experiments one after the other without shuffling numerous times between experiments. This means that the outcome of one experiment will be related to the outcome of the next one(s), particularly if you are using the same two ranks each time. For example if a K and Q are together and then you OH shuffle, they will probably still be together after the shuffle (a bit like when you have a k*y c**d and let the spectator shuffle).


All reasonable guesses, but actually none are on the mark. By "lightning fast" I mean I move the cards from my left hand to my right as if I were examining a gin rummy hand for the first time instead of actually dealing out every card and flipping them over, and while I'm no cardistry wizard I certainly wouldn't miss a card between my target cards even once, much less enough times to warp the results as much as I've seemed to. My vision sucks, but it isn't THAT bad.

I always riffle shuffle several times and never use the same values on consecutive hands.

In fact, I brought my kids into this and had them do the trials, even paid them to do so. My wife has pitched in, too. Not once has anyone scored less than 63/100. So you can see why I'm even more confused what with being shown math that runs so rampantly counter to my results but looks so solid at the same time.
R2D2
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Ok, I can't explain any of this. It looks like I understood the original problem correctly, so I can't understand how your kids got different results.

How many trials did they do? (I'm reaching here.)
RiderBacks
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I'm provisionally convinced by R2D2's explanation and answer of 48.6%. That said, I would expect that in the real-world, the percentage would be higher. If we ask John Q. Public to pick up his deck and name two values, we can expect him to pick up a deck which isn't shuffled well. It's probably left over from his last poker game. And if we ask him to shuffle, he isn't going to give it a faro or two either. He's going to overhand or riffle it once or twice in a fairly lame fashion.

If he plays poker, I bet if we looked through his deck before he shuffled, we'd find lots of standard pairs, probably often involving face cards (say from last night's poker game's full house). Most of the face cards top the list of "verbal accessibility" with the AS being at the very top of the list. So if on the spot and asked to name two values, we expect a lot will opt for the most verbally accessible cards (which, I suspect, are more likely to be next to one another in most people's decks.) Considerations such as this could push the real-world percentage much higher than the theoretically computed percentage (which isn't good).

That's a potential partial explanation for the difference between theory and practice.
Terrible Wizard
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That makes some sense.
RogerTheShrubber
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R2D2, the household results are as follows as of this writing:

Me: 558/700
Wife: 139/200
Oldest heir: 221/300
Second oldest heir: 152/200
Second youngest heir: 293/400
Youngest heir: 68/100

What puzzles me most about this is that no matter how solid the math is, it seems unthinkable that Lovell could be so far off given how much he knows about odds and how many odds he includes in other sections of the book. He didn't have the exact numbers when I wrote him, but he said it was roughly 75% and that's about what we're seeing in Shrubbery Central.
Terrible Wizard
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Weird .... This is intriguing.
RogerTheShrubber
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On Nov 4, 2015, Terrible Wizard wrote:
Weird .... This is intriguing.


My thoughts exactly. And while Riderbacks' post is logical, I've already addressed the shuffling question. Further, instructions given to the family is to never use the same two values together in any one trial run.
R2D2
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On Nov 4, 2015, RogerTheShrubber wrote:
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On Nov 4, 2015, Terrible Wizard wrote:
Weird .... This is intriguing.


My thoughts exactly. And while Riderbacks' post is logical, I've already addressed the shuffling question. Further, instructions given to the family is to never use the same two values together in any one trial run.


Wait, what do you mean by that? So if you pick 6 and 8 for the first trial then you might pick 2 and 9 for the next trial but not 6 and 8 again?

And for each trial, you shuffle, then pick two ranks (before looking at the cards) and then go through the deck?

Let's say you pick 6 and 8. If you see a 6 followed by another 6 does that count as a success? Or do you need to see a 6 next to an 8?
RogerTheShrubber
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On Nov 4, 2015, R2D2 wrote:
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On Nov 4, 2015, RogerTheShrubber wrote:
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On Nov 4, 2015, Terrible Wizard wrote:
Weird .... This is intriguing.


My thoughts exactly. And while Riderbacks' post is logical, I've already addressed the shuffling question. Further, instructions given to the family is to never use the same two values together in any one trial run.


Wait, what do you mean by that? So if you pick 6 and 8 for the first trial then you might pick 2 and 9 for the next trial but not 6 and 8 again?

And for each trial, you shuffle, then pick two ranks (before looking at the cards) and then go through the deck?

Let's say you pick 6 and 8. If you see a 6 followed by another 6 does that count as a success? Or do you need to see a 6 next to an 8?


In order:

1.) Yes, you wouldn't pick a 6 and 8 again during a series of trials, at least not together. I instructed the family to make it as varied as possible, so you might see an order like this: K-6, then 4-2, then 7-J, then Q-3, then 5-9, and so on. I don't tell them what to pick, I only tell them how, and told them not to do anything that would use either of the same values in a previous run in the next one. So no A-2, then A-3, then A-4 and so on. Eventually if you do enough trials you have to repeat somewhere, but I mix it up and have the family do the same.

2.) Yes, shuffle, pick two ranks you haven't yet used together without looking at the deck.

3.) No, a 6 followed by a 6 would not be a success. Any 6 next to any 8 in either order with no card(s) between them would. If during a single trial you have two or more instances of a 6 being next to an 8, though, it only counts once (and in fact once I see one, I stop and go to the next trial, which now that I think of it I wish I hadn't been doing, it would have been interesting to see how often I got multiple successes in the same run through a deck).
R2D2
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Ok, you're doing everything right! I can't explain this.

Does someone else want to try? Smile
Terrible Wizard
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Anyone know any math professors with an interest in magic Smile
R2D2
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I still can't get over the fact that RTS got his family to participate *hundreds* of times! I wonder if RTS and I should Skype each other so I can watch him do a trial and vice versa so we can see if we're doing anything different.

Maybe I should try it first with regular cards.

Incidentally, has anyone asked Simon L himself?
RogerTheShrubber
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On Nov 4, 2015, R2D2 wrote:
I still can't get over the fact that RTS got his family to participate *hundreds* of times! I wonder if RTS and I should Skype each other so I can watch him do a trial and vice versa so we can see if we're doing anything different.

Maybe I should try it first with regular cards.

Incidentally, has anyone asked Simon L himself?


Don't give my family too much credit here. In order to get them in on this, I have to pay the kids and barter with the wife (take some of her turns with household chores).

As I said earlier, I wrote to Simon some time ago and asked him about it. Earlier in the thread I said he answered with an approximate figure of 75%, but I just dug up the email and it seems that he actually said 7/10. The email was from January of 2013 and the exchange reads as follows:

Dear Mr. Lovell,

I must ask (this is a question that's haunted me for years) the following:

In one segment you explain a bet which in a different form you use as a card trick over the radio. You tell someone to pick two denominations at random (king and three, five and nine, etc) and shuffle the deck, and the odds say that two cards of those denominations will be found together.

My aunt taught me the same thing when I was a child and for years I've wondered what the exact odds are. Math teachers in school either didn't know or thought I was trying to pull some hustle on other students and refused to tell me. Since that level of math was never my strong suit, I'm honestly unable to do it myself.

Do you know what the odds are and how they're arrived at?

Many thanks in advance for any answer you might give. Loved your book. My father is hooked on it too.

Regards from Alexandria, VA -

Roger R*****



RE: From your book, How to Cheat at Everything

Hi Roger,

Like yourself I was taught the effect many years ago as a very young man (in my case from a hustler pal of my grandfather) and have used it ever since.

I've never bothered working out the exact odds and have just been happy that it works a lot of the time. I would, at a rough guess, from numerous performances of it, say that the the percentagess are around 7 out of 10 in my favor. That's just a guesstimation though!

Glad you like the book!

Kind regards,

Simon
simon@***********.com
RiderBacks
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May I ask what kind of shuffle you are having the fam use and, if it's a riffle shuffle, how many times you have them riffle shuffle between tries?
RogerTheShrubber
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On Nov 5, 2015, RiderBacks wrote:
May I ask what kind of shuffle you are having the fam use and, if it's a riffle shuffle, how many times you have them riffle shuffle between tries?


Sure. I haven't been overly specific with them on the riffle shuffle (that's what we all use). I'm in the habit of seven riffle shuffles between pretty much anything (deals in a game, deals for these trials, etc), but I told them to shuffle them well and then trusted them to do that (it should be noted before anyone asks that I don't supervise them or watch them when they do this, nor did I tell them what numbers could be expected). And again, I did tell them not to use the same values on consecutive deals, so I really don't think shuffles (good or bad) are having a big effect here, but as with anything else, I could be wrong.
R2D2
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Btw, if anyone wants to see a simple spreadsheet simulation of this, send me a PM and I can "share" it over google docs. Google's version of Excel isn't super-fast but you can manage a sheet with 1000 sims and it's not *too* slow. 1000 sims isn't great as far as "margins of error" go, but you can hit delete in an empty cell to reset the random numbers and see a new sim. The percentage of successes indeed hovers around the high 40s and you hardly ever see it go into the high 50s. More than 1000 sims would make the whole sheet unmanageable.

RTS, you might want to take a look (if you're spreadsheet-savvy) and tell me if it matches what you're doing.
carlyle
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Interesting thread and a fairly addictive experiment. I tried it with my wife 35 times and came out 18 times correct, 17 times wrong. We shuffled between us and varied the shuffles, the cards were always well mixed. Sometimes she'd choose the pair, sometimes I would, sometimes she'd say a value and I'd suggest the second value.

I found that it was never correct more than 3 times in a row (and that only once), I wonder about that kind of thing by the others. I'm not a betting man, but it's pretty magical when it does happen.
RogerTheShrubber
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On Nov 6, 2015, R2D2 wrote:
Btw, if anyone wants to see a simple spreadsheet simulation of this, send me a PM and I can "share" it over google docs. Google's version of Excel isn't super-fast but you can manage a sheet with 1000 sims and it's not *too* slow. 1000 sims isn't great as far as "margins of error" go, but you can hit delete in an empty cell to reset the random numbers and see a new sim. The percentage of successes indeed hovers around the high 40s and you hardly ever see it go into the high 50s. More than 1000 sims would make the whole sheet unmanageable.

RTS, you might want to take a look (if you're spreadsheet-savvy) and tell me if it matches what you're doing.


I'm probably not as spreadsheet savvy as you are, but I'd LOVE to take a look. PM coming.