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Topic: What are the odds of this (Simon Lovell "trick")? 


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.) 


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%. 


Tom Ransom provided data for this in Ted Lesley in his Paramiracles. 


[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%. [/quote] Thank you for the answer, but I meant overall  the exact odds of a 52card 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. 


[quote]On Aug 27, 2015, magicfish wrote: Tom Ransom provided data for this in Ted Lesley in his Paramiracles. [/quote] 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. 


You're right Roger. 


[quote]On Aug 29, 2015, magicfish wrote: You're right Roger. [/quote] That's what I thought but I wasn't completely sure. Thanks, magicfish. 


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 


Given your data, 81.8% is the exact odds of a 52card 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%. [/quote] Thank you for the answer, but I meant overall  the exact odds of a 52card 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. [/quote] 


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. 


Hmmm, I just convinced myself that the above post it nonsense and is underestimating the probability! So please feel free to ignore it. If you apply the approach to a 3card deck (AKQ), it gives a provably wrong answer  back to the drawing board. 


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. 


[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 [/quote] 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.). 


I asked this same question in an old thread on the 'magical equations' forum (search for it under the title "Ninetynine 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. 


[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 "Ninetynine 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. [/quote] 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. 


48.6% sound very low to me also, but in matters like this, it's always wise to follow Tomas's lead. :) But I'm still trying to figure out the flaw in WilburrUK's thinking. He had convinced me before he retracted his answer. 


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 


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. 


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. 


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. 


Ok this just in: I computed it analyticly and the answer is indeed 0.486. If anyone's interested I can give the mathematical details, but basically I had to write down a big recurrence relation. Nothing too elegant. 


Elegant or not, it's all miles above my ability to figure it out. I'd love the details if you can provide them without going to too much trouble. I'd also be interested in the Monte Carlo simulation you described. 


Here goes... you actually don't need to know any fancy math for this. You just need some basic probability. But it gets tedious. Suppose we want to know the probability of having a queen next to a king somewhere in the deck. (This is what I'll refer to as a "success".) Define the following functions: P(q,k,n) = probability of having a queen next to a king given that you have n cards left in the deck, q of which are queens, k of which are kings, and that the previous card you dealt was neither a queen nor a king. In other words, P is the probability of success given a certain mix of cards remaining. Note that n is the number of remaining cards, not the total number of cards. Let Q(q,k,n) = probability of success given that the previous card dealt was a queen. The variables are the same as for the function P. Notice that capital Q is a function and little q is a variable. Let K(q,k,n) = probability of success given that the previous card dealt was a king. The variables are the same as for the function P. Now the easy part: Figure out the base cases. P(0,k,n) = 0, i.e. if you have no queens left, you have a zero probability of success (since the previous card was neither a Q nor K.) P(q,0,n)=0, same reason as above. P(q,k,0)=0, (there are no cards left!) P(q,k,1)=0, i.e. you can't succeed if there’s only one card left. Q(q,0,n)=0, i.e. you need at least one king remaining (but not a queen – the previous card dealt was a queen after all) Q(q,k,0)=0, i.e. you need at least one card in order to succeed from here. K(0,k,n)=0, i.e. you need at least one queen remaining. K(q,k,0)=0, i.e. you need at least one remaining card. Now the moderately tricky part. Define the functions recursively. First let’s do Q(q,k,n). If the card you just dealt was a queen and you have n cards left, what are the ways you can succeed from here? Possibility 1: The next card you deal is a king. Probability of that is k/n. Possibility 2: The next card you deal is a queen, and then you succeed with the remaining n1 cards. Probability of the next card being a queen is q/n, and the probability of success after that is Q(q1,k,n1). Overall probability: q/n * Q(q1,k,n1) Possibility 3: The next card is neither a queen nor a king. You then have n1 cards left, and you succeed with the remaining n1 cards. The probability of an indifferent card being next is (nqk)/n and the probability of success after that is P(q,k,n1) (since you’d then be starting “fresh”). Overall probability: (nqk)/n * P(q,k,n1). Total probability of success given that you just dealt a queen = Q(q,k,n) = k/n %2B q/n * Q(q1,k,n1) %2B (nqk)/n * P(q,k,n1) Yes, this is a recursive function, but all the recursive evaluations involve smaller parameter values, and the whole computation bottoms out eventually (because of the base cases). If you've gotten this far, then you’re past the worst of it. You now define K(q,k,n) similarly are you get: K(q,k,n) = q/n %2B k/n * K(q,k1,n1) %2B (nqk)/n * P(q,k,n1) This also follows by symmetry. (The role of the kings and queens are interchangeable.) Finally, we define P(q,s,n). Recall that this is the probability of success from a new deck of n cards, or equivalently from a deck with n cards remaining but where the previous card was neither a king nor a queen. Here are the three cases to consider: Possibility 1: The next card you deal is a king and then you succeed from there. Probability of that is k/n * K(q,k1,n1) Possibility 2: The next card you deal is a queen and then you succeed from there. Probability of that is q/n * Q(q1,k,n1) Possibility 3: Next card is indifferent. Just like possibility 3 from earlier: (nqk)/n * P(q,k,n1) Therefore: P(q,k,n) = k/n * K(q,k1,n1) %2B q/n * Q(q1,k,n1) %2B (nqk)/n * P(q,k,n1) Now you just write a computer program that codes up these three functions and you evaluate P(4,4,52). The one complication that comes up is that the recursion branches into a *huge* tree. Each function call results in two other function calls, and n only drops by 1 each time! I think it's basically intractable unless you shorten the computation somehow. Here's how I fixed that: I cached the values of K, Q, and P so that I never had to evaluate these for the same parameters twice. Huge, huge savings. (I suppose this is equivalent to doing the whole thing with dynamic programming, for those of you who are familiar with that.) Look at the first term of P: k/n * K(q,k1,n1). Clearly if k is zero, there's no need to evaluate K(q,k1,n1). You can make this optimization in every instance of a recursive call, i.e. before you recurse, check the coefficient to see if it’s zero. This results in a small speedup. Finally, the formulas for Q and K are symmetric. So if you fill in the cache for the value of (for instance) Q(3,2,10) then you can also fill in the value for K(2,3,10). This also gives you a small speedup. Once you make these optimizations, the whole things runs really fast, and it spits out 0.486. I realize this looks long and complicated, but it's really just a long sequence of simple ideas. Let me know if anything here is unclear.  You also asked about the Monte Carlo sim. For this, I really just used a brute force approach. For each trial, I randomly picked positions for the kings and queens, making sure that they were distinct. Then I checked to see if any of the king positions differed from any of the queen positions by 1. Repeat 1 million (or 10 million) times. I can post code if you’re really curious. P.S. If you're wondering how I got a plus sign to appear despite the BBCode parser, I just typed percent then 2 then capital B. 


Arghh.... using percent2B worked in preview mode (it showed up as a plus sign). So if you're having trouble understanding my formulas, mentally convert that to a plus sign. Why does this have to be so difficult? 


This is much easier to understand in pdf form (see attached). 


R2D2, thanks very much for all of this. The math is still miles above my head, but I really appreciate being able to see it. Cheers, Roger 


No problem, Roger. And if you're ever in NYC I'd be happy to meet up and take you through it. 😀 


I'll probably be in NYC once or twice between now and the end of the year, but I'd be lying if I said I'd be able to follow the math. When I went to high school we were allowed to drop math after one year and I was dumb enough to do just that. Did the absolute minimum on math in college, too. One of my bigger regrets, especially thinking at the time how great it was that I didn't have to do more. 


There is a way of calculating this without recursion but it's not easy. Let C(n,r) be n!/(r!(nr)!). Let p be the plus sign and * be multiplication. Ignoring suits, there are C(52,4) ways to place the Qs and then C(48,4) ways for the Ks. So there are C(52,4) * C(48,4) = 52677670500 possibilities in total. It is easier to first work out the number of cases where a Q and a K are NOT adjacent. If the four Qs are in the deck, there are up to 8 adjacent spots for the Ks. Suppose there are 8  Z adjacent spots where Z is in 0,1,...,7. For example, if the four Qs are in the first four places then Z = 7 because there is only one adjacent spot. If there are Z spots, then there are C(40 p Z,4) possibilities where the Kings will NOT be adjacent to any Q. Now the tricky part. How many possible positions for the Qs are there for each Z ? Suppose A is the number of pairs of queens, so A is in 0,1,2,3. So for two pairs or for threeofakind, A = 2, and for fourofakind A = 3. Suppose B is the number of times there is a single card (exactly one card) between two Qs; also, add one if a Q is in the 1st position and add one again if a Q is in the 52nd position. This means that Z = 2*A p B. For a fixed Z, the number of positions for the Qs is C(3, A) * C(5A, B) * C(44 p A, 4  A  B) where the sum is over the pairs (A,B) such that Z = 2*A p B. The C(3, A) term corresponds to the choices of paired Qs, the C(5A, B) term corresponds to the choices of single cards between Qs (or Qs at the 1st or 52nd pos). The C(44 p A, 4  A  B) term is the others. Now multiply this by C(40 p Z, 4), sum over Z = 0,...,7, and you get 27091923270. The fraction 27091923270/52677670500 reduces to 300684703/585307450. So the answer is one minus this, which is 284622747/585307450 = 0.486279. 


Corrections: "If there are Z spots" should be "if there are 8Z spots" "the number of positions for the Qs is" should be "the number of positions for the Qs is the sum of" Simulating is probably a better idea though... 


[quote]On Oct 26, 2015, alecStephenson wrote: Simulating is probably a better idea though... [/quote] I disagree. Your solution rocks! I think you won the thread. Very clever approach. 


Thanks R2. One more (hopefully!) typo correction: 27091923270 should be 27061623270 in both cases. For some reason I typed 9's rather than 6's. 


[quote]On Oct 26, 2015, R2D2 wrote: [quote]On Oct 26, 2015, alecStephenson wrote: Simulating is probably a better idea though... [/quote] I disagree. Your solution rocks! I think you won the thread. Very clever approach. [/quote] Seconded. Proof trumps simulation in my book. 


So are the actual odds less than fiftyfifty? Sorry for being dumb ... 


Yes, less than 50%. It's 48.6%. Not a dumb question; there was a lot to wade through! 


Sounds too risky to do on the radio ...c 


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 :) 


[quote]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 :) [/quote] 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? 


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. 


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). 


[quote]On Oct 30, 2015, R2D2 wrote: [quote]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 :) [/quote] 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? [/quote] Yes. With no cards between them. 


[quote]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. [/quote] 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. 


[quote]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). [/quote] 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. 


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.) 


I'm provisionally convinced by R2D2's explanation and answer of 48.6%. That said, I would expect that in the realworld, 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 realworld 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. 


That makes some sense. 


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. 


Weird .... This is intriguing. 


[quote]On Nov 4, 2015, Terrible Wizard wrote: Weird .... This is intriguing. [/quote] 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. 


[quote]On Nov 4, 2015, RogerTheShrubber wrote: [quote]On Nov 4, 2015, Terrible Wizard wrote: Weird .... This is intriguing. [/quote] 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. [/quote] 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? 


[quote]On Nov 4, 2015, R2D2 wrote: [quote]On Nov 4, 2015, RogerTheShrubber wrote: [quote]On Nov 4, 2015, Terrible Wizard wrote: Weird .... This is intriguing. [/quote] 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. [/quote] 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? [/quote] 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: K6, then 42, then 7J, then Q3, then 59, 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 A2, then A3, then A4 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). 


Ok, you're doing everything right! I can't explain this. Does someone else want to try? :) 


Anyone know any math professors with an interest in magic :) 


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? 


[quote]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? [/quote] 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: [i]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 [/i] 


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? 


[quote]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? [/quote] 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. 


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 superfast 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 spreadsheetsavvy) and tell me if it matches what you're doing. 


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. 


[quote]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 superfast 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 spreadsheetsavvy) and tell me if it matches what you're doing. [/quote] I'm probably not as spreadsheet savvy as you are, but I'd LOVE to take a look. PM coming. 


[quote]On Nov 6, 2015, carlyle wrote: ...but it's pretty magical when it does happen. [/quote] It really is. No matter how many times I do this, even if I've gotten a couple misses in a row, for me there's still a sense of "holy crap" when it happens. 


[quote]On Nov 6, 2015, RogerTheShrubber wrote: [quote]On Nov 6, 2015, carlyle wrote: ...but it's pretty magical when it does happen. [/quote] It really is. No matter how many times I do this, even if I've gotten a couple misses in a row, for me there's still a sense of "holy crap" when it happens. [/quote] I'm going to disagree. I see nothing magical about this at all. It is obviously just a probabilitybased thing, and, as we've learned, the odds aren't even in your favor! I would expect a reasonably intelligent layman to think to themselves, "Oh, so this event occurs more frequently than I would have thought. That's interesting." I would not expect them to think, "That's magical." May the odds be ever in your favor. ;) 


Not sure if anyone still cares, but if you want to do a bunch of trials at home then you can speed things up as follows: after you shuffle and go through the deck looking for a king next to a queen (for instance), record your result and then immediate go through again and look for a jack next to a ten, and so on. In other words if you pick a bunch of pairs of ranks, you can check each pair against a single shuffled deck. And then shuffle (really well) and keep going. I hope that made sense. 


[quote]On Nov 12, 2015, RiderBacks wrote: [quote]On Nov 6, 2015, RogerTheShrubber wrote: [quote]On Nov 6, 2015, carlyle wrote: ...but it's pretty magical when it does happen. [/quote] It really is. No matter how many times I do this, even if I've gotten a couple misses in a row, for me there's still a sense of "holy crap" when it happens. [/quote] I'm going to disagree. I see nothing magical about this at all. It is obviously just a probabilitybased thing, and, as we've learned, the odds aren't even in your favor! I would expect a reasonably intelligent layman to think to themselves, "Oh, so this event occurs more frequently than I would have thought. That's interesting." I would not expect them to think, "That's magical." May the odds be ever in your favor. ;) [/quote] Good point, but that's why I said that FOR ME there's still a sense of "holy crap" when it happens. I'm aware that others may not feel the same. 


[quote]On Nov 27, 2015, R2D2 wrote: Not sure if anyone still cares, but if you want to do a bunch of trials at home then you can speed things up as follows: after you shuffle and go through the deck looking for a king next to a queen (for instance), record your result and then immediate go through again and look for a jack next to a ten, and so on. In other words if you pick a bunch of pairs of ranks, you can check each pair against a single shuffled deck. And then shuffle (really well) and keep going. I hope that made sense. [/quote] Thanks, it does make perfect sense, but I still do it the oldfashioned way. Habits which have lasted decades are uncomfortable to break. 