*A loving tribute, a thank-you and a card trick: all from the heart.*

My Dad died earlier this month: I don’t yet know when or if the tears will stop. He was a big man and he leaves a big hole, but filled with happy memories. I can’t exactly say I modelled myself on Dad: that was a bit beyond me. But I did follow his example where I could in the basics of honesty and hard work: painfully slowly at times perhaps but I hope I got there in the end. We were different, but each proud of the other I think.

Of course, I now wish I’d managed to spend more time with him in the last few years. The past year alone, he suffered strokes, heart problems, survived sepsis, countless other infections, pneumonia, a collapsed lung and saw off Covid-19. But the last illness was just one too many although, as always, he fought it to the end. He was just short of 90 and lived a full and worthwhile life: you can’t ask for anything more.

This isn’t the place for a poetic eulogy. But, even just in an academic sense, the older I’ve got, the more I’ve realised just how much I owe to Dad. I was never really a gifted student, despite what people might think. And I proved many times over the years the ease with which I could get into trouble of one sort or another. If it hadn’t been for the start that Dad gave me, I’d be in a very different place now.

No-one from our family had ever been to university – or even really thought about it – and Dad often jokingly claimed to be barely able to read and write. That was an exaggeration but he was certainly no scholar. He grafted all his life one way or another. But he wanted the best for his kids and decided he was going to teach me reading, writing, arithmetic and a fair bit more before I went to school. He did, by about the age of three. I don’t remember not being able to. That start made school tolerable for me and kept me out of the worst of mischief. It set me up at each stage to (sometimes only just) reach the next one. When it eventually did get hard, I finally had the self-confidence to sort it out for myself. None of that would have happened without Dad.

And it didn’t stop there. I still had a few screw-ups left in me and Dad (and Mum of course) were always there in times of trouble. And they were so pleased for both my sister, Vanessa, and me when it did go right. I love them all loads. This is a tribute of sorts – one in keeping with the flavour of these pages …

*So, here we go …*

**‘Malcolm’s Trick’**

Dad taught me a card trick when I was young. Stuff like this wasn’t really his thing so I imagine someone showed him at work and he came home and taught it to me while he remembered it. He did pretty much the same with *chess* over a longer period of time and so many other things that he held onto just long enough to pass them on. They stayed with me and developed long after Dad had probably forgotten them. That was the way of it: he always wanted the best for his family but never cared much for himself.

I’ll outline the trick Dad’s way before we go anywhere else with it. It’s in three simple stages and it’s sort of up to you just how much of it you explain to the person you’re showing it to as you go. (Don’t worry: that’ll make more sense as we go through it.) Anyway, it goes roughly like this …

**First stage**

Take a full, shuffled deck of cards, face down, and deal the first one, face up, on the table …

Ignore the suit; it’s only ever the card *values* that matter for this trick. Now, using the value of this card as your starting point, you’re going to ‘count’ from here up to ‘King’, dealing a card face up on the pile for each number you count through. So, if our starting card is an eight, we count, ‘Eight’ when we see the first card, then, ‘ Nine’ as we turn over the next card and put it on the pile. It makes no difference whatsoever to this process what the new card actually is – we ignore that! So on top of the eight goes …

… and so on … Repeating the process, from this starting card of eight, we count ‘Eight’, ‘Nine’, ‘Ten’, ‘Jack’, ‘Queen’, ‘King’:

… and stop with a ‘finished’ pile of cards, which we then turn over …

And we restart the process with another pile …

Here we just count ‘Jack’, ‘Queen’, ‘King’ …

and turn this new pile over …

We keep doing this, making new piles, until we’ve run out of cards. The ‘extreme’ cases for each starting card are: an Ace (counting as One), in which case we count all the way through ‘Ace’, ‘Two’, … ‘King’, and turn over quite a large pile; and a ‘King’, in which case we’re done before we really start, so we just count ‘King’ and turn over the single card to form the smallest of piles.

Most of the time this process won’t ‘come out’ exactly, by laying the last card of the deck as we count ‘King’ for a particular pile. It will occasionally but, more usually, we’ll be trying to make another pile but the deck will run out before we get to ‘King’. In this case, we simply collect the cards from the aborted pile back up and keep them in the ‘deck’ that we’re going to use in the next stage. The odd time it does work out exactly, we still *notionally* have this ‘deck’, it just starts off with no cards in it.

**Second stage**

Ask the person you’re showing the trick to, either one-at-a-time or all-at-once, to point to *all but three* of the piles. As they point to each pile in turn, pick it up and add the cards to the deck – always face down, without looking at any of the card faces. You now have just three piles left and a larger deck …

Next, ask the other person to turn over the * top* card from each of

*any*two of the three piles. (That’s the top card only;

*don’t*turn over the whole pile.)

**Stage three**

Now, with the rest of the deck in your hands, you’re going to ‘count off’ three sets of cards; you can either just quietly transfer them from one hand to the other, as you count, or deal them more openly to the table – still face down. (It all depends on how you’re presenting the trick to the other person: how much you want to explain it to them – or keep from them – as you go along.)

The three sets of cards you ‘count off’ are (1). The value of the *first upturned card*, then (2). The value of the *second upturned card*, and finally (3). *Another ten cards*. So, in the above example, we count off *eleven* (for Jack), then *five*, then *ten*.

We then count how many cards there are *left* in the deck and, *Hey Presto*, that number *should* match the value of the final card when we turn over the top card of the last pile. So, if we’ve counted off eleven, five and ten cards, and that then leaves us with *six* cards in the deck, then the card we turn over from the last pile should be a *six*!

__Well, it should work but you’ve probably realised how important it is to count accurately at all stages of the process; it’s easy to go wrong!__

Now, depending on how much you explain what you’re doing (the counting in stages one and three) as you go along, the trick can be presented as a mathematical curiosity (*why* does it work?), a simple feat of memory (because, in truth, you *did see* all the cards as they were originally dealt) or, with enough obfuscation, just *magic*! (We’ll develop a way to remove the ‘memory’ angle later.)

*And that’s Dad’s trick! I always liked it, partly I think because it didn’t rely on sleight of hand or any other magician’s dexterity. The numbers just worked out somehow – and I could follow the simple rules. Over the years, I learned, and forgot, many other card tricks but this one stuck with me and I wheeled it out occasionally to someone who hadn’t seen it. As already mentioned, following the same basic stages, it can be presented in different ways depending on the audience.*

But, perhaps strangely for someone in my line of work, I’d never really given much thought to *how* the trick *worked*. It was obvious that some basic mathematics were behind it, and would explain it, and that there would be a simple logical proof buried not too deeply in there, but I’d never found the time to look for it; until a few years ago – when Dad had previously become poorlier and I was thinking about him, and all things connected to him, a lot. And this is how I started thinking about Dad’s trick …

OK, a few things were trivial. In each pile, only the starting card mattered, of course, in determining the size of the pile. (And this would become the top card of each face-down pile when they were turned over.) But that applied to *all* piles equally – however many there were. So Dad’s example, where we quickly took back all but three piles into the deck (without bothering with their respective starting/top cards at all) was a special case of some sort of general one. In this particular case, irrespective of how many piles we’d managed to lay down initially, any calculation to be done could be done purely on the basis of the three that we eventually exposed the top cards for: the rest was diversion.

And once you know you’re dealing with a *general* model, a rock-solid way to get to grips with it is to think first about the very *simplest* one. Here, that’s not Dad’s three piles, but just *one* …

And that’s pretty easy to think about. Assuming (as we already have really) that Ace = *1*, Jack = *11*, Queen = *12* & King = *13*, then if the value of the starting card is *C,* the number of cards in the pile built up from *C* is *14-C*. (Try it. Starting with an Ace gives *14-1 = 13* cards, a King gives *14-13 = 1* card, and everything else works the same way in between.) Or, equivalently, when we’re done with making the piles, the top card of a pile with *14-C* cards will be a *C*. Then, if we’re just considering that single pile (starting from *C*) then there will be *52* minus this number (*14-C*) left in the rest of the deck: that’s *52 – (14-C)* or *38 + C*. So how does that help?

Well it means that the cards in the deck can be counted off in two sets: *38* of them, and *C* of them. In other words, if we count off *38*, then however many there are left will be *C*, which will be the number of the *starting* card for the pile – the *top* card when we turn it over. For example …

… takes us to …

So the very simplest version of the trick is to deal just one pile: counting *C, C+1, C+2, …, 12* (Queen), *13* (King), (or as much of it as the first card lets you), then turn them over, count *38* off the remainder of the deck and whatever’s left will be *C*. Turn over the top card and there it is: a *C*! For our little example, that’s …

Problem is, that probably doesn’t look too impressive! Depending on how you’re presenting the trick it either doesn’t take much to *remember* one card or, as a mathematical exercise, even if people don’t exactly reproduce the analysis we’ve managed here, it’s going to be pretty obvious what the *general* principle behind it is. The larger the starting/top card, the smaller the pile, so the more left in the deck … and vice versa.

So, like all good tricks, the simple essence that makes it work has to be hidden behind extra – if ultimately unnecessary – layers of diversion. As we’ll see, Dad’s version is based on exactly the same principle but the counting of the deck is camouflaged by breaking the calculation up into steps as the significant piles are reduced from three to two to one. (In fact, this works so well, we’ll eventually extend it to a more sophisticated version.)

So, we *could* repeat this individual analysis for the case of two piles, then Dad’s three, etc., but maybe we’ve got enough of a feel for this now to consider the *general* case? Let’s try …

Suppose, in the first stage, we manage to lay down *p* piles (before we run out of cards), with the starting/top card of each pile *1, 2, …, p* being *C _{1}, C_{2}, …, C_{p}* respectively. Then the number of cards in each pile

*i*(

*1≤i≤p*) will be

*14-C*and the number of cards left in the deck:

_{i}(if you’re comfortable with this ‘summation’ notation, or just

*52 – (14-C _{1}) – (14-C_{2}) – … – (14-C_{p})* long-hand if you’re not.)

which is the same as

* *(*52 – 14p + (C _{1}+C_{2} + … +C_{p})*)

This is the general formula (for the number of cards left – *at any point* – in the deck with *p* piles on the table) that drives *any* version of the trick, however we might want to present it. The critical thing to note is that the terms on the left (*52 – 14p*) can be calculated just by counting the *number of piles* and the compound term on the right is just the *total of all the starting/top cards*. That separation turns out to be pretty useful (and if we think about it a bit, not entirely surprising). To take Dad’s version as an example, once we’ve taken back into the deck the piles we’re not bothering with, we’ve *p = 3*, and the number of cards in the deck will be

Let’s assume that we’re *always* trying to *guess* *C _{1}*. (This is OK because, although we won’t generally know until the last step which pile/card we’re

*actually*trying to guess, the overall summation always works the same whichever way round we do individual additions, and we can always renumber the piles in the maths so we make it turn out this way –

*‘without loss of generality’*or

*‘wlog’*as the mathematicians call it.) Then, taking these

*10 + C*cards in the deck, if we count off

_{1 }+ C_{2}+ C_{3}*C*cards, then

_{2}*C*cards, then

_{3}*10*cards, we’re left with

*C*cards, which is the answer we want.

_{1}*And that’s how Dad’s trick works – and why it works. Nice one, Dad!*

But, in the well-established tradition of Dad giving the head-start and it then being carried on, we can extend his version to a general offering, with *any number of piles*, presented pretty much *any way we want* …

So let’s revisit that formula for the number of cards in the deck with *p* piles; in fact, let’s call that number *d _{p}*:

(*d _{p} = 52 – 14p + (C_{1}+C_{2} + … +C_{p})*)

and note some simple principles:

- The formula works at
*any*point we want it to. Whenever there are*p*piles in the table, there will be this number of cards in the deck. But also … - The formula holds true as we
*add*(or more likely,*remove*) piles. Suppose, starting from*p*piles (and a deck of*d*cards), we remove (_{p}*wlog*) pile*p*(with starting/top card*C*). Then, by induction, the number of cards in the deck becomes the old deck plus this new pile, which is:_{p}

*d _{p}* + (14 –

*C*)

_{p}*= d*

_{p-1 }(OK, yes, in a sense, this is obvious if we consider the piles as independent entities – as in the previous point – but being able to recalculate ‘on-the-fly’ like this turns out to be useful so it’s worth emphasising.)

- There are some practical limits on the value of
*p*. The*minimum*number of piles we could deal out initially would be four: starting with the lowest values: precisely four Aces, or other small values then not completing the fifth pile (although we*can*have one, two or three piles if we then remove some). The*maximum*number of piles comes from starting piles with the highest possible values: that’s four Kings, then four Queens, four Jacks, four Tens, then running out of cards after two piles starting with Nines. (Try it!) So that’s*4 + 4 + 4 + 4 + 2 = 18*. So a theoretical maximum for*p*is*18*(although, in practice, dealing out more than*10*piles is unusual). - We can, of course, extract
*C*from the summation above and rearrange the formula for the number of cards in the deck to give an_{1}*explicit*formula for the card we’re trying to ‘guess’:

(*C _{1} = d_{p} – 52 + 14p – (C_{2}+C_{2} + … +C_{p})*)

(The summation from *i=2* now represents the ‘rest of’ the piles, without the *C _{1}* pile.)

- Finally, since the number of piles (
*p*) is fairly constrained, we could make this a little easier on the eye by writing it out long-hand for each case. We can also then lose the*p*itself because that’s implicit and the number of cards in the deck becomes just*d*. So:*1*pile:*C*(as before)_{1}= d –**38***2*piles:*C*_{1}= d –**24**– C_{2}*3*piles:*C*(Dad’s version)_{1}= d –**10**– C_{2}– C_{3}*4*piles:*C*_{1}= d +**4**– C_{2}– C_{3}– C_{4}*5*piles:*C*_{1}= d +**18**– C_{2}– C_{3}– C_{4}– C_{5}*6*piles:*C*_{1}= d +**32**– C_{2}– C_{3}– C_{4}– C_{5}– C_{6}*7*piles:*C*_{1}= d +**46**– C_{2}– C_{3}– C_{4}– C_{5}– C_{6}– C_{7}*8*piles:*C*_{1}= d +**60**– C_{2}– C_{3}– C_{4}– C_{5}– C_{6}– C_{7}– C_{8}*9*piles:*C*_{1}= d +**74**– C_{2}– C_{3}– C_{4}– C_{5}– C_{6}– C_{7}– C_{8}– C_{9}*10*piles:*C*_{1}= d +**88**– C_{2}– C_{3}– C_{4}– C_{5}– C_{6}– C_{7}– C_{8}– C_{9}– C_{10}- … and so on, adding
to the figure in bold for each extra pile.*14*

In other words, to calculate the top card of *any* pile (we can rearrange the piles in the maths to make *C _{1}* whichever one we want it to be, remember), we simply (a) count the number of cards in the deck, (b) add or subtract a

**certain number**to it (dependent on how many piles there are), then (c) subtract the values of

*all*the other top cards.

Before we go any further, let’s quickly check this by extending the example we’ve used before. But this time, let’s ‘cheat’ and have a look at *all* of those top cards …

Suppose we wanted to calculate the value of the top card of the middle pile (before it was turned over). There are *9* piles so we use the formula:

*C _{1} = d + 74 – C_{2} – C_{3} – C_{4} – C_{5} – C_{6} – C_{7} – C_{8} – C_{9}*

giving, in this case, *C _{1} = 1 + 74 – 8 – 11 – 13 – 5 – 9 – 10 – 12 – 6 *

* = 1*

… the Ace in the middle! It works!

So, the only pieces of information we have to actually *remember* are the critical values (in **bold**) that apply for each different number of piles. But, really, we don’t even have to remember *all* of those because we could instead just fix on one (say ** 32** for

*6*piles) and work up and down in steps of

*14*from there. (There can be no more than

*18*piles and more than

*10*is rare, remember. Also, as we’ve already seen, we can force a particular number of piles if we want to by just taking some away.) However we choose to do it, we just need to know that key starting number; we then add on the number of cards in the deck and (one at a time) subtract the value of each of the other top cards. The value we’re left with is then the value of the final top card.

Once you’ve got this basic pattern, you can deliver the trick however you like, explaining more or less of what you’re doing, being more or less open about counting the cards and largely working with as many piles as you like. You can do all the work yourself or invite the other person to do some of it. You can even turn your back at key moments to add some extra ‘how did you do that?’ factor into it. Whatever you like, really: you (or even the other person) just have to count carefully. Here are just two of many possible variants …

**Variant One**

This variation removes any possibility of you simply remembering cards (because you never get to see them until they’re turned over) but does hint at the underlying maths. It goes like this:

- Explain to the other person how to build (count out) the piles (up to ‘King’) as in the ‘first stage’ at the start of this post. (You’ll have to trust them to get this right; perhaps do the first one for them as an example.)
- Turn your back while they do this. (If they have any queries about what to do, you can still answer them generally without looking.)
- When all the piles are made and turned over, you can turn round again and take back the deck. (Remember, you’ve not seen
*any*of these cards – except perhaps the first pile, if you did that as a demo.) - Look to see how many piles there are and work out the critical starting number. Covertly count the number of cards in the deck and add to the critical number to start your ‘running value’.
- For all but the last pile (the one you’re going to guess), ask them to turn over the top card of each pile. As they do that, subtract each top card value from your running value.
- When all but the last pile have had their top cards turned over, your running value will be the remaining top card. Make this prediction, turn it over and
*Hey Presto!*

An advantage of using as many piles as possible here is that there will never be very many cards left in the deck to count secretly. However there will be more subtractions to do in your head.

**Variant Two**

This is one of my favourite variations as it both hides the maths (as much as possible) but still effectively removes the ‘you just remembered that’ element …

- Build the piles yourself as in the ‘first stage’. Don’t explain the counting process though; try to make it look as random as possible. (It can sometimes look more random/haphazard if you quickly deal all the piles out first then turn them all over at the end.)
- Turn your back and ask the other person to move the piles around (so you won’t be able to remember what’s where). Insist each pile stays ‘intact’ though.
- Turn round again and ask them to point at some piles to remove – ‘maybe about half’ – no need to be precise (so you now don’t know which initial piles are left –
*or*where they are – so there’s no possibility of this being a memory exercise). - Take all the removed piles back into the deck without looking at any cards. Surreptitiously count the cards in the deck.
- As before, work out the critical starting number for the number of piles
*left*and add (or subtract if less than four piles – see the list above) it to/from the number of cards in the deck to start your ‘running value’ - For all but the last pile (the one you’re going to guess), ask them to turn over the top card of each pile. As they do this, subtract each top card value from your running value. (I like to make up some daft story here about the cards talking to each other: anything to distract from any calculations that may be going on.)
- When all but the last pile have had their top cards turned over, your running value will be the remaining top card. Throw in a bit more nonsense narrative, make this prediction, turn it over and
*Hey Presto!*

In this variant, with fewer meaningful piles, there are fewer subtractions on the fly but a larger number of cards in the deck to count without letting on that’s what you’re doing.

You can pretty much work through the earlier example to trace how variant one works. So, here’s a worked example of variant two: I’m actually now dealing the cards as I type …

(Remember, no audible counting: make it look random.) These are the initial piles as they’re dealt (with the start card left just visible for our convenience) …

It so happened this time that I used the last card (the Five of Hearts) as counted ‘King’ on the seventh pile so there are no cards initially in the deck. Now, turn the piles over …

Turn your back while they move the piles around then ask them to take ‘about half’ of them away. Return these cards to the deck and quietly count them.

There are now four piles left (and you’ve no idea which or where) and *23* cards in the deck. The critical number for four piles, in the list above, is (*add*) *4*. So *23 + 4* gives *27* as your running value. Suppose the other person wants you to guess the bottom right top card. Then, expose each of the others in turn (making up a nice nonsense story as you go) and subtract each value from your running total.

*27 – 4 = 23.*

*23 – 11 = 12.*

*12 – 6 = 6*. Burble out some more distracting narrative, predict that the final card will be a 6 and turn it over …

*Wow!*

And that’s it. That’s the trick … or tricks. And that’s my tribute to Dad – well, here anyway. Yes, it’s daft and, yes, there’s more and better things I can say – and will say – elsewhere. But this just seems fitting here.

So, give it a go if you get the chance. Practice it and remember the key bits if you can. See if you can find someone to impress with ‘Malcolm’s Trick’.

*And then, when you can, raise a glass to Dad please. A good man.*

*Malcolm Valentine Grout, 14th February 1931 – 4th December 2020*

December 13th, 2020 at 12:20 pm

Great read and a lovely tribute to your Dad.

May 26th, 2021 at 1:19 am

thumbs up. Sorry to hear about your Dad passing. Great memories.

September 29th, 2021 at 10:20 pm

What a lovely tribute to your Dad, a very engaging tale which kept me reading, despite the blur of mathematical equations which I must admit I scrolled past!

January 30th, 2023 at 4:39 pm

Sorry to hear about your Dad, great tribute and lovely to see that he passed this on to you.