The Down Under Deal

One card down onto the table, the next under the packet, round again until one card is left. Which card that is was settled before you started, by how many you were holding.

The deal is old and the name is literal: cards go down to the table and under the packet, and it is also called the Australian deal for the same joke. It turns up inside a great many self-working tricks, usually as the thing that quietly gets rid of fifteen cards while the audience watches a fair-looking procedure.

Which card survives, for any packet

Pick a size. Both answers below are produced by dealing that packet through, one card at a time, rather than by a formula.

The rule you can do in your head

Find the largest power of two that fits inside your packet — 1, 2, 4, 8, 16, 32 — and take it away. Double what is left over. That is the position that survives, counting from the top.

Twenty cards: sixteen fits, four is left, and the eighth card survives. Fifty-two cards: thirty-two fits, twenty is left, and the fortieth survives.

And if nothing is left over, the bottom card survives. That is the case worth memorizing, because it is the one you can build a trick on. Eight, sixteen, or thirty-two cards dealt down-under always leaves the card that started at the bottom, no matter what the cards are or who does the dealing.

Starting with under instead

Put the first card under and deal the second one down, and everything shifts by one. The surviving position becomes double-the-leftover plus one, and on a power of two it is the top card rather than the bottom.

So the two deals never agree on any packet larger than a single card, and they cannot: starting with down can only ever leave an even position, and starting with under can only ever leave an odd one. Between them they give you either end of a sixteen-card packet, and the choice is made by which action you take first.

How to perform it

1. Count out sixteen cards. Any sixteen, from a deck somebody else has shuffled. Counting them into a pile reverses them, which is worth knowing about before it surprises you.
2. Get their card to the bottom. The easiest honest way is to let them look at the face-down packet's bottom card and remember it. Nothing is switched and nothing is forced — the card is genuinely theirs.
3. Hand them the packet and let them do the dealing. One card down to the table, one card under, and repeat. You do not touch it. That is the whole reason this deal is worth having: the procedure looks like a mixing process and is the opposite of one.
4. One card is left. It is theirs. Because sixteen is a power of two, it could not have been anything else.

If you would rather their card started on top, tell them to begin with the under instead of the down. Same packet, same deal, other end.

Why does the down under deal work?

Because one pass through a packet of even size throws away exactly half of it and leaves the other half in order.

Deal sixteen cards down-under and the eight cards that went under are still in your hand, still in the same relative order, and the card that began at the bottom is still at the bottom — it went under on the last action of the pass rather than down onto the table. Now you are holding eight, and the same thing happens. Then four, then two, then one. The bottom card is never the one being dealt down, so it survives every pass, and when the packet is a power of two the passes come out exactly even every time.

When the packet is not a power of two, the first pass is uneven and the deal effectively starts over on a smaller packet with a shifted starting point. That shift is what the doubling accounts for.

This is the Josephus problem with every second position removed. The problem is usually told about people standing in a circle, and a packet of cards dealt one down and one under is the same loop: the cards on the table are the ones taken out, and the cards going under are the ones passed over.

Using this as a Josephus problem calculator

The selector above is one. Set the packet size to the number of people in the circle and read off the surviving position — and because the answer is produced by dealing the packet through one card at a time rather than by evaluating a formula, it is a simulator as much as a calculator.

Choose the under-first answer if you want the textbook figure. The classical problem counts person one, removes person two, and carries on around the circle, so the first position survives its first count — which is the card going under, not down. That is why under-first gives 2L + 1, the standard result, where L is however much the circle exceeds the largest power of two inside it.

Down-first removes position one immediately instead, which is the same circle entered one step later, and it gives 2L. Both are dealt out in full below, so the two conventions can be compared on any size rather than argued about.

Questions people ask

What is the down under deal?

A way of dealing a packet of cards. The top card goes down onto the table, the next card goes under the packet, and you repeat until one card is left in your hand.

It is also called the Australian deal, because the cards go under. Nothing is hidden and nothing is sleight of hand: which card is left is decided entirely by how many you started with.

Which card survives the down under deal?

Take the largest power of two that fits inside your packet and subtract it from the packet size. Double what is left, and that is the position that survives, counting from the top.

Twenty cards: sixteen fits, four is left over, and the eighth card survives. If nothing is left over, the packet is a power of two and the bottom card survives.

Why does a power of two always leave the bottom card?

Because one pass through a packet of even size removes exactly half of it and leaves the other half in the same order, with the bottom card still on the bottom.

Halve sixteen and you have eight, then four, then two, then one, and the bottom card is never the one being dealt down. It survives every pass, so it survives the deal.

What is the difference between the down under deal and the under down deal?

Which action you start with. Down first leaves position 2L, where L is the amount by which the packet exceeds the largest power of two inside it. Under first leaves 2L plus one.

So the two deals never agree on any packet bigger than a single card, and down first can only ever leave an even position while under first can only ever leave an odd one.

Is the down under deal the same as the Josephus problem?

It is the same arithmetic. The Josephus problem asks who is left when every second person in a circle is removed, and dealing one down and one under removes every second card from a loop in exactly that way.

The cards on the table are the ones removed, and the cards going under are the ones passed over.

Harrison Alley

Hobbyist · Writes Sleightless

A hobbyist, not a professional. I write up the self-working end of card magic, because it is the door I would point a beginner at, and every trainer here is played through and checked before it ships.

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The 21 card trick is the other deal whose answer depends only on the packet size, and it has a calculator of its own: three piles instead of one down and one under, and the card lands in the middle rather than at an end.

The binary card trick runs on powers of two as well, from the other direction — there they build a number up rather than cut a packet down.