There's an old scam in which a victim is offered a free choice of any five of ten face-down cards, to serve as a poker hand, the other five being retained by the person offering the cards. When the resulting poker hands are compared, it turns out that the victim has the losing hand.
This can be repeated as often as is wished. This principle, often referred to as the "Ten Card Poker Deal," goes back at least as far as Card Control (2nd ed., 1947) by Arthur Buckley.
Another version sees the victim being permitted to win a few times at first, in order to instill a false sense of confidence, perhaps before the game is played for money. Then, the winds of fortune mysteriously change.
For one performance only, the cards could be offered openly, face up, one at a time. A high-valued card such as Ace is an early contender for attention, it being assumed that the victim won't be able to resist its fatal charms.
The secret is very simple. As hinted at above, there is a single dud among the ten cards---in isolation it may appear to be valuable---such that whoever gets it loses without a doubt. The other nine cards are carefully planned in advance of course, but the person offering the card choices just has to make sure that the victim gets stuck with the dud.
We've been deliberately vague about how the cards are offered to the victim and then selected; various methods can be used. Perhaps the dud's back is subtly marked, so that it can easily be tracked, and steps are taken to unload it ASAP.
Such not-so-honest approaches are far from the spirit of mathemagical effects, so in what follows we instead offer three ingenious and fair-seeming ways to arrive at the desired card distribution. These methods have been explored in previous Card Colms, and can be used here to control who wins on each round.
The required ten cards are: three sets of three of a kind, for instance three 3s, three 8s, and three Kings, together with the dud, which is any non-matching card---below we'll settle for a Jack---which is known as the Jonah card. Whoever gets the Jonah card loses, without fail.
The suits here can be arbitrary; for that matter the relative values play no role either, as they are never compared.
Before we show why success or failure here always boils down to control of the Jonah card, we need to be clear as to the ranking of the various possible poker hands, i.e., selections of five cards from a standard deck.
There are 52!/(5!47!), or roughly 2.6 million such selections, which is so many that even if you were to play enough poker in your life to encounter 100,000 hands, you'd still only have seen less than 5% of them.
The relative scarcity (and hence value) of some of the commonly desired Poker hands is reflected in this chain, starting with the rarer hands:
flush> straight > three of a kind > two pairs > one pair.
The rankings are transitive, so that for instance, a flush beats two pairs, but four of a kind beats both.
Returning to the Jonah card issues, let's suppose that the cards in question are: 3♣, 3♠, 3♦, 8♥, 8♣, 8♦, K♥, K♠, K♦ and J♠.
First focus on the hand with the Jonah card; it's not difficult to see that one of the following three cases must hold.
Case 1: This hand also contains three of a kind, and one other card, as shown in the top row here. The bottom row is the other hand.
Figure 1: Typical losing and winning hands in Case 1
Then the other hand wins, being a full house containing three of a kind and a matching pair. There are 54,912 five-card hands with just three of a kind, compared to only 3744 full houses. The hands are something like what is shown above.
Case 2: The hand with the Jonah card also contains two pairs, as shown in the top row here. The bottom row is the other hand.
Figure 2: Typical losing and winning hands in Case 2
Then the other hand still wins, containing three of a kind and two unmatched cards. There are 123,552 hands with two pairs, compared to only 54912 hands with just three of a kind. The hands look something like what is shown above.
Case 3: The hand with the Jonah card also contains one pair and two non-matching cards, as shown in the top row here. The bottom row is the other hand.
Figure 3: Typical losing and winning hands in Case 3
Then the other hand wins one more time, as it contains two pairs. There are 1,098,240 hands with just one pair, compared to 123,552 hands with two pairs. The two hands look something like those above.
In conclusion: the trick, so to speak, is to control who gets the Jonah card.
Here are three clever ways to control who gets the Jonah card, while appearing to randomise how the cards are distributed.
- In the April 2006 Card Colm, we explained Bill Simon's 64 Principle in general, and two months later applied it to the control of ten cards for two poker hands in Better Poker Hands Guaranteed, specifically in "Better Poker Hands with Bill Simon." The probabilistic considerations there can be ignored however; we know what we're dealing with here, and there's only one card to control.
- It was also in the June 2006 Card Colm, that we introduced a Position Parity method of controlling card distribution for two poker hands, in "Martin Gardner's Coins to Cards Effect," and "Better Poker Hands with Martin Gardner."
- In the October 2008 Card Colm, we explored the Monge Shuffle, which is an in-hand way to mix cards.
As seen in "Better Poker Hands With Gaspard Monge" there, it's especially interesting for a packet of ten cards, because the cards in positions two, five and eight just keep cycling around, and the one in position four doesn't move at all. So the ten cards can be subjected to as many Monges as the victim asks for, and assuming the Jonah card started in one of the four key positions just mentioned, it's still in one of those slots. The losing hand can be teased out using a spelling routine as suggested in that 2008 column, but a much easier option is available.
- When using Bill Simon's 64 Principle, simply start with the Jonah card in any of positions three to six from the top of the packet of ten cards. Plenty of innocent-looking in-hand shuffling can be done which preserves that key fact, before launching into the "free-choice" phase.
- To implement the Position Parity method, pay attention to which end of the row of ten the selections start from, and who picks first. That way, you control everything just by knowing if the Jonah card starts in an even- or odd-numbered position. Additional shuffling of various types can be done which won't change the outcome, for instance, experiment with running off---into a waiting hand---either an even or odd number of cards, and putting them back at the face or behind the remainder. (Start with alternating Black and Red cards to get a feel for the possibilities.)
- It's easiest of all to Monge Shuffle repeatedly: arrange it so that the Jonah card is in the position four at the outset. It stays there throughout any number of Monges, so that when the victim is satisfied that the cards are mixed, you can either say, "You take the top five, I'll take the rest, let's see how we did," or deal into two piles, alternating, making sure the victim gets the second one. You could even add more layer of deception in the dealing situation; asking for one of the piles to be indicated, saying, ``You want me to have that pile? Fine," if it's the first one, and "That's the pile you want? Okay," if it's the second one.