“A card force is one of any number of methods used in close-up magic to apparently offer a subject a free or random choice of card, when in fact the magician knows in advance exactly which card will be chosen. This can then be revealed later in the trick.”
Premise: We have a set of three cards, all of which are known to us. We attempt to force a specific card on the unsuspecting participant by instructing them to randomly point at one of three cards, which are all laying face down. Theoretically, they have a one in three chance of picking the correct card randomly. If they point at the desired card, we immediately instruct them to flip it, effectively “forcing” the card on them in one try. If this works on the first try, the trick will be especially impressive. However, if it doesn’t work on the first try, we still have a fall-back method:
If they point at one of the other cards, instead of telling them to flip the card, we pretend that they’re playing a game of elimination and we simply remove the card, instructing them to point again. If they point at the next incorrect card, we instruct them to remove it, leaving one final card: the correct one.
Potential downside: If they point at the desired card on the second step, then the “force” fails,
because when we remove it, the end card will actually be the incorrect one.
Cards used: Ace of Hearts (the desired card), Queen of Clubs, and 8 of Spades
Approximate Results After 2000 Simulations:
Times ended on Ace of Hearts: 1322 (0.661)
Times ended on Queen of Clubs: 339 (0.1695)
Times ended on 8 of Spades: 339 (0.1695)
Odds of picking the right card during the first step: 1/3. If incorrect card is chosen on first step, we pretend it’s an elimination game and remove the card, leaving only two cards, the desired card, and the incorrect card. At this point, you might think there is an even 50/50 chance they will choose the correct card, but in reality, the odds of them choosing and eliminating the desired card are still 1/3 because it hasn’t been touched. The odds of them choosing and eliminating the second incorrect card, however, have increased to 2/3. This is unintuitive, but the simulation shows it to be true.
Because of the fact that we know which cards are which, we can effectively double our seemingly low 33% odds all the way to 66% simply by using this card force method.
The only unsolved problem is, how do you handle a dead end where someone chooses the wrong card, and then the right card on the second step?