Little Fibs
You choose any two of six cards and say only their total. From that one number, both cards are named, suits and all.
Watch the lectureLecture 48 · Little Fibs: two cards from their sum
Try it
Mix the six cards, choose any two while I look away, and tell me only their total. Every choice is yours.
Why it works
Addition usually loses information. A total of 8 could be 1 + 7, 2 + 6 or 3 + 5, if all those cards were in play. This packet was chosen so that nothing is lost. Its values 1, 2, 3, 5, 8, 13 are the first six distinct Fibonacci numbers: each one is the sum of the two before it.
Two different cards can be chosen from six in 6 × 5 ÷ 2 = 15 ways, few enough to check every one. Step through the table below. Each row collects the pairs that share the same larger card. Watch where each row ends, and where the next one begins. Every pair that includes the king totals at least 13 + 1 = 14; every pair without it totals at most 8 + 5 = 13. The same split happens at every card.
Each row of totals starts above where the row before it ended, so no total ever has two explanations.
The decoder reads the table backwards. A total’s row belongs to the largest value strictly below it; subtracting that value gives the column. The word strictly matters at 13, which is 5 + 8: taking the king itself would leave 0, and 0 is not a card.
Decoding and checking are separate jobs. A total of 12 lands in the eight’s row, but 12 − 8 = 4, and there is no 4. That is not a second answer. It means the addition went wrong, so ask for a recount and never invent a pair.
One more piece of information arrives before the total: exactly two different cards were chosen. The king alone is worth 13 and the three alone is worth 3, but single cards are not allowed choices, so they never compete with 5 + 8 or 1 + 2.
Challenge
Edit the packet. First break the code by making two different pairs share a total. Then find another six-card packet, values ace to king, whose fifteen pair totals are all different. There are exactly three. The three-card and four-card settings show where uniqueness stops, and where it comes back.
Perform it yourself
Set-up
- Take out the ace of clubs, two of hearts, three of spades, five of diamonds, eight of clubs and king of hearts. The packet is shown openly; there is no hidden stack.
- Agree on values before anyone chooses: the ace counts 1 (not 11), the king counts 13 (not 10), and the others count as printed.
- Learn the rule in step 4 below, or keep the fifteen-line table on a slip.
Performance
- Name all six cards and their values, so the spectator knows exactly what is available. Let them mix as thoroughly as they like; order plays no part.
- Turn away. The spectator chooses any two different cards, adds their values (on paper if they like), checks the addition and announces only the total: no values, no suits.
- Check that the total is possible: 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 18 or 21. If it is not, say: “That total is not possible under these rules. Please check that there are two cards, with ace one and king thirteen, and add again.” Keep the cards hidden and never invent a pair.
- Decode: take the largest value strictly below the total; the remainder is the other card. For 13, take 8, not 13: the cards are the five of diamonds and the eight of clubs.
- Name both cards before they are shown, then have them revealed.
- Have both cards returned and all six mixed again before each repeat. The same pair may come up again.
Exactly two different cards, every time. One card or three cards change the problem: with three, 1 + 2 + 13 and 3 + 5 + 8 both make 16.
| Total | Cards | Total | Cards |
|---|
Going deeper
Three cards collide; four cards do not
Allow three cards and there are twenty possible choices. Exactly one total repeats: 1 + 2 + 13 = 3 + 5 + 8 = 16. Both are genuine three-card choices, so no decoder can split them from the total alone. They are complements, using all six cards between them; the whole packet totals 32, and 16 is half of it.
Four cards are decodable again, through the two cards left behind. If the chosen four total S, the missing pair totals 32 − S, and pair totals are unique. Four cards totalling 21 leave a pair totalling 11, which is 3 + 8, so the four are 1, 2, 5 and 13.
This is not Zeckendorf’s theorem
Zeckendorf’s theorem says every positive whole number is a sum of Fibonacci numbers from 1, 2, 3, 5, 8, … in exactly one way, provided no two neighbours in the sequence are used. It is tempting to quote it here, but its conditions do not match the trick: a spectator may choose neighbours such as 5 and 8.
Little Fibs asks a different question, with the number of cards fixed at two. Under Zeckendorf’s rules, 13 is the single value 13; under the pair rule, 13 is 5 + 8. Both answers are right, because the allowed objects differ. A uniqueness claim always needs its rules stated.
Why the shortcut is exact for Fibonacci values
Call the larger chosen value b and the total s. The total is more than b, because the other card is positive. It is at most the next value in the sequence, because the other card is at most the value just before b, and those two add to the next value. So b is the largest available value strictly below s.
The same argument works for any longer packet that continues the same sequence: 1, 2, 3, 5, 8, 13, 21, …. Other packets can have unique pair sums without the shortcut. In the challenge, 1, 2, 3, 7, 10, 13 has fifteen different totals, yet the largest value below 17 = 7 + 10 is 13.
Other packets, repeated values and suits
The design requirement is only that sums of two different available values never repeat; a Fibonacci recurrence is one way to achieve it. The five values 1, 2, 4, 6, 10 give ten different totals. The four values 1, 2, 3, 4 fail: 1 + 4 = 2 + 3 = 5.
We never choose the same card twice, so 1 + 3 = 2 + 2 is no threat: there is only one card of value 2. Rules that allow repeated summands make a stronger requirement. And if two cards shared a value, the total could still identify values while failing to identify suits.
Sources
Colm Mulcahy, Mathematical Card Magic: Fifty-Two New Effects, Chapter 4, “Little Fibs”, pp. 89–92 (the chapter continues to p. 103 with the triple collision, the complement idea and other value sets). The original effect camouflages a prepared stock inside a larger deck; this version uses the six-card packet openly, so the mixing needs no control.
Colm Mulcahy performs the trick for Numberphile: Little Fibs.
From episodes 48 and 49 of Magic & Math with Cards.