Missing dollar riddle
Although the wording and specifics can alter, the puzzle runs along these lines:
Three people check into a hotel room. The clerk says the bill is $30, so each guest pays $10. Later the clerk realizes the bill should only be $25. To rectify this, he gives the bellhop $5 to return to the guests. On the way to the room, the bellhop realizes that he cannot divide the money equally. As the guests didn't know the total of the revised bill, the bellhop decides to just give each guest $1 and keep $2 as a tip for himself. Each guest got $1 back, so now each guest only paid $9, bringing the total paid to $27. The bellhop has $2. And $27 + $2 = $29 so, if the guests originally handed over $30, what happened to the remaining $1?
The misdirection in this riddle is at the end of the description, where a bunch of unrelated totals are added together, and the listener assumes these numbers should add to 30. There is, in fact, no reason this sum should add to 30. The exact sum mentioned in the riddle is computed as:
SUM = $9 (payment by Guest 1) +
$9 (payment by Guest 2) +
$9 (payment by Guest 3) +
$2 (money in bellhop's pocket)
The trick here is to realize that this is not a sum of the money that the three people paid originally, as that would need to include the money the clerk has ($25). This is instead a sum of a smaller amount the people could have paid ($9 * 3 people = $27), added with the additional money that the clerk would not have needed had they paid that smaller amount ($27 paid - $25 actual cost = $2). Another way to say this is, the $27 already includes the bellhop's tip. To add the $2 to the $27 would be to double-count it. So, the three guests' cost of the room, including the bellhop's tip, is $27. Each of the 3 guests has $1 in his pocket, totalling $3. When added to the $27 revised cost of the room (including tip to the bellhop), the total is $30.
To obtain a sum that totals to the original $30, every dollar must be accounted for, regardless of its location.
Thus, the sensible sum that we really desire is this one:
$30 = $1 (inside Guest pocket) +
$1 (inside Guest pocket) +
$1 (inside Guest pocket) +
$2 (inside bellhop's pocket) +
$25 (hotel cash register)
This sum does indeed come out to $30.
To further illustrate why the riddle's sum does not relate to the actual sum, we can alter the riddle so that the discount on the room is extremely large. Consider the riddle in this form:
Three people check into a hotel room. The clerk says the bill is $30, so each guest pays $10. Later the clerk realizes the bill should only be $10. To rectify this, he gives the bellhop $20 to return to the guests. On the way to the room, the bellhop realizes that he cannot divide the money equally. As the guests didn't know the total of the revised bill, the bellhop decides to just give each guest $6 and keep $2 as a tip for himself. Each guest got $6 back: so now each guest only paid $4; bringing the total paid to $12. The bellhop has $2. And $12 + $2 = $14 so, if the guests originally handed over $30, what happened to the remaining $16?
Now it is more obvious that the question is silly. One cannot simply add a bunch of payments together and expect them to total an original amount of circulated cash.
More economically, money is accounted by summing together all paid amounts ( liabilities ) with all money in one's possession ( assets ). That abstract formula holds regardless of the relative perspectives of the actors in this exchange.
- The guests of the hotel paid $27, but also have $3 among their pockets at the story's end. Their assets are $3, and their liabilities are $27 ( $30=27+3 ) Thus the original total is accounted.
- From the perspective of the hotel clerk, the hotel has $25 in assets and lost $5 in liabilities ($30=25+5).
- From the perspective of the bellhop, his assets are $2, and his liabilities are $3 to guests and $25 to the register at the desk. ($30 = 2+3+25).
There are many variants of the puzzle. Professor David Singmaster's Chronology of Recreational Mathematics suggest these type of mathematical misdirection puzzles descended from a problem in an 18th-century arithmetic, Francis Walkingame's Tutor's Assistant which was published, and republished, from 1751 to 1860 where it appeared on page 185, prob. 116 in this form, "If 48 taken from 120 leaves 72, and 72 taken from 91 leaves 19, and 7 taken from thence leaves 12, what number is that, out of which, when you have taken 48, 72, 19, and 7, leaves 12?" Singmaster adds, "Though this is not the same as the withdrawal problems below, the mixing of amounts subtracted and remainders makes me think that this kind of problem may have been the basis of the later kind."
An 1880 misdirection is given as "Barthel sees two boxes at a jeweller's, priced at 100 and 200. He buys the cheaper one and takes it home, where he decides he really prefers the other. He returns to the jeweller and gives him the box back and says that the jeweller already has 100 from him, which together with the returned box, makes 200, which is the cost of the other box. The jeweller accepts this and gives Barthel the other box and Barthel goes on his way. Is this correct?"
A model more similar in style to the modern version was given by Cecil B. Read in his 1933 Mathematical fallacies. His puzzle produces an extra dollar. A man puts $50 in the bank. Then on subsequent days he withdrew $20 leaving $30; then withdrew $15 leaving $15; withdrew $9 leaving $6; withdrew $6 leaving $0. But $30 + $15 + $6 = $51. Where did the extra dollar come from?
The actual solution to this riddle is to add correctly (correct time, correct person and correct location) from the bank point of view which in this case seems to be the problem: 1) First day: 30$ in the bank + 20$ owner already withdrew = 50$ 2) Second day: 15$ in the bank + (15$ + 20$ owner already withdrew)= 50$ 3) Third day: 6$in the bank + (9$ + 15$ + 20$ owner already withdrew) = 50 $
From the owner point of view the correct solution is this: 1) First day: 20$ owner already withdrew + 30$ in the bank = 50$ 2) Second day: 20$ owner already withdrew+15$ owner already withdrew + 15$in the bank = 50$ 3) Third day: (20$ owner already withdrew + 15$ owner already withdrew + 9$ owner already withdrew) +6$in the bank = 50$
The solution appears very obvious if the owner withdraws every day only 10$ from 50$. To add up 40+30+20+10 using the same pattern from above would be too obviously wrong (result would be 100$).
The answer to the question: „Where did the extra dollar come from?” From consecutively adding the bank rest from three different days. This way is correct only if the money owner withdraws every day exact half of the money. Than it will add up. (25$ + 12,5$ +6,25$) + 6,25$ = 50$
Another entry from 1933, R. M. Abraham's Diversions and Pastimes (still available in a Dover version) poses a slightly different approach with this problem from page 16 (problem 61). "A traveller(sic) returning to New York found that he had only a ten dollar postal money order, and that his train fare was seven dollars. The ticket clerk refused to accept the money order, so the traveller went across the road to a pawn shop and pawned it for seven dollars. On his way back to the station he met a friend, who, to save the traveller the trouble of returning to redeem the money order, bought the pawn ticket from him for seven dollars. The traveller then bought his ticket and still had seven dollars when he got to New York. Who made the loss?" David Darling in his The Universal book of Mathematics, credits this as an earlier version of the three men in a hotel version above.
Even more similar is the English, The Black-Out Book by Evelyn August in 1939; What happened to the shilling?, pp. 82 & 213. Three girls each pay five shillings to share a room. The landlord refunds 5 shillings via the bellboy, who gives them each one and keeps two.
And one more from the same theme appears in an Abbot and Costello routine in which Abbot asks Costello for a fifty dollar loan. Costello holds out forty dollars and says, "That's all I have." Abbot responds, "Fine, you can owe me the other ten."
- The Universal Book of Mathematics: From Abracadabra to Zeno's Paradoxes Publisher: Wiley, 2004 ISBN 9780471270478