Missing dollar riddle
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 bellboy $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 key to understanding the riddle intuitively is to realize that, while each man did pay $9, he did not pay $9 to the hotel. Each one paid $8⅓ to the hotel ($8⅓ × 3 = $25), and $⅔ to the bellhop ($⅔ × 3 = $2), for a total of $27 spent by the group. $25 + $2 = $27, and so does $9 × 3.
The initial payment of $30 is accounted for as follows: the clerk takes $25, the bellhop takes $2, and the guests get a $3 refund. It adds up. After the refund has been applied, we only have to account for a payment of $27. Again, the clerk keeps $25 and the bellhop gets $2. This also adds up.
There is no reason to add the $2 and $27; the $2 is contained within the $27 already. Thus the addition is meaningless (mixing cost and cash). Instead the $2 should be subtracted from the $27 to get the revised bill of $25.
This becomes clear when the initial and net payments are written as simple equations. The first equation shows what happened to the initial payment of $30:
- $30 (initial payment) = $25 (to clerk) + $2 (to bellhop) + $3 (refund)
The second equation shows the net payment after the refund is applied (subtracted from both sides):
- $27 (net payment) = $25 (to clerk) + $2 (to bellhop)
Both equations make sense, with equal totals on either side of the equal sign. The correct way to get the bellhop's $2 and the guests $27 on the same side of the equal sign ("The bellhop has $2, and the guests paid $27, how does that add up?") is to subtract, not add:
- $27 (final payment) - $2 (to bellhop) = $25 (to clerk)
This question is very simple when thought of in a different manner. The hotel receives $30 from the guests and gives $5 to the bellhop to return to the guests. $30 minus $5 equals $25. $3 is given to the guests and the bellhop keeps $2. Thus, $3 plus $2 equals $5, and the remaining $25 is in the possession of the hotel.
This is clearly not a paradox, and involves only the switching of subtraction for addition. Each patron has paid $9 for a total of $27. The storyteller adds the $2 that the bellhop pilfered, but he should have subtracted the $2 to make a total of $25 paid. So 3 × $9 = $27, which accounts for the $25 room and the $2 given to the bellhop.
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 a 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 missing dollar come from?
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