Splitting the Restaurant Bill

The problem

Alice and Brian go out to lunch. Brian drinks only water with his sandwich, but Alice has a glass of wine and a more expensive sandwich. After the meal, Alice leaves her credit card on the table and goes to the restroom. When the waiter comes, Brian gives him his own and Alice’s credit cards and tells him to split the $50 bill evenly. The waiter does so and charges each card $25. Alice comes back and objects that since her portion of the bill came to $30 and Brian’s to $20, she should pay more. How much cash should she give Brian so that she ends up paying her fair share?


When I presented this problem to a mathphobic writer friend, she came up with the answer within a few seconds — $5. A bit more probing and she devised a general rule. Alice should pay Brian half the difference between the two portions. Alice pays Brian (30 – 20)/2 = $5. Now the cost of the meal to Alice is $25 (on her credit card) plus $5 cash to Brian, which comes to $30. And Brian is out $25 (on his credit card), but he gets $5 cash from Alice, so his net cost comes to $20.

General rule

A and B evenly split a bill totalling t = a + b. Assume a is greater than b. Then A should pay B half the difference between a and b. That is (a – b)/2.

Why does this always work? A has paid half the total bill, (a + b)/2 dollars, on her credit card. But she should be paying a dollars. So she owes the balance, a – (a + b)/2, to someone. Is there a simpler way of writing that down? We take a deep breath and do some secondary school algebra.

a - \frac {(a+b)} {2} = \frac {2a} {2} - \frac {(a+b)}{2} = \frac {2a-a-b}{2} = \frac{a-b}{2}

You may have done that operation in a less laborious manner. I put in the extra steps for those whose algebra is shaky. What we have done is to show that Alice owes (a-b)/2 dollars. That’s half the difference between her share and Brian’s share.

What about B? He has paid (a + b)/2 dollars. How much more is half the bill — (a + b)/2 — than the b dollars he should be paying? Well,

\frac {a + b} {2} - b = \frac {a+ b - 2b} {2} = \frac {a-b} {2}

In summary, the amount A owes — (a – b)/2 — is exactly the amount owed to B. So A simply pays this amount to B.

Painful elaboration of the obvious (PEOTO)

There are three reasons for this painful elaboration of the obvious.

First, I wanted to demonstrate that the formula will work no matter what a and b are.

Second, remember that I asked you to assume that a is greater than b. Of course if b is greater than a, you can just relabel a as b, A as B and so on. But I want a system that works without having to do that.

Third, I want to devise a method that will work for any number of lunchers. In short, I want to generalize the problem.

General Problem

n people go out to lunch. Person1’s portion of the bill comes to d_1 dollars, person2’s portion comes to d_2 dollars, and so on. So the total of the bill comes to

t = d_1 + d_2 + d_3 + \dotsb + d_n

dollars. The ellipsis (…) tells you to fill in the missing terms for yourself using the pattern of the first terms.

The lunchers split the bill evenly on their credit cards. How should they arrange payment among themselves so that each ends up paying the right amount? Remember — the total amount paid by those who owe must equal the amount collected by the rest.

Answer next week.

About aharmlessdrudge

Way back during the late Bronze age -- actually it was the 1950s -- all of us in high school had to take a vocational test to determine our interests and, supposedly, our future careers. I cannot remember the outcome, but I do recall one question that gave me pause. "If you were to win a Nobel prize, would it be in literature or in physics?" I hesitated over the question: although I enjoyed mathematics and science more than English class, I did have a couple of unfinished (and very bad) novels hidden away at home. I cannot remember what I chose back then, but the dilemma followed me to university, where I switched from mathematics to English and -- after a five-year stint in journalism -- back to mathematics. I recently retired as a professor of statistics. Retirement. What a good chance to revive my literary ambitions. I have finished a novel -- more about that in good time -- and a rubble of drafts of articles about mathematics and statistics is taking up space on my hard disk.
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