ONE costs $1

TWO costs $2

ELEVEN costs $11

The math teacher wants to buy TWELVE. How much does TWELVE cost?

Watch the video for a solution.

**The World’s Easiest Math Word Problem – Can You Solve It?**

Or keep reading.

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**Answer To The World’s Easiest Math Word Problem**

If you guessed $12 without thinking, then you’re correct! This *is* the easiest word problem.

This puzzle involves a clever anagram. The letters on the left side of the equals sign can be re-arranged to form the letters on the right side:

TWO PLUS ELEVEN = ONE PLUS TWELVE

Interestingly the mathematical equation is also true. And both anagrams have the same number of letters, 13, which is coincidentally the correct answer to each sum! One more thing: 2 + 11 = 1 + 12 is also an anagram of numerals.

The word problem can be solved by noting TWELVE has the letters of ELEVEN and TWO minus the letters for the word ONE. This implies the cost of TWELVE is the cost of ELEVEN plus TWO minus the cost of ONE. Thus,

cost(TWELVE) = cost(ELEVEN) + cost(TWO) – cost(ONE)

cost(TWELVE) = 11 + 2 – 1

cost(TWELVE) = 12

So this is the easy answer to the easiest word problem: the word TWELVE costs $12.

**The formal algebra problem with extra details**

We do not know how much each letter costs, but we can readily solve for the word twelve. Let’s use a lowercase letter variable to indicate the cost of a particular uppercase letter. Then we have:

*o* + *n* + *e* = 1

*t* + *w* + *o* = 2

*e* + *l* + *e* + *v* + *e* + *n* = 11

The last equation can be re-written by grouping like variables:

3*e* + *l* + *v* + *n* = 11

We want to solve for TWELVE, which is:

*t* + *w* + 2*e* + *l* + *v*

After experimentation, we can find the relationship:

*t* + *w* + 2*e* + *l* + *v* = (3*e* + *l* + *v* + *n*) + (*t* + *w* + *o*) – (*o* + *n* + *e*)

Substituting the known costs then leads to:

*t* + *w* + 2*e* + *l* + *v* = 11 + 2 – 1 = 12

**Sources and further reading**

http://blog.tanyakhovanova.com/2017/03/the-hidden-beauty/

https://twitter.com/neiltyson/status/144594005852831744

http://old.qi.com/talk/viewtopic.php?t=34230&start=0&sid=3f281a47a14fd758ff5566d0ca5c3584

https://mathjokes4mathyfolks.wordpress.com/2011/05/06/anagram-game/

https://wordsmith.org/board/ubbthreads.php?ubb=showflat&Number=196932

]]>image by Ron Cogswell. CC BY 2.0

The Chicago Cubs won their first World Series championship since 1908. To commemorate the end of a 108 years drought, they are gifting 1,908 employees with extravagant championship rings that have 108 diamond studs. The rings could fetch $50,000 or more, according to a Chicago Sun-Times story.

This is an extremely lavish celebration–the rival Chicago White Sox gave a ring with 432 diamond studs when they won in 2005. Furthermore, since gifts are taxable, the Cubs are even paying for the taxes of some employees, so this gift comes without any out of pocket cost.

(A fun tangent here: in theory, paying the tax could also be considered a gift, which would then also be taxable. And if the Cubs paid that, that would also be another gift, which would be taxable. This creates a kind of pyramiding tax which is calculated as the sum of an infinite geometric series. For a tax rate *t*, we can readily calculate the total tax = (gift)*t*/(1 – *t*). But to encourage leases where the lessor pays taxes, and simplify the calculation, the IRS stops the pyramid after one step–after taxes are paid on a gift, no further taxes are incurred).

The organization faces a game theory dilemma of gift giving. The organization can either give a generous gift or stingy one, and a gift recipient can either honor the gift or sell it. Knowing an employee would sell the gift, the Cubs would probably rather give a stingy gift. So employees will claim they will never sell the gift in order to get a generous gift. But once an employee gets a generous gift, the employee could readily turn around and sell it. (Players will pay taxes and will be allowed to sell the rings, but presumably they have enough wealth they will not immediately sell them for money).

This suggests there are two Nash equilibria to this game: (stingy, sell gift) and (generous, do not sell). Which outcome is achieved depends on whether the Cubs believe many employees will sell the gift.

The Cubs want to give a generous gift, but they cannot legally bind employees from selling it. So the Cubs have taken an unusual step: they have added a catch that they can always re-purchase a ring for $1. The re-purchase clause also applies to any future owners of the ring.

This is an example of a right of first refusal contract. The person with the right has privileges on the asset before an owner can sell it–for example, a sports team may be able to offer a matching contract before a player signs with another team.

The Cubs contract aims to diminish the resale value of the ring, because who would pay for a ring which can be re-purchased for only $1? Amazingly, what the Cubs are doing has legal precedence with the Oscars, which I detail below.

**The Oscar is not really yours**

Who really won Best Picture last year? Arguably all Oscars still belong to the Academy Award which is kind of loaning them out.

I used to think actors could do whatever they wanted with the award. But it turns out the Academy Award does not want its Oscar awards reputation to be sullied with secondary sales (the Academy apparently keeps the exclusive right to tarnish its reputation).

There is a stipulation that award winners must first offer the award to the Academy for a token price of $1.00 before offering it for sale (the price used to be $10). The stipulation holds even if the Oscar is gifted to someone else or passed to descendants.

Is this even a legal contract? An article on Quartz explains several court cases have ruled in favor of the Academy. Some of the court cases seem to rest on bizarre logic.

For example, the Academy started the right of first refusal contract in 1950. Mary Pickford won an award in 1930 for best actress that never had the stipulation. She later received an honorary award in 1975, and her husband received one in 1985. After she passed, she requested her best actress Oscar be sold with the proceeds given to her charity. The Academy blocked her heirs from selling the award, saying that by accepting the 1975 award, and because she had been a lifelong member of the Academy, the right of first refusal retroactively applies to the 1930 award. The jury ruled in the Academy’s favor and the heirs could not sell the Oscar. The Academy had tried to offer $50,000, and even $200,000 to her charity to avoid the lawsuit, but as the award was likely worth $800,000 the heirs tried and lost in court.

**A game of trust**

The Cubs are using the same kind of contract for their championship rings. And if an employee is desperate for money, the organization says it will try to help in other ways. The Cubs likely cannot make any exceptions for selling a ring or else that opens up the possibility for more sales.

The Cubs are footing the bill for the gift and many employees will be happy to receive it. But the re-purchase clause adds a strange note to the celebration. While employees receive rings with great sentimental value, the rings have little value as a store of wealth because, unlike other memorabilia, the rings have little resale value. An employee really desperate for money might try to skirt the restriction by selling it for its individual pieces, but that would presumably not be allowed, and it would also fetch a much lower value.

So ultimately, even this solution does not escape the game theory dilemma. The Cubs have essentially assumed employees might resell the gifts, and so the result is not unlike the (stingy, sell the gift) equilibrium. And that is the strange way a championship ring with 108 diamonds studs is like a stingy gift. Such is life for the once lovable losers.

**Sources**

“Cubs are handing out World Series rings — but there’s a catch.” Chicago Sun Times. Stefano Esposito and Mitchell Armentrout. 19 April 2017.

http://chicago.suntimes.com/sports/cubs-are-hand-out-world-series-rings-but-theres-a-catch/

“Cubs ask non-players to sign doc discouraging sale of World Series rings.” ESPN.com, Jesse Rodgers.

http://www.espn.com/mlb/story/_/id/19189422/chicago-cubs-ask-non-players-sign-agreement-discouraging-sale-world-series-rings

“You never really win an Oscar, you just get to borrow it.” Quartz. Ritchie King. 22 February 2013.

https://qz.com/55895/you-never-really-win-an-oscar-you-just-get-to-borrow-it/

Alice starts at point A, and each second, walks one edge right or up (if a point has two options, each direction has a 50% chance) until Alice reaches point B.

At the same time, Bob starts at point B, and each second he walks one edge left or down (if a point has two options, each direction has a 50% chance) in order to reach point A.

What is the probability Alice and Bob meet during their random walks? (Meet means occupy the same point at the same time).

What is the probability for an *n* by *n* grid? What is the limiting probability as *n* goes to infinity?

Watch the video for a solution.

**Can You Solve The Probability Two Random Walkers Meet? A Strange Way To Approximate Pi**

Or keep reading.

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**Answer To The Probability Two Random Walkers Meet**

The first step is to consider the possible positions of Alice and Bob after *k* steps. When Alice is *k* steps from A, Bob is also 10 – *k* steps from A. The only points at which they could meet are when *k* = 10 – *k* which has a solution of *k* = 5 steps. These points are along the diagonal of the grid as pictured below.

Now we consider the number of ways they can meet at the possible meeting points. Some points can be reached by multiple paths. Hence, we will count the number of ways to reach each of the points.

In a given step, Alice can either move up (U) or right (R). In the first step, there are two possible paths, U or R, so there is 1 way each to reach the points adjacent to point A.

In the second step, Alice can move U or R again. Thus there are 4 paths: UU, UR, RU, and RR. Notice the paths UR and RU are 2 ways to reach the same point. We can write out the number of paths by adding the numbers from adjacent vertices.

In the third step, Alice can move U or R again. There are 8 total paths. Rather than writing them out, we can count the number of paths to a point by adding the numbers in connecting edges.

At this point, we can also notice a pattern: the 8 paths after 3 steps are the paths in the binomial expansion of (U + R)^{3}. The numbers we have been writing on the points are numbers from Pascal’s triangle, and they are equal to the binomial coefficients.

So let us go ahead and write the number of paths for the points after 4 and 5 steps. These could also be calculated as the binomial coefficients for the expansion of (U + R)^{4} or (U + R)^{5}.

By symmetry, Bob has a similar distribution for the number of paths from B to the possible meeting points. As Bob can move down (D) or left (L) in a given step, these could also be calculated from the expansion of (D + L)^{5}.

Now let’s calculate the probability.

**Probability calculation**

How many paths are there from A to a possible common meeting point? For a random walker, 2 possible paths for a step, so after *k* steps there are 2^{k} possible paths. After 5 steps Alice has 2^{5} paths, and so does Bob. thus, there are (2^{5})(2^{5}) = 4^{5} total paths they could both take after 5 steps.

Furthermore, we have determined the point that is *j* units up from A and 5 – *j* units right from A has a number of paths equal to the binomial coefficient C(5, *j*), where C(*n*, *k*) = *n*!/[*k*!(*n* – *k*)!]. The same is true for Alice and Bob, so the number of ways that both meet at to a particular point is the square of a binomial coefficient.

As Alice and Bob can meet at any of the 6 points along the main diagonal, the total number of paths in which they could meet is the sum of the squares of the 6 binomial coefficients:

1^{2} + 5^{2} + 10^{2} + 10^{2} + 5^{2} + 1^{2} = 1024

This needs to be divided by the total number of paths they could take, which is 4^{5}.

1024/2056 = 63/256.

The probability they meet is approximately 24.6%.

**Generalizing to n by n board**

On an *n* by *n* board, Alice and Bob could possibly meet at the points that are *n* steps away–along the diagonal of the grid.

As each walker has 2 choices in a step, there are 2^{n} possible paths in *n* steps for each walker, making for a total of (2^{n})(2^{n}) = 4^{n} paths.

The number of paths for each walker to a point also follows a binomial distribution. The point that is *j* units up from A and *n* – *j* units right from A has a number of paths equal to the binomial coefficient C(*n*, *j*). The same is true for Alice and Bob, so the number of paths where they pick the same point is the square of each binomial coefficient. There are *n* + 1 points at which they could meet, and the number of ways they can meet is the sum of the squares of the binomial coefficients:

There is a nice simplification to this formula. First we use the property C(*n*, *j*) = C(*n*, *n* – *j*), and we write the product as:

Now imagine we have a group of *n* women and *n* men. How many ways are there to form a group of *n* people? The above sum counts this. The first term is the number of ways to have 0 women and *n* men, the next term is the number of ways to have 1 woman and *n* – 1 men, and so on.

But we also know the number of ways to pick *n* people from a group of 2*n* is equal to C(2*n*, *n*). Thus, the two ways to count are equal, and we have derived the identity:

We now divide the number of ways they could meet by the total number of paths 4^{n}:

Pr(meet) = C(2*n*,*n*)/4^{n}

**What happens as n goes to infinity?**

The term C(2*n*, *n*) is known as the central binomial coefficient, and it has a rather interesting limit. As *n* goes to infinity, it is approximately 4^{n}/√(π *n*).

Using this approximation, we have our probability is related to pi!

We could therefore approximate pi as follows. Simulate two random walkers on a large *n* by *n* grid.

As the probability approaches 1/√(π *n*), we can estimate by:

(1/Pr(meet)^{2} *n*) ~ π

Before you actually try to code this, the formula has very poor accuracy. Even on a 1000×1000 grid, it will only approximate pi to 2 decimal places accurately.

Nonetheless, it’s somewhat surprising that π appears in a problem about two random walkers on a grid!

Links for asymptotic formula

https://en.wikipedia.org/wiki/Central_binomial_coefficient

https://en.wikipedia.org/wiki/Stirling%27s_approximation

http://math.stackexchange.com/questions/58560/elementary-central-binomial-coefficient-estimates

http://planetmath.org/asymptoticsofcentralbinomialcoefficient

]]>William Spaniel, of GameTheory101, asked why United didn’t consider using game theory to solve the problem.

Thanks @united for giving me a lesson plan! pic.twitter.com/5hFi1OHFt5

— William Spaniel (@gametheory101) April 11, 2017

If you can’t read the tweet, here is the scheme. United could have asked each passenger to write down how much they valued being on the flight. The person with the lowest value is bumped and then paid the second-lowest value.

This is an example of a Vickrey auction (second-bid auction), and it can be proven to be efficient–the logic is explained below.

So why didn’t United use such a procedure, and why do we rarely see Vickrey auctions in practice? I’ll review the theoretical advantage of the procedure and then mention a few practical issues.

But not all is lost. Since 2011 Delta has employed a very similar auction that has led to fewer involuntarily bumped passengers. I explain this system and how it gets around the problems of the Vickrey auction below.

**Theoretical advantages of a Vickrey auction**

A Vickrey auction involves everyone submitting a bid for an item, in secret, and then awarding the item to the highest bid. That person pays the price of the second highest bid.

When bumping passengers, the “bids” are the amounts each passenger is asking for compensation. From United’s perspective, this is a cost, so each bid is a negative number. The passenger asking for the least amount of money is the lowest negative number (so this is the largest bid from United’s perspective). So the Vickrey auction works the same way: the passenger asking for the least money gets awarded the amount of the passenger asking for the second least amount.

In theory, there is no incentive to for any passenger to lie. Each passenger should honestly write down how much they value being on the flight. Why is that?

Obviously you would not want to write too small of a number, or else you might be bumped and compensated less than you value the flight.

Similarly, you do not want to inflate your bid. If you write down an absurdly high value (say $100,000), when you actually valued it at $1,000, then there will be times you might have been bumped and paid more than $1,000.

The amount you bid exactly determines when you get bumped: if you bid your value, you will only get bumped if you can get paid at least your value.

Every passenger has a weakly dominant strategy to write down their honest value.

Furthermore, if all passengers do this, then the person with the lowest value is bumped. This is economically the most efficient outcome as the person most willing to de-board is identified and compensated.

So this idea should work. But why don’t we see it in practice more?

**Practical issues**

Two auction theory professors, Lawrence M. Ausubel and Paul Milgrom, write about The Lovely but Lonely Vickrey Auction:

“Despite the enthusiasm that the Vickrey mechanism and its extensions generate among economists, practical applications of Vickrey’s design are rare at best.”

So why don’t more people use a second-bid auction?

One problem is people may not trust the seller or auctioneer to be truthful. Imagine you wrote $1,000 for your value as the lowest, and United is supposed to pay you $10,000 which is the next lowest. But United is the only one that sees all the values, so it could make up that $1,050 was the next highest value and stiff you on compensation. In other auctions, a seller might organize shill bids to artificially raise the selling price. If people do not trust the auction, then there is no reason to bid honestly, which ruins some of the point of a Vickrey auction.

A second issue is bidders can collude. For example, if enough people on the flight wrote down absurdly large values like $1 million, then United might get stuck with paying a ridiculously large bill.

This second issue is a particularly worrisome possibility in practice. In theory, second price auctions should raise the same amount as first price auctions: in the first price auction bidders “shave” their bids to about what the next highest value would be. But in practice, sellers are scared of low bids in auctions, and Vickrey auctions may generate less money than first price auction where people overbid irrationally (in a first price auction, a seller benefits when a single person drastically overbids–in a second-price auction the overbid is negated if the next highest bid is reasonable).

Vickrey auctions are interesting theoretically, but they are not so easy to implement in practice. That’s why Delta implemented a slightly different auction.

**A modified auction from Delta**

Delta airlines does a kind of blind auction: when a flight is overbooked, it asks passengers during check-in how much they are willing to be compensated to be put on a later flight. The instructions indicate that Delta accepts lower bids first.

This is a first-price auction–people offer an amount and they actually get that amount. Passengers can offer a fair value and if Delta accepts the offer they get that amount–there is no worry Delta will manipulate the bids. This takes care of one concern of the Vickrey auction.

Second, passengers will find it harder to collude as each passenger bids secretly while checking in.

As an added benefit, the bidding process happens before people are on the flight, so there is an efficiency to bumping passengers earlier and allowing them time to re-arrange travel plans.

(Incidentally, what is your best strategy for this auction? Let’s say you would be willing to get bumped for $1,000 minimally. If you accept this offer, you don’t profit from winning, you just break even. If you ask for an absurdly high amount, on the other hand, you are unlikely to win at all. What you want to do is pad your offer to be slightly lower than the next lowest person’s offer. This way you get your offer, and you profit as much as you can.)

Has Delta’s auction method worked? Per PBS, in 2014 Delta bumped 3 passengers involuntarily per 100,000 versus American bumping 5 and United bumping 11. This translates into thousands of passengers fewer per year, and fewer chances for a controversy when bumping a passenger involuntarily.

United is likely going to try something to fix its reputation and avoid another incident. Perhaps United will also use the lessons of auction theory and strive for a more efficient outcome for everyone.

]]>If we exclude one parent, who is 40, the average age drops to 15. How many children are in the family?

Watch the video for a solution.

**Can You Solve The Number Of Children Riddle? **

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**Answer To The Family Average Age Riddle**

There are no tricks to this problem, but many people do not get the correct answer because they either solve for the total number of family members (and forget to subtract to count only children), or they incorrectly set up the equations (by, say, forgetting there are 2 parents plus the children).

Let *c* denote the number of children. As there are 2 parents, there will be *c* + 2 total family members.

Let *s* denote the sum of the ages of everyone in the family. We first know the average age of the entire family is 20, which means:

*s*/(*c* + 2) = 20

If we exclude a parent aged 40 from the average, the numerator gets reduced by 40, and the denominator gets reduced by 1. So the new average age of 15 is denoted by the following equation:

(*s* – 40)/(*c* + 1) = 15

Now we need to solve for *c*. We can solve each equation for *s* to get:

*s* = 20(*c* + 2)

*s* = 20*c* + 40

*s* – 40 = 15(*c* + 1)

*s* – 40 = 15*c* + 15

*s* = 15*c* + 55

We can set these two expressions for *s* equal to each other and then solve:

20*c* + 40 = 15*c* + 55

5*c* = 15

*c* = 3

Thus, there are 3 children in the family.

**Source**

Adapted from a problem on Math StackExchange

http://math.stackexchange.com/questions/2162668/average-of-age-in-a-family