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Math puzzles and tricks |
Suppose the distance between two cities is 1000 miles. In zero-wind, a plane can travel at 500 mph. Will it take longer to make the round trip with no wind, or a direct 100 mph tailwind in one direction, and equal head wind the other way?
— Kevin from Portland, OR
In zero-wind it will take 2 hours each way, for a total of 4 hours. With the tailwind, the plane will travel at 600 mph, making the trip in 1000/600 = 1.667 hours. With the headwind, the plane will travel at 400 mph, taking 1000/400 = 2.5 hours. So, in the wind, the total time is 4.167 hours, or 10 minutes longer.
This just goes to show that it is dangerous to average averages. You can't say the average rate of a trip is 500 mph, if it is 400 mph one way and 600 mph the other, because the 400 mph leg is over a longer period of time.
If this isn't intuitive, consider a 500 mph wind. The plane would take 1 hour only with the wind, but it would stay in place the other way, taking forever. January 12, 2009
If all thirty major league baseball teams were of equal talent, then what would be the probability that any given team from a specified division would make the playoffs? Clearly the current rules favor the National League Central and disadvantage the American League West.
Why don't teams in the American League West complain about this inequity? The differences here to me are not trivial. Since making the playoffs is big money for a team, I'm surprised that there aren’t more complaints from the NL Central. I would also be a little miffed, as a fan of any of these six teams, to know my team is getting the short end of the stick.
I can't imagine that I'm the first one to ever notice this? Does MLB compensate these teams in some way?
— Ben
For the benefit of other readers, there are two leagues, each with three divisions, in Major League Baseball. Each division has five teams, except the American League West, with 4, and the National League Central, with 6. Every year, in both leagues, the three division leaders, and a Wild Card team, make the playoffs. The Wild Card team is the team with the best win/loss record in the League, not counting the three division leaders. There are some tie-breaking rules, which I won't get into, and assume they are resolved randomly.
Indeed, the American League West has a big advantage, and the National League Central has a big disadvantage, all other things being equal. I do not know of any compensating rules. Nor do I know the reason for this imbalance. Before 1998 there were only two divisions. In 1998, Major League Baseball added two new teams, the Tampa Bay Devil Rays and Arizona Diamondbacks. They also increased the number of divisions from four to six, and added the Wild Card rule. However, why they didn't balance the leagues, I have no idea. The best solution to this inequity, in my opinion, would be to move the Houston Astros to the American League West. Some may say Houston is not west enough, but the Texas Rangers are also in that division.
I posted the answer and solution to the probability question at my companion site, mathproblems.info, as problem number 200.
p.s. Since I posted this column, one reader wrote that the reason for the imbalance was to keep the number of teams in each league an even number. This allows every team to play on a given day, and keep the play within the division. However, this I don't buy as an excuse. In 2008 the regular season consisted of 162 games per team, played over 185 days (not counting the all-star game day, and a day on each side). So, each team played 0.8757 games per day. Of those 162 games, 18 are played against teams in the opposite division, and 144 in the same division. I suggest that with balanced divisions of 15 teams each, on any given day 12 teams play within their own league. Over 185 days, that will accomplish 185 × (12/15) = 148 games. In the other 37 days, schedule 14 interleague games, for a total of 162 games. So, the only change will be a reduction in the number of interleague games per team from 18 to 14. It seems to me, most fans oppose interleague games to begin with, including me.
p.p.s. Another reader wrote to say that my system would not accommodate the baseball traditions of keeping every team playing on Saturdays and Sundays, and designating interleague play to only certain times of the seasons. Okay, fair points. However, if tradition is so important in baseball, why introduce interleague games at all? Personally, I value fairness over tradition. Put me in charge of baseball scheduling, and I’ll no only balance the leagues, but keep every team playing on weekends. However, it would come at the expense of the days off being clumped together. Maybe the easier thing to do would be to add two more teams. My hometown of Las Vegas will be the first to volunteer to be one of them. December 8, 2008
I need help with a puzzle called "Eternity II." The prize for solving the puzzle is a staggering $2,000,000, a considerable amount of money to me.
Here's a link to an interview, including the game maker himself,
Christopher Monckton (former advisor to Margaret Thatcher, among many things). The game is obviously not really about gambling at all, but despite this
fact, maybe you could add a word or two on your web page about it.
The game maker brags about the puzzle to be insolvable, in that link given above. I'm starting to think that he's actually right, and that he himself is the only one who will eventually become rich from selling that (ridiculous but fascinating) game. How would you, being a mathematician and all, go about solving this type of puzzle? — Robert
I hope you're happy; I’ve been obsessed by this puzzle for the last month or so. I was lucky (or perhaps unlucky) to find the 256-piece puzzle at the local Borders book store, but I had to buy the four clue puzzles on eBay, from a guy in Australia.
I wrote a program that can easily solve the four clue puzzles. It solved the 72-piece clue puzzle #4 in less than a second. The way I did it was with a simple brute-force recursive program. I mapped out a path on the board, starting with the border. At each position, the program looped through all the unused pieces, looking for one that fit. If it found one, it moved to the next square, if it didn’t, it moved back a square.
I have had two computers crank away at the 256-piece $2 million puzzle for weeks, and neither have come anywhere close. I tend to agree with what the creator said in that video, that if you hooked up ten million of the world’s fastest computers, they still might not find the solution by the death of the universe. You would think I would have heeded his warning before starting, but in the face of a good puzzle, all consideration for practical use of my time goes out the window.
I have lots of ideas for shortcuts, but even if they sped up my program by a factor of a billion, it still probably wouldn't help. I'm going to be extremely impressed if anybody solves this thing. What really nags at me is I feel there is some undiscovered branch of mathematics that could solve puzzles like this easily. Until then, I think glorified trial and error is the best we can do to solve it. Today's computers are simply too slow, and the number of combinations too vast, for that to have much of a chance of success.
November 17, 2008
Thank you for your entertaining collection of math puzzles. My
girlfriend and I came up with this variation on the pirate puzzle ( problem #85). What if all the pirates are of equal rank, and in each round the proposer of the division is chosen by lot? In this variation, assume that each pirate's highest priority is to maximize his expected amount of coins received. I have what I think is the solution, but perhaps you'd like to try your hand at it first. Thanks again.
— Jon S
You're welcome. If there are only two pirates left, then the one chosen to make a suggestion has no hope, because the other pirate will vote no. The one drawn will get zero, and the other all 1000. So, before the draw, the expected value with two pirates left is 500 coins.
At the three pirate stage, the drawn pirate should suggest giving one of the other pirates 501, and 499 to himself. The one getting 501 will vote yes, because it is more than the expected value of 500 by voting no. Before the draw, with three pirates left, you have a 1/3 chance each of getting 0, 499, or 501 coins, for an average of 333.33.
At the four pirate stage the drawn pirate should choose to give 334 to any two of the other pirates, and 332 to himself. That will get him two 'yes' votes from the pirates getting 334 coins, because they would rather have 334 than 333.33. Including your own vote, you will have 3 out of 4 votes. Before the draw, the expected value for each pirate is the average of 0, 334, 334, and 332, or 1000/4=250.
By the same logic, at the five pirate stage, the drawn pirate should choose to give 251 to any two pirates, and 498 to himself. Unlike the original problem, it isn't necessary to work backwards. Just divide the number of coins by the number of pirates, not including yourself. Then give half of them (rounding down) that average, plus one more coin. September 29, 2008
Three logicians are playing a game. Each must secretly write down a positive integer. The logician with the lowest unique integer will win $3. If all three have the same number, each will win $1. The logicians are selfish, and each wishes to maximize his own winnings. Communication is not allowed. What strategy will each logician follow?
— Matthew from Fort Wayne, IN
We've been given a challenge at work -- just for fun, and none of us can work it out. A farmer has 5 trailers full of sheep. Four of the trailers contain sheep weighing 39kg and the 5th trailer contains sheep weighing 40kg. All of the sheep are identical. He goes to the market. He wants to find out which of the trailers contains the sheep weighing 40kg, and he can only use the large weighing scales once!!! How does he do it? Please help, it is driving us all mad at my work place -- it's a vet's!! – Becca
Take one sheep from trailer 1, two from trailer 2, three from trailer 3, four from trailer 4, and zero from trailer 5. If all the sheep weighed 39 kg then the total weight would be 39 * 10 = 390 kg. However 0 to 4 sheep are one kg heavier. If the total weight is 391, then there is one heavy sheep on the scale; thus it must have come from trailer 1. Likewise, if the total weight is 392, then there are two heavy sheep on the scale, which must have come from trailer 2. In the same manner a weight of 393 means the heavy sheep are in trailer 3, a weight of 394 means the heavy sheep are in trailer 4, and a weight of 390 means the heavy sheep are in trailer 5. March 29, 2007
A test consists of 10 multiple choice questions, each with 5 possible answers, 1 of which is correct. To pass the test a student must get 60% or better on the test. If a student randomly guesses, what is the probability that the student will pass the test? – Kirk from Canton
The probability of exactly 6 correct is combin(10,6)×0.26×0.84 = 0.00550502.
The probability of exactly 7 correct is combin(10,7)×0.27×0.83 = 0.00078643.
The probability of exactly 8 correct is combin(10,8)×0.28×0.82 = 0.00007373.
The probability of exactly 9 correct is combin(10,9)×0.29×0.81 = 0.00000410.
The probability of exactly 10 correct is 0.210 = 0.00000010.
Adding the probabilities for 6 to 10 correct, the probability of at least six correct is 0.00636938.
March 5, 2007
I'm trying to compare the cost of replacing an old refrigerator now in order to save on electrical costs, vs. waiting until it dies to replace it. I can calculate how much cheaper it is to run the new fridge vs. the old one: $37/yr., that's easy. But how do I factor in the cost of the new fridge? Say the new fridge costs $425. I can't say that *all* of that $425 is a new expense, because I'll have to replace the old fridge *someday*, if not now, so I'll have that new-fridge expense at some point anyway. Let's say that a typical fridge lasts 14 years and my old fridge is 9 years old, so if I replaced it now I'd be replacing it in 5 years. I tried to make a two-column table, comparing the cost of keeping the current fridge for 9 years and then replacing it, vs. replacing it now, but I didn't know how to make an apples-to-apples comparison because I didn't know for how far into the future to consider the costs, and because the fridges are replaced in different years. How do I compare the economics of replacing now vs. replacing later? By the way, this isn't for my own situation, because my current fridge is probably 30 years old. It's for, uh, a friend. - Spanky McBluejay from Austin, TX
If you keep the current fridge then in five years you will have spent an extra $37*5 = $185 on electricity compared to a new one. If you replace it now you’ll be out $425 but assuming linear depreciation after five years it will still be worth $425*(9/14) = $273.21. So you will have lost $425*(5/14) = $151.79 due to depreciation. So the cost of depreciation of the new fridge is less than the additional electricity expense of keeping the old one, so I favor buying a new one now.
July 31, 2006
How does this work?
- Grab a calculator. (you won't be able to do this one in your head)
- Key in the first three digits of your phone number (NOT THE AREA CODE)
- Multiply by 80
- Add 1
- Multiply by 250
- Add the last 4 digits of your phone number
- Add the last 4 digits of your phone number again.
- Subtract 250
- Divide number by 2
Do you recognize the answer? - Chris M. from Las Vegas
Let’s call the first three digits in your phone number x, and the last four y. Now let’s see what I have at each step.
- Ready!
- X
- 80x
- 80x+1
- 250*(80x+1) = 20000x+250
- 20000x+250+y
- 20000x+250+2y
- 20000x+250+2y-250 = 20000x+2y
- (20000x+2y)/2 = 10000x+y
So that is of course going to equal your phone number. We need the 10000x to move the prefix four places to the left, and then we add on the last four digits.
June 9, 2006
You are in a boat with a rock, on a fresh water lake. You throw the rock into the lake. With respect to the land (shore), does the water level increase, decrease, or stay the same?
My co-workers think that the water level will stay the same. - David
The water level relative to the shore will decrease. Inside the boat the rock is pressing down on the canoe and thus pushing up the water around it. The amount of water displaced is equal in weight to that of the rock. For example, a 10 pound rock will displace 10 pounds of water upward. When the rock is thrown overboard the weight will not matter but rather the volume of the rock. So the rock will push upward an amount of water equal in volume to the rock. The mass of a rock is greater than that of water so the rock displaces more water pushing down on it than in it. So the level of the lake will be higher with the rock in the canoe than at the bottom of the lake.
May 10, 2006What do you think of the Bible Code? - Vince from Manila
I would put those behind it on the same level as those selling get rich quick gambling schemes. The mathematically ignorant taking advantage of the mathematically ignorant.
Nov. 22, 2005
My question is about a problem that is known as the "two envelope paradox". You are on a game show. In front of you are 2 envelopes, each containing an unknown amount of cash. You are told that 1 envelope has twice as much money as the other. You are now asked to choose an envelope. You choose one. It contains $50,000. Now you are told that you can keep the envelope you picked, or swap for the other one. Should you swap? Knowing ahead of time that you could swap, then it doesn't matter, as you would just choose the envelope you ultimately want. But because you only find out about swapping after you choose an envelope, then the original selection and the option to swap are 2 independent events, correct? That said, when deciding to swap or not, the other envelope contains either twice as much or half as much as what you currently have. So it has either $100K or $25k. Since there is a 50% chance of either occurring, the Expected Value of the other envelope is $62,500. Generically speaking, if we let x = the amount you originally selected, then the other envelope's EV is 1.25x. Therefore it is always correct to swap. Is this correct? Thank you. - Derek from Boston
I'm very familiar with this problem. I address it on my web site of math problems, problem number 6. There I address the general case, including not looking in the first envelope at all. However to answer your question we can not ignore the venue of where the game is taking place. You said it was a "game show." On most game shows $50,000 is a nice win. Few contestants on the Price is Right ever make it that high. I would guess that fewer than 50% of players on Who Wants to be a Millionaire get that high. Meanwhile wins of $25,000 are not unusual on game shows. Cars are won routinely on the Price is Right, which have values of about $25,000. The $32,000 level is a common win on Who Wants to be a Millionaire. The average win on Jeopardy per show is roughly $25,000. The great Ken Jennings averaged only $34,091 over his 74 wins. So, my point is that $50,000 is a nice win for a game show, and $100,000 wins are seen much less often that $25,000. Thus as a game show connoisseur it is my opinion that the other envelope is more likely to have $25,000 than $100,000. So I say in your example it is better to keep the $50,000. It also goes to show you can never assume the chances that the other envelope has half as much or twice as much are exactly 50/50. Once you see the amount and put it in the context of the venue it is being played you can make an intelligent decision on switching, which throws the 1.25x argument out the window.
Nov. 2, 2005
The radius of the circle is 1. The triangle is
equilateral. Find the area of each colored region.
I don't want to blow the answer for those who want to
solve it for themselves. For the answer and solution
visit my other web site mathproblems.info,
problem 189. Aug.
28, 2005
How does this work: www.1800gotjunk.com/genie/?
Let's express your number as 10t+u. You are
asked to subtract each digit, leaving you with 10t+u-t-u
= 9t, a number divisible by 9. Note how all the numbers
divisible by 9 have the same item, which is the one the
genie predicts. July 28,
2004
A friend sent me this,
I was wondering if there was a formula as to how this works.
Often these mind reading number puzzles work
because an interesting mathematical oddity. If the sum of
digits of a number is divisible by 9 then the number it
self is divisible by 9. Let's try it on the phone number
of the Las Vegas Tropicana (702-739-2222). The sum of
digit is 7+0+2+7+3+9+2+2+2+2 = 36. 36 divides evenly by
9, so 702739222 must also be divisible by 9. Here is a
proof of this.
- Let n bet any integer. Express n as
d0*1 + d1*10 +
d2*100+ d3*1000+ ... +
dn*10n, where dn is
the first digit, dn-1 is the second, and so
on.
- n = [d0 + d1 +
d2 + ... + dn ] +
[d1*9 + d2*99+
d3*999+ ...+ dn*999Ö9 ( a
number with n nines)]
- n = [d0 + d1 +
d2 + ... + dn ] +
9*[d1*1 + d2*11+
d3*111+ ... dn*111...1 (a number
with n ones)]
- 9*any integer is evenly divisible by 9. So if
d0 + d2 + d2 + ... +
dn , or the sum of digits, is divisible by
9, then the entire number must be divisible by 9.
Now that we have that proof out of the way we can look
at this magic trick. The problem asks you to pick any
number. Then rearrange the digits to make a second
number. Then subtract the smaller number from the larger
number.
The answer is always going to have a sum of digits
divisible by 9. Why? For every digit in the original
number it appears somewhere else in the other number.
Going one set of digits at a time, changing all the other
numbers to zero, we could boil down each set as +/-
n*[10x - 10y] (where
x>=y and n is the digit) = +/-n *10y *
(10x-y - 1) = 10y * (a number
composed of only nines) = a number divisible by 9.
Let's look at an example. Let the original number be
1965. Scramble it up to get 6951. 6951 - 1965 =
6*(1000-10) + 9*(100-100) + 5*(10-1) + 1*(1-1000) = 6*990
+ 9*0 + 5*9 + 6*-999. Note that each part is divisible by
9, thus the number you get after subtracting must also be
divisible by 9, and finally the sum of digits is also
divisible by 9.
The trick then asks you to circle a number except 0
and enter the sum of all the other digits. The program
then only needs to add a number to the number you entered
so that the sum is divisible by 9. For example if you
said the sum of your digits was 13 then you must have
circled a 5, because 13+5 = a number divisible by 9.
The reason you can't circle a zero is because if you
did and then entered a number already divisible by 9 then
the program wouldn't know whether you circled a 0 or a 9.
Sept. 26, 2003
Eight golfers went to a new course. The caddy master
put 8 bags on four carts at random. The golfers put 8 marked
golf balls in a hat. The balls were thrown in the air. The 2
closes balls to each other were partners. In every case the
partners' golf bags were already on the same cart. What is
the probability of that the golf bags were paired up
correctly before the throw?
The formulaic answer for the number of
combinations would be
combin(8,2)*combin(6,2)*combin(4,2)/fact(4) = 25*15*6/24
= 105. Another way to solve the number of combinations
would be to take one golfer at random. There are 7
possible people to pair him with. Then pick another
golfer at random from the six left. There are 5 possible
people to pair him with. Then pick another golfer at
random from the four left. There are 3 possible people to
pair him with. So the number of combination is 7*5*3 =
105. Thus the answer is 1 in 105.
May 5, 2003
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