So You Think You've Got Problems

by Alex Bellos

  He then offers you the option of sticking with door No. 1, or switching to door No. 3.

  Is it in your advantage to make the switch?

  The question was devised in 1975 by Steve Selvin, a young statistics professor at the University of California, for a lecture course he was giving on basic probability. The first time he asked it, all hell broke loose. ‘The class was divided into warring camps,’ he remembers. Do you stick or do you switch?

  Perhaps the most common initial response is that it makes no difference to switch. If there are two doors left, and we know that the car has been randomly placed behind one of them, it’s intuitive to think that the odds of the car being behind either one are 1 in 2.

  Wrong! The correct answer is to switch.

  In fact, when you switch, you double your odds of getting the car. If you stick, your chances of winning the car are 1 in 3. There were three closed doors and you had a 1 in 3 chance of choosing correctly. Your odds don’t change when Monty Hall opens another door, because he was always going to open another door regardless of your choice.

  If you switch, however, your chances of winning the car rise to 2 in 3. When you choose your first door, as I said in the previous paragraph, you have a 1 in 3 chance of choosing correctly. In other words, there is a 2 in 3 chance that the car is behind one of the other doors. When Monty Hall reveals that one of the doors you did not choose has a goat behind it, there is now only a single choice remaining for the ‘any other door’ option. Therefore, the probability that this remaining door is the one with the car behind it is 2 in 3.

  If you are finding this confusing, or counter-intuitive, you are not alone. In 1990, the problem appeared in Marilyn vos Savant’s column in the large-circulation US magazine Parade. She said switch. Her article provoked a deluge of about 10,000 letters, almost all disagreeing with her. Almost a thousand of these letters came from people with a PhD after their names. The New York Times reported on the storm on its front page.

  The Monty Hall problem requires careful thought. As does its sister puzzle …

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  THE MONTY FALL PROBLEM

  You have made it to the final round of Let’s Make a Deal. In front of you are three doors. Behind one door is an expensive car; behind the other two doors are goats. Your aim is to choose the door with the car behind it.

  The host, Monty Hall, says that once you have made your choice he will open one of the other doors to reveal a goat. (Since he knows where the car is, he can always do this. If the door you choose conceals the car, he will choose randomly between which of the other two doors to open.)

  The game begins and you pick door No. 1.

  Monty Hall walks to the doors, stumbles, falls flat on his face and accidentally, at random, opens one of the two doors that you didn’t pick. It’s door No. 2, which reveals a goat.

  Monty Hall dusts himself off and gives you the option of sticking with door No. 1, or switching to door No. 3.

  Is it in your advantage to make the switch?

  Monty Hall’s fame has spawned many similar ‘stick or switch’ problems.

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  RUSSIAN ROULETTE

  You are tied to a chair and your crazed captor decides to make you play Russian roulette. She picks up a revolver, opens the cylinder and shows you six chambers, all of which are empty.

  She then puts two bullets in the revolver, each in a separate chamber. She closes the gun and spins the cylinder so you do not know which bullet is in which chamber.

  She puts the gun to your head and pulls the trigger. Click. You got lucky.

  ‘I’m going to shoot again,’ she says. ‘Would you like me to pull the trigger now, or would you prefer me to spin the cylinder first?’

  What’s your best course of action if the bullets are in adjacent chambers?

  What’s your best course of action if the bullets are not in adjacent chambers?

  Congratulations, you survived. You didn’t blow your brains out, although I hope they have been thoroughly massaged over the course of the book.

  This chapter concerned probability, an area that underlies much of modern life – from finance to statistics to the frequencies of buses – and one in which it is easy to fall into traps. The more we are aware of how probability can mislead us, the better the decisions we will be able to make in real life. In fact, my aim in the whole of this book has been to present you with puzzles from which there is something to learn, whether a new concept, a clever strategy, or just a simple surprise. I have wanted to kindle creativity, encourage curiosity, hone powers of logical deduction, and share a sense of fun.

  Most problems in life don’t come with a set of solutions. Thankfully, mine do. And here they are.

  ANSWERS

  Tasty Teasers

  Number conundrums

  1)

  2)

  3)

  4)

  3 x 4 = 12

  13 x 4 = 52

  54 x 3 = 162

  12 = 3 + 4 + 5

  2 x 6 = 3 + 4 + 5

  3 + 6 = 4.5 x 2

  9⁄12 + 5⁄34 + 7⁄68

  5)

  The first step is to eliminate 5 and 7. If 5 was in the grid, the multiplication(s) including the 5 would be divisible by 5. But the multiplication(s) without the 5 would not be divisible by 5, since no other number between 1 and 9 is divisible by 5. Likewise with 7.

  6)

  The puzzle zoo

  ANIMAL PROBLEMS

  1 THE THREE RABBITS

  2 DEAD OR ALIVE

  See now the four lines. ‘Tally ho!’

  We’ve touch’d the dogs, and away they go!

  3 GOOD NEIGHBOURS

  The rabbit was not harmed because it was already dead when the dog picked it up.

  4 A FERTILE FAMILY

  The single doe has more than 28 trillion descendants, which is about 300 times the estimated total number of humans that have ever lived.

  Now to how we get there.

  For the first six months of her life, the doe is not fertile. Let M(n) be the total number of does at the end of month n. We have

  M(1) = M(2) = M(3) = M(4) = M(5) = M(6) = 1

  Finally, at the end of the seventh month, she produces six kits, three of which are does, and she produces three more does per month until the end of the year. Note that we can write each monthly total in terms of the total of the previous month.

  M(7) = 1 + 3 = 4

  M(8) = M(7) + 3 = 7

  M(9) = M(8) + 3 = 10

  M(10) = M(9) + 3 = 13

  M(11) = M(10) + 3 = 16

  M(12) = M(11) + 3 = 19

  In the second year, she will start to have grandchildren. In the first month of the second year, she adds another three does, and her first litter produces three each. That’s four batches of three added to the previous month’s total.

  M(13) = M(12) + 4 x 3 = 31

  In the second month of the second year, she adds another three does, and her first and second litters also produce another three. That’s seven batches of three.

  M(14) = M(13) + 7 x 3 = 52

  M(15) = M(14) + 10 x 3 = 82

  M(16) = M(15) + 13 x 3 = 121

  You may have spotted a pattern by now – the number of new does is the monthly total from six months previously, multiplied by 3. In other words:

  M(14) = M(13) + 3 x M(8)

  M(15) = M(14) + 3 x M(9)

  M(16) = M(15) + 3 x M(10)

  Rewritten as a formula we have M(n) = M(n – 1) + 3M(n – 6). The M(n – 1) part is the running total from the month before, and the 3M(n – 6) gives us three new does for every doe that was alive six months ago, which is necessary and sufficient for it to be fertile now. Since the life span of a rabbit is seven years, or 84 months, we want to know M(84).

  Calculating this by hand will take too long. Put it in a computer and out will pop 14,340,818,086,651.

  But this is not the answer, yet. We need to subtract 1 (the rabbit whose descendants we are trying to calculate) and multiply by
2, since for every doe born there is also a buck.

  The final answer is 28,681,636,173,300 descendants.

  A bunny bonanza!

  5 A BUNCH OF HOPS

  The rabbits were a clue. The answer is 55, the ninth term in the Fibonacci sequence.

  Like many questions of this type, we embark on our solution by simplifying the problem and looking for a pattern. With only two lilies, there is only one way to get from the first to the last lily: a single jump of 1. With three lilies there are two ways: either two single jumps or a double jump; and with four there are three ways: three singles, a single followed by a double, or a double followed by a single.

  Now consider five lilies, as shown opposite. The frog’s first jump will be to either A or B. The total number of ways to get to the last lily is therefore the number of ways from A plus the number of ways from B, which is 2 + 3 = 5.

  Continuing this line of thought, the number of ways to jump across six lilies is the number of ways to jump across four of them plus the number of ways to jump across five, which is 3 + 5 = 8. The answer for seven lilies is 5 + 8 = 13, for eight the answer is 8 + 13 = 21, for nine it’s 13 + 21 = 34, and for ten it’s 21 + 34 = 55.

  This recursive process – starting with 1, 2 and 3, and generating the next term in a sequence by summing the last two terms – produces the Fibonacci sequence.

  6 CROSSING THE DESERT

  The four Bedouin tribesmen – let’s call them A, B, C and D – depart together, each loaded with five days’ worth of water.

  At the end of the first day, they each have four days of water left. A gives a day’s worth of water to each of his three colleagues, leaving him with a single day’s worth of water.

  On the second day A returns home (since he is a day away from there, and he has a day’s worth of water), while B, C and D carry on. At the end of the second day the three of them are now two days out, each with four days of water left. B gives a day’s worth of water to the other two, leaving him with two days’ worth of water, and the others at capacity with five days’ worth.

  On the third day B now returns home (which he is able to do since he has two days’ worth of water, and the start is two days away) and C and D carry on, reaching a point three days away from the start, with both carrying four days’ worth of water. C now gives D a day’s worth of water.

  On the fourth day, C returns home (which he is able to do since he has three days’ worth of water and home is three days away), while D finally reaches the camp, where he delivers the package. D has four days’ worth of water left, which is just what he needs to travel back home.

  7 SAVE THE ANTELOPE

  You set off with a full load of five gallons (that’s a full tank and four full canisters) and use a gallon to drive 100 miles. Call this point A. Fill the tank, leave the remaining three canisters by the side of the road, and return to base.

  Set off again with five gallons and use a gallon to drive the 100 miles to A. Pick up a canister, fill the tank, and with a full load drive another 100 miles. Call this point B. Leave two canisters by the side of the road here. You have two canisters left, and this is enough to get you the 200 miles back to base.

  So far you have taken 10 gallons from base, and have two canisters at A and two canisters at B. The petrol in these four canisters will let you drive for 400 miles, meaning that you only need another four gallons to do the round trip.

  Set off for the final time with four gallons: one in the tank and three in canisters. Here is one way you can now split the journey: at A, fill the tank with one canister and take on the other. You now have a full load which will get you the 300 miles to the antelope and a further 200 miles back to B. Once you are back at B, take the two remaining canisters, which is just enough for the 200 miles home.

  8 THE THIRTEEN CAMELS

  The 17 camel puzzle is a lesson about how easily numbers can be used to mislead. The apparent contradiction comes from assuming that the fractions demanded by the father – 1⁄2, 1⁄3 and 1⁄9 – add up to 1. But they don’t. Written as fractions with lowest common denominator 18 they are 9⁄18, 6⁄18 and 2⁄18 which add up to 17⁄18. When the children are given 9, 6 and 2 camels they are not being given a half, a third and a ninth of the total inheritance, as the puzzle suggests. They each receive slightly more. What they do receive, however, is a number of camels that are in the proportions set out by the father – 1⁄2:1⁄3:1⁄9 – which when multiplied by 18 are the same as 9:6:2.

  The 13 camel problem is a twist on the 17 camel problem in which the wise woman is a bit more provocative. Rather than loaning them a camel, she temporarily takes one away.

  The woman hears the children’s gripe and says she can solve it if she can confiscate one of their camels. The children reluctantly agree, leaving them with only 12 camels left. She gives 6 to the eldest, which is half the total, and she gives 4 to the middle one, which is a third of the total.

  That leaves only 2 camels, which she gives to the youngest child. To these camels she adds the one that she confiscated, meaning that the youngest now has 3 camels, a quarter of 12.

  The three children thus receive 6, 4 and 3 camels each.

  If you read the question carefully, the man does not say clearly that he wants his children to receive a half, a third and a quarter of the total number of camels. A fair interpretation is that he wants the children to have camels in the proportions 1⁄2:1⁄3:1⁄4, or 6⁄12:4⁄12:3⁄12, which is indeed the solution the wise woman presents.

  9 CAMEL VS. HORSE

  Ada says: ‘Get on the other person’s animal!’

  10 THE ZIG-ZAGGING FLY

  The simple way to solve this problem is not to think about the distance travelled, but about the time spent travelling. If two cyclists 20 miles apart are cycling at 10 miles an hour, they will meet in one hour. The fly, which is flying at 15 miles an hour, will therefore have travelled 15 miles.

  11 THE ANTS ON A STICK

  Carlos, our anty-hero, falls off last after exactly 100 seconds.

  The arbitrary-sounding distances 38.5cm, 65.4cm and 90.8cm were a sign that you need to think laterally about this puzzle. No puzzle that passes as entertainment is going to require algebra or arithmetic with numbers like 38.5 and 65.4. (Of course, it would be possible to work out the answer by calculating the positions, but that would be messy.)

  The piece of insight that makes this an elegant puzzle is this: think about what happens when two ants collide and both walk back the way they came. If you blur your eyes, this is equivalent to those two ants walking past each other. In other words, this puzzle can be treated as if six ants, each on their own track, are walking to the end of the stick. Another way to visualise it is this: imagine that each ant is carrying a leaf, and that on each collision they exchange leaves. The leaves will be moving at 1cm per second in a unique direction. The leaf that takes the longest to fall off the edge is the one that starts with Aggie. So, it will take 100 seconds – 1m at 1cm per second – until the last ant drops.

  Now to the identity of the ant in question. Let’s continue thinking about the leaves. Four leaves will fall off the right side (because leaves only move in one direction, and four of them are moving to the right). The last leaf to fall off on the right side is the one that started on Aggie, and we can deduce that the ant carrying it at that point must be the ant that started fourth from the right. Stand up Carlos, that’s you. It has to be him because ants can’t walk past each other, so the order of the ants on the stick cannot change. The fact that we don’t know his position to start with is irrelevant.

  12 THE SNAIL ON THE ELASTIC BAND

  During the first second, the snail crawls 1cm.

  In other words, the snail travels 1⁄100,000 of the length of the 1km-long elastic band.

  The elastic band stretches instantly to 2km. During the following second, the snail travels another 1cm, which now represents 1⁄200,000 of the length of the elastic band. After the third second the length of the band is 3km, so the 1cm trave
lled by the snail represents 1⁄300,000 of the length of the elastic band. And so on.

  If we add up all these fractions, we get the following term:

  This number describes the snail’s progress after n seconds, expressed as a fraction of the total length of the elastic band.

  If you got this far, congratulations. To finish this proof you need some background knowledge. The term in the brackets, (1 + 1⁄2 + 1⁄3 + 1⁄4 + … + 1⁄n), is also known as the ‘harmonic series’, and it gets bigger and bigger the more terms you add, eventually exceeding any finite number. (In my discussion of the jeep problem I mentioned the series 1 + 1⁄3 + 1⁄5 + …, which also increases beyond any finite number. If this series increases to infinity, so must the harmonic series. Proving the divergence of the harmonic series is not hard, and a proof can be found easily online.)

  We know, therefore, that there exists a number N such that the term (1 + 1⁄2 + 1⁄3 + 1⁄4 + … + 1⁄N) exceeds 100,000. After N seconds the expression above describing the snail’s progress will be more than 1, which means that the snail will have reached the end of the elastic band. It will take a while! N seconds is longer than the age of the universe. And after N stretches, the elastic band will be so long it will not even fit in the universe.

  13 ANIMALS THAT TURN HEADS