So You Think You've Got Problems
Page 17
If you removed C, the other counter in the same box is white.
In other words, in two out of three equally likely outcomes, the remaining counter is black.
So the chance of the remaining counter being black is 2 in 3, or 66.7 per cent.
109 THE DICE MAN DIET
Monday.
You have a 1 in 6 chance of eating your first pudding today. In order for your first pudding to be eaten tomorrow, you must not eat pudding today (a 5 in 6 chance) and roll a 6 tomorrow (another 1 in 6 chance). A 5⁄6 chance and a 1⁄6 chance combined is less than a 1⁄6 chance, so it’s more likely that your first pudding will be today than tomorrow. Following that logic, pudding on the day after tomorrow becomes even less likely than pudding tomorrow, and so on.
110 DIE! DIE! DIE!
The bet is not in your favour. In fact, this game is chuck-a-luck, a casino game often played with the three dice in a spinning cage.
To show that the house always wins, imagine that six players bet £100 on each of the six possible outcomes. The house takes in £600.
If all three of the dice land on the same face, the house pays back £400 (£300 to the winner plus the stake).
If two dice land on the same face, the house pays back £500 (£200 to one winner, £100 to another winner, plus their two stakes).
If all three land on different faces the house pays back £600 (three £100 prizes plus the three original stakes).
In other words, the house never loses. Which means that, overall, over time, the punters never win.
111 THE PHONEY FLIPS
The second sequence is the one I made up.
Let’s look at the similarities: both sequences contain an equal number of heads and tails, which is what we might expect from a random set of flips.
Now to the differences. Look at the largest runs in each sequence. The first sequence has a run of five Ts and a run of four Hs. The largest runs in the second sequence, however, are three Ts and three Hs. While a run of five Ts might seem unlikely in a sequence of 30 flips – and therefore evidence of human interference – in fact, an occurrence is more likely than not. Randomness throws up this type of cluster, or ‘coincidence’, all the time.
The second sequence gives itself away as man-made if you count the number of times the flips alternate between heads and tails. If all the flips are random, you would expect a head (or tail) to be followed by a tail about 50 per cent of the time and followed by a head about 50 per cent of the time. So, in a run of 30 flips, you would expect the results to alternate about 14 or 15 times. In the first sequence, the results alternate between heads and tails 14 times, while in the second sequence the results alternate 18 times. The second sequence therefore appears more biased, and thus less random, than the first.
112 JUST FOUR KIDS
At first glance, you might think that half boys and half girls is more likely. Each child has a 50/50 chance of being a boy or a girl, so with four siblings you would expect, on average, two to be boys and two to be girls. Sloppy thinking!
Let’s look at the 16 equally probable combinations of four siblings (where B is a boy and G is a girl):
BBBG GGGB GGBB GBBG
BBGB GGBG BBGG BGGB
BGBB GBGG GBGB BBBB
GBBB BGGG BGBG GGGG
Once you tot them up, you’ll see that only six possible combinations have two boys and two girls, but eight have three of one and one of the other.
So the most likely outcome will be three of one and one of the other.
113 THE BIG FAMILY
Mrs Brown’s strategy is likely to result in a smaller family.
Let G = a girl and B = a boy. There are four equally likely ‘doublets’ of two children born one after the other: BB, BG, GG and GB.
The question states that Mr Brown wants to stop when they get a BB, and that Mrs Brown wants to stop when they get a GB.
In other words, the problem boils down to working out whether BB or GB is more likely to come first in a random sequence of Bs and Gs.
In a family of two children, the sequence BB and BG appear with equal likelihood.
However, in a family of more than two children, GB is more likely to appear before BB. This is because in any sequence of Bs and Gs, a BB is always preceded by a GB except when the first two children are BB. In other words, as soon as a girl is born, you are guaranteed that GB will come before BB.
Since there is only a 1 in 4 chance that the first two children are BB, the chance that GB appears first is 3 in 4.
In other words, when it gets to the point that one of them wants to stop having children, it is more likely to be Mrs Brown (75 per cent chance) than Mr Brown (25 per cent chance).
114 PROBLEMS WITH SIBLINGS
[1] ⅓
Problems involving probabilities are often easier to understand when expressed as expected frequencies. Imagine that 4,000 two-children families are chosen at random. We’d expect the following frequencies for each equally likely combination:
Older child Younger child Frequency
Boy Girl 1,000
Girl Boy 1,000
Boy Boy 1,000
Girl Girl 1,000
In all occurrences of boy-girl, Albert will circle the first line. That’s 1,000 cases.
In half of the occurrences of boy-boy, Albert will circle the first line. That’s 500 cases.
So, the top line is circled in 1,500 cases, and in only 500 of them are both children boys. The chance of two boys is therefore 1 in 3.
[2] No.
Here we have a lovely paradox, which captures one of the reasons that this type of problem causes lots of anguish and disagreement. We know from above that Albert has a 1 in 3 chance of having two children. All the journalist has done is communicate what Albert wrote on the form. So if readers were aware of the rules that governed how Albert filled in the form, they would deduce that Albert has a 1 in 3 chance of having two boys. Yet if readers assume that Albert has been randomly selected from all two-children families in which the older child is a son, then we know from the preamble to the puzzle that the chance of his having two boys is 1 in 2.
The moral is: in order to avoid any ambiguity we need to know how the information presented was arrived at.
[3] ½
If Beth is thinking of the older child, there is a 1 in 2 chance the other child is a girl. If she is thinking of the younger one, there is also a 1 in 2 chance that both are girls. So the overall chance must be 1 in 2.
[4] No it doesn’t, because of the three possible combinations – GG, GB and BG – all are still equally likely.
[5] ½
Remarkably, this situation produces a different answer from the previous one. Imagine I asked this question 100 times. On 50 occasions I would have asked ‘is your older child a girl?’, and a yes to this answer would have given 25 instances in which two children are girls. On the remaining 50 occasions I would have asked ‘is your younger child a girl?’ and a yes would have given another 25 instances of two girls. So, in 50 of 100 instances there are two girls, meaning that the chances are 1 in 2.
115 THE GIRL BORN IN AN EVEN YEAR
Let’s approach this problem by looking at frequencies, as we did in the answer to the previous question. Imagine selecting 400 two-children families at random. We’d expect the following frequencies:
Older child Younger child Frequency
Boy Girl 100
Girl Boy 100
Boy Boy 100
Girl Girl 100
We know that babies are equally likely to be born in odd or even years. If ‘girlODD’ and ‘girlEVEN’ refer to girls born in odd and even years, the table becomes:
Older child Younger child Frequency
Boy GirlODD 50
Boy GirlEVEN 50
GirlODD Boy 50
GirlEVEN Boy 50
Boy Boy 100
GirlEVEN GirlEVEN 25
GirlEVEN GirlODD 25
GirlODD GirlEVEN 25
GirlODD GirlODD 25
> The total number of families with at least one girlEVEN is 50 + 50 + 25 + 25 + 25 = 175. Of these families, 75 of them have two girls.
So the probability of having two girls, given that one is a girlEVEN, is 75⁄175 = 3⁄7, or about 43 per cent.
*
Now to the Tuesday-boy problem, in which a parent has a boy born on a Tuesday. We can draw up a similar table of frequencies. Imagine selecting 196 two-children families at random. (It will become clear why I chose that particular number). We’d expect the following frequencies:
Older child Younger child Frequency
Boy Girl 49
Girl Boy 49
Boy Boy 49
Girl Girl 49
Each equally likely combination has a frequency of 49. When dealing with days of the week it is helpful to have a number divisible by 7. Now if ‘boyT’ is a boy born on a Tuesday and ‘boyNT’ is a boy born on any other day, we can expand the table to:
Older child Younger child Frequency
BoyT Girl 7
BoyNT Girl 42
Girl BoyT 7
Girl BoyNT 42
BoyT BoyT 1
BoyNT BoyT 6
BoyT BoyNT 6
BoyNT BoyNT 36
Girl Girl 49
The number of families with a boy born on a Tuesday is 7 + 7 + 1 + 6 + 6 = 27. Of these families, 13 have two boys, so the probability of having two boys, given that one is born on a Tuesday, is 13⁄27, which is about 48 per cent.
Let’s summarize what we have discovered.
When a two-child family has ‘at least one boy’, the chances of two boys is 33 per cent.
When a two-child family has ‘at least one boy who was born in an even year’, the chance of two boys is 43 per cent.
When a two-child family has ‘at least one boy who was born on a Tuesday’ the chance of two boys rises to 48 per cent.
As more details are specified about the boy, the chance that the family has two boys gets closer to 50 per cent.
If the question was to specify enough information to make it absolutely clear who that boy is, then the chances of two boys would be exactly 50 per cent. Likewise, if we state that the boy is the older (or younger) son, we are also making it absolutely clear who he is, and the chance of two boys is also 50 per cent.
116 THE TWYNNE TWINS
The most likely position is in first place. Each position has an equally likely probability of having one of the twins. But when a twin is in first position, it is always the first, whereas when further back there is a chance it is the second twin. Let’s run the numbers. Imagine that there are only three students: the Twynne twins, and an Other. There are three equally likely ways the students can line up, from left to right:
TTO
TOT
OTT
In 2/3 cases the first twin is at the front of the queue, and in 1/3 cases the first twin is in second place. (The first twin can never be in last place.)
With two Others, the six equally likely ways the four students can line up are:
TTOO
TOTO
TOOT
OTTO
OOTT
OTOT
In 3/6, or 1/2, cases the first twin is at the front of the queue; in 2/6, or 1/3 cases the first twin is in second place; and in 1/6 cases the first twin is in third place. If you were to do the same with more and more Others, you would always discover that the most likely position for the first Twynne is first place. Likewise, the second twin is most likely to be in last place.
(In general, in a class of n people, the chance of the first twin being in first place is n – 1 divided by the sum of the numbers from 1 to n – 1. So in a class of 30 students the chance of a twin being in the first position is 29⁄435.)
117 A JAB OF MMMR
The answer is (2, 5, 5, 6, 7) and (3, 4, 5, 5, 8).
How do we get there? We know the median is 5, so we can fill in our first value:
(X, X, 5, X, X)
If the range is 5, the highest and lowest values will be one of the following pairs: (0, 5), (1, 6), (2, 7), (3, 8), (4, 9) or (5, 10).
We can eliminate some of these pairs by considering the mean. If the mean is 5, the sum of all the numbers divided by 5 must equal 5, in other words, the sum of all the numbers must be 25. If the range is (0, 5) or (1, 6), then the numbers will always sum to less than 25. Likewise, if the range is (4, 9) and (5, 10), the numbers will always sum to more than 25. So we have two possible sets:
(2, X, 5, X, 7) and (3, X, 5, X, 8)
In the former case the two unknowns must sum to 11. We can eliminate 4 and 7, since this would make 7 the mode. So, these unknowns must be 5 and 6.
In the latter case the two unknowns must sum to 9. We can eliminate 3 and 6 since that would make 3 the mode. So these unknowns must be 4 and 5.
118 LIES AND STATISTICS
Imagine that everyone in the junior school has a grade E, and that everyone in the upper school has a grade C.
If the upper school has one more pupil than the junior school, then the median will be a C.
If everyone who is an E improves to a D, and everyone who is a C improves to a B, and two new pupils whose grade is D or lower join the junior school, everyone’s grades will have improved and the median will now be D.
119 THE LONELINESS OF THE LONG-DISTANCE RUNNER
Using only the information presented in the question the best estimate is 251 runners.
By looking out of the window you have confirmed that there are at least 251 runners. If there are exactly 251, then the probability that you would have seen this runner is 1⁄251. If there are 252 runners, then the probability you would have seen this runner is 1⁄252, which is a tiny bit less likely. The estimate that makes your guess the most likely option is that there are 251 runners.
Of course, you may have extra information that can help you make a more informed estimate, such as knowledge of your local race that you picked up from the radio or friends, or an assumption that races tend to have a round number of contestants. But without any of this information, the best guess is 251.
Statisticians call this method of deduction ‘maximum likelihood estimation.’
120 THE FIGHT CLUB
Beast, Mouse, Beast.
It’s a counter-intuitive answer, since your best course of action is to fight the strongest opponent twice.
If you had to win a single bout, you would choose to go against Mouse. He is the weaker fighter, less good than you, and it’s likely you would win.
Similarly, if you were fighting three bouts, and you wanted to win as many as possible, you would choose to fight Mouse – the weakest – twice and Beast only once.
However, the question asks about maximising your chances of winning two fights in a row. In this case, it is in your best interests to choose the option where you fight the tougher opponent twice, because if you have to win two bouts in a row, you must win the middle one. Your best course of action, therefore, is to choose the easiest adversary for the middle bout, which gives you two chances to beat the harder opponent in the other bouts.
We can prove this more rigorously using the numbers given.
If the probability of beating Beast is 2/5 and of beating Mouse is 9⁄10, we can draw up the following table. We work out the probabilities of each possible combination of winning two in a row, and add them up.
In the first combination the total odds are 126⁄250, which is just over half, so you are more likely than not to win two bouts in a row. In the second combination the odds are under half, so you are more likely to fail to win two bouts in a row.
The moral of the story is that winning the most battles is not always the best way to win the war.
121 TYING THE GRASS AND TYING THE KNOT
It is more likely.
Once the top ends are joined in pairs, the blades can be arranged as so:
In order to make a loop, end A can be joined to any end except B, meaning that there is a choice of four out of the five options. Once
A is joined to a permitted end, say C, then B can join to anywhere except D, since this will close a loop of four blades of grass. In other words, there are two out of three options for B. Let’s say B is joined to E. This means that the final pair must be F and D. So the chances of creating a loop are 4⁄5 x 2⁄3 = 8⁄15, which is just over half.
122 THE THREE SLIPS OF PAPER
Let’s say the three numbers are A, B and C, where A > B > C.
The best strategy is to turn over two slips. If the second slip you turn over has a number that is bigger than the first, choose the second slip, but if the second slip has a lower number than the first, choose the remaining slip. This increases your chance of picking the slip with the highest number to 1 in 2.
We can see this by looking at the table below. If Slip 1 is the first slip you choose, and Slip 2 the second, the table lists the six equally likely arrangements of A, B and C, together with what your choice will be. In three out of the six arrangements you will choose the highest number.
Slip 1 Slip 2 Slip 3 Choice
A B C C
A C B B
B A C A
B C A A
C A B A
C B A B
The solution for the problem when there are only two slips of paper is similar to the solution with three slips, only now we must randomly generate a number for an imaginary third slip. Here’s how it works.
Let A and B be the numbers on the slips, with A > B.