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 2018-05-02 
Every morning just before 7am, one of the Today Programme's presenters reads out a puzzle. Yesterday, it was this puzzle:
In a given month, the probability of a certain daily paper either running a story about inappropriate behaviour at a party conference or running one about somebody's pet being retrieved from a domestic appliance is exactly half the probability of the same paper containing a photo of a Tory MP jogging. The probability of no such photo appearing is the same as that of there being a story about inappropriate behaviour at a party conference. The probability of the paper running a story about somebody's pet being retrieved from a domestic appliance is a quarter that of its containing a photo of a Tory MP jogging. What are the probabilities the paper will (a) run the conference story, (b) run the pet story, (c) contain the jogging photo?
I'm not the only one to notice that some of Radio 4's daily puzzles are not great. I think this puzzle is a great example of a terrible puzzle. You can already see the first problem with it: it's long and words and very hard to follow on the radio. But maybe this isn't so important, as you can read it here after it's been read out.
Once you've done this, you can re-write the puzzle as follows: there are three news stories (\(A\), \(B\) and \(C\)) that the newspaper might publish in a month. We are given the following information:
$$\mathbb{P}(A\text{ or }B)=\tfrac12\mathbb{P}(C)$$ $$1-\mathbb{P}(C)=\mathbb{P}(A)$$ $$\mathbb{P}(B)=\tfrac14\mathbb{P}(C)$$
To solve this puzzle, we need use the formula \(\mathbb{P}(A\text{ or }B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\). These Venn diagrams justify this formula:
Venn diagrams showing that \(\mathbb{P}(A\text{ or }B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\).
Using the information we were given in the question, we get:
\begin{align} \tfrac12\mathbb{P}(C)&=\mathbb{P}(A\text{ or }B)\\ &=1-\mathbb{P}(C)+\tfrac14\mathbb{P}(C)-\mathbb{P}(A\text{ and }B)\\ \mathbb{P}(C)&=\tfrac45(1-\mathbb{P}(A\text{ and }B)). \end{align}
At this point we have reached the second problem with this puzzle: there's no answer unless we make some extra assumptions, and the question doesn't make it clear what we can assume. But let's give the puzzle the benefit of the doubt and try some assumptions.

Assumption 1: The events are mutually exclusive

If we assume that the events \(A\) and \(B\) are mutually exclusive—or, in other words, only one of these two articles can be published, perhaps due to a lack of space—then we can use the fact that
$$\mathbb{P}(A\text{ and }B)=0.$$
This means that \(\mathbb{P}(C)=\tfrac45\), \(\mathbb{P}(A)=\tfrac15\), and \(\mathbb{P}(B)=\tfrac15\). There's a problem with this answer, though: the three probabilities add up to more than 1.
This wouldn't be a problem, except we assumed that only one of the articles \(A\) and \(B\) could be published. The probabilities adding up to more than 1 means that either \(A\) and \(C\) are not mutually exclusive or \(A\) and \(B\) are not mutually exclusive, so \(C\) could be published alongside \(A\) or \(B\). There seems to be nothing special about the three news stories to mean that only some combinations could be published together, so at this point I figured that this assumption was wrong and moved on.
Today, however, the answer was posted, and this answer was given (without an working out). So we have a third problem with this puzzle: the answer that was given is wrong, or at the very best is based on questionable assumptions.

Assumption 2: The events are independent

If we assume that the events are independent—so one article being published doesn't affect whether or not another is published—then we may use the fact that
$$\mathbb{P}(A\text{ and }B)=\mathbb{P}(A)\mathbb{P}(B).$$
If we let \(c=\mathbb{P}(C)\), then we get:
\begin{align} \tfrac12c&=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\\ &=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A)\mathbb{P}(B)\\ &=1-c+\tfrac14c-\tfrac14(1-c)c\\ \tfrac14c^2-\tfrac32c+1&=0. \end{align}
You can use your favourite formula to solve this to find that \(c=3-\sqrt5\), and therefore \(\mathbb{P}(A)=\sqrt5-2\) and \(\mathbb{P}(B)=\tfrac34-\tfrac{\sqrt5}4\).
In this case, our assumption appears to be more reasonable—as over the course of a month the stories published by a paper probably don't have much of an effect on each other—but we have the fourth, and probably biggest problem with the puzzle: the question and answer are not interesting or surprising, and the method is a bit tedious.
                        
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@Stefan: It's possible that that was intended but it's not completely clear. If that was the intention, I stick to my point that it's odd that the other story can be published alongside these two.
Matthew
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Doesn’t the word ‘either’ in the opening phrase make A and B explicity exclusive?
Stefan
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 2018-03-23 
This morning, I heard about Arnie Ellis on the Today programme. Arnie is the first baby boy to be born in his family in five generations, following ten girls. According to John Humphrys, there is a 20,000-to-1 chance of this happening. Pretty quickly, I started wondering where this number came from.
After a quick Google, I found that this news story had appeared in many of today's papers, including the Sun and the Daily Mail. They all featured this 20,000-to-1 figure, which according to The Sun originally came from Ladbrokes.

What is the chance of this happening?

If someone is having a child, the probability of it being a girl is 0.5. The probability of it being a boy is also 0.5. So the probaility of having ten girls followed by a boy is
$$\left(\tfrac12\right)^{10}\times\tfrac12=\frac1{2048}.$$
If all 11 children were siblings, then this would be the chance of this happening—and it's a long way off the 20,000-to-1. But in Arnie's case, the situation is different. Luckily in the Daily Mail article, there is an outline of Arnie's family tree.
Here, you can see that the ten girls are spread over five generations. So the question becomes: given a baby, what is the probability that the child is male and his most recently born ten relatives on their mother's side are all female?
Four of the ten relatives are certainly female—Arnie's mother, grandmother, great grandmother and great great grandmother are all definitely female. This only leaves six more relatives, so the probability of a baby being in Arnie's position is
$$\left(\tfrac12\right)^{6}\times\tfrac12=\frac1{128}.$$
This is now an awful lot lower than the 20,000-to-1 we were told. In fact, with around 700,000 births in the UK each year, we'd expect over 5,000 babies to be born in this situation every year. Maybe Arnie's not so rare after all.
This number is based on the assumption that the baby's last ten relatives are spread across five generations. But the probability will be different if the relatives are spread over a different number of generations. Calculating the probability for a baby with any arrangement of ancestors would require knowing the likelihood of each arrangement of relatives, which would require a lot of data that probably doesn't exist. But the actual anwer is probably not too far from 127-to-1.

Where did 20,000-to-1 come from?

This morning, I emailed Ladbrokes to see if they could shed any light on the 20,000-to-1 figure. They haven't got back to me yet. (Although they did accidentally CC me when sending the query on to someone who might know the answer, so I'm hopeful.) I'll update this post with an explanaation if I do hear back.
Until then, there is one possible explanation for the figure: we have looked at the probability that a baby will be in this situation, but we could instead have started at the top of the family tree and looked at the probability that Beryl's next ten decendents were girls followed by a boy. The probability of this happening will be lower, as there is a reasonable chance that Beryl could have no female children, or no children at all. Looking at the problem this way, there are more ways for the situation to not happen, so the probability of it happening is lower.
But working the actually probability out in this way would again require data about how many children are likely in each generation, and would be a complicated calculation. It seems unlikely that this is what Ladbrokes did. Let's hope they shed some light on it...
                        
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@Steve Spivey: Nothing
Matthew
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Any response from Ladbrokes yet?
Steve Spivey
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 2018-03-22 
Back in 2014, I worked out the cost of filling an official Panini World Cup 2014 sticker book. Today, the 2018 sticker book was relased: compared to four years ago, there are more stickers to collect and the prices have all changed. So how much should we expect it to cost us to fill the album this time round?

How many stickers will I need?

There are 682 stickers to collect. Imagine you have already stuck \(n\) stickers into your album. The probability that the next sticker you buy is new is
$$\frac{682-n}{682}.$$
The probability that the second sticker you buy is the next new sticker is
$$\mathbb{P}(\text{next sticker is not new})\times\mathbb{P}(\text{sticker after next is new})$$ $$=\frac{n}{682}\times\frac{682-n}{682}.$$
Following the same method, we can see that the probability that the \(i\)th sticker you buy is the next new sticker is
$$\left(\frac{n}{682}\right)^{i-1}\times\frac{682-n}{682}.$$
Using this, we can calculate the expected number of stickers you will need to buy until you find a new one:
$$\sum_{i=1}^{\infty}i \left(\frac{682-n}{682}\right) \left(\frac{n}{682}\right)^{i-1} = \frac{682}{682-n}$$
Therefore, to get all 682 stickers, you should expect to buy
$$\sum_{n=0}^{681}\frac{682}{682-n} = 4844 \text{ stickers}.$$

How much will this cost?

You can buy the following [source]:
First of all you'll need to buy the starter pack, as you need an album to stick everything in. This comes with 31 stickers; we should expect to buy 4813 more stickers after this.
The cheapest way to buy these stickers is to buy them in multipacks for 15p per sticker. This gives a total expected cost of filling the sticker album of £725.94. (Although if your local newsagent doesn't stock the multipacks, buying 80p packs to get these stickers will cost you £774.07.)

What if I order the last 50 stickers?

If you'd like to spend a bit less on the sticker book, Panini lets you order the last 50 stickers to complete your album. This is very helpful as these last 50 stickers are the most expensive.
You can order missing stickers from the Panini website for 22p per sticker's sticker ordering service for the 2018 World Cup doesn't appear to be online yet; but based on other recent collections, it looks like ordered stickers will cost 16p each, with £1 postage per order.
Ordering the last 50 stickers reduces the expected number of other stickers you need to buy to
$$\sum_{n=0}^{631}\frac{682}{682-n} = 1775 \text{ stickers}.$$
This reduces the expected overall cost to £291.03. So I've just saved you £434.91.

What if I order more stickers?

Of course, if you're willing to completely give up on your morals, you could order more than one batch of 50 stickers from Panini. This raises the question: how many should you order to minimise the expected cost.
If you order the last \(a\) stickers, then you should expect to pay:
The total expected cost of filling your album for different values of \(a\) is shown in the graph below.
The red cross shows the point at which the album is cheapest: this is when the last 550 stickers are ordered, giving a total expected cost of £120.18. That's another £170.85 I've saved you. You're welcome.
Still, ordering nearly all stickers to minimise the cost doesn't sound like the most fun way to complete the sticker book, so you probably need to order a few less than this to maximise your fun.

What about swaps?

Of course, you can also get the cost of filling the book down by swapping your spare stickers with friends. In 2016, I had a go at simulating filling a sticker book with swapping and came to the possibly obvious conclusion that the more friends you have to swap with, the cheaper filling the book will become.
My best advice for you, therefore, is to get out there right now and start convinding your friends to join you in collecting stickers.
                        
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@RM: It would slightly reduce the expected cost I think, but I need to do some calculations to make sure.

I'm not certain whether they guarantee that there are no duplicates in a pack or just it's very unlikely... Follow up post looking into this coming soon
Matthew
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Helpful article!! My head always go into a spine with probability, but does the fact that you would not get a double within a pack affect the overall strategy or cost? My head says this is an another advantage of buying 30 packs over the 5.
RM
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 2015-10-08 
This post also appeared on the Chalkdust Magazine blog. You can read the excellent second issue of Chalkdust here, including the £100 prize crossnumber which I set.
From today, the National Lottery's Lotto draw has 59 balls instead of 49. You may be thinking that this means there is now much less chance of winning. You would be right, except the prizes are also changing.
Camelot, who run the lottery, are saying that you are now "more likely to win a prize" and "more likely to become a millionaire". But what do these changes actually mean?

The changes

Until yesterday, Lotto had 49 balls. From today, there are 59 balls. Each ticket still has six numbers on it and six numbers, plus a bonus ball, are still chosen by the lottery machine. The old prizes were as follows:
RequirementEstimated Prize
Match all 6 normal balls£2,000,000
Match 5 normal balls and the bonus ball£50,000
Match 5 normal balls£1,000
Match 4 normal balls£100
Match 3 normal balls£25
50 randomly picked tickets£20,000
The prizes have changed to:
RequirementEstimated Prize
Match all 6 normal balls£2,000,000
Match 5 normal balls and the bonus ball£50,000
Match 5 normal balls£1,000
Match 4 normal balls£100
Match 3 normal balls£25
Match 2 normal ballsFree lucky dip entry in next Lotto draw
One randomly picked ticket£1,000,000
20 other randomly picked tickets£20,000

Probability of Winning a Prize

The probability of winning each of these prizes can be calculated. For example, the probability of matching all 6 balls in the new lotto is $$\mathbb{P}(\mathrm{matching\ ball\ 1})\times \mathbb{P}(\mathrm{matching\ ball\ 2})\times...\times\mathbb{P}(\mathrm{matching\ ball\ 6})$$ $$=\frac{6}{59}\times\frac{5}{58}\times\frac{4}{57}\times\frac{3}{56}\times\frac{2}{55}\times\frac{1}{54}$$ $$=\frac{1}{45057474},$$ and the probability of matching 4 balls in the new lotto is $$(\mathrm{number\ of\ different\ ways\ of\ picking\ four\ balls\ out\ of\ six})\times\mathbb{P}(\mathrm{matching\ ball\ 1})\times\\...\times\mathbb{P}(\mathrm{matching\ ball\ 4})\times\mathbb{P}(\mathrm{not\ matching\ ball\ 5})\times\mathbb{P}(\mathrm{not\ matching\ ball\ 6})$$ $$=15\times\frac{6}{59}\times\frac{5}{58}\times\frac{4}{57}\times\frac{3}{56}\times\frac{53}{55}\times\frac{52}{54}$$ $$=\frac{3445}{7509579}.$$ In the second calculation, it is important to include the probabilities of not matching the other balls to prevent double counting the cases when more than 4 balls are matched.
Calculating a probability for every prize and then adding them up gives the probability of winning a prize. In the old draw, the probability of winning a prize was \(0.0186\). In the new draw, it is \(0.1083\). So Camelot are correct in claiming that you are now more likely to win a prize.
But not all prizes are equal: these probabilities do not take into account the values of the prizes. To analyse the actual winnings, we're going to have to look at the expected amount of money you will win. But first, let's look at Camelot's other claim: that under the new rules you are more likely to become a millionaire.

Probability of winning £1,000,000

In the old draw, the only way to win a million pounds was to match all six balls. The probability of this happening was \(0.00000007151\) or \(7.151\times 10^{-8}\).
In the new lottery, a million pounds can be won either by matching all six balls or by winning the millionaire raffle. This will lead to different probabilities of winning on Wednesdays and Saturdays due to different numbers of people buying tickets. Based on expected sales of 16.5 million tickets on Saturdays and 8.5 million tickets on Wednesdays, the chances of becoming a millionaire on a Wednesday or Saturday are \(0.0000001398\) (\(1.398\times 10^{-7}\)) and \(0.00000008280\) (\(8.280\times 10^{-8}\)) respectively.
These are both higher than the probability of winning a million in the old draw, so again Camelot are correct: you are now more likely to become a millionaire...
But the new chances of becoming a millionaire are actually even higher. The probabilities given above are the chances of winning a million in a given draw. But if two balls are matched, you win a lucky dip: you could win a million in the next draw without buying another ticket. We should include this in the probability calculated above, as you are still becoming a millionaire due to the original ticket you bought.
In order to count this, let \(A_W\) and \(A_S\) be the probabilities of winning a million in a given draw (as given above) on a Wednesday or a Saturday, let \(B_W\) and \(B_S\) be the probabilities of winning a million in this draw or due to future lucky dip tickets on a Wednesday or a Saturday (the values we want to find) and let \(p\) be the probability of matching two balls. We can write $$B_W=A_W+pB_S$$ and $$B_S=A_S+pB_W$$ since the probability of winning a million is the probability of winning in this draw (\(A\)) plus the probability of winning a lucky dip ticket and winning in the next draw (\(pB\)). Substituting and rearranging, we get $$B_W=\frac{A_W+pA_S}{1-p^2}$$ and $$B_W=\frac{A_S+pA_W}{1-p^2}.$$
Using this (and the values of \(A_S\) and \(A_W\) calculated earlier) gives us probabilities of \(0.0000001493\) (\(1.493\times 10^{-7}\)) and \(0.00000009736\) (\(9.736\times 10^{-8}\)) of becoming a millionaire on a Wednesday and a Saturday respectively. These are both significantly higher than the probability of becoming a millionaire in the old draw (\(7.151\times 10^{-8}\)).
Camelot's two claims—that you are more likely to win a prize and you are more likely to become a millionaire—are both correct. It sounds like the new lottery is a great deal, but so far we have not taken into account the size of the prizes you will win and have only shown that a very rare event will become slightly less rare. Probably the best way to measure how good a lottery is is by working out the amount of money you should expect to win, so let's now look at that.

Expected prize money

To find the expected prize money, we must multiply the value of each prize by the probability of winning that prize and then add them up, or, in other words,
$$\sum_\mathrm{prizes}\mathrm{value\ of\ prize}\times\mathbb{P}(\mathrm{winning\ prize}).$$
Once this has been calculated, the chance of winning due to a free lucky dip entry must be taken into account as above.
In the old draw, after buying a ticket for £2, you could expect to win 78p or 83p on a Wednesday or Saturday respectively. In the new draw, the expected winnings have changed to 58p and 50p (Wednesday and Saturday respectively). Expressed in this way, it can be seen that although the headline changes look good, the overall value for money of the lottery has significantly decreased.
Looking on the bright side, this does mean that the lottery will make even more money that it can put towards charitable causes: the lottery remains an excellent way to donate your money to worthy charities!
                        
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 2014-04-11 
As well as letters games, the contestants on Countdown also take part in numbers games. Six numbers are chosen from the large numbers (25,50,75,100) and small numbers (1-10, two cards for each number) and a total between 101 and 999 (inclusive) is chosen by CECIL. The contestants then use the six numbers, with multiplication, addition, subtraction and division, to get as close to the target number as possible.
The best way to win the numbers game is to get the target exactly. This got me wondering: is there a combination of numbers which allows you to get every total between 101 and 999? And which combination of large and small numbers should be picked to give the highest chance of being able to get the target?
To work this out, I got my computer to go through every possible combination of numbers, trying every combination of operations. (I had to leave this running overnight as there are a lot of combinations!)

Getting every total

There are 61 combinations of numbers which allow every total to be obtained. These include the following (click to see how each total can be made):
By contrast, the following combination allows no totals between 101 and 999 to be reached:
The number of attainable targets for each set of numbers can be found here.

Probability of being able to reach the target

Some combinations of numbers are more likely than others. For example, 1 2 25 50 75 100 is four times as likely as 1 1 25 50 75 100, as (ignoring re-orderings) in the first combination, there are two choices for the 1 tile and 2 tile, but in the second combination there is only one choice for each 1 tile. Different ordering of tiles can be ignored as each combination with the same number of large tiles will have the same number of orderings.
By taking into account the relative probability of each combination, the following probabilities can be found:
Number of large numbersProbability of being able to reach target
00.964463439
10.983830962
20.993277819
30.985770510
40.859709475
So, in order to maximise the probability of being able to reach the target, two large numbers should be chosen.
However, as this will mean that your opponent will also be able to reach the target, a better strategy might be to pick no large numbers or four large numbers and get closer to the target than your opponent, especially if you have practised pulling off answers like this.
Edit: Numbers corrected.
Edit: The code used to calculate the numbers in this post can now be found here.
                        
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@Francis Galiegue: I've pushed a version of the code to https://github.com/mscroggs/countdown-...
Matthew
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@Francis Galiegue: Sadly, I lost the code I used when I had laptop problems. However, I can remember what it did, so I shall recreate it and put it on GitHub.
Matthew
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If you could, I'd love to have the code you used to do this exhaustive search?

I'm a fan of the game myself (but then I'm French, so to me it's the original, "Des chiffres et des lettres"), but for the numbers game, this is pretty much irrelevant to the language and country :)
Francis Galiegue
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