# One-in-a-million

Suppose you can play on either of 2 slot machines:

1. Slot machine A pays out with probability one in a million.
2. Slot machine B pays out with probability one in 10.

Are you more likely to get a payout with one million attempts with slot machine A or with 10 attempts on slot machine B?

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I was prompted to think about this question by the following tweet, which includes both the answer and the relevant calculations.

So, there’s a bigger probability (0.65) that you’ll get a payout from 10 spins of slot machine B than from a million spins of slot machine A (probability 0.63).

Hopefully, the calculations above are self-explanatory. But just in case, here’s the detail. Suppose you have N attempts to win with a slot machine that pays out with probability 1/N.

1. First we’ll calculate the probability of zero payouts in the N spins.

2. This means we get a zero payout on every spin.

3. The probability of a zero payout on one spin is one minus the probability of a win: 1 – 1/N.

4. So the probability of no payout on all the spins is

$(1-1/N)^N$

5. And the probability of at least one payout is

$1- (1-1/N)^N$

As explained in the tweet, with N=10 this gives 0.65 and with N=1,000,000 it gives 0.63. The tweet’s author explains in a follow-up tweet that he was expecting the same answer both ways.

But as someone in the discussion pointed out, that logic can’t be right. Suppose you had one attempt with slot machine C which paid out with probability 1. In other words, N=1 in my example above. Then, of course, you’d be bound to get a payout, so the probability of at least one payout is 1. So, although it’s initially perhaps surprising that you’re more likely to get a payout with 10 shots at slot machine B than with a million shots at slot machine A, the dependence on N becomes obvious when you look at the extreme case of slot machine C.

Footnote: What does stay the same in each case however is the average number of times you will win. With N shots at a slot machine with win probability 1/N, you will win on average once for any choice of N. Sometimes you’ll win more often, and sometimes you may not win at all (except when N=1). But the average number of wins if you play many times will always be 1.

# Sleight of hand

A while ago I sent a post with a card trick which I explained had a statistical element to it, and asked you to try to work out how it was done. Thanks to those of you who wrote to me with variants on the correct answer.

The rules of the game were that my assistant, Matteo, chose a card at random hidden from me. It happened to be a 5 in the video. I then turned the cards over one at a time and Matteo had to play a counting game. Once he reached the 5th card, he noted its value, which was a 10. So he then counted another 10 cards in the sequence, noted the value of that card, and so on until we ran out of cards. Matteo had to remember the final card in his sequence before the cards ran out, which turned out to be the eight of diamonds. My task as the magician was to predict what Matteo’s final card was, which I did successfully.

Now, there are 2 reasons why this is a statistical card trick.

1. It doesn’t always work. It does so with a reasonably high probability, but depending on the  configuration of the cards once they are shuffled, won’t always. I’ll be honest: we had to remake the video several times, but that was always due to my incompetence in explaining the trick and not because it ever failed. Still, it won’t always work.
2. The second reason it’s a statistical trick is in its execution. The way it works is that I also play the same counting game as Matteo, but starting with the value of the first card I turn over, which happened to be a 10. So, we’re both playing the same counting game but from different starting points. Matteo’s starting point is 5, mine is 10. Although we start from different places, it turns out to be quite likely – though not certain – that the counting sequences we follow will overlap at some point. And once they do overlap, we are then following exactly the same sequence and so will arrive at the same final card.

Technically, the sequences of cards Matteo and I are both following are called Markov chains. These are sequences of random numbers such that in order to understand what the next card might be I only need to know the value of the current card, without knowing the past sequence that took me to the current state. In other words, when Matteo has to start counting 10 cards, it doesn’t matter how he got to that position, just that that’s where he currently is. And I also generate my own Markov chain. With an unlimited number of cards in a pack, the mathematical properties of Markov chains would guarantee that our sequences  meet at some point, after which we would be following exactly the same sequence, leading me to have the same final card as Matteo. With just 52 cards in a pack, there’s no guarantee, which is why the trick won’t always work.

The fact that the trick might not work is a little undesirable, but you can increase the chances by counting picture cards as 1 rather than 10. This forces the sequences to change card more often, which increases the chances of our two sequences overlapping.

Markov chains are actually really important building blocks for modelling in many areas of Statistics, which is one reason why I used posted the card trick to the blog. I’ll use a future post to explain this though.

# Massively increase your bonus

In one of the earliest posts to the blog last year I set a puzzle where I suggested Smartodds were offering employees the chance of increasing their bonus, and you had to decide whether it was in their interests to accept the offer or not.

<They weren’t, and they still aren’t, but let’s play along>.

Same thing this year, but the rules are different. Eligible employees are invited to gamble their bonus at odds of 10-1 based on the outcome of a game. It works like this…

For argument’s sake, let’s suppose there are 100 employees that are entitled to a bonus. They are told they each have the opportunity to increase their bonus by a factor of 10 by playing the following game:

• Each of the employees is randomly assigned a number between 1 and 100.
• Inside a room there are 100 boxes, also labelled 1 to 100.
• 100 cards, numbered individually from 1 to 100, have been randomly placed inside the boxes, so each numbered box contains a card with a unique random number from 1 to 100. For example, box number 1 might contain the card with number 62; box number 2 might contain the card with number 25; and so on.
• Each employee must enter the room, one a a time, and can choose any 50 of the boxes to open. If they find the card with their own number in one of those boxes, they win. Otherwise they lose.
• Though the employees may discuss the game and decide how they will play before they enter the room, they must not convey any information to the other employees after taking their turn.
• The employees cannot rearrange any of the boxes or the cards – so everyone finds the room in the same state when they enter.
• The employees will have their bonus multiplied by 10 if all 100 of them are winners. If there is a single loser, they all end up with zero bonus.

Should the employees accept this game, or should they refuse it and keep their original bonuses? And if they accept to play, should they adopt any particular strategy for playing the game?

Give it some thought and then scroll down for some discussion.

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A good place to start is to calculate the probability that any one employee is a winner. This happens if one of the 50 boxes they open, out of the 100 available, contains the card with their number. Each box is equally likely to contain their number, so you can easily write down the probability that they win. Scroll down again for the answer to this part:

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There are 100 boxes, and the employee selects 50 of them. Each box is equally likely to contain their number, so the probability they find their number in one of the boxes is 50/100 or 1/2.

So that’s the probability that any one employee wins. We now need to calculate the probability that they all win – bearing in mind the rules of the game – and then decide whether the bet is worth taking.

In summary:

• There are 100 employees;
• The probability that any one employee wins their game is 1/2;
• If they all win, their bonuses will all be multiplied by 10;
• If any one of them loses, they all get zero bonus.

Should the employees choose to play or to keep their original bonus? And if they play, is there any particular strategy they should adopt?

If you’d like to send me your answers I’d be really happy to hear from you. If you prefer just to send me a yes/no answer, perhaps just based on your own intuition, I’d be equally happy to get your response, and you can use this form to send the answer in that case.

This is a variant on a puzzle pointed out to me by Fabian.Thut@smartodds.co.uk. I think it’s a little more tricky than previous puzzles I’ve posted, but it illustrates a specific important statistical issue that I’ll discuss when giving the solution.

# Who wants to win £194,375?

In an earlier post I included a link to Oscar predictions by film critic Mark Kermode over the years, which included 100% success rate across all of the main categories in a couple of years. I also recounted his story of how he failed to make a fortune in 1992 by not knowing about accumulator bets.

Well, it’s almost Oscar season, and fabien.mauroy@smartodds.co.uk pointed me to this article, which includes Mark’s personal shortlist for the coming awards. Now, these aren’t the same as predictions: in some year’s, Mark has listed his own personal favourites as well as what he believes to be the likely winners, and there’s often very little in common. On the other hand, these lists have been produced prior to the nominations, so you’re likely to get better prices on bets now, rather than later. You’ll have to be quick though, as the nominations are announced in a couple of hours.

Anyway, maybe you’d like to sift through Mark’s recommendations, look for hints as to who he thinks the winner is likely to be, and make a bet accordingly. But if you do make a bet based on these lists, here are a few things to take into account:

1. Please remember the difference between an accumulator bet and single bets;
2. Please gamble responsibly;
3. Please don’t blame me if you lose.

If Mark subsequently publishes actual predictions for the Oscars, I’ll include a link to those as well.

Update: the nominations have now been announced and are listed here. Comparing the nominations with Mark Kermode’s own list, the number of nominations which appear in Mark’s personal list for each category are as follows:

Best Picture: 1

Best Director: 2

Best Actor: 1

Best Actress: 2

Best supporting Actor: 3

Best supporting Actress: 1

Best Score: 2

In each case except Best Picture, there are 5 nominations and Mark’s list also comprised 5 contenders. For Best Picture, there are 8 nominations, though Mark only provided 5 suggestions.

So, not much overlap. But again, these weren’t intended to be Mark’s predictions. They were his own choices. I’ll aim to update with Mark’s actual predictions if he publishes them.

# And the winners are…

In previous posts I discussed the Royal Statistical Society’s ‘Statistic of the Year’ award. I’m now grateful to Richard.Greene@smartodds.co.uk for having pointed out that the winners for 2018 have now been announced. They are as follows:

International award: 90.5%

UK award: 27.8%

Before reading any further you might like to have a quick guess at where those statistics derive from and why they might have been selected.

Actually, the two statistics have contrasting motivations: one is pretty depressing, while the other is a cause for some optimism. Maybe this balance was intentional. The 90.5% is the proportion of plastic waste that has been produced and not recycled. The 27.8% is the peak percentage of all electricity produced in the UK due to solar power on 30 June this year, which actually made it the largest single form of energy production in the UK on that particular day.

You can find a fuller explanation of the awards here. This also includes a list of ‘highly commended’ nominations. I guess my favourite is 16.7%, which is the proportional reduction in the number of Jaffa Cakes in a McVities’ Christmas tube due to shrinkflation. This is the process whereby manufacturers hide actual price increases by reducing the volume of a product – often by stealth – while keeping the price the same. Who knows why so many manufacturers should suddenly be adopting this strategy?

Meanwhile, remember to take note of any potential nominations for the 2019 awards.

# Christmas quiz

I mentioned in a previous post the Royal Statistical Society (RSS), which is the UK’s foremost organised body of statisticians. In addition to its role in promoting and publishing all-things statistical, it is also famous for one other thing: its annual Christmas quiz, which is widely considered to be one of the toughest quizzes around. It’s been going for 25 years and is famous enough that it gets reported in full in the Guardian.

Though produced by the RSS, the questions have nothing to do with statistics, and not much mathematics either. That said, the questions do require a good general knowledge, logical thinking and a capacity to approach problems laterally; skills that are useful for statisticians. My personal total score for the quiz over the last 5 years or so is zero.

So, the 2018, 25th anniversary, edition of the quiz is now available here. If you like a good challenge you might enjoy having a go at it. Good luck, and remember that you can’t possibly do worse than me. I’ll post a link to the solutions once they are available.

Just to give you some idea of the types of questions you’re likely to face in the quiz, here’s a question from the 2017 edition:

POLYMERISATION
If 5 is IHNTOBBTTAS, and 10 is IDAATINELR, what will 20 be?

Can you get the answer? There’s a clue in the question title. Once you’ve had enough, scroll down for the solution.

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SOLUTION:

The polymer £5 note introduced in 2016 by the Bank of England features the quote “I have nothing to offer but blood, toil, tears and sweat” (initial letters IHNTOBBTTAS), while the new polymer £10 note features the quote “I declare after all there is no enjoyment like reading!” (initial letters IDAATINELR). The polymer £20, due for release in 2020, will feature the quote “Light is therefore colour” – so the answer is “LITC”.

Obviously I failed to find this solution in much the same way as I failed to find any solution to any of the questions in each of the quizzes for the last five years. I didn’t look at the quizzes in previous years, but you might extrapolate my more recent scores to get a reasonable estimate of what my score would have been if I had.

If you want further practice, you can find the complete 2017 version of the quiz here and the solutions here.