Written by Colin+ in probability, statistics.

You don’t see many run-down, out-of-business casinos, which should serve as a tip-off: almost nobody beats the house in the long run. In this article, I’ll show you – using the binomial distribution and the normal distribution – why that is.

Sure, you might get a few people who come out ahead by luck or guile – check out The Newtonian Casino or Bringing Down the House for a couple of examples, both of which highlight the amount of subterfuge needed to avoid getting your backside kicked out, or your head kicked in.

But this article’s not about subterfuge, it’s about random variables, which I’m sure you’ll agree is a much more exciting topic. (If you don’t agree, buy one of the books in the last paragraph and I’ll get a few pence in kickbacks, hooray!)

How do the casinos always win? Well, it’s largely down to two things:

- For each bet you win, the casino pays slightly less than ‘fair’ odds
- The Law Of Large Numbers says that, if you run an experiment often enough, the mean of the outcomes will converge onto the ‘true’ mean of the experiment

Quick note: The Law Of Large Numbers – which is actual, proper maths – isn’t the same thing as the Law Of Averages – which isn’t; the law of averages says ‘I’ve been losing a lot recently, so I’m bound to win soon to make up for it’, as if there’s some kind of God of the roulette wheel who’d notice you were losing and say ‘oops! my bad, let me balance that out.’ There isn’t.

Now, as it happens, winning and losing over a certain number of spins on the roulette wheel is a perfect example of a binomial distribution – the probability of winning is fixed, the spins are independent, and you have a fixed number of trials. One nice thing about the binomial distribution is that if you do enough trials – more than 30 or so – you can model it as a normal distribution, which is a bit easier to work with.

I’m going to talk about red and black here, but you can do the same experiment with any of the bets available on the roulette wheel.

On a European wheel, there are 18 black and 18 red numbers, plus a green 0 which is what makes the casino its money. If you were to bet one chip on red 37 times, you’d expect to win 18 times and lose 19 – on average, you lose one chip every 37 spins, or about 2.7%. That doesn’t sound like much, does it?

So, let’s imagine we do 1,000 spins. That’s plenty for us to use the normal assumption, with the mean and variance you can look up in the tables: the mean number of wins you’ll get is np, which is 1000 × 18/37, about 486. The variance is np(1-p), which is about 250 – making the standard deviation 15.8. So, how likely are you to come out ahead over 1000 spins?

You’re going to need a z-score. To come out ahead, you’re going to need to win 501 times, and the z-score is just how many standard deviations you are above the mean: $\frac{x – \mu}{\sigma} = \frac{501 – 486}{15.8} = 0.918$ (I’ve used the accurate numbers there rather than the rounded ones).

You look up that number in the table and see you have an 82% chance of losing – so you’d win something that you might reasonably expect to be a 50-50 chance a little more than one time in 5.

This is how the casinos make their money. They make tens of thousands of bets a day, each with a tiny built-in advantage – if the number of spins was a million, your z-score would be up to 27 (check it!), which is so far off the table as to mean ‘impossible’ – I’m guessing your chances of coming out ahead are about 10^{-70} – it’s not going to happen.

**EDIT:** Ah, Abramovitz and Stegun to the rescue. $log_{10}{\Phi(27)} = -160.13139$ (on page 972 of my edition), meaning $\Phi(27) = 7.38 \times 10^{-161}$. Three interesting things there:

- I was off by a factor of $10^{90}$, which is probably the wrongest I’ve ever been about anything;
- I was correct to about 70 decimal places, which is probably the rightest I’ve ever been about anything inexact;
- it’s still not going to happen.

The casino, meanwhile, happily rakes in its large number of tiny profits and turns them into a huge one.

## William Inches

HI

A football match has 3 possible results H home win A-away win and D -a draw .

To Guarantee the results of 4 matches correctly needs 81 tries.3x3x3x3

Reducing this to a guarantee of 3 wins in 4 can be done in 9 coulumns of 4 numbers .e.g.

111222333

123123123

231312123

231123312

Using the first two as a guide can you suggest a way to have a reasonable chance of success of 1 or 2 correct in the other 2 -3 and 4 with odds on offer of 2/1 ?

## Colin

Sorry, William, I don’t quite understand your question. To guarantee two wins, you could use three columns: 1111, 2222 and 3333.

## Carlosa

Hi Colin,

What is the table you refer to in the sections a thousand spins and a million spins?

## Colin

Good question, Carlosa! Something like this!

## Carlosa

Hey Colin,

I was just wondering what you think of progressive betting systems in general. If you do not think they work (in general, not just a specific one), could you please explain why they are not the most profitable option for a player?

## Colin

How you stake at a standard roulette wheel does not make a difference: on average, every bet you make loses money, so in the long term you will lose money. Adopting different strategies can give you better chances of being ahead in the short term, but often have large downside risk (something like a martingale will fairly reliably win you a small amount each time – but when you go on a bad run, your stake size increases rapidly, until you hit limits imposed by the casino or by your wallet. When you eventually lose, you lose very badly).