By Winterland

Wind yourself back to school days when the summers were long and at least in my day the most dangerous thing in the world was Richard Hadlee or Dennis Lillie, depending on your preference for the shores of the Tasman. You’re stuck in a stuffy maths class and the teacher is drilling in the virtues of the normal distribution, the wonders of mean and standard deviation for some banal variable such as heights of your classmates. You were told that 95% of heights fall within 2 standard deviations each way of the mean, which consigned the dwarf in the front row into jockey-dom and the guy in the back figuring life as a hoop shooter. Now just over the page in the text was the binomial distribution, the cases there involved much more interesting results of coin tossing, dice rolling, but before you got there they closed shop on you lest little Dennis stray from his destiny of filling Toorak with another dentist. You see the back story for the binomial distribution and its application was a bit edgy for the school board to unleash it on you.

Most punters will be familiar with probability e.g. a horse paying $2 in a land where all things are fair and the market is efficiently priced has a probability of winning of 0.5, likewise a horse at $10 has a probability of 0.1 of winning. Now with a perfectly efficient market (without any takeout) if you enter into a long series of bets at any selected odds with level stakes you have an expected average gain/loss of zero, i.e. you are most likely to just retain your bankroll but the ride may be a little bumpy in getting there. How bumpy is where the binomial distribution comes in.

Let’s look at p = 0.5, akin to your horse paying evens. You make n bets, let’s say 100 bets. Your expected number of successes is np = 100 x 0.5 = 50 wins. Your expected number of failures is n(1-p) = 100 x (1-0.5) = 50 losers. The binomial distribution shows you, in a relative sense, how many times if you repeated this 100 bet run (a lot) you would end up with 0 wins all the way through to 100 wins, with the peak being 50 as most likely.

It’s about 1000 years of work to do it and chart it so some clever non-dentist came up with a shortcut:

Providing np is more than about 5, and n(1-p) also more than about 5 (here both are 50) then the distribution can be approximated by the normal distribution with mean = np, and standard deviation σ = √(np(1-p))
i.e. sigma = the square root of (number of bets x winning probability x losing probability).

In this case np = 50 and σ = 5.

Ok put the flame to another nicotine stick because here’s where the chicken gets fried.

If you were paying attention in maths you would know that 68% of your results end up within one standard deviation of the mean, 95% within two standard deviations and 99.7% within 3 standard deviations.

Hence if you back 100 horses at evens in a fair market, 68% of the time you will end up with between 45 and 55 winners, 95% of the time (being a useful standard) you will end up with between 40 and 60 winners.

Backing 100 horses at 10’s gives an expected np = 10, σ = 3.

Hence 95% of the time you would end up with between 4 and 16 winners.

So what, you’ve got me here but all I have is a budding lung condition?

Ok so if you create your own ratings, track your bets and their success/failure, and group them into price intervals then you can see whether you may have a rating problem and at what odds.

If you had 100 bets that you rated at close to $2, paying hopefully at least $2 in the market and you scored 43 winners, no big deal, keep rating as you do. If you got less than 40 winners then you are outside that 95% expected range and it may be worth thinking about whether you should be rating these at more like $2.20 or more and see how things progress.

Now that’s more useful than relative heights of your classmates, why don’t they teach this?