How it works

Here’s how it works (using my real world example):

Taking a look at the odds of matches for round 14, I can calculate the potential gaurenteed profit I will make under a few assumptions.

My aim is to match the profit I would get if I staked the free bet on my preferred team. For example if I put the $100 free bet on a team at 1.40 odds, my potential profit is $40. So I want to determine what amount of money I need to put on the opposing team to match this profit. Then I want to determine if my overall profit is positive. If it is then its a good match to hedge on.

So lets take a look at how this works on the actual match I staked on.

I chose to use my $100 free bet to stake on the Broncos playing at home to beat Canberra in round 14.

The Broncos were at odds of 1.33 while Canberra were at odds of 3.45

If I use the $100 free bet to stake on Broncos my potential profit is $33 ($100*1.33 - $100).

If I want to ‘match’ this profit by betting on Canberra I would need to put $13.75 on Canberra to win. ($13.47*3.45 - $13.47 = $33)

So now; given I have placed a $100 (free) bet on Broncos and 13.75 bet on Canberra then I am gaurenteed to profit $19.53. Yup that’s right, no matter who wins I get $19.53 profit. Now I don’t even have to watch the game and I can go off and buy that ‘special’ something I’ve always wanted….

But in all seriousness at least its a profit, and I only had to outlay $13.47 to gaurentee it.

You can double check the math if you like:

I put $13.47 on Canberra to win @ 3.40 (total actual outlay $13.47).

I put $100 (free bet) on Broncos to win @ 1.33 (total actual outlay = $0.00 because its a free bet)

If Broncos win I get (1.33*100-100)-$13.47 = $19.53

If Canberra wins I get (3.4*13.47)-$13.47 = $19.53

So there you go! As demonstrated you can use that free bet to guarantee yourself a profit (I actually did)

Math

Here’s the step by step guide (with formula) to determine how much to stake on the opposing team, and how much overall profit you will make using this strategy. You can use the amount of overall profit to determine which match is the best value to hedge on

1. Determine the:

• free bet amount (s1)

• odds of the team which free stake will be used (o1)

• odds of the opposing team (o2)

2. Calculate the expected profit (p) from the free bet stake

• p = s1*o1 - s1

3. Calculate the amount of money needed to bet on the opposing team in order to match the expected profit from the free bet stake (s2)

• s2 = p/(o2-1)

4. Calculate the total expected profit (tp) from the hedge betting

• tp = p-s2

5. Use the hedging strategy to stake on the game with the highest total profit (tp).

Using my real world example:

s1 = $100

o1 = 1.33 (Odds on Broncos)

o2 = 3.4 (odds on Canberra)

p= $100*1.33-100 = $33

s2 = 33/(3.45-1) = $13.47

tp = $33- $13.47 = $19.53

This was the highest tp of all matches in round 14 (at the time of betting) so was my preferred hedging stake.

If we wanted to quickly look at the best free bet hedging opportunities we can tabulate and graph to get a bit of an idea of how this works.

Assuming a $100 free bet and a completely fair market free of vig/overround; lets tabulate and plot the total potential profit of odds from 1.0 through to 2.0:

We observe that in a fair market the maximum profit of $25 is achieved when you put a $100 free bet on odds of 1.5 and $25 on the opposing team at odds of 3.00. Profit declines from this maximum as you move lower towards 1.2 or higher towards 2.0. At odds of 2.00 no profit is achieved. While this is the theoretical maximum in a fair market, the reality is that the ‘best’ value hedging opportunity will vary based on the vig/overround offered for the match, and the vig/overround bias. To refresh your memory a bookmakers market is never fair as the probabilities they offer on each team will add up to over 1.00.

For example, in a fair market assuming even probability (50%) that each team will win, then the odds offered should be 2.00 on each team.

In the bookmakers market when the odds are ‘even’ instead of being set at 2.00 you will observe that they might be at 1.92.

For example, the opening odds for the round 14 match up between the Sharks and Cowboys were 1.92 each. This is equivalent to an implied chance of 52% chance that each team will win. This adds up to 104% so the bookmaker has a 4% overround.

So unfortunately you can’t use 1.50 exclusively, you will need to do the calculation to determine which match has the highest profit potential like I did.

Lets take the odds of the round 14 matches* and see how the profit is affected by the vig/overround. We will also order the odds lowest to highest.

NB these odds are slightly different to the odds determined at the time of my stake - these are just the latest odds I had in my database

We can see that odds of 1.5 no longer represents the maximum profit but rather, maximum profit is achieved at odds of ~1.42. You can also see that the vig/overround does actually vary slightly between matches which does play into the potential profit. So the bottom line is that to maximize your profit you need to make sure that you do your math in the market you want to apply your hedging stake in.

Finally, although I said that this hedging strategy guarantees you a profit… there is a slight caveat. In the event of a draw I would have lost my free bet and also lost my $13.47…. however draws are a relatively rare beast..

Also as a final disclaimer in reality I had to put $14 on Canberra to win, as I wasnt able to put fractional bets on.. so my actual profit was very slightly lower

Hedging

Tables and graphs to assist in intepreting hedging oppertuities

OK, so now we have identified the opportunities which have existed and we understand how to calculate the potential profit margins; how do you easily keep an eye on any hedging opportunities after you have put down a stake?

Probably the easiest way to do this is to visualize in a graph and/or refer to a table rather than scribbling down calculations all the time. If you are code eccentric, I’ll show you how to quickly whip these visualizations up in R programming language (otherwise just keep my table/graph handy).

The first table/graph we will look at is a break even hedging opportunity. This is our baseline. A break even hedging opportunity is one where; if you stake a specified amount on the opposing team you are gaurenteed to break even. You might want to do this if you fear that the team you placed (maybe a large) bet on really doesn’t stand a chance to win and so you are looking to ‘opt out’ without loosing money (if given the opportunity)

Lets quickly review a few formulas so we know how to calculate this. What we want to calculate is the o2 under a break even scenario (margin = 1, profit margin = 0)

We know that given:

o1 = Decimal odds of initial stake

o2 = Decimal odds of the dependent outcome to compare

The margin (m) = 1/o1 + 1/o2 and the profit margin (pm) = 1/m - 1

Therefore to calculate o2 for m=1 we re-arrange the formula/s so that:

o2 = o1/(o1*m-1)

Or if we want to know o2 given a desired profit margin:

o2 = o1/((1/pm+1)-1)

So for example if we have put down a stake at odds of 1.90 we would need to wait until the opposing teams odds are at least 2.11 to be able to hedge to break even

o2 = 1.90/(1.90*1-1) = 2.11

We can quickly and easily generate a reference table using this knowledge. All we will do is generate a table showing the odds ‘o1’ of the team we may have bet on and the corresponding value of the the opposing odds ‘o2’. The value of ‘o2’ is what the odds must be on the opposing team in order to break even when hedging:

The way to read the table is pretty simple. o1 represents the odds of the initial stake and o2 is the odds that the opposing team must reach in order to break even in hedging. So you can see that when o1 is 1.3 the opposing team odds must reach 4.33 before you have a hedging opportunity and so on. We also include ‘m’ which is the margin, and ‘p’ which is the profit margin. Since this is the break even odds, ’p is obviously 0.

An alternative way to visualize this is in a graph. You could plot o2 against o1 and read off the o2 value required directly from the graph:

As an example you can read from the graph that when ‘o1’ is 1.5, ‘o2’ must be 3.00:

Since we might want to do better than just break even, we can expand on the above and generate a table and graph with multiple options. Lets create a table showing the odds that o2 have to be in order to generate profit margins from 0-30%

We observe that when our staking odds are 1.3 that to break even the odds of o2 have to be 4.33 to break even. To generate a profit margin of 30% the odds of o2 have to be impossibly high (infinite). For a more reasonable comparison, using starting odds (o1) of 2.0, we see that the odds range for a profit margin of 0-30 are 2.00-3.71.

Finally we can plot this table data if we want a visual graph to pick values from:

Again using o1 of 2.00, we can read across the y-axis and see that to break even we need an o2 of 2.00, to generate a profit of 20% we need an o2 of 3.00 etc.

Arbitrage

As I indicated in the introduction, this article focus’ more on Hedging, and if you want more detail on arbitrage I encourage you to read any of the articles I reference in the introduction. So briefly:

Arbitrage is very similar in concept to Hedging (and in fact you can determine the margin and profit margin of an arbitrage opportunity in the exact same way), however the difference is that arbitrage opportunities occur similtaniuosly or in very close space of time. As an example:

Lets just say that you observe that a bookmaker ‘A’ has the following (even) odds for two teams:

o1: 1.9

o2: 1.9

You also see that at the same time bookmaker ‘B’ has different odds for the same team/s:

o1: 1.6

o2: 2.35

Using the Hedging opportunity table, you see that when o1 is 1.9, then you can generate a profit margin of 5% when o2 = 2.35.

Knowing this, you will observe that if I placed a stake on ‘o1’ with bookmaker ‘A’ (at 1.9 odds), and immidiatly after/ similtaneuosly also stake on ‘o2’ at Bookmaker ‘B’ then I will guarantee myself a 5% profit. I have exploited the differing odds that bookmakers have offered and guaranteed myself a profit no matter the outcome of the match!

You can double check the math:

Assume a $100 stake on o1 at Bookmaker A.

Now calculate the stake I need to place on o2 at bookmaker ‘B’ to guarantee my 5% profit

Stake 2 = $100*(1.9/2.35) = $80.85

Now determine the guaranteed profit amount:

If o1 win: 1.9*100 - 180.85 = $9.15

If o2 win: 2.35*80.85 - 180.85 = $9.15

9.15 as a percentage of 180.85 (9.15/180.85) equals 5%

So we can see that the calculations are the same as for Hedging, and you can use the tables/graphs to also identify potential arbitrage opportunities. Fair warning however, Arbitrage opportunities are even less frequent then hedging opportunities and last for a very brief period of time. In order to take advantage of Arbitrage opportunities you need accounts with very many bookmakers, be able to place bets very quickly and additionally would need some sort of ‘automated’ notification system to alert you of any opportunity which arose.

Calculating multibet odds & probability

Determining the value of a multibet is a simple exercise in which you multiply the implied odds of each match together (compute the product) and compare this to the product of your perceived odds. If your odds are shorter, then it would be a ‘good value’ (albeit long odds) bet. As an example, lets take the recent details of the multibet I placed for round 8 of the 2016 NRL season. In the table below, the team_descr is the team I tipped to win and the bookie_odds_stake were the odds at the time I placed the bet:

If we calculate the product of the perceived odds we get 5.12 and if we calculate the product of the bookie odds we get 5.85. Because 5.12 is less than 5.85 it is a ‘good value’ bet. Just a reminder, the product of odds is simply all the odds multiplied together.

As an example the product of my perceived odds is:

1.33 x 1.42 x 1.51 x 1.37 x 1.31 = 5.12

If we look at it from a probability perspective, the perceived probability of this multibet paying off is 1/5.12 which is 19.5%. The bookmakers implied probability is 1/5.85 which is 17.1%. So I perceive that there is a greater chance of this event occurring than the bookmakers, therefore I deem it a good value bet. Of course we must also consider that the probability of it paying off (according to our own perceived probability) is only 19.5%.

Value of ‘cash back’ promotional offer/s

Now lets take a look at the scenario where a ‘cash back’ offer exists.

Specifically, for multibets, Sportsbet currently has a promotion whereby if you place a 5 leg multibet (on S2016 NRL matches) and only 4 of your tips actually win then you receive cash back (up to $50). So how do I tell if that’s good ‘value’ or not?

Well, I can tell you straight up that simply receiving your cash back on 4 of 5 legs is never good value because your perceived odds will be the product of 4 legs winning vs the cash back value which is equivalent to odds of 1.00. As an example the range of perceived odds for any 4 combinations of my 5 leg multi is between 3.39 and 3.9 and the odds offered (the cash back) is only 1.00. This is obviously negative expected return. So the ‘value’ of a cash back offer is really determined by your panchant for the overall risk of the multibet.

As an example in this 5 leg multi I can tell myself:

I perceive I have around (at best) a ~30% (1/3.39) chance of at least getting my cash back and about a 20% chance of a return of 5.8. So if I bet $50 I have around a 1 in 3 chance of getting my money back and about a 1 in 5 chance of a $290 return ($240 profit). In this case I took the ‘punt’ but at least I understood my chances. In this case it paid off!

As you can see multibets are high ‘risk’ high reward, and the promotional offer does not represent good value for the odds at time of potential cash back, however provides a ‘psychological’ buffer for the punter (which is why they offer them).

In all, given this sort of promotional offer it is worth understanding your chances of money back after 4 legs so you can put the multibet into perspective.

For example if you determine you have a 1 in 4 chance of money back and a 1 in 5 chance of a ‘big’ payout you might be less inclined to bet. Conversely if (in the case that one of your legs is very long odds) you might calculate you have a 1 in 2 chance of money back and a 1 in 4 chance of a big payout. In this case the higher chance of cash back might be enough for you to take a ‘punt’

Determing the odds of 4 of 5 betting legs

The odds of 4 of 5 of the multibet legs are simply calculated in the same way as previous (its just the product of the odds of 4 of the matches). If you assume that your top 4 ‘highest’ probability matches are most likely to win, just calculate the product of these odds and this is the (perceived) odds of cash back. For my multi, the highest probability games were games 1,2,3 and 8 and if I compute the product of these I get 3.3 which is equivalent to 29.5%. So I perceive I have a 29.5% chance of getting my money back.

It gets a little more tricky if you want to evaluate the range of odds for any four bets winning. To do this you have to compute the odds for every combination of odds for the 4 of 5 matches. Combination math is a little bit off topic (if your interested see http://stattrek.com/online-calculator/combinations-permutations.aspx#examples) but when you have 4 objects from a set of 5 objects there are 5 possible combinations of odds.

The 5 possible combinations of perceived odds odds expressed as a matrix looks like:

##      [,1] [,2] [,3] [,4] [,5]
## [1,] 1.33 1.33 1.33 1.33 1.42
## [2,] 1.42 1.42 1.42 1.51 1.51
## [3,] 1.51 1.51 1.37 1.37 1.37
## [4,] 1.37 1.31 1.31 1.31 1.31

Now If we compute the product of these combinations (re-ordered from lowest to highest) we observe the range of odds of getting our cash back after 4 legs:

## [1] 3.39 3.60 3.74 3.85 3.91

As probabilities this looks like:

## [1] 0.295 0.277 0.268 0.260 0.256

So if I assume that any combination of 4 legs is possible, the chances of getting my cash back is between 25.6% and 29.5%

Remember of course this all assumes my perceived probabilities are accurate 😉

Conclude

So now you know how to calculate the probability of a multibet based on your perceived odds and determine if its a good value bet. In the event of ‘cash back offer’, you will understand that the odds at time of cash back will never be favorable; however you will know its worth calculating the odds of to determine if the chances of cash back are favorable enough to factor into your psychology of bet placement. Finally you understand that multibets are high risk (long odds) high reward (high return) bets.

The data set

Of course what you need first and foremost is a decent data set of bookmaker odds. Luckily I have a sample up on my website which can be downloaded for such analysis. Some of the data manipulation is already done for you such as normalising the odds, calculating the implied probability, vig and overround, labeling true positive, false positive events etc. A couple of quick notes on the data set:

• There is only one odds column (bookie_back_odds) and this represents the average of multiple bookmaker odds close to closing of bets.

• Odds are the decimal odds of a team wining

• The probability is simply calculated using 1/Odds

• Normalised odds/probabilities are just weight averaged to 100%

The key variables of interest will be:

• win_lose (binary indicator if the team won (1) or lost (0))

• bookie_back_prob_norm (bookies probability that a team will win normalised to 100%)

• bookie_label (binary indicator of the bookies prediction if the team will win or lose)

• TP (True positive results (where predicted win and actual result was a win))

• FP (False positive results (where bookie predicted a win and result was a loss))

OK, lets get started!!

b <- read.csv ("http://maxwellai.com/wp-content/uploads/2016/03/nrl_2009_2015_bookmaker_odds.csv", header = TRUE)

colnames(b)

##  [1] "match_id"              "match_team_id"        
##  [3] "season"                "round_no"             
##  [5] "game_no"               "team_descr"           
##  [7] "team_against_descr"    "score"                
##  [9] "score_against"         "home_away"            
## [11] "win_margin"            "win_lose"             
## [13] "bookie_back_odds"      "bookie_back_prob"     
## [15] "bookie_back_odds_norm" "bookie_back_prob_norm"
## [17] "overround"             "vig"                  
## [19] "bookie_label"          "TP"                   
## [21] "FP"                    "TN"                   
## [23] "FN"

str(head(b, n=5))

## 'data.frame':    5 obs. of  23 variables:
##  $ match_id             : int  1 1 2 2 3
##  $ match_team_id        : int  1 2 4 3 5
##  $ season               : int  2009 2009 2009 2009 2009
##  $ round_no             : int  1 1 1 1 1
##  $ game_no              : int  1 1 2 2 3
##  $ team_descr           : Factor w/ 16 levels "Brisbane Broncos",..: 1 10 7 14 4
##  $ team_against_descr   : Factor w/ 16 levels "Brisbane Broncos",..: 10 1 14 7 12
##  $ score                : int  19 18 17 16 18
##  $ score_against        : int  18 19 16 17 10
##  $ home_away            : Factor w/ 2 levels "A","H": 2 1 2 1 2
##  $ win_margin           : int  1 0 1 0 8
##  $ win_lose             : int  1 0 1 0 1
##  $ bookie_back_odds     : num  1.69 2.15 1.4 2.93 1.42
##  $ bookie_back_prob     : num  0.592 0.465 0.714 0.341 0.704
##  $ bookie_back_odds_norm: num  1.6 2.03 1.33 2.78 1.35
##  $ bookie_back_prob_norm: num  0.56 0.44 0.677 0.323 0.669
##  $ overround            : num  0.057 0.057 0.056 0.056 0.053
##  $ vig                  : num  0.054 0.054 0.053 0.053 0.05
##  $ bookie_label         : int  1 0 1 0 1
##  $ TP                   : int  1 0 1 0 1
##  $ FP                   : int  0 0 0 0 0
##  $ TN                   : int  0 1 0 1 0
##  $ FN                   : int  0 0 0 0 0

bp <- subset(b, b$bookie_label == 1)
#add precision column
bp$precision <- sum(bp$TP)/(sum(bp$FP) + sum(bp$TP))
#report the overall precision
round (max(bp$precision), digits=2)

## [1] 0.65

Bookmakers probability distribution

Now let’s have a look at how the the bookmakers probabilities compare between the false positives (when they picked a team to win and they lost) and the true positives. This is going to tell us how well calibrated the probabilities are and will be useful for determining appropriate staking strategies

Fist let’s take a quick look at the summary statistics for the normalised probabilities for true positive events and false positive events:

For true positive events:

bTP <- subset(bp,bp$TP ==1)
#save third quartile for use later
bpTPq <-quantile(bTP$bookie_back_prob_norm, c(.75)) 

print (summary(bTP$bookie_back_prob_norm))

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.5010  0.5870  0.6470  0.6561  0.7070  0.9380

For false positive events:

bFP <- subset(bp,bp$FP ==1)
#save third quartile for use later
bpFPq <-quantile(bFP$bookie_back_prob_norm, c(.75)) 
print (summary(bFP$bookie_back_prob_norm))

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.5010  0.5630  0.6180  0.6267  0.6740  0.9120

And lets take a look at that visually in a simple box-plot:

p2 <- ggplot(bp,aes(factor(TP),bookie_back_prob_norm))
p2 <- p2 + geom_boxplot()
plot(p2)

Great! we can see straight way that the probabilities are well calibrated because the mean of the probability of the false positive events is lower than the mean of the probability of the true positives. This means that even though the bookies have mis-predictions (false positives) their implied probability is lower for these events (which is really good! (for them – not so much for the punters)).

We also observe that the third quartile for false positives is 0.67 (equivalent to 1.49 odds) and that there are a large number of true positive >0.67. Now this is important; this means that when the bookmaker gives odds odds on which are >0.67 implied probability (<1.49 odds) then in the overwelwing majority of cases, the bookmaker will accurately pick the winner. We can actually calculate the bookmakers precision in this region Which is 75%!!. This means if you bet on anything with odds <1.49, then you should win ~75% of you bets:

bp75 <- subset(b, b$bookie_back_prob_norm> bpTPq)
#add precision column
bp75$precision <- sum(bp75$TP)/(sum(bp75$FP) + sum(bp75$TP))
#report the overall precision
round (max(bp75$precision), digits=2)

## [1] 0.75

If you want to look at it from an odds perspective, we can create the same box-plot; but just with odds on the y axis instead of the probability. Note that the plot should be reversed because higher (IE longer odds means lower probability). I personally prefer looking at it from a probability point of view but in some scenarios it might be good to look from and odds point of view. Lets see:

p3 <- ggplot(bp,aes(factor(TP),bookie_back_odds_norm))
p3 <- p3 + geom_boxplot()
plot(p3)

Yes the box-plots are reversed but you can make the same inference as above. That if odds on are <1.49 then in the majority of cases the bookie will be correct!

But what does it all mean??

So what does all this mean for us?? Well for one the bookmaker is going to be damn hard to beat! They have well calibrated probabilities a precise predictive model, and on top of this have a little lea-way due to the overround/vig. But like I said in the introduction all is not lost; just in doing this simple exercise we have learnt a great deal about the bookmakers. And we can try use this analysis to help us out in our own ‘informed’ gambling. So lets review what we know and think about how we can use it logically.

Ok, so we know that the bookmaker predicts winners correctly about 75% of the time when odds on are <1.47. This means that if we pick games when the odds are <1.47 then we also have a 75% chance of a return on a bet… but does it mean we will make a profit? We should just be able to place bets on anything less than odds of <1.47 and get a return right?

Lets see. I will quickly make a simulated bet with the logic of staking on everything which is >0.67 in probability (equivalent to <1.47 odds):

#create a new dataframe using bp for simple simulated staking
bss <- bp
#create a 'stake' column and just set it to a fixed staking value
bss$stake <- 10
#find the 3rd quantile of the bookie probability
bssFPq <-quantile(bFP$bookie_back_prob_norm, c(.75)) 
bssTPq <-quantile(bTP$bookie_back_prob_norm, c(.75)) 
#create stake column using logic
bss$stake <- ifelse(bss$bookie_back_prob_norm > bssFPq , bss$stake,0)
#now simulate profit/loss on the stake amount
bss$profit <- ifelse (bss$TP == 1, (bss$stake*bss$bookie_back_odds-bss$stake),-bss$stake)
#create a total profit/loss cloumn
bss$profit_total <- sum(bss$profit)
#now create a cumulative profit for graphing
bss$profit_cum <-  round(cumsum(bss$profit),2)

#print the amount of profit we would make
print(max(bss$profit_total))

## [1] -284.6

Bummer!! that’s -$-284.6, so we actually made a loss. Seems its not so simple after all!! So what went wrong here? Well, there must be enough outliers in the False positive region to offset our small profits.

Lets take a look at how many false positives there are which are > 0.67

sum(ifelse(bss$bookie_back_prob_norm >bssFPq & bss$FP ==1, 1,0))

## [1] 120

So there a 120 of these occurrences. Ok so that means straight up we are going to lose $1,200 ($10 stake * the amount of false positives).

Now lets count the number of occurrences of True positives which are >0.67 probability

sum(ifelse(bss$bookie_back_prob_norm >bssFPq & bss$TP ==1, 1,0))

## [1] 338

Ok, so there are 338 of these. That seems like a lot more, but obviously not enough.. lets see why. The average of the odds for the bookmaker above >0.67 probability is 1.28:

mean(subset(bp$bookie_back_odds,bp$bookie_back_prob_norm >bssFPq))

## [1] 1.276114

So on average our profit on all our bets above >0.67 is (on average) going to be $946:

# (1.34 * stake * number of bets) - stake * number of bets
(1.28 * 10 * 338) - (10 * 338)

## [1] 946.4

So we can see straight away that we are going to make a loss because this value ($946) is smaller than than the value from false negatives at the same threshold ($1,200)

We can use this knowledge to create a simple formula to check if we can profit from any sort of ‘short odds’ betting. We can use:

((TPd*TPb)-TPb) - FPb

where

• TPd is the average decimal odds of the true positives above/within the threshold
• FPd is the number of False positive bets above/within the threshold
• TPb is the number of True positive bets above/within the threshold

So we could reduce our previous staking simulation to: (1.28*338)-338 - 120

((1.28 *338)-371)-131

## [1] -69.36

Which is -69. Since this is negative we aren’t going to make any sort of money staking like this because the number of false postives outwieghs the number of true postives times the odds of the true postives

Right! so is there anything that might work? Well when we looked at the box plot there looked to be as many outliers in the False positive range above 0.67 as there were True positives in this range. So what about if we just use the range between the 3rd quartiles (75th percentiles) of the false positives and true positives?

p2b <- ggplot(bp,aes(factor(TP),bookie_back_prob_norm))
p2b <- p2b + geom_boxplot()
p2b <- p2b + geom_hline(aes(yintercept=bpFPq), col = "coral", linetype = "dashed", size = 0.8)
p2b <- p2b + geom_hline(aes(yintercept=bpTPq), col = "coral", linetype = "dashed", size = 0.8)
plot(p2b)

Sounds like a pretty simple strategy, maybe to good to be true so lets quickly check how it performs. Just using a really simple fixed staking calculation on the data using the logic that we will bet $10 on every game which has an implied probability of between ~0.67 and ~0.71:

#create a 'stake' column and just set it to a fixed staking value
bp$stake <- 10

#save 75 percentile (3rd quartile of true postives and false postives)
bpTPq <-quantile(bTP$bookie_back_prob_norm, c(.75)) 
bpFPq <-quantile(bFP$bookie_back_prob_norm, c(.75)) 
#set up basic staking logic 
  #if the bookie back odds are between 3rd q of the FP and 3rd Q of the TP then we are going to stake $10 otherwise we are not going to stake ($0)
bp$stake <- ifelse(bp$bookie_back_prob_norm > bpFPq & bp$bookie_back_prob_norm < bpTPq, bp$stake,0)
#now simulate profit/loss on the stake amount
bp$profit <- ifelse (bp$TP == 1, (bp$stake*bp$bookie_back_odds-bp$stake),-bp$stake)
#create a total profit/loss cloumn
bp$profit_total <- sum(bp$profit)
#now create a cumulative profit for graphing
bp$profit_cum <-  round(cumsum(bp$profit),2)

#print the amount of profit we would make
print(max(bp$profit_total))

## [1] 42.2

Yay!!!! All that hard work and we got $42.00!! On my way to money town oh yeah!!

Ahem… back to reality. Lets chart that baby up and see what it looks like..

p4 <- ggplot(bp, aes(match_id,profit_cum))
p4 <- p4 + geom_path(aes(col=profit_cum), size = 0.5)
p4 <- p4 + geom_hline(aes(yintercept=0.0), col = "red", linetype = "dashed", size = 0.8)
plot(p4)

Not so pretty… BUT what really stands out is that profit never dropped below 0… now that is interesting…

So it appears that betting in this region is profitable (just). Lets just explore a little bit further. The total amount staked across all seasons was $42:

sum(bp$stake)

## [1] 1500

So if we made $42 then return on investment (ROI) is ~3%:

(sum(bp$profit)/sum(bp$stake))

## [1] 0.02813333

That is we invested $1,500 over the term and ended up with $1,542 (profit $42) for a total return of…. ~3%… amazing.. (I am being sarcastic, BUT making any profit as you will come to know is actually quite a challenge)

So how many bets did we actually make?

sum(ifelse(bp$stake ==10, 1,0))

## [1] 150

184 bets! That’s not many…, how many games did we forgo betting on…

sum(ifelse(bp$stake ==0, 1,0))

## [1] 1242

1,242! Ok so we ended up betting and profiting small on a low number of games, but came out on top. What was our staking accuracy??

Its the number of bets we placed and had profit over the total number of bets we placed:

(sum(ifelse(bp$profit >0, 1,0)))/
  (sum(ifelse(bp$stake >0, 1,0)))

## [1] 0.7466667

74%!! well that’s pretty damn good!

Recap

Ok lets just recap what we have learned.

• If we use the bookmakers normalized probability to predict the outcome of matches then we would have achieved a 64% precision across seasons 2009-2015

• If we stake on matches within the noramilsed probability of the 3rd quartile of the true positives and false positive results then we can achieve small (~3% ROI) and hence constitutes a safe bet

• Using a simple staking threshold to back strong favorites did not achieve profit because the number of false positives outweighed the number of true positives times the odds.

Summary

While brief, the above is a simple review of how good the bookmakers are at predicting winners (65% accuracy across 2009-2015) and gives brief in-site into the structure of the bookmakers odds. It shows that the bookmakers probabilities are well calibrated (the mean of the false positives is lower than the mean of the true positives) and additionally shows that there is a theoretical window of short odds betting which can make a very small, long term return on investment (~3%).

Now its time to explore any strategies which might make a higher ROI!! Stay tuned!!