​Every year we have an office sweepstake for the Cheltenham horse racing festival. Like most sweepstakes, this one attempts to remove the skill, allowing everyone to take part without necessarily needing to know much about horse racing.

In case you’re not familiar with a Sweepstake, here’s a simple example of how one based on the World Cup might work:

• Entry costs £5
• Each player randomly selects a team
• The player who picked the team that goes on to win, gets all the money

This allows anyone to enter and have an equal chance of winning (the chance is equal prior to drawing, but not equal once the draw has been made), otherwise everyone would just want Spain or Germany, or whichever team is currently considered the favourite.
 
Note that in order for this to work properly, you need to ensure that the team draw is not carried out until everyone who wants to play has put their money in – otherwise you introduce some optionality and people can then decide whether to enter based on the teams that are still left. i.e. if you know that Germany and Spain have already been picked then there is less value in entering the competition.
 
The rules for our Cheltenham sweepstake were as follows:
 
The Rules
 
The festival comprises 7 races a day for 4 days, for a total of 28 races.
 
The sweepstake costs £20 to enter, and the winnings are calculated as follows:
 
3rd place in competition = 10% of pot
2nd place in competition = 20% of pot
1st place in competition = 70% of pot
 
Each participant picks one horse per race each morning. Points are then calculated using the following scoring system:

• 1st place – 5 x odds
• 2nd place – 3 x odds
• 3rd place – 1 x odds

 
A running total is kept throughout the competition, and the winner is determined after the final race.
 
The odds the scoring are based on are set using the odds printed in the Metro Newspaper on the morning of the races.
 
(as an example, for a horse which has odds of 11/2 in the Metro - if the horse then places 1st, if we selected this horse, we would win (1+11/2)*5 = 32.5 points)
 
Initial thoughts
 
Any set of betting odds can be converted to an implied probability of winning, these would be the odds which over the long run would cause you to breakeven if the race were repeated multiple times with each horse winning a proportion equal to its probability of winning.

Because the scoring in our sweepstake is based on the betting odds, using implied probabilities derived from the odds to help select our horses ends up cancelling itself out (which was the intention when designing the rules). The implied probability can be calculated as one over the odds As an aside, the bookie is indifferent to whether this is the correct probability of winning, they structures the odds purely on the ratio of how their customers are betting. They then structure the odds so that they make money on each race, regardless of which horse wins, for an explanation of this, see my post on creating a Dutchbook:
​
​www.lewiswalsh.net/blog/archives/12-2017
 
Here is an example showing how we would calculate the implied probabilities using some made up odds:
 

​

We can then use the implied probabilities we just calculated to see what would happen if each horse finished in the first three positions. Once we have done this, we can then calculate the Expected Value of betting on each horse:

We see that the payout varies for each horse, but the EV is the same. This is by design, the intention is that it should not matter which horse you bet on, the sweepstake rules equalise everyone.

So what can we - an actuary who knows nothing much about horse racing - do? I don’t have any special knowledge that would allow me to select a horse which would beat the odds listed in the Metro, we appear to be at an impasse.

I could attempt to build a model of which horse will win, and then select my horse based on that, but unless it proves to be more accurate than the Metro odds, I might as well just pick at random. Furthermore, if I could build such a model, then I could just start betting actual money. This probably shows you what a difficult problem this is! There's no such thing as free money. It would be a cool project to try, and it’s something I’ve been meaning to attempt for a while, but that’s best saved for another day.

Attempt 1 - Metro vs pre-race odds

My first thought was that we can exploit the difference in odds between those published in the Metro in the morning, and the latest odds published closer to the start of the race. It seems reasonable to assume that the odds just before the race should be more accurate than the metro odds. There will have been time for additional information to be included in the more up to date odds, e.g. the weather is worse than expected therefore horse x is expected to be slower than usual. Since the payout will be based on the Metro, we will then be able to maximise our EV by exploiting this differential. 

Our table will end up looking something like this:
​

 
We see that we have very small variations in the EVs for some of the horses. It looks like according to this analysis Horse 3 would be the best selection as it has the highest EVs for 1st, 2nd, and 3rd. Based on this strategy, we would then go through each race and select the horse with the highest EV.

Is this what I did? No, for a couple of reasons. The biggest issue was that the Metro did not publish odds in the morning for all races, meaning we couldn’t use the Metro, and therefore the rules of the sweepstake were amended to use the official pre-race odds to calculate the payout instead. This meant there was only one set of odds used, and our edge disappeared!
​
Even if we had used this method, there was a more fundamental issue - the margins we ended up with were tiny anyway. The Metro vs pre-race odds did not swing wildly, meaning that even selecting the horse with the highest EV were only marginally better than picking at random.

So, was there an alternative?

Attempt 2 - 2nd and 3rd place odds
​
My next attempt at an exploitative strategy was based on the insight that the payout multiplier for 2nd and 3rd place was based on the odds of the horse coming 1st, rather than the odds of the horse coming 2nd or 3rd.

The expected value of a horse was not quite as I calculated above, it was actually:
$$EV = P(1)*p_1 + P(2)*p_2 + P(3)*p_3$$
Above, we were using the implied probability of the horse coming first as a proxy for the probability it would come second and third. This is not the same, and some betting websites do allow you to bet on whether a horse will come 2nd or 3rd.

For websites that do not allow you to bet directly on this, then we may still be able to calculate it from the odds for whether a horse finishes in the top 2 or 3 places. We just need to subtract out the implied probability of coming 1st from the probability of coming in the top 2, and then subtracting this out from coming top 3 etc.

​I therefore added some more columns to my table above, corresponding to the probability of the horses coming 2nd and 3rd, and then used this to calculate the EV instead.

We see that the final column, Total EV, now has quite different values for each horse. In this case Horse 15, Seddon has an EV of 11.72. The favourite horse on the other hand - number 7  - only has an EV of 6.2. The intuitive explanation of this is that the probability of Seddon coming first is very low – this is the reflected in the long odds of 67, this then gives us a large multiplier, but the odds of the horse coming second or third are actually relatively less far out – the fact that it is not the favourite actually increases the odds of it coming in a position which is not 1st. But then if does come in 2nd or 3rd, we would still apply the same large multiplier for the odds of it coming 1st. This then gives us our 2nd edge – we can gain a positive EV by focusing.

As a thought experiment, imagine we have a race with three horses – horse A is a clear favourite, horse B is an average horse, and horse C is clearly the weakest. By betting on horse C – the odds of it winning should be very low, so the multiple should be very high, but then this multiple will be applied even if it comes 2nd or 3rd, which is exactly where it is expected to finish.

This therefore suggests our next potential strategy – select horses which maximise our EV using the implied probabilities of the horses coming 2nd, 3rd etc.

So is this what I did? Well kind of....

The issue with this approach is that typically the horses that provide the best EV also have very long odds. In the race analysed above, our horse has an EV of 11.7, but it only has a 7% chance overall of coming in the top 3. In race two for example, the horse with the best EV actually only had a 2.7% chance of coming in the top 3. Since there are only 28 races in total, if each horse we selected only had a 2.7% chance of coming in, then the probability of us getting 0 points overall in the entire competition would then be:
​
$(1-2.7 \%)^{28} = 48 \%$​

So there is roughly a 50% chance we will get 0 points!

Alternatively, if we selected the favourite every time, we could expect it to come top 3 almost every time, and thereby guarantee ourselves points most races, but it also has the lowest EV.

So we have what appears to be a risk vs reward trade off. Pick outsiders and give ourselves the highest EV overall, or pick the favourites thereby reducing our overall EV but also reducing our volatility.

This leads us  neatly to attempt 3 - trying to think about how to maximise our probability of winning the competition rather than simply maximising EV for each race.
​
Attempt 3 - EP curves
​
From the work above, we now have our model of the position each horse will finish in each race – using the 1st, 2nd, and 3rd implied probabilities - and we have the payout for each horse – using the odds of the horse coming 1st. We can then bring our selection of horse and these probabilities together in a summary tab and simulate our daily score stochastically using a Monte Carlo method. To do this we just need to turn the implied probabilities into a CDF, and lookup the value of each position and repeat 10,000 times. The output for this then ends up looking like the following, where the value is the number of points we will for a given race.

So we see that across this sample of 20 simulations, most days we to end up with 0 points overall, but a few days have very high scores. So far so good!

The next step is to set up an EP table of the following, which looks like the following:

The EP table gives us the probability of exceeding various scores in the competition based on our horse selections. In this case, we see that there is a 1 in 20 chance of getting 453 points or greater in the day.

This is useful even on its own – when I was deciding which horses to bet on, I simply played around with the selections until I got an EP table I was comfortable with. My reasoning was quite basic – I decided I wanted to maximise the 1 in 20 value. I wanted to give myself something like a 1/4 chance of winning the whole competition and a 3/4 chance of getting very few points. Since there were four days of races, dividing the 1/4 by another 4 suggests we should be looking at maximising the 1 in 20 level (I admit this reasoning was a bit rough, but it seemed to serve its purpose)

The insight here is that the payout structure of the sweepstake is such that coming in the top 3 is all that matters, and in particular coming 1st is disproportionately rewarded. To see this, we can think of the daily selection of horses as attempting to maximise the EV of the overall prize rather than the EV of our overall score - maximising the EV of each race is only a means to this end. So we are actually interested in maximising the following:

$0.7 * P(1st) + 0.2 + P(2nd) + 0.1 * P(3rd)$

Which will largely be dominated by P(1st), given the $0.7$ factor.
​
This is largely the strategy I went for in the end.

Attempt 4 - Game theory?

I’ve brushed over one difficulty above; in order to maximise our prize EV we need to consider not just which strategy we should take, but how this strategy will fare against the strategies that other people will take. If everyone is maximising their 1 in 20 return period then there’s little point us doing exactly the same. Luckily for me, most people were doing little more than picking horses randomly. We could then formalise this assumption, and come up with a numerical solution to the problem above.

To do this, we would simulate our returns for each day across 10,000 simulations as above, but this time we would compare ourselves against a ‘base strategy’ of random selection of horses and we would simulate this base strategy for the approximately 40 people who entered.

Each simulation would then give us a ranking we would finish in the competition, Here is an example of what that would look like:

​And we could then convert this into an EP table which would look like the following:

So we see that if we select these horses, we end up having somewhere between a 1 in 10 and a 1 in 5 chance of winning the competition.

Now that we have all of this set up, we can then optimise our horse selection to target a particular return period.

I didn’t actually end up setting up the above for the sweepstake, but I suspect it would have been an improvement on my approach

Attempt 5 - Multiple day?

There is a further refinement we can make to the above. We have so far only really been focusing on maximising our chance of winning by using a fixed strategy throughout the competition. But there is no reason we have to do this. After the first day, we should really be including the current scores of each competitor as part of our calculation of our ranking. i.e. if person 1 had a great day and now has 200 points but we had a bad day and still have 0 points, by accounting for this, the model should automatically increase our volatility i.e. start picking horses with longer odds so as to increase the chance of us catching up. If on the other hand, we had a really good first day and are now in the lead, the model should then automatically reduce our volatility and start selecting the favourites more often to help safely maintain our lead.
​
How did it work in practice?

I ended up placing 2nd, and taking 20% of the prize pot which was great! I was behind for most of the competition but then pulled back on the last day when a 66/1 came in 1st place, and I picked up 330 points off a single race. This may look miraculous, but is exactly how the model is supposed to work. 

Does that have any applicability to gambling generally? Unfortunately not, basically all of the work above is based on exploiting the specific scoring system of the sweepstake. There's no real way of apply this to gambling generally.

If you have ever generated Random Variables stochastically using a Gaussian Copula, you may have noticed that the correlation of the generated sample ends up being lower than the value of the Covariance matrix of the underlying multivariate Gaussian Distribution. For an explanation of why this happens you can check out a previous post of mine:
www.lewiswalsh.net/blog/correlations-friedrich-gauss-and-copula.

It would be nice if we could amend our method to compensate for this drop. As a quick fix, we can simply run the model a few times and fudge the Covariance input until we get the desired Correlation value. If the model runs quickly, this is quite easy to do, but as soon as the model starts to get bigger and slower, it quickly becomes impractical to run it three of four times just to get the output Correlation we desire.

We can do better than this. The insight we rely on is that for a Gaussian Copula, the Pearson Correlation in the generated sample just depends on the Covariance Value. We can therefore create a precomputed table of Input and Output values, and use this to select the correct input value for the desired output. I wrote some R code to do just that, we compute a table of Pearson's Correlations obtained for various Input Covariance values when using the Gaussian Copula.

a <- library(MASS)
library(psych)

set.seed(100)

m <- 2
n <- 10^6

OutputCor <- 0
InputCor <- 0

for (i in 1:100) {
  
  sigma <- matrix(c(1, i/100,
                    i/100, 1), 
                  nrow=2)
  z <- mvrnorm(n,mu=rep(0, m),Sigma=sigma,empirical=T)
  
  u <- pnorm(z)
  
  OutputCor[i] <- cor(u,method='pearson')[1,2]
  InputCor[i] <- i/10
  
}

OutputCor
InputCor

Here is a sample from the table of results. You can see that the drop is relatively modest, but it does apply consistent across the whole table.

Here is a graph showing the drop in values:
​

Updated Algorithm

We can then use the pre-computed table, interpolating where necessary, to give us a Covariance value for our Multivariate Gaussian Distribution which will generate the desired Pearson Product Moment Correlation Value. So for example, if we would like to generate a sample with a Pearson Product Moment value of $0.5$, according to our table, we would need to use $0.517602$ as an input Covariance. We can test these values using the following code:

a <- library(MASS)
library(psych)

set.seed(100)

m <- 2
n <- 5000000
sigma <- matrix(c(1, 0.517602,
                  0.517602, 1), 
                nrow=2)
z <- mvrnorm(n,mu=rep(0, m),Sigma=sigma,empirical=T)
u <- pnorm(z)

cor(u,method='pearson')

​Analytic Formulas

I tried to find an analytic formula for the Product Moment values obtained in this manner, but I couldn't find anything online, and I also wasn't able to derive one myself. If we could find one, then instead of using the precompued table, we would be able to simply calculate the correct value. While searching, I did come across a number of interesting analytic formulas linking the values of Kendall's Tau, Spearman's Rank, and the input Covariance..

All the formulas below are from Fang, Fang, Kotz (2002) Link to paper: 
www.sciencedirect.com/science/article/pii/S0047259X01920172

The paper gives the following two results, where $\rho$ is the Pearson's Product Moment

We can then use these formulas to extend our method above further to calculate an input Covariance to give any desired Kendall Tau, or Spearman's Rank. I initially thought that they would link the Pearson Product Moment value with Kendall or Spearman's measure, in which case we would still have to use the precomputed table. After testing it I realised that it is actually linking the Covariance to Kendall and Spearman's measures.

Thinking about it, Kendall's Tau, and Spearman's Rank are both invariant to the reverse Guassian transformation when moving from $z$ to $u$ in the algorithm. Therefore the problem of deriving an analytic formula for them is much simpler as one only has to link their values for a multivariate Guassian Distribution. Pearson's however does change, therefore it is a completely different problem and may not even have a closed form solution.

As an example of how to use the above formula, suppose we'd like our generated data to have a Kendall's Tau of $0.4$. First we need to invert the Kendall's Tau formula:

$$ \rho = sin ( \frac{ \tau \pi }{2} ) $$

We then plug in $\rho = 0.4 $ giving:

​We can then test this value with the following R code:

a <- library(MASS)
library(psych)

set.seed(100)

m <- 2
n <- 50000
sigma <- matrix(c(1, 0.587785,
                  0.587785, 1), 
                nrow=2)
z <- mvrnorm(n,mu=rep(0, m),Sigma=sigma,empirical=T)
u <- pnorm(z)
cor(z,method='kendall')

Which we see gives us the value of $\tau$ we want. In this case the difference between the input Covariance $0.587785$, and the value of Kendall's Tau $0.4$ is actually quite significant.

It's the second week of your new job Capital Modelling job. After days spent sorting IT issues, getting lost coming back from the toilets, and perfecting your new commute to work (probability of getting a seat + probability of delay * average journey temperature.) your boss has finally given you your first real project to work on.

You've been asked to carry out an annual update of the Underwriting Risk Capital Charge for a minor part of the company's Motor book. Not the grandest of analysis you'll admit, this particular class only makes up about 0.2% of the company's Gross Written Premium, and the Actuaries who reserve the company's bigger classes would probably consider the number of decimal places used in the annual report more material than your entire analysis. But you know in your heart of hearts that this is just another stepping stone on your inevitable meteoric rise to Chief Actuary in the Merger and Acquisition department, where one day you will pass judgement on billion dollar deals in-between expensive lunches with CFOs, and drinks with journalists on glamorous rooftop bars.

The company uses in-house reserving software, but since you're not that familiar with it, and because you want to make a good impression, you decide to carry out extensive checking of the results in Excel. You fire up the Capital Modelling Software (which may or may not have a name that means a house made out of ice), put in your headphones and grind it out.

Hours later you emerge triumphant, and you've really nailed it, your choice of correlation (0.4), and correlation method (Gaussian Copula) is perfect. As planned you run extracts of all the outputs, and go about checking them in Excel. But what's this? You set the correlation to be 0.4 in the software, but when you check the correlation yourself in Excel, it's only coming out at 0.384?! What's going on?

Simulating using Copulas

The above is basically what happened to me (minus most of the actual details. but I did set up some modelling with correlated random variables and then checked it myself in Excel and was surprised to find that the actual correlation in the generated output was always lower than the input.) I looked online but couldn't find anything explaining this phenomenon, so I did some investigating myself.

So just to restate the problem, when using Monte Carlo simulation, and generating correlated random variables using the Copula method. When we actually check the correlation of the generated sample, it always has a lower correlation than the correlation we specified when setting up the modelling.

My first thought for why this was happening was that were we not running enough simulations and that the correlations would eventually converge if we just jacked up the number of simulations. This is the kind of behaviour you see when using Monte Carlo simulation and not getting the mean or standard deviation expected from the sample. If you just churn through more simulations, your output will eventually converge. When creating Copulas using the Gaussian Method, this is not the case though, and we can test this.

I generated the graph below in R to show the actual correlation we get when generating correlated random variables using the Copula method for a range of different numbers of simulations. There does seem to be some sort of loose limiting behaviour, as the number of simulations increases, but the limit appears to be around 0.384 rather than 0.4.

The actual explanation

First, we need to briefly review the algorithm for generating random variables with a given correlation using the normal copula.

Step 1 - Simulate from a multivariate normal distribution with the given covariance matrix.

Step 2 - Apply an inverse gaussian transformation to generate random variables with marginal uniform distribution, but which still maintain a dependency structure

Step 3 - Apply the marginal distributions we want to the random variables generated in step 2

We can work through these three steps ourselves, and check at each step what the correlation is.

The first step is to generate a sample from the multivariate normal. I'll use a correlation of 0.4 though out this example. Here is the R code to generate the sample:

a <- library(MASS)
library(psych)

set.seed(100)

m <- 2
n <- 1000
sigma <- matrix(c(1, 0.4,
                  0.4, 1), 
                nrow=2)
z <- mvrnorm(n,mu=rep(0, m),Sigma=sigma,empirical=T)

​

And here is a Scatterplot of the generated sample from the multivariate normal distribution:
​

We now want to check the product moment correlation of our sample, which we can do using the following code:
​

cor(z,method='pearson')

Which gives us the following result:

> cor(z,method='pearson')
     [,1] [,2]
[1,]  1.0  0.4
[2,]  0.4  1.0

So we see that the correlation is 0.4 as expected. The Psych package has a useful function which produces a summary showing a Scatterplot, the two marginal distribution, and the correlation:

Let us also check Kendall's Tau and Spearman's rank at this point. This will be instructive later on. We can do this using the following code:

cor(z,method='spearman')

cor(z,method='Kendall')

Which gives us the following results:

> cor(z,method='spearman')
          [,1]      [,2]
[1,] 1.0000000 0.3787886
[2,] 0.3787886 1.0000000

> cor(z,method='kendall')
          [,1]      [,2]
[1,] 1.0000000 0.2588952
[2,] 0.2588952 1.0000000

Note that this is less than 0.4 as well, but we will discuss this further later on.

We now need to apply step 2 of the algorithm, which is applying the inverse Gaussian transformation to our multivariate normal distribution. We can do this using the following code:

u <- pnorm(z)

We now want to check the correlation again, which we can do using the following code:

cor(z,method='spearman')

Which gives the following result:

> cor(z,method='spearman')
          [,1]      [,2]
[1,] 1.0000000 0.3787886
[2,] 0.3787886 1.0000000

​Here is the Psych summary again:

u is now marginally uniform (hence the name). We can see this by looking at the Scatterplot and marginal pdfs above.

We also see that the correlation has dropped to 0.379, down from 0.4 at step 1. The Pearson correlation measures the linear correlation between two random variables. We generated normal random variables, which had the required correlation, but then we applied a non-linear (inverse Gaussian) transformation. This non-linear step is the source of the dropped correlation in our algorithm.

We can also retest Kendall's Tau, and Spearman's at this point using the following code:

cor(z,method='spearman')

cor(z,method='Kendall')

This gives us the following result:

> cor(u,method='spearman')
          [,1]      [,2]
[1,] 1.0000000 0.3781471
[2,] 0.3781471 1.0000000

> cor(u,method='kendall')
          [,1]      [,2]
[1,] 1.0000000 0.2587187
[2,] 0.2587187 1.0000000

Interestingly, these values have not changed from above! i.e. we have preserved these measures of correlation between step 1 and step 2. It's only the Pearson correlation measure (which is a measure of linear correlation) which has not been preserved.

Let's now apply the step 3, and once again retest our three correlations. The code to carry out step 3 is below:

x1 <- qgamma(u[,1],shape=2,scale=1)
x2 <- qbeta(u[,2],2,2)

df <- cbind(x1,x2)
pairs.panels(df)

The summary for step 3 looks like the following.

This is the end goal of our method. We see that our two marginal distributions have the required distribution, and we have a correlation between them of 0.37. Let's recheck our three measures of correlation.
​

cor(df,method='pearson')

cor(df,meth='spearman')

cor(df,method='kendall')

> cor(df,method='pearson')
          x1        x2
x1 1.0000000 0.3666192
x2 0.3666192 1.0000000

> cor(df,meth='spearman')
          x1        x2
x1 1.0000000 0.3781471
x2 0.3781471 1.0000000

> cor(df,method='kendall')
          x1        x2
x1 1.0000000 0.2587187
x2 0.2587187 1.0000000

So the Pearson has reduced again at this step, but the Spearman and Kendall's Tau are once again the same.

Does this matter?

This does matter, let's suppose you are carrying out capital modelling and using this method to correlate your risk sources. Then you would be underestimating the correlation between random variables, and therefore potentially underestimating the risk you are modelling.

Is this just because we are using a Gaussian Copula? No, this is the case for all Copulas. Is there anything you can do about it? Yes, one solution is to just increase the input correlation by a small amount, until we get the output we want.

A more elegant solution would be to build this scaling into the method. The amount of correlation lost at the second step is dependent just on the input value selected, so we could pre-compute a table of input and output correlations, and then based on the desired output, we would be able to look up the exact input value to use.   

Recently I've been reading The Mathematics of Poker (2009, Bill Chen and Jerrod Ankenman) and I came across an interesting idea that I thought I'd write about. For me, understanding how to analyse this situation really gets to the heart of how I think about poker. I'd love to spend more time playing and studying poker but it's such a time-sink and I don't really have the oodles of free time it would require, but every now and again I'll still open up a poker book and read about something, this is one of those interesting topics I was reading about hopefully you find it as interesting as I do.

Calling a shove pre-flop in heads up

The scenario being analysed in the book is a relatively common situation, particularly in online poker where people are more inclined to shove than in real life games. The question is:

How should we analyse whether to call when we have a moderately strong hand against an opponent who has gone all-in pre flop.

Let's set up an example so that we have something concrete to talk about. Say we have pocket Kings pre flop, and our opponent goes all-in, how should we decide whether to call? Obviously without knowing our opponent's hand there is no 100% correct answer. There is  however one very useful way of analysing the situation.

Equity against a range

We need to ask ourselves - what cards would are opponent go all-in with, and how does my current hand fare against that range? i.e. we need to calculate our equity against our opponent's range. We are adding another layer of stochastic uncertainty on the event, instead of trying to guess what hand our opponent has (which is almost impossible) we are trying to guess what kind of hands he might go all-in with (which is very much possible). We then take this extra level of uncertainty and calculate the correct mathematical way to proceed.

On the one extreme, let's suppose that based on our read of how our opponent is playing, we might think that they would only go all-in with very strong hands, in this case just pocket Aces. We would then be a 4:1 underdog if we call with Ks, and we should definitely fold. (In order to calculate this we could use any standard poker calculator like the following)

www.cardplayer.com/poker-tools/odds-calculator/texas-holdem

One the other hand, suppose we know for a fact that our opponent has not looked at their cards at all but has still decided to go all in. In this case we should definitely call. The only cards we will be behind are pocket Aces, which make up a small fraction of the possible hands that our opponent could shove with, and we will be ahead or equal against all other possible hands. Therefore we would have a positive EV when calling.

What if our read on our opponent's range is somewhere in between though?

What we need to do is calculate our equity against each individual hand in our opponent's range, and then calculate the probability of our opponent having a given hand from that range. That is to say, in order to calculate the conditional expectation, we need to calculate the unconditional expectations against each hand and then multiply by the conditional probability of our opponent having that hand, given our belief about our opponent's range. 

Numerical Example

Let's go back to our numerical example, and suppose that we have pocket Kings, and we put our opponent on either Pocket Aces, Kings, Queens, or Jacks. All of these hands are equally likely, so there is a 25% chance of our opponent having each hand. We can look up our equity against each hand (after you've been playing for a while, you naturally start to memorise hand equities anyway)

Our probability of winning is then:

$$P(A) * P(beating A) + P(K)*P(beating K) + P(Q)*P(beating Q) + P(J) * P(beating J)$$
Putting in our values:

$$ 0.25*0.2 + 0.25*0.5 + 0.25*0.8 + 0.25*0.8 = 0.575.$$

​We therefore see we have a positive expectation against this range, and should call.

No one actually thinks like this in real games?

It is a general misconception that professional poker is a game where players are trying to guess exactly what hand their opponent has, are constantly trying to bluff each other, or trying to pick up on subtle tells or signs that their opponent is or isn't bluffing.

The more mundane truth is that poker is ultimately a game of imperfect information, where the better player is the one who can correctly interpret the betting information their opponent is giving them, and can then quickly and accurately make the kind of judgements described above during a game.

Obviously poker players are not carrying out these calculations in their head to multiple decimal places in real time, what they will do though is review their hands after a game, calculate exactly what they should have done, and try to build up an intuition as to what the correct answer is, so that in the middle of a game they can quickly make decisions.

Software to Analyse this situation

Is there an easy software based method way of calculating our equity against a range? After I did a quick google there are a few programs that offer this type of analysis. For example:

combonator.com/
www.power-equilab.com/

More interestingly though, I also found the following open source software, that can be adapted to carry out this type of analysis:

github.com/zekyll/OMPEval

At some point, I might try to use this code to set up a page on this website to let people analyse this situation.

The Reinstatement Premium payable following a loss to an Excess of Loss contract, is related to the ceded loss to the contract by a simple formula. Therefore, it seems reasonable that we should be able to come up with a simple formula relating the price charged for the Excess of Loss contract to the price charged for a Reinstatement Premium Protection (RPP) cover. I was in a meeting last week with two brokers who were trying to do just this. They had come up with an indicative price for an Excess of Loss programme and were trying to use this to price the equivalent RPP cover.

At the time I didn't have an answer for them, and when I did a quick Google, nothing came up. When thinking about it subsequently though, there are a couple of easy approximate methods we can use. Below I discuss three different methods for how you can price an RPP cover, two of which do not require any stochastic modelling assuming you already know the price of the Excess of Loss layers.

Let's quickly review what we mean by all these terms so that we are starting from the same point.

What is a Reinstatement Premium?

If you already understand how a Reinstatement Premium works, then feel free to skip this section.

Most Excess of Loss contracts will have some form of reinstatement premium. This is a payment from the Insurer to the Reinsurer to reinstate the protection in the event of a loss. In the London market, most contracts willl have either $1$, $2$, or $3$ reinstatements and generally these will be payable at $100 \%$.

What is a Reinstatement Premium Protection Cover?

The reinstatement premiums can be quite a large proportion of the overall premium paid for the Excess of Loss reinsurance. This is especially the case for lower level, working layers. Furthermore, the Reinstatement Premium will be payable when the insurer has just suffered a loss. From the point of view of the insurer, this additional payment comes at the worst possible time. The Insurer is being asked to fork over another large premium to the Reinsurer, just after having suffered a loss.

To address some of these concerns, Reinsurers developed a product called a Reinstatement Premium Protection cover (RPP cover). This cover pays the Insurer's Reinstatement Premium for them, which gives the insurer further indemnification in the event of a loss. Here's an example of how it works in practice:

Let's suppose we are considering a $5m$ xs $5m$ Excess of Loss contract, there is one reinstatement at $100 \%$ (written $1$ @ $100 \%$), and the Rate on Line is $25 \%$. The Rate on Line is just the Premium divided by the Limit. So here, the Premium can be found by multiplying the Limit and the RoL:

$$5m* 25 \% = 1.25m$$

So we see that the Insurer will have to pay the Reinsurer $1.25m$ at the start of the contract. Now let's suppose there is a loss of $7m$. The Insurer will recover $2m$ from the Resinsurer, but they will also have to make a payment  to cover the reinstatement premium of: $\frac {2m}  {5m} * (5m * 25 \% ) = 2m * 25 \% = 0.5m$ to reinstate the cover. So the Insurer will actually have to pay out $5.5m$. The RPP cover, if purchased by the insurer, would pay the additional $0.5m$ on behalf of the insurer

Now that we know how it works, how would we price the RPP cover?

Three methods for pricing an RPP cover

Method 1 - Full stochastic model

If we have priced the original Excess of Loss layer ourselves using a Monte Carlo model, then it should be relatively straight forward to price the RPP cover. We can just look at the expected Reinstatements, and apply a suitable loading for profit and expenses. This loading will probably be broadly in line with the loading that is applied to the expected losses to the Excess of Loss layer, but accounting for the fact that the writer of the RPP cover will not receive any form of Reinstatement for their Reinsurance.

What if we do not have a stochastic model set up to price the Excess of Loss layer? What if all we know is the price being charged for the Excess of Loss layer?

Method 2 - Simple formula

This was the situation I was in when I was asked to price the RPP cover last week. The broker had come up with some very approximate pricing for a programme of Excess of Loss layers. This pricing was driven by the burning cost, and other commercial factors rather than any actuarial modelling. The brokers wanted to come up with an approximate price for the RPP cover, just based on the price of the Excess of Loss layer. The two should be related, as they pay out dependant on the same underlying losses. So what can we say?

If we denote the Expected Losses to the layer by $EL$, then the Expected Reinstatement Premium should be:
$$EL * ROL $$
To see this is the case, I used the following reasoning; if we had losses in one year equal to the $EL$ (I'm talking about actual losses, not expected losses here), then the Reinstatement Premium for that year would be the proportion of the layer which had been exhausted $\frac {EL}  {Limit} $ multiplied by the Deposit Premium $Limit * ROL$ i.e.:

$$ RPP = \frac{EL}  {Limit} * Limit * ROL  = EL * ROL$$

Great! So we have our formula right? The issue now is that we don't know what the $EL$ is. We do however know the $ROL$, does this help?

If we let $DP$ denote the deposit premium, which is the amount we initially pay for the Excess of Loss layer and we assume that we are dealing with a working layer, then we can assume that:

Which is our first very simple formula for the price that should be charged for an RPP. Was there anything we missed out though in our analysis? 
​
Method 3 - A more complicated formula:

There is one subtlety we glossed over in order to get our simple formula. The writer of the Excess of Loss layer will also receive the Reinstatement Premiums during the course of the contract. The writer of the RPP cover on the other hand, will not receive any reinstatement premiums (or anything equivalent to a reinstatement premium). Therefore, when comparing the Premium charged for an Excess of Loss layer against the Premium charged for the equivalent RPP layer, we should actually consider the total expected Premium for the Excess of Loss Layer rather than just the Deposit Premium.

What will the additional premium be? We already have a formula for the expected Reinstatement premium:

$$EL * ROL $$

Therefore the total expected premium for the Excess of Loss Layer is the Deposit Premium plus the additional Premium:

$$ DP + EL * ROL $$

This total expected premium is charged in exchange for an expected loss of $EL$.

So at this point we know the Total Expected Premium for the Excess of Loss contract, and we can relate the expected loss to the Excess of Loss layer to the Expected Loss to the RPP contract. 

i.e. For an expected loss to the RPP of $EL * ROL$, we would actually expect an equivalent premium for the RPP to be:

$$ RPP =  (DP + EL * ROL) * ROL $$

This formula is already loaded for Profit and Expenses, as it is based on the total premium charged for the Excess of Loss contract. It does however still contain the $EL$ as one of its terms which we do not know.

We have two choices at this point. We can either come up with an assumption for the profit and expense loading (which in this hard market might be as little as only be $5 \% - 10 \%$ ). And then replace $EL$ with a scaled down $DP$:

$$RPP =  \frac{DP} {1.075} * ( 1 + ROL) * ROL $$

Or we could simply replace the $EL$ with the $DP$, which is partially justified by the fact that the $EL$ is only used to multiply the $ROL$, and will therefore have a relatively small impact on the result. Giving us the following formula:

$$RPP =  DP ( 1 + ROL) * ROL $$

Which of the three methods is the best?

The full stochastic model is always going to be the best in my opinion. If we do not have access to one though, then out of the two formulas, the more complicated formula we derived should be more accurate (by which I mean more actuarially correct). If I was doing this in practice, I would probably calculate both, to generate some sort of range, but tend towards the second formula.

That being said, when I compared the prices that the Brokers had come up with, which is based on what they thought they could actually place in the market, against my formulas, I found that the simple version of the formula was actually closer to the Broker's estimate of how much these contacts could be placed for in the market. Since the simple formula always comes out with a lower price than the more complicated formula, this suggests that there is a tendency for RPPs to be under-priced in the market.

This systematic under-pricing may be driven by commercial considerations rather than faulty reasoning on the part of market participants. According to the Broker I was discussing these contracts with, a common reason for placing an RPP is to give a Reinsurer who does not currently have a line on the underlying Excess of Loss layer, but who would like to start writing it, a chance to have an involvement in the same risk, without diminishing the signed lines for the existing markets. So let's say that Reinsurer A writes $100 \%$ of the Excess of Loss contract, and Reinsurer B would like to take a line on the contract. The only way to give them a line on the Excess of Loss contract is to reduce the line that Reinsurer A has. The insurer may not wish to do this though if Reinsurer A is keen to maintain their line. So the Insurer may allow Reinsurer B to write the RPP cover instead, and leave Reinsurer A with $100 \%$ of the Excess of Loss contract. This commercial factor may be one of the reasons that traditionally writers of an RPP would be inclined to give favourable terms relative to the Excess of Loss layer so as to encourage the insurer to allow them on to the main programme and to encourage them to allow them to wrte the RPP cover at all.

Moral Hazard

​One point that is quite interesting to note about how these deals are structured is that RPP covers can have quite a significant moral hazard effect on the Insurer. The existence of Reinstatement Premiums is at least partially a mechanism to prevent moral hazard on the part of the Insurer. To see why this is the case, let's go back to our example of the $5m$ xs $5m$ layer. An insurer who purchases this layer is now exposed to the first $5m$ of any loss. But they are indemnified for the portion of the loss above $5m$, up to a limit of $5m$. If the insurer is presented with two risks which are seeking insurance - one with a total sum insured of $10m$, and another with a total sum insured of $6m$, the net retained exposure is the same for both risks from the point of view of the insurer. By including a reinstatement premium as part of the Excess of Loss layer, an therefore ensuring that the insurer has to make a payment any time a loss ceded to the layer, the reinsurer is ensuring that the insurer keeps their financial incentive to not have losses in this range. 

By purchasing an RPP cover, the insurer is removing their financial interest in losses which are ceded to the layer. There is an interesting conflict of interest in that the RPP cover will almost always be written by a different reinsurer to the Excess of Loss layer. The Reinsurer that is writing the RPP cover is therefore increasing the moral hazard risk whichever Reinsurer has written the Excess of Loss layer. Which will almost always be business written by one of the Reinsurer's competitors! 

Working Layers and unlimited Reinstatements

Another point to note is that this pricing analysis makes a couple of implicit assumptions. The first is that there is a sensible relationship between the expected loss to the layer and the premium charged for the layer. This will normally only be the case for 'working layers'. These are layers to which a reasonable amount of loss activity is expected. If we are dealing with clash or other higher layers, then the pricing of these layers will be more heavily driven by considerations beyond the expected loss to the layer. These might be capital considerations on the part of the Reinsurer, commercial considerations such as 

Another implicit assumption in this analysis is that the reinstatements offered are unlimited,. If this is not the case, then the statement that the expected reinstatement is $EL * ROL$ no longer holds. If we have limited reinstatements (which is the case in practice most of the time) then we would expect the expected reinstatement to be less than or equal to this.
​

I was attempting to set up a Loss Model in VBA at work yesterday. The model was a Compound-Poisson Frequency-Severity model, where the number of events is simulated from a Poisson distribution, and the Severity of events is simulated from a Severity curve.

There are a couple of issues you naturally come across when writing this kind of model in VBA. Firstly, the inbuilt array methods are pretty useless, in particular dynamically resizing an array is not easy, and therefore when initialising each array it's easier to come up with an upper bound on the size of the array at the beginning of the program and then not have to amend the array size later on. Secondly, Excel has quite a low memory limit compared to the total available memory. This is made worse by the fact that we are still using 32-bit Office on most of our computers (for compatibility reasons) which has even lower limits. This memory limit is the reason we've all seen the annoying 'Out of Memory' error, forcing you to close Excel completely and reopen it in order to run a macro.

The output of the VBA model was going to be a YLT (Yearly Loss Table), which could then easily be pasted into another model. Here is an example of a YLT with some made up numbers to give you an idea:

It is much quicker in VBA to create the entire YLT in VBA and then paste it to Excel at the end, rather than pasting one row at a time to Excel. Especially since we would normally run between 10,000 and 50,000 simulations when carrying out a Monte Carlo Simulation. We therefore need to create and store an array when running the program with enough rows for the total number of losses across all simulations, but we won't know how many losses  we will have until we actually simulate them.
​
And this is where we come across our main problem. We need to come up with an upper bound for the size of this array due to the issues with dynamically resizing arrays, but since this is going to be a massive array, we want the upper bound to be as small as possible so as to reduce the chance of a memory overflow error.

Upper Bound

What we need then is an upper bound on the total number of losses across all the simulations years. Let us denote our Frequency Distribution by $N_i$, and the number of Simulations by $n$. We know that $N_i$ ~ $ Poi( \lambda ) \: \forall i$. 

Lets denote the total size of the YLT array by $T$. We know that $T$ is going to be:

$$T = \sum_{1}^{n} N_i$$
​We now use the result that the sum of two independent Poisson distributions is also a Poisson distribution with parameter equal to the sum of the two parameters. That is, if $X$ ~ $Poi( \lambda)$ , and $Y$ ~ $Poi( \mu)$, then $X + Y$ ~ $Poi( \lambda + \mu)$. By induction this result can then be extended to any finite sum of independent Poisson Distributions. Allowing us to rewrite $T$ as:

$$ T \sim Poi( n \lambda ) $$

$$ T \sim N ( n \lambda , n \lambda ) $$
Remember that $T$ is the distribution of the total number of losses in the YLT, and that we are interested in coming up with an upper bound for $T$.

Let's say we are willing to accept a probabilistic upper bound. If our upper bound works 1 in 1,000,000 times, then we are happy to base our program on it. If this were the case, even if we had a team of 20 people, running the program 10 times a day each, the probability of the program failing even once in an entire year is only 4%.

I then calculated the $Z$ values for a range of probabilities, where $Z$ is the unit Normal Distribution, in particular, I included the 1 in 1,000,000 Z value.

We then need to convert our requirement on $T$ to an equivalent requirement on $Z$.

$$ P ( T \leq x ) = p $$ 

If we now adjust $T$ so that it can be replaced with  a standard Normal Distribution, we get:

Now replacing the left hand side with $Z$ gives:

Hence, our upper bound is given by:

Dividing through by $n \lambda $ converts this to an upper bound on the factor above the mean of the distribution. Giving us the following:

We can see that given $n \lambda$ is expected to be very large and the $Z$ values relatively modest, this bound is actually very tight.

For example, if we assume that $n = 50,000$, and $\lambda = 3$, then we have the following bounds:

So we see that even at the 1 in 1,000,000 level, we only need to set the YLT array size to be 1.2% above the mean in order to not have any overflow errors on our array.


References
(1) Proof that the sum of two independent Poisson Distributions is another Poisson Distribution
math.stackexchange.com/questions/221078/poisson-distribution-of-sum-of-two-random-independent-variables-x-y

(2) Normal Approximation to the Poisson Distribution.
stats.stackexchange.com/questions/83283/normal-approximation-to-the-poisson-distribution