Tag Archives: lottery

Decision Theory, Economics, Maths, Microeconomics, Probability, Risk

From Simpson’s paradox to pies

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Today, I wanted to publish a post on economics, and decision theory. And probability too… Those who do follow my blog should know that I am a big fan of Simpson’s paradox. I also love to mention it in my
econometric classes. It does raise important questions, that I do relate to multicolinearity, and interepretations of regression models, with multiple (negatively correlated) explanatory variables. This paradox has amazing pedogological virtues. I did mention it several times on this blog (I should probably mention that I discovered this paradox via Marco Scarsini, who did learn me a lot of things, in decision theory and in probability). For those who do not know this paradox, here is an example that Marco gave in one of his talk, a few years ago. Consider the following statistics, when healthy people entered in some hospital

hospital total survivors deaths survival
rate
hospital A 600 590 10 98%
hospital B 900 870 30 97%

while, when sick people entered in the same hospitals

hospital total survivors deaths survival
rate
hospital A 400 210 190 53%
hospital B 100 30 70 30%

Somehow, whatever your health situation, you should choose hospital A. Now, if we agregate

hospital total survivors deaths survival
rate
hospital A 1000 800 200 80%
hospital B 1000 900 100 90%

i.e. without any doubts, people should choose hospital B.

Actually, Simpson’s paradox is called Simpson’s paradox because Colin Blyth named it that way in 1972, in his paper entitled on Simpson’s paradox and the sure-thing principle (an economic article in a statistical journal), that can be downloaded from http://www.stat.cmu.edu/~fienberg/…. He found this paradox in a paper published in 1951 by Edward Simpson, even if other papers actually did mention it earlier. The most popular application is probably admission at Berckley’s graduate studiesprograms, and sex bias, see Bickel, Hammel & O’Connell (1975), that can be downloaded from http://www.unc.edu/~nielsen/…. I also mentioned a geometric interpretation of this paradox a few years ago on my blog, which is so simple to understand that the paradox is no longer a paradox actually, since on the example above, we had

and

while

With symbolic notations, one can have at the same time

and

with also

as shown on the graph below

There should be connection between Simpson’s paradox and the ecological fallacy (which is an issue I recently discovered and that I found extremely interesting, related again to difficulties of interpreting
regressions). But that’s another story. My point today is that Colin Blyth did mention another nice paradox, that is related, this time, to stochastic orderings. The idea is the following. Consider the three spinners drawn below (imagine some arrows in those circles)

  • spinner A: no matter where the arrow stops, the gain is 3,
  • spinner B: 56% chances to gain 2, 22% chances to gain 4, and 22% chances to gain 6,
  • spinner C: 51% chances to gain 1, 49% chances to gain 5.

Instead of spinners, it is also possible to consider three different lotteries,

You play against a friend, you pick a spinner, while the friend picks another. Everyone flick his arrow, the highest number wins (no matter the difference). Let us compute the odds. First case, A against B, from
A’s perspective

B-2 B-4 B-6
A-3 56%
+1
win		 22%
-1
lose		 22%
-3
lose

In that case, A has 56% chance of beating B. Second case, A against C, from A’s perspective,

C-1 C-5
A-3 51%
+1
win		 49%
-2
lose
In that case, A has 51% chance of beating C. Third (an final) case, B against C, from B’s perspective. Assuming independence between the spinners, joint probabilities can easily be computed,
C-1 C-5
B-2 28.56%
+1
win		 27.44%
-3
lose
B-4 11.22%
+3
win		 10.78%
-1
lose
B-6 11.22%
+5
win		 10.78%
+1
win
In that case, B has 61.78% chance of beating C. So, if we try to summarize,
  • A is the best choice, since it beats both with – always – more than 50% chance,
  • C is the worst choice, since it is beaten by both with – always – more than 50% chance,
Now, assume that you play not against one friend, but two friends. An everyone picks a different spinner. Let
us compute the odds, one more time. First case, A against B and C, from A’s perspective
B-2
C-1		 B-2
C-5		 B-4
C-1		 B-4
C-5		 B-6
C-1		 B-6
C-5
A-3 28.56%
+1
win		 27.44%
-2
lose		 11.22%
-1
lose		 10.78%
-1
lose		 11.22%
-3
lose		 10.78%
-3
lose
In that case, A has 28.56% chance of beating B and C. Second case, B against A and C, from B’s perspective,
A-3
C-1		 A-3
C-5
B-2 28.56%
-1
lose		 27.44%
-2
lose
B-4 11.22%
+1
win		 10.78%
-1
lose
B-6 11.22%
+3
win		 10.78%
+1
win
In that case, B has 33.22% chance of beating A and B.Third (an final) case, C against A, from C’s perspective,
A-3
B-2		 A-3
B-4		 A-3
B-6
C-1 28.56%
-2
lose		 11.22%
-3
lose		 11.22%
-5
lose
C-5 27.44%
+2
win		 10.78%
+1
win		 10.78%
-1
lose

In that case, C has 38.22% chance of beating A and B. So, if we try to summarize, this time

  • C is the best choice, since has (strictly) more than 1/3 chances to win, which the highest probability
  • A is the worst choice, since has (strictly) less than 1/3 chances to win, which the lowest probability

Odd isn’t it ? Now, is there an interpretation of that paradox ? Yes, Martin Gardner, in his paper on induction and probability, mentioned the case of drug testing. The value we had with the spinner is the health level, rated from 1 to 6. Thus, taking drug A, you always get an average health level of 3. With drug C, on the other hand, you get either very sick (level 1) or very well (level 5). Consider now a doctor who wants to maximize the patient’s chance of being well. If only pills A and C are available, then the doctor should choose A. This is what we’ve seen in the first part. Assume that now a company delivers a third pill, called drug B. Then the doctor should find C more interesting…. Odd, isn’t it ?

Colin Blyth gave a more amusing application. Assume that you like to go to the restaurant, and you like get a dessert there. Dessert A – the apple pie – is the average one, with a standard level, that you rank 3 (on a scale from 1 to 6). Dessert C – the cheese cake – can either be awfull (ranked 1) or delicious (ranked 5). You’d better go for the apple pie if you want to maximize the probability of not being disappointed (i.e. maximizing your “best chance” according to Colin Blyth, but I guess it can be interpreted as regret minimization too). Now assume that dessert B – the blueberry pie – is available (with ranks given by the spinner). Then you should go for the cheese cake. I let you imagine the discussion that you can have, then, with your favorite waitress

– Hi Mr Freakonometrics, do you want a piece of apple pie ? (yes, actually she also comes frequently on my blog, and knows me from my pseudo…)

– Probably. But actually, I was wondering if you did have your blueberry pie today ?

– Yes, in fact we do….

– Great, in that case, I’ll go for the cheese cake.

She’ll probably think that I am freak… so I hope she’ll come and read my post, to understand that, actually, it does make a lot of sense to go for what was supposed to be my worst case.

Computer, Probability, Randomness, Statistics

Cursed numbers ?

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In Lost, Hugo “Hurley” Reyes played the numbers 4, 8, 15, 16, 23 and 42 at the lottery, and ended up winning the $114-million jackpot. And over the ensuing weeks, everyone around him seems to suffer increasingly bad luck: Hurley’s grandfather dies of a heart attack, his brother’s wife walks out, his mother breaks her ankle while the house Hurley bought her goes up in flames, and Hurley himself is falsely arrested.
Anyway, last week (here) 4 numbers (out of 6) appeared at the lottery in LA. As pointed out by Xian (here), the odds were not that small, i.e. it is a 1‰ chance,

http://freakonometrics.blog.free.fr/public/perso/probaloto.png
Hence, with one lottery per week, the return period is 16 years. Note this percentage is very close to what we did observe on the French lottery (below the statistics in ‰, from here, in a zip file),

> loto=read.table("loto.csv",dec=",",header=TRUE,sep=";")
> ntirage=nrow(loto)
> loto=loto[51:ntirage,]
> ntirage=nrow(loto)
> N=as.matrix(loto[,c("boule_1","boule_2","boule_3",
   "boule_4","boule_5","boule_6")])
> P=rep(NA,nrow(N))
> for(s in 1:nrow(N)){
+ P[s]=sum(N[s,1]%in%c(4, 8, 15, 16, 23, 42)+
+          N[s,2]%in%c(4, 8, 15, 16, 23, 42)+
+          N[s,3]%in%c(4, 8, 15, 16, 23, 42)+
+          N[s,4]%in%c(4, 8, 15, 16, 23, 42)+
+          N[s,5]%in%c(4, 8, 15, 16, 23, 42)+
+          N[s,6]%in%c(4, 8, 15, 16, 23, 42))
+ }
> table(P)/nrow(N)*1000
P
         0          1          2          3          4
435.732113 405.366057 137.271215  19.966722   1.663894

But what about the full sequence…? Imagine that in France, at the official lottery, the exact sequence played by Hugo appears. What a coincidence.The probability that the sequence appears, assuming that there are 48 possible numbers in the lottery, is

http://freakonometrics.blog.free.fr/public/perso/lotto2.png
i.e. the expected number of draws we need before seeing that sequence for the first time is almost a billion.
Now if we look at all official lotteries around the world, say 100 per week, what is the probability to see Hurley’s sequence shows up – at least once – in 25 years (assuming that after 25 years, no one will remember Lost and those cursed numbers) ? It looks like it is a 1% chance…
http://freakonometrics.blog.free.fr/public/perso/proba-lost.png
So let us wait and see…

Computer, Probability, Statistics

Comments on probabilities

2 Comments

The only thing I remember from courses I had in probability a few years ago is that we also have to clearly defined the event we want to calculate the probability. On the Freakonomics blog, last week, the Israeli lottery was mentioned (here, see also there where I mentioned that, and odds facts from the French lottery),

Yesterday, Andrew Gelman claimed (here) that there was a probability error… Well, since Andrew is really a statistician (and a good one… while I am barely an economist), I tried to do the maths…. and to understand where the error was coming from…

Since 6 numbers are drawn out of a pool of numbers from 1 to 37, the total number of combination at each lottery is

https://perso.univ-rennes1.fr/arthur.charpentier/latex/loto01.png
> (n=choose(37,6))
[1] 2324784

Over 8 lotteries (since there are two draws per week, we can assume there 8 draws per month), the probability of no identical draws is

https://perso.univ-rennes1.fr/arthur.charpentier/latex/loto02.png

Here is the R code for those who want to check, again,

> prod(n-0:7)/n^8
[1] 0.999988

Each month, the probability of “coincidence” (I define “coincidence” the event “over 8 draws, at least two times, we obtained the same 6-uplet” or more precisely (as mentioned here) “over one calendar month, at least two times, we obtained the same 6-uplet“) is

> (p=1-(prod(n-0:7)/n^8))
[1] 1.204407e-05

The occurrence of a coincidence each month as a Geometric distribution, with probability p. And it is classical, following Gumbel’s definition (here), to consider 1/p, called the “return period“, i.e. the number of months we have to wait until we observe a coincidence (i.e. a repetition in the same month), since for a geometric distribution

https://perso.univ-rennes1.fr/arthur.charpentier/latex/loto03.png
> 1/p/(12)
[1] 6919.034

Here, the (expected) return period is 6919 years.

From my point of view, this is “the incident of six numbers repeating themselves within a calendar month”, and this is an event of once in 6919.034 years. On the other hand the median of a geometric distribution is

https://perso.univ-rennes1.fr/arthur.charpentier/latex/loto04.png
> -log(2)/log(1-p)/(12)
[1] 4795.88

which means that we have 50% chance to get such a coincidence over 4796 years.

Of course, if instead of looking at a longer period, say 100 draws, i.e. one year (here I define “coincidence” the event “over 100 draws, at least two times, we obtained the same 6-uplet“), we have in red the expected return period, and in blue the median of the geometric distribution,

> M=E=rep(NA,100)
> for(i in 2:100){
+ p=1-exp((sum(log(n-0:(i-1)))-i*log(n)))
+ E[i]=1/p/(100/i)
+ M[i]=-log(2)/log(1-p)/(100/i)
+ }
> plot(1:100,E,ylim=c(0,10000),type="l",col="red",lwd=2)
> lines(1:100,M,col="blue",lwd=2)
> abline(v=8,lty=2)
> points(8,E[8],pch=19,col="red")
> points(8,M[8],pch=19,col="blue")

or below of a log-scaled version

As Xi’an did (here), assume now that there is a lottery over 100 countries. Here I define “coincidence” the event “over k lottery draws over 100 around the world, at least two times, we obtained the same 6-uplet“, and then the previous graph becomes (with on the x axis the level of k)

Here I have a 12% chance if we consider probability to have identical numbers over a month…

But here, we can have one 6-uplet in Israel, and the other one in Egypt, say… If we want to get the same 6-uplet in the same country, the graph is now

i.e. each month there is a chance over one thousand…

> i=8
> p=1-exp((sum(log(n-0:(i-1)))-i*log(n)))
> 1-(1-p)^100
[1] 0.001203689

Note: actually, Xi’an mentioned that the probability that this coincidence [of two identical draws over 188 draws] occurred in at least one out of 100 lotteries (there are hundreds of similar lotteries across the World) is 53%! And I got the same,

> 1-(1-P[188])^100
[1] 0.5305219
Computer, Statistics

Lottery, and martingales

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I recently got a comment on a post I published one year ago, here, about the fact that in September 2009, on the 6th and the 10th, the 6 same numbers came out at the lottery, in Bulgaria (but  I do not understand the question: the author of the comment ask about the order the numbers came out…)
Xi’an published also a post on that topic, there, since last week, the same thing happened in Israel.
All that reminded me a discussion I had with a colleague about another post (here) where I mentioned that I found a strange distribution of numbers in the French lottery (the old one actually). For those who want to check, all historical events are here, in a zip file. My colleague was wondering if I found the martingale to win the lottery…

First, I do not like that term, since martingale is something different from a mathematical point of view… Second, let us look if it would have been possible to make some money… (free lunch ?)

> loto=read.table("D:\\loto.csv",dec=",",header=TRUE,sep=";")
> ntirage=nrow(loto)
> loto=loto[51:ntirage,]
> ntirage=nrow(loto)
>   N=as.matrix(loto[,c("boule_1","boule_2","boule_3","boule_4","boule_5","boule_6")])
> n=as.vector(N)
> length(n)
[1] 28848
> (TN=table(n))
n
1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20
607 576 571 618 579 598 608 582 588 590 562 577 577 580 591 630 558 567 594 608
21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38  39  40
578 562 579 583 574 589 602 572 550 598 604 582 545 646 597 618 599 636 609 588
41  42  43  44  45  46  47  48  49
576 589 577 585 618 596 560 571 604

So, it might look nice, but we have to compare that distribution with the one we should have with “independent” draws. It is not possible to look at a discrete uniform distribution: the six numbers are not independent. Each day, the 49 balls are back in the urn, but within a day, we do not have independent draws (it is a sample without replacement of balls). Hence, with 4808 lottery draws, each number cannot be obtained more than 4808 times. So, let us use monte carlo techniques to  look at the theoretical distribution,

> M=matrix(NA,49,1000)
> for(s in 1:1000){
+ B=NA
+ for(i in 1:ntirage){B=c(B,sample(1:49,size=6,replace=FALSE))}
+ B=B[-1]
+ M[,s]=sort(table(B))
+ }
> q50=function(x){quantile(x,.5)}
> Q50=apply(M,1,q50)
> lines(1:49,Q50,col="red",lwd=2)
> q10=function(x){quantile(x,.1)}
> Q10=apply(M,1,q10)
> q90=function(x){quantile(x,.9)}
> Q90=apply(M,1,q90)
> polygon(c(1:49,49:1),c(Q10,rev(Q90)),col="light blue",border=NA)
> lines(1:49,Q10,col="red",lty=2)
> lines(1:49,Q90,col="red",lty=2)
> lines(1:49,Q50,col="red",lwd=2)
> points(1:49,sort(TN),pch=19,type="b")

Looking at the graph, it looks like some numbers appeared too frequently, especially the ones that did not appear frequently (bottom left). So, since I have removed the last 50 draws, let us see if we could have used that information, somehow…

> nb=names(sort(TN))
> loto=read.table("D:\\loto.csv",dec=",",header=TRUE,sep=";")
> loto=loto[1:50,]
> N=as.matrix(loto[,c("boule_1","boule_2","boule_3","boule_4","boule_5","boule_6")])
> n=as.vector(N)
> TN=table(n)
> TN[nb]
> barplot(TN[nb])

Unfortunately, numbers that came out too frequently over 4800 draws did not appear that frequently of the last 50. Playing top number might not have been a great strategy.

(numbers that came out frequently are on the right, while those we did not see much are on the left)… What about worst numbers: if I had decided to play the 6 that did not come out very frequently (we’ve seen earlier that they should have appeared even less, actually), would it have been interesting ? As we can see, our top 2 numbers were numbers that did not appear frequently earlier (29 and 47 appears respectively 10 and 11 times over 50 draws)….
Over 50 draws of 6 balls, the expected frequency of 6 given number is around 36.7,..

> S=rep(NA,10000)
> for(s in 1:10000){
+ B=NA
+ for(i in 1:50){B=c(B,sample(1:49,size=6,replace=FALSE))}
+ B=B[-1]
+ S[s]=sum(B%in%(1:6))
+ }
> mean(S)
[1] 36.7694

But here for the top 6, we have

> z=TN[nb]
> sum(rev(z)[1:6])
[1] 29

i.e. the top 6 appeared 29 times over 50 draw of 6 balls (which looks low) and for the worst 6, it is a bit higher,

>  sum(z[1:6])
[1] 38

If we look at the theoretical density of the frequency of 6 given number, we have

i.e. our worst 6 is a nice average (in green) while top 6 did not appear frequently this time (here in blue) ! So we could not have used that information….
Anyway, if some of you are interesting using statistics to get a free lunch, with the nouveau loto, I did not see any strange pattern (data can be downloaded here).

I am terribly sorry, but I cannot help anyone winning at the French Lottery….