Showing posts with label mathematics. Show all posts
Showing posts with label mathematics. Show all posts

Tuesday, January 12, 2010

The Fundamental Theorem Of Betting

I used two popular websites to get betting odds for the today's match between Egypt and Nigeria. Here are some odds from two example sites:

Site 1: Egypt (2.90) - X (3.20) - Nigeria (2.35)
Site 2: Egypt (2.65) - X (3.10)- Nigeria (2.65)


Which site of the two should be preferred over the other? Generally, how to compare the attractiveness of different betting odds? In what follows, I am presenting what I call as the "fundamental theorem of betting".

Consider just the single game "Egypt vs Nigeria" which has 3 possible turnouts: "Egypt wins", "nobody wins" and "Nigeria wins". Let us also consider that we have a budget of Δ dollars and that we split this sum to all three turnouts, betting Δ1, Δ2 and Δ3 respectively. Finally, for every event, we expect to win some money, say W1, W2 and W3. In general, we are just interested in winning more money than we spent ( Δ dollars), so for every event that might take place (e.g. "Egypt wins" ) we might want W1 > Δ or W2 > Δ or W3 > Δ. But we might want to ask: Can all of these inequalities hold? The obvious answer is no. Otherwise, the booker would go bankrupt. Next, we have to ask: What values do the W's satisfy? It turns out that this is a linear function, with coefficients determined by the betting odds of the booker!

Consider that we have a budget Δ and we want to split this in all k possible turnouts. Thus, we will have to pay Δ1, Δ2,Δ3,..., and Δk respectively. For every turnout i the booker provides a betting odd Mi. This means that we will win Wi = Mi * Δi, in case of this turnout. Finally, every profit has a specific ratio of the whole budget Δ. This is the real profit since we are interested in making more money than we spent. For every turnout the profit will be denoted as: Ai = Wi / Δ. One can easily see that:

(A1/M1)+(A2/M2)+...+(Ak/Mk) = 1

For clearer inspection:

This is what one could call the "Fundamental Theorem Of Betting".

Remember that the μ's are the booker's betting odds and the α's are the ratios of profit over the overall budget. For example, if we invested $200 in total and Α1 = 2.0 then if turnout #1 happened we would win 200x 2.0 = $400 and so on. When an α is greater than 1 this means that we won more than we invested in total and so we are interested in establishing at least one α greater than 1.0. This means that it would be useful to consider one single quantity:

M = (1/M1)+(1/M2)+(1/M3)+...+(1/Mk)

The following cases now exist:
1) M > 1.0
If this is the case then the booker is going bankrupt. Players can construct a set of α's such that all of them are greater or equal to 1.0. This means that every player either wins more than he spent in total or he gets his money back (zero loss)

2) M = 1.0
In this border case, the players can set all α's to 1.0 and never suffer a loss. This is not very meaningful and players might try to disturb the ratios by increasing and decreasing some of the ratios.

3) M < 1.0
This is expected to be much more frequent in real world betting odds. Players will have to decide how to distribute their bets.

What's more important, the quantity M can be used as a metric of how attractive the betting odds are. The greater the quantity M is, the better the odds are for the players. Going back to the example in the beginning we have:

Site 1: M = (1/2.90)+(1/3.20)+(1/2.35) = 1.0828
Site 2: M = (1/2.65)+(1/3.10)+(1/2.65) = 1.0773


One can see with a single arithmetic comparison that Site 1 gives slightly better margins for profits (1.0828 is bigger) and it should be preferred to Site 2. Even within the same book house, players should prefer games with higher M factors than others. The first step to a complete "Theory of Betting" has been made!

Sunday, March 2, 2008

Google Interviews Part V - Just Two More Questions

This is the last part of posts describing my own experience interviewing with Google. You may find the first interview here and then follow the traces until this last one.

Time had already passed (almost a month) after three successive interviews with Google and this last one was most probably the critical one. I did not prepare for this as much as the previous. The whole thing tires you up and at some point it seems better to relax and concentrate on the psychology rather than the technical stuff.

I was only asked two technical questions, and since my language of choice was Java the interviewer wanted to see some Java code for the answers. In summary, this interview was perfect. While in all previous interviews, mostly the first and the second, I had some important flaws in my performance, but this one was different. And all this because I was playing in my field.

Google at some point asks for a CV. However, the way I see it, Google and other major software companies search always for something unconventional in the candidate's application. You could say of course you know 5 or 10 programming languages, or that you can build an internet spider in 1 minute, however there might be some knowledge you have, that is somewhat rare and hence you should mention. In my case, this is Number Theory.

For me, number theory is a passion that has not passed over time. It was love in the first sight, when I became aware of it, maybe at the age of 15 or so, and from that time it was the field mathematics that I was really feeling comfortable with. Anyway, it is not hard to see why it has been named as the "Queen Of Mathematics". First, it is very easy to grasp the basics (primes, divisors etc) since they are inherently tractable by the human brain. After all, every human being starts my learning how to count, which is the first step of the true intelligence, aka understanding that 5 ice creams and 5 cars, share something in common: that they are 5. This helps solving silly cases like, if I ate 5 ice creams and then I ate 1 more, how many did I eat? There is no need to think of ice creams, nor the eating process. All you have to do is to construct the proper model for this situation: use natural numbers and solve 5+1.

Another great thing with number theory, is that your 'lab' can be a blank piece of paper. You can argue that 15 is divisible by 3, but all you need is a paper to perform the division. While a physicist might say that in light speeds some hypo-atomic elements, called mambojambions, are created, he still needs a gigantic, CERN-like laboratory to test his theories.

Anyway, I myself was raised in a Pythagorean culture(numbers are holy) and so I claimed in my CV to know some number theory stuff. Although I was a little worried if I would be able to defend such claims (after I would be talking to Google), it soon proved to be a wise decision. Both in phone and on site interviews with Google, there was no better time for me than doing number theory. And to much of my surprise, and despite that Google's name is inspired by a natural number (or a unit of measure if you wish), all the 'Googlers' that I spoke and met with, did lack elementary knowledge on the field. This time, (I assume) the interviewer had dug in my profile and wanted to see for himself. Every question and conversation was related to Number Theory.

So, the first question was about sets: "Create a function, called powerSet(), that will output the power set of the input set." The power set in Algebra theory is the set of all subsets of a set (no..bull-set!) If a set has N elements then the power set will have 2^N elements. So if a set is denoted by {a,b} with a,b as elements then the power set is { {},{a},{b},{a,b} }.
The {} is the empty set. Note that all elements of the power set are sets. Recursion can help for solving this problem. If for example the powerSet function can produce power sets for sets that have at most N elements how should we solve for the N+1 case?

If we denote sets with capital letters and the initial set as A, we can use the following algorithm
1. define B set initially empty
2.
a = extract one element from A
3. add powerSet(A) to B
4. for every set in B, say X, add a, and then add it to B
5. return B

Making the proper checks, this yields a function that can produce the power set of a set. For the Java case you can use the java.util.Set interface to write the code. I was asked to write the Java code but not asked any further questions, like complexity, runtime and so on. So we moved to the next and final question.

This was: "Create a function, called findZeros(), that will compute the number of zeros in the decimal representation of n!". Now this is interesting and despite seemingly easy it has subtle points. For example, 5! = 1.2.3.4.5= 120 has one zero and 10! = 1.2.3.4.5.6.7.8.9.10 = 3628800 and has two zeros. Now the obvious answer is to produce this C-style pseudo code:

int findZeros(int n) {
z<-0;
N<- n!

while ( N%10==0) { z++; N/=10;}
return z;
}

It seems right but has a terrible flaw: it fails at very 'small' values of n, in our 4-byte world, for n=15. Using some Java code like below, we can verify this case.

int n=1,i=2,m=1;
while(n>=m) {
m=n; n*=i;
System.out.println("Previous value "+m+" Current value "+n+" Counter "+i);
i++; }

System.out.println("OOPS! Overflow!");

So, if our code works only for 14 cases, it is pretty much useless. We should do better than that.
For number theory geeks, like myself, this means show time! The equation below is called the Legendre's formula
and basically computes the factorization of
n!. The exponent in which prime p is 'participating' in the factorization is the sum in the figure. Now, we should use it for our case. 10 is the product 2x5. So, the exponents of 2 and 5 in the factorization of n! defines how many zeros we get. For example, 10! = 1.2.3.4.5.6.7.8.9.10 = (2^8).(3^3).(5^2).7 So 2 is found 8 times and 5 is found 2 times in the factorization, hence if we pair up the 2's and 5's we get two zeros. Now, there are obviously more 2's than 5's in the expansion so finally we have to answer at the question:
What is the exponent of 5 in the factorization of n!?
Based on the reasoning above this is equal to the number of zeros of n!.
Based on Legendre's formula we have to calculate:
[n/5]+[n/5^2]+[n/5^3]+... to infinite. This is done for example by the following Java code

calculateZeros(int n) {
int s=0,r,p=5;
while((r=(n/p)) !=0)

{s+=r;
p*=p;}

System.out.println(n+"! has "+s+" zero"+((s==1)?"":"s"));
}

For example, 25 = 1.2.3....23.24.25 = 15511210043330985984000000. Running the above code will produce the message 25! has 6 zeros. Instead of messing with a number with 26 digits (25!) you only have to deal with the number 25. Why?

In the sum S(n) = [n/5]+[n/5^2]+[n/5^3]... the first term, [n/5], computes how many numbers up to n are multiples of 5. For the 25 case there are 5 of them: 5,10,15,20,25. The second computes all multiples of 25, i.e. only one. The third and all others are zero. A number that in its factorization, the prime 5 is in the power of k, e.g. 10 = 2.5 and k=1 or 25 = 5^2 and k=2, participates in the first k terms of the sum by adding one to each term. Hence its overall contribution is k, hence the sum S(n) calculates the power of 5 in the factorization of n!, and therefore the number of zeros in the decimal representation, Q.E.D.

That was it! There is no other way to handle such big values and the solution can deal with really gigantic numbers without explicit calculation. The interviewer had elementary number theoretic knowledge but was able to follow the reasoning and we had a very nice conversation after that. He seemed to be fond of such mathematical tricks as the one we dealt with.

In summary, I think this was the interview question that booked me the plane ticket to fly to the Google site. I was very satisfied after we hang up and I was impatient for the outcome but deep inside I was sure it was an ideal interview session and that my chances were great. The next morning I received a phone call from Google inviting me for the on site interviews. Mission was accomplished.

This post ends a series of posts in which I wrote about my phone interviews with Google along with many interview questions. For the on site interviews I am bound to an NDA so maybe I will post them encrypted! That's all for now about Google interview questions. Until, I come back for the on site experience remember a small tribute to Number Theory:
"Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk" (Numbers were created by good Lord. Everything else is human's work-Leopold Kronecker)

Tuesday, February 5, 2008

Google Interviews Part IV

This is the 3rd interview I had with Google. You can find the previous and the questions asked here:
* First Google interview
* Second interview

Update: Fourth Google Interview

In the 3rd interview I talked with a woman software engineer from Mountain View. As usually it lasted for about 45 minutes but there was a surprise waiting at the last question..But let's take things from the beginning.

The first question was a hard test for my memory: "How do you test your code?" This is a fair one for software engineers. But not for my style. I personally like things that are quick and dirty. For larger projects I use some of my own class libraries that employ some kind of logging, measure method execution time and so on. But saying "Well, I use my own class libraries." I thought would not be a good answer. Nor a professional one.

So talking about Java, I made the mistake of mentioning the JUnit framework. I had used for some time (long ago) but the time had passed so I had forgotten even the basics. And of course the interviewer started the questions (How the JUnit handles exception and others) To tell you a secret I had already opened in my browser a JUnit site (some kind of tutorial I guess) but this didn't help at all. So, don't do it. It will only make things worse. Trying to think logically didn't help either. The interviewer kept pushing, until I 'broke' and admitted that I couldn't answer. That was it. We moved to the next question.

All next questions were really typical, the kind you find in tech interview sites:
* "What is a Unix kernel?"
* "What is the difference between interfaces and abstractt classes?"
* "What is the difference between threads and processes?"
* "What is inheritance, polymorphism and encapsulation?"
* "What is overriding and what is overloading?"

I think there is no need to elaborate further on that. You most probably have come across these concepts and have your own mind.

Which brings us to the last question. Actually it was a puzzle including programming with handicaps, i.e. with small processor, low memory etc. The puzzle was this:
"You have 1000 integers. All are less than 1000 and greater or equal to 1. Among them, 999 are distinct and there is one that is found twice. How can you find the duplicate?" To this I gave an answer I think is optimal. The idea is simple. If you denote the duplicate number by d and the sum of all the integers by S then the following equation holds:
S-d= 499500 since S-d is the sum 1+2+3...+999 which is equal to 499500 by applying the formula 1+2+3...+n = n*(n+1)/2
Now the good thing is that we can be able to find the duplicate even we have capacity for one integer, or when reading from a stream and so on. We can optimize even if we apply modulus to the first equation: d = (S-500) (mod 1000) Now if d is equal to a positive number mod1000 then this is the duplicate, otherwise the duplicate is the negative part plus 1000. For example is 1 was the duplicate, then d=1(mod 1000) and the duplicate had to be the 1. If the duplicate was 600, then d=-400(mod 1000) so the duplicate had to be -400+1000=600. This means we do not need to have storage for integer (int type) but only the byte type is enough.

While fairly easy to grasp the interviewer had difficulties in understanding how this would work, so she asked for an example when we have 10 integers. I replied this would transform the equation to d =S-45 but then the counter question was disappointing: "How did you compute 45?" It is quite obvious however that I had to compute 1+2...+9 which is equal to 45 when applying the formula that Gauss found when he was 8. But the interviewer started computed 'by hand' the sum, adding the numbers from 1 up to 9, which left me speechless for some seconds. I mentioned that there was a formula for that but she didn't listen, still being concentrating on her computations. I didn't bother elaborating on the modulus idea since obviously would not give me any more credit.

After that, the interview had finished. I didn't ask anything, saying something like 'I have many questions but I do not wish to spend any more of your time' and we ended the conversation. In overall the last incident was really awkward. To that time I believed that all Google engineers had a good mathematical background. The engineer that I spoke with, did lack elementary skills. So in my eyes, the myth had been destroyed and it is a good advice to anybody, not bother berating himself more than he should. If you already knew the formula for the sum of consecutive integers, you have a good reason to feel more confident.

Update: Go to the last Google phone interview