Can I Count on You?

It's as easy as 1, 2, 3

There can hardly be anything simpler than learning to count. Most of us have mastered it before we started elementary school. Similarly we all knew our a-b-c's before we graduated kindergarten. Yet any attempt to combine the two, leading to the dreaded realm of algebra, and most people's brains turn into jello, or so they say. Well, it doesn't have to be that way, and if you keep an open mind and read this article through to the end, you may discover that algebra is not only easy, but fun too.

Let's start with something simple, like counting. One, two, three, four, five ... not so hard is it. Suppose we start adding these numbers together, as follows:
1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 = 10
1 + 2 + 3 + 4 + 5 = 15
Nothing too profound here, but suppose I asked you for the sum of the first 100 numbers. In other words the sum of 1 + 2 + 3 + (keep going from 4 to 98) + 99 + 100. (In the future I'll write ... instead of (keep going from 4 to 98.)) This was an arithmetic problem for young Carl Friedrich Gauss, the most famous mathematician of all time, when he was in first grade. He immediately responded with 5,050 the correct answer. Now this was in the days before pocket calculators, even though that probably wouldn't help you much anyway. I usually make a mistake after typing in 3 numbers into a calculator, imagine trying to add up a hundred numbers. So how did he do it? We all know that 2 + 3 = 3 + 2 right? When you add a bunch of numbers together, the order in which you add them doesn't matter. Well, he noticed that if you rearrange the sum as follows 1 + 100 + 2 + 99 + 3 + 98 + 4 + 97 +... + 50 + 51, that every two terms add up to 101, and there are 50 of them. Check it out, 1+100 = 101, 2+99 = 101, and 3+98=101. Since there were 100 numbers to start with, there are now 50 = 100/2 pairs of numbers. So the total has to be 101 * 50 = 5,050. Simple isn't it, but pretty clever for a six year old. Are you still with me? All we have done so far is add up the same list of numbers, but in a different order. Now lets go a little further. Suppose we don't want to say exactly how many numbers we should add up. How can you do that?

Oh no, Algebra!

By using a letter instead of number. Let's use the letter N. So we would ask, what is the sum of the following sequence: 1 + 2 + 3 + 4 + ... + N? Can it possibly be that simple? Yes! All we have to remember is that N stands for a number. It could be 5, it could be 100, it could be anything. Now what is the number that is one less than N? Why N-1 of course. How about the number that is 2 less than N? Try N-2 on for size. Let's rewrite the sequence now adding in a few more terms on the end:
1 + 2 + 3 + 4 ... + (N-2) + (N-1) + N = ?
All I've done is explicitly show you the last few terms of the sum "symbolically." Instead of ending with 98 + 99 + 100 or 998 + 999 + 1000, I've used the letter N to represent the last number. Now let's apply the profound notion that 2 + 3 = 3 + 2 and rearrange the sequence like Gauss did. We get
(1 + N) + (2 + N-1) + (3 + N-2) + ...
We can write (2 + N - 1) as (N + 2 - 1) = (N+1). Also we can write (3 + N - 2) as (N + 3 - 2) = (N+1). Each of these terms adds up to N+1, and there are exactly N/2 of them, so what is the total? Well, it must be (N+1) times N/2, or if we write it the way mathematicians do, N*(N+1)/2 Now plug a number for N into this expression, and we have the sum of the first N numbers. If N is 5 you get 5 * 6 / 2 = 15 correct? If N is 100 you get 100*101/2 = 50*101 = 5,050 just like Gauss. It is a lot easier to calculate N*(N+1)/2 than it is to add up N numbers, yet the answer is always the same. This idea of letting letters stand for numbers is the heart of algebra, and allows us to express very complicated ideas in a very simple way. Now why didn't they tell you it was that easy in high school?

A Raw Deal

Let's look at another example, perhaps much more practical for you poker players out there. We all know that in regular poker, flushes (all cards the same suit) are harder to get and hence beat straights (all cards in a sequence.) But what does it mean, that flushes are harder to get than straights, and how can we be sure that it is true? Well, to answer these questions we have to learn how to count. We need to count how many different flushes there are, and how many different straights there are. A simpler question is how many different 5 card poker hands are there? The answer isn't obvious, but if you can count you can get there. Let's start with a simpler problem, how many 1 card hands are there? Well there are 52 cards in a deck, unless you're playing with my friend Dean, and then there are 48 cards in the deck and 4 aces up his sleeve, but let's ignore that and stick with the 52 cards. When someone gives you a card, it could be any one of the 52. Thus there are 52 possible one card poker hands. No problem so far. How many 2 card poker hands are there? Well the first card you get can be any one of 52 cards, and the second card you get can be any one of 51 cards, since you already have one card in your hand. So there are 52*51 = 2652 different 2 card hands. But wait, it doesn't matter if you have an King-Jack or Jack-King, so the real number if "different" poker hands is 52*51 / 2, since the order doesn't matter. Okay, now let's continue with a 3 card poker hand. The first card can be any one of 52, the second any one of 51, and the third any one of 50. Again order doesn't matter, King Jack Deuce is the same as Deuce King Jack, they're both losers. We can think of having 3 slots for our cards. The first card we get can go into any one of the three slots, the 2nd card can go into any one of the two remaining slots, and the last card has to go in the last slot. Thus we can arrange these 3 cards in 3*2*1 = 6 different ways, and they are always the same poker hand. Thus with 3 cards there are (52*51*50) / (3*2*1) different poker hands. We can see where this is heading. With 5 cards we will get (52*51*50*49*48) / (5*4*3*2*1) = 2,598,960. Mathematicians call this thing the number of combinations of N things taken K at a time, such as 52 cards taken 5 at a time. The formula for this is N * (N-1) * (N-2) .. (N-K) / (K * (K-1) * ... 3 * 2 * 1) Just plug in a number for N and a number for K and you have how many combinations there are. Let's get back to counting flushes now. In order to have a flush, your first card can be anything, thus you have 52 choices. Your second card can be any one of 12 cards, since it now must be in the same suit as your first card. Your third card can be any one of 11, and so on so the number of flushes is 52*12*11*10*9/(5*4*3*2*1) = 5148. So you can see that only 5148/2,598,960 or .00198 or about 2 in a thousand is your chance of getting a flush in five cards. I'll leave it to you to figure out how many different straights there are in five cards, and you'll see there are more straights than flushes, which makes them easier to get.

What's behind door number 2?

This leads us to the Monty Hall problem, which you may have heard about, but in case you haven't it is one of my favorites. Suppose you are playing Let's Make a Deal, and you can pick any one of 3 curtains. One curtain has a wonderful prize, and the other two have junk. Monty knows what is behind each curtain, and after you choose, say curtain number one, he shows you that behind curtain number two is a three legged goat. He now gives you the opportunity to "stay with curtain number one, or switch to curtain number three." What should you do? Well, believe it or not, you should always switch. Why? If you stay with curtain 1, you have a one in three chance of winning the prize, after all there is one good prize out of three. The fact that you know what is behind curtain two doesn't change that. On the other hand if you switch your odds double to two out of three. Let's analyze the situation. There are 3 possibilities of how to arrange the prize and goats behind the curtains. Case 1 is (Prize, goat, goat), case 2 is (goat, Prize, goat) and case 3 is (goat, goat, Prize). If you stay with door number one, you will only win in case number 1, which happens one third of the time. If you use the always switch strategy, you will win in cases 2 and 3, or two thirds of the time. So you should always switch. If this seems odd to you, imagine the following. Suppose there were 100 curtains, and after you pick one, say curtain number 1 again, Monty reveals what is is behind 98 of the other curtains, all of which have goats. Now only curtain number 1, the one you chose, and some other curtain that Monty hasn't revealed yet are left. My guess is that at this point you would be begging Monty to switch. After all, you could have just as well have chosen any of the 98 curtains Monty revealed as well as your own, and discovered it had a goat behind it, while, for some mysterious reason, Monty has not disclosed what is behind the sole remaining curtain other than yours. If he gives you the opportunity to switch, your chance of winning a prize is 1/100 of you stay, and 99/100 if you switch. I can hear the audience members now screaming SWITCH! SWITCH! I was in the Pacific Pearl offices last week, and Mike told me that he prefers my articles to crossword puzzles, but both require the same amount of effort out of his gentle readers. I don't really like crossword puzzles, but I do like puzzles in general. Here is one of my favorites. Can you think of a four letter word that ends in "eny?" Yes, there are only 26 possibilities, and no it isn't an obscure word, but a common everyday word that you have heard many times. Nevertheless, this is a fiendish puzzle and I leave you to work it out. If you get desperate, you can always resort to a dictionary. Enjoy.

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