7 Drawing objects from an urn: What actually happens during a lottery?

Slide 0

Hello and welcome to a new episode in which we again talk about math. Once more we will deal with random experiments; we will again draw objects from an urn (or at least do something very similar). This time we won’t replace the objects and we are also not interested in the order. Surely, you are familiar with the lottery game “6 of 49.” This is the prototype for the actions that we will take today.

Slide 1

Amara is a young woman who loves traveling. And each time she has the same problem: What should she pack in her suitcase? She has 12 dresses. Let’s assume that four of them will fit in her suitcase. That means there are  

12 • 11 • 10 • 9

possibilities, right?

Slide 2

Not so fast ... packing a suitcase doesn't work that way. Ultimately, it doesn’t make a difference whether she packs the green dress first and then the red dress or whether this order is transposed.

It won’t surprise you that in this case as well, we must first consider things systematically.

Slide 3

Stated very simply: Here, the order of the result is really not important.

Slide 4

We will simplify our lives a little in order to understand the basic idea and watch Lene pack her suitcase. She has only five dresses – easy to manage – of which three should go into her suitcase.

For the first dress there are five possibilities, for the second dress four, for the third dress three. We start initially with 5 • 4 • 3 possibilities and that is – stated differently and with our mathematical repertoire – exactly 5! divided by (5-3)! = 2! possibilities.

Slide 5

However, now we have counted some combinations multiple times and in this case that is all combinations of three dresses in the same colors. However, there are exactly 3 • 2 • 1 = 3! duplicates; therefore, you have to eliminate all permutations. In this case, eliminate means divide, because of the mentioned 3! = 6 possibilities, only exactly one of each remains. Is this clear?

Slide 6

Let’s think about this again using another example. Surely you know that you play skat with 32 cards. Ten cards are dealt to each of the three players. How many possibilities exist for such a hand with ten cards?

Clearly, the product of the numbers 32, 31, 30, 29, 28, 27, 26, 25, 24, and 23 are again the starting point, thus the number 32! divided by (32-10)! This time we must eliminate all possibilities for ten identical cards in another order and those are 10! possibilities. And as just mentioned, eliminate means dividing by 10!.

You then end up with 64,512,240 possibilities, an impressively large number.

Slide 7

And as we already said, the popular “6 of 49” lottery follows precisely this pattern.

We start with 49 • 48 • 47 • 46 • 45 • 44 possibilities and divide by 6!, thus by the number of permutations of six numbers.

The number 49! divided by (49-6)! and 6! – that is more than 15 million possibilities.

Slide 8

What have we done so far? Well, we have observed three situations that do not have a lot in common at first glance. We have selected three of five dresses, ten of 32 cards, and lastly six of 49 numbers.

From a mathematics perspective, however, we have looked at more or less the same situation. Specifically, we determined the quantity of all subsets of three elements from a set of five elements, the quantity of all subsets of ten elements from a set of 32 elements, and the quantity of all subsets of six elements from a set of 49 elements.

In a nutshell, each time we determined quantities of very specific subsets. And in the process we actually always followed the same pattern.

Slide 9

We have observed a very typical situation in combinatorics, that is, combination without repetition and without taking the order into account.

The quantity of subsets with k elements from a set with n elements (and that applies to 0 ≤ k ≤ n and n≠ 0 ) is n! divided by (n-k)! and k!.

Slide 10

Using this formula, you can now very easily determine how many possibilities Amara has for packing her suitcase. You remember. Amara has 12 dresses and would like to pack four of them in her suitcase.

Then she has 12! divided by (12-4)! and by 4! possibilities, a total of 495 possibilities. That’s enough for more than a year of a suitcase packed with different contents.

Slide 11

Would you like to calculate for yourself? Then here is a selection of exercises that unfortunately are all associated with testing situations.

Slide 12

A new symbol exists for the term n! divided by (n-k)! and k!. You write n above k and draw large parentheses around them. This term is called a binomial coefficient and is read as “n choose k.”

In this term, n is a natural number, as is k , but k can also be zero. And k is at most equal to n.

Slide 13

The name comes from the fact that the binomial coefficient occurs in binomial terms. Do you remember? That is

(a+b)² = a² +2ab + b²       

                        (a+b)³ = a³ +3a²b + 3ab² + b³

                               (a+b)⁴ = a⁴ +4a³b + 6a²b² + 4ab³ + b⁴

                               (a+b)⁵ = a⁵ +5a⁴b + 10a³b² + 10a²b³  + 5ab⁴ + b⁵

for all real numbers a and b.

Slide 14

These are the coefficients very systematically. They form a Pascal’s triangle.

Slide 15

You could also write the coefficients in the following way. Verify this for low numbers.  

Slide 16

That’s all for today. Thanks for being here, thanks for your interest, and until the next time.