8 From counting to probability. The important role of combinatorics.

Slide 0

A warm welcome to everyone. Let’s talk about math again. Today we will discuss an introduction into the mathematical concept of probability. In particular we will see how combinatorics and probabilities can be linked to one another. However, we will also see an example where this is not possible in principle and we have to rely on an actual experiment.

Slide 1

Let’s reflect on the development of the probability concept. The focus is on random experiments. As the name already says, the outcome of such an experiment cannot be predicted. When we flip a coin or roll a die, we don’t know the outcome in advance. However, we do know that only certain outcomes are possible, for example, head or tails or one of the numbers from 1 to 6.

In detail we can also specify what we’re interested in. For example, there are games in which the only concern is whether or not a 6 turns up. The German game “Mensch ärgere dich nicht,” similar to the American game “Trouble,” is one example you are probably familiar with.

We group the possibilities in a set of outcomes that is usually symbolized using the Greek letter Ω.

Slide 2

I know that sometimes a Greek letter like the basically harmless Ω can trigger anxiety. Let’s look as examples, and then everything will become quite simple and clear.

Let’s assume we are rolling a normal die. We might be interested in which of the numbers between 1 and 6 come up. Our set Ω₁ would then include exactly these numbers.

Let’s look again at the game “Mensch ärgere dich nicht.” At the beginning, we’re interested in two results, specifically 6 or a number other than 6. You can see the possibilities in set Ω₂.

And of course it might be that you want to roll an uneven or even number. This example is shown in set Ω₃.

Even though this may appear abstract, there is an important consequence. In the first example, we have six possibilities that, in the long run, should all occur equally often. In the second example, there are two possibilities, of which one is significantly more probable. Ask the children about this! And in the third example, there are also only two possibilities, but they should be equally distributed.

In principle, it doesn't matter whether you flip a coin or draw a ball from an urn. It is always possible to define which outcomes are of interest in a certain situation.

Slide 3

A soccer match is also a random experiment. There are three possible outcomes, namely team 1 wins, team 2 wins, or the teams end in a tie. Don’t tell me that this doesn't apply to teams like Bayern Munich or Real Madrid. Even they have occasionally unexpectedly lost against seemingly harmless opponents. How boring soccer would be without this possibility.

The basic idea of the random experiment also shows up here: There are clear results known in advance. However, the result that will occur cannot be influenced and is not known before the experiment is conducted.

Slide 4

What do we understand by probability? Let’s try to answer this question as we listen to Bem and Favour. We will consider three situations in which they talk about likelihood.

Favour’s sister wanted to stop by this afternoon. Now it is almost 6 p.m. and she isn’t there yet. “She probably forgot the date,” says Bem.

“Let’s eat without her and flip a coin to decide who will wash the dishes,” Favour suggests. “At least then I have a 50% chance of not having to do it.”

“I have a thumbtack here,” says Bem. “Let’s use it. If it lands on its side, then I will do the dishes; otherwise, you will.”

Slide 5

The first situation involves a guess. You cannot likely explain Favour’s sister mathematically because she could have various reasons for failing to appear and the reason cannot be evaluated in principle.

Slide 6

The second situation is simple. You can theoretically assume that a coin has an equal chance of coming up heads or tails after being flipped. And instead of “chance,” we speak of “probability.”

Slide 7

The third situation is also not complicated. A thumbtack can land on its head or side and we have to determine the frequencies experimentally for this. However, there is no theoretical assumption for the result, but the situation is basically manageable and suitable for the mathematical term of probability.

Slide 8

The counting procedures of combinatorics are important tools for determining probabilities.

Let’s use examples to consider how we can systematically go from counting to probability. This time we will first draw balls of different colors from an urn.

Slide 9

We will experiment with various urns, always using the same method: We will take a ball out, note down the color, return the ball to the urn, and take another out. Urn 1 contains two balls, one blue and one red. Urn 2 contains two blue balls and two red balls. Urn 3 contains three blue balls and two red balls.

Slide 10

Urn 1 is very simple and in principle there are three results: You can draw two red balls, two blue balls, or one red and one blue ball.

BUT: There are two possibilities for the third case, thus first red and then blue or first blue and then red. For the two other cases, there is only one possibility each of the four total possibilities.

Slide 11

We can clearly go from these combinatoric observations to the term of probability. We use “p” for “probability” and set p(two red) = ¼, p(two blue) = ¼, and p(one blue and one red) = 2/4 = ½. That’s reasonable, right? Here we’re assuming that when drawing a ball without looking, each ball has the same chance. Instead of “chance,” we say “probability.”

Slide 12

Let’s look at urn 2. The first time, we can take out one of the four balls. And now we overlook the fact that they cannot all be distinguished from one another. The second time is the same; we draw one of the four balls again. In total there are 4 • 4 = 16 possibilities. We draw two blue balls four times, two red balls four times, and one blue and one red ball eight times in any order.

Slide 13

Not surprisingly, we come up with the same probability as for urn 1. It is p(two red) = 4/16 = ¼, p(two blue) = 4/16 = ¼, and p(one red and one blue) = 8/16 = ½.

Slide 14

We can also easily handle the last urn using counting. As a reminder: We assume that each of the five balls in urn 3 can be drawn with equal probability. If you simply count and don’t think structurally, this of course doesn’t change the result at all. There are a total of 5 • 5 = 25 possibilities, whereby in 2 • 2 = 4 cases both balls are red, in 3 • 3 = 9 cases both balls are blue, and in 3 • 2 + 2 • 3 = 12 cases one ball is red and one ball is blue (and here the sequence doesn't matter).

Slide 15

The probabilities are set accordingly for urn 3 as well. It is p(two red) = 4/25, p(two blue) = 9/25, and p(one red and one blue) = 12/25. And anything else would run counter to common sense, right?

Slide 16

Let’s summarize some facts that became clear in the examples. Everything is simple, but it will be important, not only today but also in principle when working with probability.

First, we draw either a red or a blue ball in each instance. What else? In other words: We know all possible outcomes.

We are then able to assign a theoretical probability to each combination of balls. We contend that this is reasonable.

Finally, we worked with three different urns. The experiment setups were basically identical; we drew two balls in succession. The underlying possibilities were also identical because you could draw two red balls, two blue balls, or one red and one blue. Although the probabilities differ, they always remain constant in one respect:

     p(two blue) + p(two red) + p(one blue and one red) = 1.

Of course, because altogether there were 8 of 8 possibilities, 16 of 16 possibilities, or 25 of 25 possibilities, or in general, n of n possibilities. If you write this as fractions, you have 8/8, 16/16, 25/25, or n/n, and the result is always equal to 1.

Slide 17

But what do we do about the thumbtack? There is no theoretical assumption in this case. So I really tossed the thumbtack 1,000 times: It landed on its side 367 times and on its head 633 times.

This suffices for an empirical probability: p(head) = 633/1,000 and p(side) = 367/1,000.

But I fear that this applies only to this box of thumbtacks and my tossing method (10 thumbtacks in a dice cup). The results on the Internet show rather a 30% chance for landing on its head.

Slide 18

We can summarize in this case as well and come to similar basic considerations as for the three urns with red and blue balls.

In all cases, the tossed thumbtack lands on its side or head. So here we also know what the possible outcomes are.

We can assign a probability to these two outcomes; they were determined empirically in this case.

It doesn't make a difference whether or not this probability is confirmed through more tosses. This always applies:

     p(side) + p(head) = 1.

And of course this works for any 1,000 tosses.

Slide 19

That’s all for today. Thank you for being here and until next time.