#### Independence of events

The title sounds really dry. It almost makes you feel like you’re in school, doesn’t it? And if I now pronounce the word “set theory”, the book is guaranteed to end up in the oven for good. Lucky, you don’t have one? I recognise your effort.

So the purely mathematical formulation is this: For independent events, the individual probabilities of occurrence are multiplied for the total probability of occurrence. Boom! But if we express it very simply or look at the simplest example, it becomes clear relatively quickly: Two heads in a row on a coin toss: 1/2*1/2 = 1/4. That’s right, isn’t it? To illustrate, one could note the 4 possible sequences, which are all equally probable (for doubts about this, read other chapters): 1st toss number, 2nd toss number; 1st toss number 2nd toss head; 1st toss head 2nd toss number; 1st toss head 2nd toss head. So the event “1st roll heads 2nd roll heads” is one of the four possibilities, all equally likely, 1/4, true. When rolling the dice, it looks something like this. Two 3s in a row, 1/6 *1/6 = 1/36. And if we throw two dice at the same time, each number is also independent of the other. That’s the theory.

So back to the “independent events”: Maybe you have already bet yourself, possibly on football matches. If you have even made a combined bet, you will have noticed that you have to multiply all the individual odds together to calculate the odds. And the law of independent events is responsible for this.

We have established above that the ratio between odds and probability is the reciprocal value. Since we have to multiply the probabilities, we also have to multiply the odds. Many players then always combine a bunch of favourites. So Bayern, (1.30) Liverpool (1.45), Chelsea (away, 1.65), ManU(1.20), Arsenal(2.20, also away, good opponent), Inter Milan (1.50), Real Madrid(1.25) and FC Barcelona (1.40). Then you get a gigantic payout ratio of 1.30*1.45*1.65*1.20*2.20*1.50*1.25*1.40 = 21.55. So 21.5 times the money.

That sounds great, doesn’t it? Fantastic, and they all win, right? Yes, that’s where the inexorable power of mathematics strikes. The probability of ALL predictions occurring at the same time can be determined by multiplication. And that is then really in the order of 1/21. Or do we want to look at it even more closely? So let’s look at each odds with an assumed probability of occurrence. And we also assume that the oddsmaker was a good one, i.e. that his estimates were correct. Then he only pays a 1.30 if the fair odds are higher, so let’s say 1.35. That would correspond to a probability of 1/1.35 = approx. 0.74. Further, the 1.45 comes from 1.50, 1/1.50 = 0.667, the 1.65 comes from a 1.72. 1/1.72 = 0.581, the 1.20 comes from 1. 23, 1/1.23 = 0.813. Next the 2.20 comes from 2.4, 1/2.4 = 0.417, 1.50 comes from 1.57, 1/1.57 = 0.637. The 1.25 comes from 1.29, 1/1.29 = 0.775 and the 1.40 again from 1.45, 1/1.45 = 0.69.

Then, according to the rules, we multiply the probabilities of occurrence. So 0.74*0.667*0.0.581*0.813*0.417*0.637*0.775*0.69= 0.033. So the total probability is 3.3%. The correct payout for this would be 30.2, but you only get 21.55. So we get a whopping disadvantage of -28.6% for the combination bet! (Calculation under “expected value”, but here is the formula: 0.033*(21.55-1) – (1-0.033) = -0.286). So although you would only have a small disadvantage in the percentage range for the individual games, you would have a very big disadvantage on the combined bet. The reason is unfortunately this: When the odds are multiplied, not only is the chance of occurrence multiplied, but the disadvantage is multiplied at the same time. This has always been an El Dorado for bookmakers. Instead of 3-4% profit, they could sometimes calculate with up to 20% or even more profit margin. Of course, this only applied to (relatively) good bookmakers who had to deal with bad players.

For me, this circumstance was of use the other way round. Because I only ever picked out the games that gave me an advantage according to my calculations. Then I combined these games (I did this in so-called system bets. For example, you select 6 games and combine them in all possible combinations of 3, which is a total of 20 combinations). And just like the disadvantages, the advantages also multiply.

Oh, I was talking about the independence of events and set theory. So “independent” are always those events that have no intersection or, in mathematical terms, an empty intersection.

An example of dependent events would be: You roll two dice again. Now we calculate the probability for the event “at least one of the two dice shows a 6”. You calculate the probability by subtracting the counter probability of this event from1. So the event “two times no 6” is easier to calculate. One time no 6 is 5/6, 2 times no 6 is 5/6 * 5/6, which is = 25/36. Subtract the whole from 1, for the counter probability, 1 – 25/36 = 11/36 (more on this elsewhere). So the probability “at least one die shows a 6” is 11/36.

Now we consider the event “The sum of the eyes is exactly 9”. The event occurs if the first die shows a 5 and the second a 4, or the first a 4 and the second a 5. Likewise if the first a 3, the second a 6, or the first a 6 and the second a 3. That is 4 possibilities out of 36. So 4/36 = 1/9.

Now we multiply (??) the probabilities and get 1/9*11/36 = 11/324. So about 1/30 is the multiplied probability “at least one die shows a 6” and at the same time “the eye sum is 9”.

Of course, this is not true. Firstly, it is obvious that the events are not independent and secondly, we already know the probability. There are exactly 2 cases out of 36, i.e. 2/36 or 1/18, that the combined event occurs (3-6 or 6-3). In plain language, the 9 comes more often when one of the two dice shows a 6.

If I had asked how often the sum of the eyes is 5 and one of the two dice shows a 6, we could have saved ourselves any calculation effort: That would be impossible.

I once showed an Italian (my friends, by the way, and I mean that, my daughters are called Chiara and Giulia, for example) a betting slip on how to bet and play. Then I said to him that he could also combine everything to get a higher payout (I didn’t mention the reduced probability of occurrence for the time being). Then he came the next day and asked me (European Championship 92, Sweden was playing against Denmark) if he could combine “Sweden win” and “1:0 for Sweden”. That was really a great idea. The rate on Sweden winning was 2.0, the rate on 1:0 was 7.0. Combined, that’s 2*7 = 14. So in one fell swoop, he raised the rate on 1:0 from 7 to 14!

Unfortunately, the bookmaker saw through him. Whenever the score is 1:0, Sweden has also won… this is an example of maximum dependence. The amount of games that end 1:0 are completely within the amount of Swedish wins. Do you see the quantity diagram in front of you?