In order not to be exposed as a charlatan at an early stage (no doubt that I am), I must once again include a somewhat more strenuous chapter. One with some arithmetic and such. We simply have to deal with these two terms mentioned, I simply cannot spare the reader and interested sports bettor that.
So what is an or also the expected value? And I am not presenting you with the mathematical definition, but trying to explain it intuitively. So when we pick up the dice again, what do we expect to roll? Well, I don’t expect anything. Let’s see what comes up, right? Yes, correct. But what if we talk about the sum of the eyes that we expect? Well, even then, it’s the same. But if we roll the dice 100 times? Do we then expect any sum? Yes, now we are getting closer. You could now take a dice and try it out. Do you enjoy it? Go ahead. I did it too. And the result was: 15* 1, 162, 213, 154, 175 and 166. So I rolled a total of 151 + 162 + 213 + 154 + 175 + 16*6 eyes. That’s 15 + 32 + 63 + 60 + 85 + 96 = 351. If I divide that number by 100, for the 100 rolls, 351/100, I get 3.51. That’s my average rolled eye total. And what was the expected value?
If we assume that we are dealing with an absolutely balanced dice, i.e. a LaPlace experiment, the expected value would be the multiplication of the probability of occurrence by the numerical value itself, thus: 1/6 * 1 +1/6 *2 + 1/6 *3 +1/6 * 4 +1/6 *5 +1/6 *6. And the result is indeed 3.5. You can make it even simpler here, and simply add up the eyes and divide by the number of possible outcomes: So 1+2+3+4+5+6 = 21. And 21/6 = 3.5. And even my experiment stuck to this, which is pure coincidence. Or is it? Try it out. There could also be larger deviations sometimes.
Multiplying the probability of occurrence by the eye value can also be done linguistically like this: One-sixth time comes 1 and One-sixth time comes 2 and One-sixth time comes 3 and One-sixth time comes 4 and One-sixth time comes 5 and One-sixth time comes 6.
Character
The example was deliberately chosen to be simple and has certainly been known for a long time. But now I will add a very small variation to the experiment. You are given the following task: roll the dice, roll the dice, add up the eyes you roll. But if you roll a 1, everything is deleted, you have 0 eyes. Then the following question arises: Up to what eye sum should you keep rolling the dice to get as far as possible? Maybe defeat someone else?
Now it becomes much more complicated. The trick now is not to look for the right formula for the calculation (at least I haven’t learned this art, so I can’t teach it either) or even to know it, but to intuitively find the best possible simplification of the problem to make it manageable. But you may also answer intuitively to the following question, for example: You have reached 15 (i.e. you had a 3, a 5, a 4 and then a 3 again, for example). Would you now continue to roll the dice? A 1 could come up, and then everything would be gone again! So what do you do? In an emergency, calculation helps. In principle, we now ask for the expected value if you roll the dice again, and this is calculated in the same way according to the above formula, only with the slightly modified values: 1/6 * 0 + 1/6 * 17 + 1/6 * 18 + 1/6 * 19 + 1/6 * 20 + 1/6 * 21 (to check this, even with more complex problems, you should always add up the probabilities so that they add up to 1. Then you have probably not forgotten any case).
The modified numerical values result from adding 15 + rolled eyes. And in case you roll a 1 you have 0, because then the game is over, your value is fixed. You have scored a 0. So, the value that results from the above calculation in this case is 15.833.
So, if you roll the dice at 15, you score 15,833. You have the choice between 15 if you stop and 15,833 if you roll the dice. That’s not a difficult choice, is it? The answer is, you have to roll the dice. And given this reasonably handy representation of the problem, it is of course easy to find the correct answer to question 1. It turns out, as you can easily see for yourself, that you have to roll the dice at 19 (expected value is 19.167), at 20 it doesn’t matter, you get 20 whether you roll the dice or not (calculate: 1/6 *0 + 1/6 * 22 + 1/6 * 23 + 1/6 * 24 + 1/6 * 25 + 1/6 * 26, that gives exactly 20). And you absolutely have to stop at 21. The expected value at 21 is 20.833 when you roll the dice, so less than 21.
So, always calculate an expected value like this: Calculate the probabilities of occurrence for the individual events, write them down, the sum must add up to 1. And then multiply by the numerical values that come out in the random experiment, i.e. the possible outcomes.
Two envelopes
All right, and now I’ll show you an example right away, which will make even a mathematician doubt the correctness of this calculation method. But the example is only meant for entertainment:
A thought experiment: I put two envelopes in front of you. You can choose one of these envelopes. Each envelope contains a sum of money. The amounts are different. One envelope contains twice the amount of the other. You now know that much. Now you can take an envelope. You can open it and look inside. Then I will ask you if you want to keep this one or if you would rather have the other envelope. You are hesitant, I assume. Why swap? Let’s say you found 50 euros. And after studying this chapter, you naturally say to yourself, I can work it out. Let’s see if I do better by swapping or by keeping.
So you write down the figures: 50% you get half of 50, 50% you get double, that’s how it has to be. So the amount in the other envelope is either 25 euros or 100 euros. Now we can easily multiply out. In half of the cases you lose 25 euros, in the other half of the cases you win 50 euros. As an arithmetic operation, it looks like this: 0.5 * (-25 Euros) + 0.5* (+50 Euros) = +12.5 Euros (in this form of calculating the expected value, I have only taken into account the change that occurs when you swap). So, you decide to swap, sure, logical. You have duped me with the power of mathematics. 12.5 euros earned, extra.
But something seems strange, doesn’t it? Why should I suddenly be able to profit if I swap? What if I had taken the other envelope? Would I then have had to swap too?
And now it gets even better: How would you have to decide if you held the envelope unopened in your hand? You now have an amount x in your hands that is unknown to you. And the other envelope contains either the amount 2x or the amount 1/2x. So then we calculate again: 0.5* (-1/2 *x) + 0.5 * (+1x) = +1/4 x. So, if you were to swap, the opened envelope or even the unopened envelope, you would benefit 1/4 of the amount contained. So, you swap, wisely. Thank you, mathematics.
But now you hold the other envelope in your hands and greedily want to open it. I put the brakes on your eagerness for a moment, and ask you to think again for a moment whether you really wanted to open this envelope, or perhaps you would rather open the other one? You pause, how smooth is ice? The calculation yields the same result: you have to swap again. I’m going to go to sleep…
You don’t have to get grey hairs about it. The answer is obvious and plausible: it doesn’t matter which envelope you take. The only question that remains open is: How can such a simple example succeed in leading mathematics ad absurdum? And that brings us back to the field of philosophy in a flash. Or should we suddenly find a confirmation of the well-tried proverb: “A gift horse…”?
Philosophically, it is like this: you have been given any amount of money just like that, out of nothing. And whether this amount is 1/2 x or x or 2 x is absolutely irrelevant. You don’t even have a clue whether it is 1 euro or 100,000 euros. So, grab it, don’t calculate.
But how relevant this example really is has hopefully not yet occurred to you during my description, otherwise I would have nothing more to tell you. In fact, it is relevant almost on a daily basis, yes. Well, I have to let the cat out of the bag: If you also watch the quiz shows? They are interchangeable in terms of structure, but they are exciting and offer a brilliant form of evening entertainment. You see a friendly, knowledgeable or nicely parlating presenter, and usually two more or less sweating candidates. Funny questions and also interesting questions are chosen. “Yes, I know that, C) is right!” And one even educates oneself, doesn’t one?
So, let’s take Jörg Pilawa. The candidates log their winning levels. The second one at 20,000 euros. All right, a bit cautious, but the candidates have to be careful about their level of knowledge. Good, the first questions make you smile, the first prize level is reached. All the jokers are still there. The way to the second prize level is a bit more difficult. Good, one correct guess with the help of the moderator, one swap. But we made it. The next question on 30,000 euros is free, falling back is not possible. The candidates have already achieved everything they dreamed of. 20,000 euros, that was their dream. The next question is easy, almost free, they have 30,000 euros! Now the exciting question: to continue or to stop?
And here we are already at the envelopes. What’s inside, what question? Well, we can only speculate about the question. But you can still do some quick calculations: in a bad case, you can fall back to 20,000 euros (but 5 minutes ago you would have been absolutely satisfied with this amount), so from 30,000 euros to 20,000 euros, that’s a loss of 10,000 euros. And you can win 20,000 euros, from 30,000 euros to 50,000 euros, that’s the division here. So you should have a 1/3 chance of answering the next question correctly. Why first? I just want to prove it briefly and not “infer”. If you are right in 1/3 of all cases with the answer, you have earned 20,000, if you are wrong in the other 2/3 cases, you have lost 10,000. So we multiply: 1/3 *20,000 + 2/3 * (-10,000) = 0. So if your estimated chance exceeds 1/3, you would have to keep playing. And now let’s think even further: there are 4 possible answers to any question, even without any idea. So with pure guessing, you would already have a chance of 1/4 (instead of the required 1/3). But if you can still swap, the chance of having even a hint of a clue for at least one of the questions increases enormously. And then there is the fact that you can almost always rule out one answer immediately anyway. So we have the one required third. Then there is the chance that the moderator, benevolently, helps at least a little. (“How do you come up with B now? Can you explain that to me again for a moment?” I’m guessing B didn’t.) We’re over 1/3, virtually certain.
But almost everyone stops. They make a mistake. They’re scared or whatever it is. At least I haven’t seen/heard any candidates do the math.
By the way, the “equity” basically means the equivalent monetary value. So let’s assume that the candidate has a 50% chance of answering the following question correctly (swap included), then the equity would be calculated as follows: 50% of the time he solves the question, so he would win 50,000 (stop and go home) and 50% of the time he would answer incorrectly, so he would only win 20,000 (then he has to go home). His equity at this moment is therefore 0.5 * 50,000 + 0.5 *20,000 = 35,000 euros. The monetary value given away by quitting would therefore be 5,000 euros. He would rather take 30,000 than 35,000, but the 35,000 is just speculative. The probability of finding the right answer also varies from person to person, I have only estimated it, but it is quite realistic. You almost always have a small part by excluding a possible solution, another mostly by the possibility of swapping and the rest by moderator help. So 50% is realistic.
But, to make the confusion complete, even here there is a calculation error. Have you found this one too? Yes, fact. Because in the following situation he would not have 50,000, but the value that he would be entitled to if he behaved correctly. And we would still have to calculate this. Because the game goes on. “50,000, a nice sum, do you want to continue playing?” And we have to calculate again here. Did he have to use the joker or not? And now comes the jump from 50,000 to 100,000. So you would risk 30,000 here to be able to win 50,000. And so on.
And if you then calculate again and realise that you have to continue, then the reason would be that the equity is higher if you continue, i.e. higher than 50,000 euros. So in the previous calculation, the calculation for the 50,000 euro question, you would already have to use this higher figure in order to find the correct answer to the question “Should I tackle the 50,000 euro question”. Let’s take another calculation example : you estimate your chance of answering the 100,000 euro question correctly at 40%. Then we would have to use:
0.4* 100,000 + 0.6 * 20,000. And that gives 52000 euros. So your equity on the 100,000 euro question would be 52,000 euros. Stop and take 50,000 or continue and take an estimated 52,000 resulting from the outputs 20,000 and 100,000 respectively with the associated probabilities.
The correct calculation for the 50,000 Euro question, i.e. the first and original calculation, would actually look like this: 0.5 * 52,000 + 0.5 *20,000 = 36,000 Euro. Your equity even improves compared to the previous calculation, because you would probably have to continue even with the 100,000 euro question.
But the questions are getting really difficult. My brothers should not stop under any circumstances! I, on the other hand, would have to stop before that. Because with my absolute cluelessness I wouldn’t even reach the required 33% (1/3). And that, unfortunately, is not even coquetry (unless I get a football question). But at least I can see you sitting there and doing the maths for Mr Pilawa. I’m looking forward to it!
