How many throws do you really need for a double?
Yes, now we have a mathematical chapter and may also calculate undisturbed. Of course, I assume that you have long since calculated the whole thing for yourself and already know what you would have done. To play along or not to play along? And whoever answered the question with “play along” also gets my approval. The prerequisite for this, however, is that you have a lot of superfluous money. Because the correct answer is, of course, “Don’t play along”.
If this should somehow cause me to be taken aback, then I feel obliged to explain this to you in a comprehensible way. There are, of course, many ways – my children even claim “all” by now – that lead to Rome, but among them there are clever and clumsy ones. One way, of course, is to pocket a lot of money and come and see me. Then I would of course warn you, like the famous black belt holder in karate, but since you have also read the book, I would be happy to play the game. I roll the dice, you collect (or pay, in the less favourable case). My tip would then be that either the money case is empty first, or the understanding is full beforehand.
Another way is to calculate beforehand. And we can just do it together.
In the following diagram you can see quite well how probable a double is in theory. It shows all 36 combinations that two dice can show:
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 1 | 11 | 12 | 13 | 14 | 15 | 16 |
| 2 | 21 | 22 | 23 | 24 | 25 | 26 |
| 3 | 31 | 32 | 33 | 34 | 35 | 36 |
| 4 | 41 | 42 | 43 | 44 | 45 | 46 |
| 5 | 51 | 52 | 53 | 54 | 55 | 56 |
| 6 | 61 | 62 | 63 | 64 | 65 | 66 |
The main diagonal of this matrix, the one in bold, represents the set of all doubles. There are 6 throws that form a double, so 6/36 = 1/6.
In principle, four questions arise in the double-Jürgen problem: Has one side made a mistake here? If one side made a mistake, which one was it? What kind of mistake did this side make? And Gar: Why did a side make a mistake here? Here, a mistake is defined by the fact that if you did the experiment again, you would lose in the long run. So one side had an advantage and if so, which one. Every gambler naturally tries to make good bets. Any money bet on an unspecified future event is a bet. So everyone who played in the game certainly thought they were making a good bet. But who really made it?
Let’s just deal with the questions in order:
1) Did one side make a mistake here?
In all bets, money bets, there is one side that has an advantage. The advantage makes itself felt in such a way that if the experiment is repeated, one side will profit in the long run. Of course, this profit cannot be realised in a single game. You win or lose the one game. In the long run, things look quite different ( See also the chapter: “Simulations”). The advantageous side will (as a rule) prevail. In practice, it would be pure chance if one could actually find and carry out a random experiment where there would be exact equality of opportunity, i.e. neither side would benefit in the long run (even when tossing a coin for equal money, this is not guaranteed, at least in a practical experiment, see also the chapter on “How odds are created”).
Now, in numerous problems it is not quite easy (up to impossible) to find the side that has the advantage. It certainly exists, because the ideal case of equality of opportunity only exists for the somewhat comfortable mathematicians who simply assume ideal cases. Especially in the case of events in the future in which people are involved (sporting events, for example?), at some point it becomes absolutely confusing and not exactly (if at all) calculable. But more on that later. Here we will examine an experiment, a game, that is still quite clear, in which, at least theoretically, mathematics still provides unambiguous answers. One side was wrong, we have to go in search of which side it was. Which side made bad money bets?
2) Which side made the mistake?
To do this, it is now necessary to calculate the probability that the event “one double in 5 rolls” occurs. Of course, the first thing to do is to describe the formulation and the execution of the experiment exactly: Obviously, due to the problem “If a double comes in 5 throws”, not all 5 throws are always carried out. Because: As soon as a double comes, the experiment is stopped and a new one is started. Because one asks, so to speak, and here the mathematicians would be in good hands again, actually after “does at least one double come in 5 throws”. Whether there would then be two or more is no longer of interest. You stop as soon as the first double comes, if any, and start the experiment again (unless you are at the point where one side surrenders or the tournament hall closes).
All right, I’ll keep trying to approach the solution to the problem intuitively: So if you were to make a double every 6 throws and thus had 6 throws instead of 5 time, then you would “always” have a double, so to speak. Since it also makes sense to us intuitively that it certainly doesn’t always happen (take out two dice, roll six times and note whether a double comes up, then repeat the experiment), but it does happen “frequently”, the probability would pretty certainly be over 50%, wouldn’t it? The 50% seems to be calculated simply by (wrongly) adding up 1/6 + 1/6 + 1/6 = 3/6 = 1/2 = 0.5 = 50%. So at about 3 throws you are already approaching the 50% mark. And this is relevant for this problem. Equal money, 100 DM against 100 DM, would be correct and fair if there were equal chances, i.e. if one side had 50% and the other 50%.
Jürgen, however, has cheekily agreed to 5 throws. Somehow I find it at least intuitively conceivable that with 4 throws one already exceeds the 50% (which is already obvious with three attempts), which, by the way, a later calculation will also confirm. But simply saying five attempts is cheeky is the right word.
So the answer is: the party that claimed no doubles in five throws made the mistake. If there are still disbelievers, I feel compelled to do some concrete calculations.
3) How great is the probability and how great the error?
We already encounter all sorts of interesting peculiarities in such a seemingly tiny problem. So does this one: Again and again, when faced with such a question, it is helpful to make the problem as manageable as possible. Experience is often a good guide. When you have solved a few such problems, you get to know the different possibilities that can lead you to your goal and, what’s more, you learn to choose the most favourable one.
I’ll take the arduous path of “forward calculation”. If we want to know how likely a double is to come in 5 throws, it makes sense to add up the probabilities that it will come in the first, the second, the third, etc. Of course, it is indispensable to know the different possibilities. Of course, it is essential to calculate correctly. The calculation 1/6 + 1/6 (as already indicated above) does not work. For he does not come to 1/6 in the second, since he can only come in the second if he has not yet come in the first. The fact that it doesn’t come in the first is obviously the counter-chance that it will come, i.e. the counter-probability of 1/6, which is 5/6. So it comes to 1/6 in the first. In the second it comes to 5/6 * 1/6. Of course, it doesn’t come to 5/6 in the first and in these cases the second is executed, there it is 1/6 again, so together it is 5/6 * 1/6.
So we would have a probability of a double coming in two throws of 1/6 + 5/6 * 1/6. That is 1/6 + 5/36. The fraction inflated gives 6/36 + 5/36 = 11/36. If we now want to continue calculating in this way, we naturally arrive at the correct result. We then calculate that it comes in the third, if it hadn’t come before, so the counter-chance of 11/36 multiplied by the recurring 1/6. So that would be 25/36 * 1/6. Here I note it down: The probability that (at least) one double comes in three throws is 11/36 + 25/36 * 1/6. The 11/36 were the chance that it came in the first or in the second, if it hadn’t come then, so at 25/36 it comes with 1/6 then in the third. The fraction that results is then 11/36 + 25/216. Inflated again, that is 66/216 + 25/216, so in total 91/216.
This is just to prove that although it is not quite 50% after three throws, it is already getting close. 50% would be 108/216, but it is “only” 91/216.
We could continue the calculation and, if there were no errors, we would soon have the desired and correct result for 4 and subsequently also for 5 throws.
But the more unwieldy all the fractions become, the more one begins to wonder whether there might not be a simpler way? The problem one has with the forward calculation can also be described in this way: You have to keep changing the way you calculate. It is remarkable that the linguistically correct formulation also gives one a clue as to which arithmetic operation to use. And here we have to formulate it again to make it understandable: Whether a double comes in 5 throws would be, expressed for calculation, that it comes in the first throw, or, if it does not come in the first, that it then comes in the second, or, if it has not yet come in the first and in the second, that it then comes in the third and … or … and … or. Each “and” stands for an *-operation and each “or” for a +-operation. Not only is this remarkable, but also the fact that the words constantly alternate.
This results in these unwieldy terms, which represent combinations of + and * operations. In addition, the error rate inevitably increases with complicated terms.
Yes, so we achieve the manageability of the problem by simply reversing the question. How likely is it that a double will come in 5 rolls? I’ll just ask: How likely is it that no double comes in 5 rolls. Obviously, it is the opposite probability. So if we know how likely it is that no double comes in 5 throws, we immediately know how often it does come. It is the complement to 1, the counter probability, so 1-(none comes) is the chance that one comes.
But is this question really easier to answer? We formulate: If we want to know how likely it is that no double will come in 5 throws, we formulate that it is the chance that it will not come in the first and that it will not come in the second and that it will not come in the third and that it will not come in the fourth and that it will not come in the fifth. So it’s five times the conjunction “and” which is analogous to the mark. And what is more, it is quite easy to calculate the chance that he will not come in the respective attempt. It is in each case the counter-probability of it coming, that is, the counter-probability of 1/6, i.e. 5/6. If we now insert this correctly, it results in the multiplication 5/6 * 5/6 * 5/6 * 5/6. Five terms, all identical, all equally manageable, no 36ths and no 216ths.
Last little problem we have: We need to subtract the result from 1. So first we multiply out. 5/6 * 5/6 * 5/6 *5/6 *5/6 = 3125/7776. The best way to press that … into the calculator and get a percentage. This number is 40.18%. The difference to 1, that is 1 – 40.18% is 59.82% and that is our result.
The question “how likely is it to throw at least one double in 5 throws” must be answered with “59.82%”. You can see how big the advantage was that Jürgen “snuck”. He wins almost 60% and loses only 40%.
I don’t want to bore you with more mathematics, but if you win on average 6 times out of 10, then the advantage is calculated as follows: 6100 DM – 4100 DM, i.e. 6 times out of 10 you win 100 DM, 4 times you lose 100 DM, that makes 600 – 400 = 200 DM profit on 10 attempts. That would be 20 DM per attempt. This means that every 100 you bet was only worth 80 DM. One run takes about 20 seconds. 10 players. So every 20 seconds that would be 10 * 20 DM = 200 DM, per minute it would be 600 DM, per hour it would be 36000 DM.
Well, obviously I exaggerated with the half hour, but it was certainly 20 attempts. And Jürgen was not unlucky. Maybe there were less than 10 players. But there were quite a few and Jürgen really won a lot.
4) Why did this site make the mistake?
Why did people, some of them very intelligent, good and experienced players, get involved (I don’t know how many of these people had trouble sleeping the next night or slapped themselves in the face, or both)?
Well, the wording was cleverly chosen and our senses are relatively easily fooled (there are supposed to be optical illusions or other forms of sensory illusions; one always puzzles over how it was possible, but at least it succeeds, one is fooled). And intuitively you think: he can’t do it in five throws, he can only do it in six. And for the moment you don’t forget or consider the actually relevant question: With odds of 2.0, I need at least 50% to have an advantage. In this case, do I have 50% or more?
In addition, Jürgen had built up a certain image in the player scene. He was able to parley easily and wonderfully, to entertain people. And he never set great store by being considered a good player. Rather, he was perhaps considered a bit crazy, and sometimes he really did do crazy things (on purpose?). This certainly made it much easier to get the other players to play than if a recognised good player had done it, they would have suspected a snag and maybe thought again for a moment.
I just watched and marvelled. But, if I may mention it, the reason I didn’t play was because I recognised the problem and saw through it. But I didn’t call anyone’s attention to it. Why should I? Jürgen was good, did a trick, exploited his image, that’s just how it goes in business. Everyone has to look after their own money.
5) The general and recurring question
We quickly calculate how likely it is that a double will come in 4 throws. To do this, we first take the counter probability, i.e. 5/65/65/6*5/6 that the double will not come, which is 625/1296 or 48.2%. The counter probability, i.e. the chance that a double will come in four attempts, is 51.8%. So from 4 attempts onwards you would already have an advantage. As I said, intuitively I don’t actually find it astounding, since the naive calculation 1/6 + 1/6 etc. would like to make you believe that somehow.
So, try it out, you’ll be amazed at how many people get the estimation wrong.
So the general and recurring question is how many attempts at an event with a probability of occurrence of 1/n (in our case n=6, but you could also take 100, or 36, for 1/36, which would be the chance of double 6 or something similar) you need until the probability is greater than 50%, until it has occurred at least once.
Answers to this can be found in the chapter “Murphies law”.
