 Hi, I'm Zor. Welcome to Unizor Education. This lecture is part of the advanced mathematics course presented on Unizor.com and it belongs to a section about theory of probabilities. Now this is basically a very simple lecture which I would like to just basically exemplify a couple of concepts of theory of probabilities and to introduce you into a state of mind which anybody who solves the problems related to theory of probabilities should actually be. Now this state of mind is relatively simple. To evaluate the probability of something, you really have to analyze your experiment. What kind of outcomes, what kind of results of the experiments can be called elementary events. So these elementary events are supposed to be in some way symmetric. Like if you are rolling the dice, then every one result, like one on the top, two on the top, three, four, five or six, they are equivalent and they are symmetrical and they all have equal chances to occur. And that's why we assign equal probabilities to these events. And that's why each one of them has the probability of one divided by total number of these events. So that's the very important part. And a couple of examples which I have would illustrate how important it is. So let me just start with real examples. And these four examples are related to the dice games. So we will play with dice. A couple of games which some of them, you know some of them maybe you don't. Anyway, the first one is the game of craps. And I'm not talking about all the details of the game. I'm just extracting one particular aspect of the game. I know that if you throw two dice, you roll two dice and if the sum of numbers is equal to seven or 11, basically somebody won't, somebody wins. It doesn't really matter who wins, how much they bet, etc. My concern right now is that there are two dice rolled and we are calculating the sum of the numbers on the top. And there are supposed to be either seven in the sum or 11. So my question is, okay, this is an event. What's the probability of this event? So what's the probability to having seven or 11 as a sum of two numbers on the top? Okay, as I was saying before, what's extremely important in all these problems is to establish what are my elementary events. If you are rolling two dice, every combination of two numbers on one dice and on another dice represents an elementary event. These elementary events are all different, obviously, because we are talking about different pairs of numbers. So, for instance, number three and five. One dice is three, another is five. That's an elementary event. Elementary event two and two. That's another elementary event. Elementary event five and three, which is different from this one, by the way. So these are all elementary events. Now, how many of them are? Well, you have six different choices for the first number and six different choices for the second number. So basically, you have six times six, 36 different elementary events, which cover completely all the possible outcomes. So there are 36 possible results. They're all different and they're all having equal chances because we're supposed that the dice are symmetrical, right? So these are all symmetrical events, which means that the probability of each one of them is equal to 136. Now, my question is what's the probability of my event which I'm interested in, which is the event when the sum of these two numbers is either 7 or 11? Okay, well, actually in this particular case, the easiest way is just to count. So let's just consider. If you have one, two, three, four, five, six, one, two, three, four, five, six, this is the first dice and this is the second one. So when do we have seven? Seven we have one and six, two and five, three and four, four and three, five and two, and six and one. So one, two, three, four, five, six different cases when I have a sum of numbers equals to seven, six. Now, when do I have 11? I cannot have 11 if one of them is less than five, right? So it's five and six and six and five, two cases. So eight out of 36 elementary events make up my event. So all others, all other points in this table, so there are 36 different points on the crossing of all the different rows and columns, right? And I'm interested only in these six and these two. So they represent eight out of 36 elementary events, which means it's like you're calculating the area of some shape, right? So which means that the probability is equal to 836 or two-ninths. Okay, so that's the answer. With the probability of two-ninths, we will have a sum of two numbers on the tops of two dice which I am rolling equal to seven or 11. So an average out of every nine rollings of the pair of dice, I will have this winning combination. On average, doesn't mean that on every nine, I will have two. And it doesn't mean that on every 900, I will have 200 of these combinations, but somewhere close to this number I will have. So that's the first problem. The second problem is there is a game, a dice game called I wish I can pronounce it correctly. Yatsy or something like this. It's spelled like this. Yatsy or Yatsy. I don't know how to say it. Anyway, there are five dice. Actually, it's very close to poker, the card game, but it's played with dice. So we have five dice and different combinations which are in a way similar to the combinations of the poker, like a pair or two pairs or something like this. Well, are events which we are interested. And in this particular case, I'm interested in the combination which is called Full House. Very analogous to Full House in the poker game of cards, where you have three cards of one rank and two cards of another. Here as well, I have three dice out of five, have one number. Let's say five, five and five. And two other dice have also equal number, but different than this one. Let's say two and two. So this is the Full House. Question again is, what's the probability of this particular event? Now, as usually, we have to start with our elementary events. So what are elementary events? When we are throwing five dice, okay? We are rolling five dice. Each one has six different combinations from one to six, which means that the total number of all the combinations of, let's call it a quintet. I think quintet is five, right? So a quintet of dice. So every time I'm rolling a quintet of dice, I have some number, one of the six numbers on the first one, one of the six numbers on the second, etc., etc., so it's six to the fifth degree different combinations of my values on the tops of the quintet of dice, right? Six to the fifth degree. Now, all of them seems to be completely equal chance. So the probability of each elementary event must be the same. And since all of them comprise the total probability, which is always assumed to be one, we are talking about frequency, right? So what's the frequency of something happening? Well, obviously something always happens. So it's one, right? So every one of them has the probability of six to the fifth degree. Okay, so I have six to the power of five different elementary events. All of them have equal chances to occur. The probability of each one is one to the over six to the power of five. Now, question is, if we are talking about full house, how many of these elementary events make up a combination called full house? If I will find out the number of these combinations, I'll just multiply it by this n, right? So what's the n? This is a purely combinatorial problem. And as I said many times, combinatorics is a very, very important part of this course, which was preceding the theory of probabilities. And so let's just calculate how many different combinations of five numbers from one to six each are comprising the full house combination. So I have three dice with one value and two dice with another value, right? So first of all to count the combinations, I have to somehow position these three and two. Well, actually I have to position only three because the two will be left, right? So out of five, I have to choose my three positions where I will have one number. That determines the other two positions with another number. Now, what's important right now, so my positions are already fixed, okay? So I will use this as a multiplier because it doesn't really matter which positions. But let's say the position is fixed. Now for each position out of this number, for each position my choice for three numbers having the same value is basically from one to six, right? So I have six choices. Now the other two, they're not supposed to be equal to these ones, right? So I have to choose the value for the other pair from the remaining five numbers, right? So the combination of all these, the product of all these, give me the total number of different combinations. So I choose certain number of triplets out of my quintet. Then as soon as I chose a particular triplet, I have to choose what exactly the number is there and I have that many numbers to choose from. And then I have to choose the number for the remaining pair. Now this is 5, 4, 3, 1, 2, 3, which is 10. So this is 30. So my answer is, my probability is, of this event, is 30. Or if you wish 5, 6, 4, right? And I think I have, no, not 30, sorry, 300. Yeah, 300. So it's 10 times 30, right? So the probability is 50 over 6 to the 4. So I have an approximate value, actually. 0.03858, which is less than 4%, as you see. So the probability of the full house in the game of Yatsi is less than 4%, 3. Whatever percent, okay? So that's the second problem. I have four problems and the third and the fourth problems are related to the, well, this is actually a historical fact. Apparently it's written in many textbooks on theory of probabilities. I learned it very, very long ago when I was reading some kind of encyclopedia for children published in the Soviet Union in the 1950s. Anyway, it's about the French person called Frenchman. His name was Cavalier de Mer. I think it's pronounced de Mer. I'm not sure, maybe. I think it's de Mer. So the Cavalier or Chevalier de Mer. So he was a gambler, very simply. So he invented an interesting game. He said that, okay, let me throw a dice four times in a row, okay? So we have four rolls. And he's saying if out of these four rolls I will have one particular number. I think it was number one. Doesn't really matter which number. Let's say it's number one. So if number one is at least once, I win. So out of four rolls, if at least once there will be number one, he wins. And people did play with him. Now, he was a smart guy. So he decided, just think about it. I mean, the chances of having one on one roll is one six. I mean, he understood that, obviously. So approximately out of every six games, one will be on average, I will have my one, right? My number one on top. Now, then he decided, okay, if out of one game, my chances are one six, then out of four, if I roll it four times in a row, well, probably it would be four six, which is greater than 50%. And I will win. Well, guess what? He was really winning. Regardless of the fact that this is an invalid calculation and I will explain why, he was really winning. But let's do it a little bit more mathematically. So let's evaluate the problem. Well, first, let me criticize this. Why this is not right? Well, let's just think about it. If the probability is multiplied as the number of experiments is multiplied, then if we will make five experiments, it will be five six. If we will make six experiments, we will roll the dice six times. It will be six six, which is one, which means that it's like always out of six rolling of the dice, I will always have at least one time when number one will be on top, which is definitely wrong. But what if if I'm rolling it seven times or eight times, the probability goes seven six or eight six, which is greater than one, which is just impossible, right? So the whole mathematics of this is just wrong. This is not the way how it's supposed to be calculated because it definitely reaches to some absurdity. And okay, so what's the right way to calculate it? Okay, so let's just think about what's the right way to calculate the probability of having number one on top if you roll the dice four times. Okay, as usually, we have to start with the set of elementary events, right? So what's the set of elementary events when you're rolling dice four times? Well, it's a quartet, right? Quartet of numbers. Let's say two, two, five and four. And this is a quartet of numbers as well as any other quartet. So all these sets of four numbers are the results of rolling the dice four times. Now, obviously they are all equally probable. Number of them is six to the fourth degree, right? Six choices for this, for this, for this and for this. So I have to multiply them. So that's the number of elementary events. They have all equal chances of occurrence, so the probability of each of them is equal to one over six to the fourth. So all I have to find out is how many of these elementary events make up my event to have at least one of these numbers to be equal to one? Well, what's actually easier is to count how many of these do not contain number one. Because if they do contain it can be one number one or two number ones or three number ones, I mean it's all separate. It's easier to calculate no number ones, all right? Okay, no number ones means that I have only choices from two to six for the first, from two to six for the second, two to six for the third and two to six for the fourth position, right? So I have only five choices for each position. So number of events is five to the fourth. These are all events which do not contain number one among them, right? So if I will subtract from the total number of quartets, the number of quartets where the number one is absent, then I will get the number of quartets where there is at least one number one, okay? Now, if I would divide it by a total number of events, that would be my relative frequency. That would be my probability of having my event, or one minus five six to the fourth, or again I have a number calculated 0.5177. Now, so as you see, the probability is not the one which demerer calculated, which was four six as he was thinking. Well, this is two-thirds. This is much smaller, this is much smaller than this one, but what's important is it's still greater than 50 percent and that's why he was still winning. Now, he was encouraged by that and now we go to the problem number four. Problem number four is about the continuation of his saga, his history. You see, he was winning too much and the problem is people stopped playing with the game. They kind of realized that something is wrong with these games whatever he is offering. So, they stopped playing with him and he was actually trying to persuade them to play, so he invented another game. Well, what's the other game? Okay, he said, you know what, instead of rolling a dice four times, I will roll a pair of dice 24 times. Now, his logic was exactly the same as before. He was thinking, okay, my pair of dice have 36 different combinations, right? Six on one, die and six on another die, so it's six by six, it's 36 different combinations. So, the probability of every one particular combination is 136. Now, I'm interested, he said, in one particular combination of two numbers, two ones in this particular case on the top, number one on the top of the first die and number one on the top of the second die, all right? So, he's interested in one particular combination, one-one on two dice and if you will repeat it 24 times according to his logic, it would be 2436, which is basically the same two-thirds, the same as before. So, it's from his perspective, it's the same problem as before, the same kind of rules as before and he would probably be winning the same as before. Well, guess what? He started losing and he lost a lot. Enough for him to persuade his acquaintance, his name was Blaise Pascal, very famous French mathematician. He actually wrote a letter to Blaise Pascal and basically saying, hey, something is wrong, I don't know what's wrong, could you help me? Well, Pascal did certain calculations and he also got involved another guy called Pierre Fermat, another very famous mathematician who is actually the author of the great Fermat theorem. So, they were actually exchanging letters, etc. And basically, that was the beginning of the whole theory of probabilities, believe it or not, or at least that's what legend says. Well, anyway, obviously Pascal did the correct calculations. So, what are the correct calculations? Let's just now do it the way how to do it properly. So, number one, what is a set of elementary events which we are talking about? Well, if you're throwing a dice, a pair of dice 24 times, the elementary event is 24 pairs. There are 24 pairs of them. So, the probability, well, the number of pairs actually is 36, right, 6 and 6, but the number of pairs, if you're repeating this 24 times, would be 36 to the 24th degree, right? 36 choices for this, for this, for this, for this, and 24 different things. So, it's 36 times, 36 times 36. So, we get 36 to the 24th different elementary events, which means that the probability of each one is this. So, this is a probability of concrete sequence of 24 pairs of numbers, where each number is from 1 to 6. Fine, great. Now, the same, exactly the same reasoning as before. What are those pairs which make up the event we need? So, what are those events, sorry, elementary events which make up? So, out of these, we need these sequences where there is at least one combination called 1-1, right? Again, it's easier to calculate those guys which do not have 1-1. So, 1-1 is only one pair out of 36. If we don't want this pair to have been represented in this sequence, then we have to choose for each place, and we have 24 places. We have to choose anything but 1-1, which means any other of 35 combinations. And obviously, the number of those is 35 to the 24th power, right? So, in each place out of these 24, we can have a pair, but any pair is good, but 1-1, which means other 35 pairs are fine. So, we have to subtract them and divide the result by 36 to the 24th. So, what will be here is 36 to the 24 minus 35 to the 24th, which is the same as 1 minus 35 36 to the 24th power. And guess what? This is equal to 0.4914, which is less than 50%, and that's why the mayor was losing money. So, again, what's very important in this particular set of problems, whenever you are trying to solve this problem, first of all, think about the set of elementary events. These are the smallest, so to speak, representatives of your experiment, out of which, combining them together, you can make up any event which you're interested in. Now, what's important is that these elementary events must be, must have the same chances to occur. They must be symmetrical. That's how you choose them. So, that's why, for instance, in the last problem, any combination of 24 pairs of numbers basically has the same chance to occur as any other combination. They are completely symmetrical, and that's why I have chosen these. And from these elementary events, now you can put together them into any event which you are interested in, like in this particular case, event is at least one of those 24 pairs equals to one one. All right. So, I do suggest you to go to unizord.com and review this lecture again. Try to solve these problems just yourself, first, and see if you get the same results as I'm getting here. Well, that's it for today, and good luck.