 Hi, I'm Zor. Welcome to a new Zor education. I would like to present to you another series of probability problems. I call it advanced probability problems. This is lecture number two in this series. They're not really that advanced. They're basically, again, based on combinatorics and your ability to count different combinations. If combinatorics present certain problems for you, I do suggest you to go through the previous chapter dedicated to combinatorics of this course. The lecture is presented on Unizor.com. That's part of the advanced course of mathematics for teenagers. It's a free course which I have started with only one purpose actually. To introduce you to the process of creation, process of solving certain problems, process of looking for the way from point A to point B. You don't really know what is the way. You just have to look for the way. That's the purpose of solving problems in mathematics. In my personal view, that's the purpose of mathematics as it is studied in school. Whenever you're going to a higher level of education, there are some specialized parts of mathematics used in certain professions, maybe something like this. But as far as the general mathematics, I do think that solving problems is the most important part, not teaching you one formula or another. Formulas probably will be forgotten anyway, but the ability to solve problems will remain with you forever. Alright, so these problems are dedicated to the game of poker. Well, basically the games are the beginning of the theory of probabilities. The theory of probabilities actually started as a response to certain game problems. In this particular case, we're talking about poker, and it's not really the game of poker. It's rather I'm using the terminology of poker to introduce you to certain mathematical problems. Now, very similar problems were introduced as part of the combinatorics. Now, I'm basically repeating the same thing, but in this case, it's based on theory of probabilities. And that's what I'm going to do. Assume you have a standard deck of 52 cards. Four different suits, you know what they are, right? It's spades, hearts, diamonds, and clubs, let's say. Actually, there are different names for the different suits and different languages for different people, but I'm using this, which is relatively standard in the United States and Europe, I guess. Now, every suit has cards of different ranks. The ranks are from 2 to 10, Jack, Queen, King, and Ace. 13 ranks in each suit, right? So that makes it 13 times 452. Now, the dealer gives you five cards. Now, these five cards can actually form certain combinations. And what I'm going to do right now is to calculate the probability of certain combinations. Right up front, without changing any cards, just pure combinations, randomly chosen five cards out of the deck of 52 cards. Okay, just described. So, the combination number one is four of a kind. Now, what is four of a kind? If you have five cards, then four of a kind means you have four different cards of the same rank. Let's say you have nine of spades, nine of hearts, nine of diamonds, nine of... Sorry, nine of clubs. And the fifth card can be anything, basically. Any other card. Let's say it's Ace of spades. It doesn't really matter. So, this is the example of a combination which is called four of a kind. Now, what's the probability to have this combination? Let's just think about it. Well, again, let's approach this from the theory standpoint. When we're talking about probability, and this is definitely an example of a discrete probability, because there are only finite number of outcomes, what is our sample space? The experiment is we get five cards out of the deck of 52. And this is a completely random experiment, and when I'm saying random without specifying how random, that actually means all the different outcomes have exactly the same probability. So, what is the number of outcomes? Well, if I'm choosing five cards out of 52, or I'm giving five cards out of 52, the number of combinations is this. So, we have a set of five cards out of 52 is an elementary event. And we have this many elementary events, number of combinations of five cards out of 52, which means that the probability of each particular combination of five cards is this. Now, the event which we are interested in is having four cards of the same rank and the fifth something else. So, the question is how many different combinations of elementary events, each of them having this probability, constitute our event? Well, that's actually very simple to calculate. First, we have to choose which is the rank of these four cards, because if we choose the rank, it completely defines the four cards out of these five. Let's say we choose nine, or we choose eight, or we choose two. So, how many different choices? We have 13 different ranks, so we have 13 different choices for the rank of this card. And that completely defines four cards out of five, because there are only four suits, right? Now, the fifth card can be anything. Now, if I have chosen four, how many choices do I have for the fifth card? Well, 52 minus four, we have 48, which means that with each of these 13 choices for my four cards, I have 48 cards for the choice of the fifth card. And so, this number is the number of elementary events, number of sets of five cards, which constitute my event. And since each of them has this probability, the total probability of my event, which is four of a kind, this is the probability of four of a kind. Next, next is full house. Now, full house, sometimes it's called 3 plus 2. Now, what it means is we have three cards of one particular rank. Let's say nine diamond, nine clubs and nine hearts, for instance. And two cards, completely different rank, but the same rank among themselves. Let's say it's a king of spades and king of diamond. So, three of one rank, three and two of another rank. That's what makes a full house. All right, now, exactly the same logic. The total number of different elementary events, as we were saying before, is the same. It's always the same for all our problems. And the probability would be one over this, obviously, of each elementary event. Now, the question is how many elementary events constitute this particular event? Well, let's just think about it. First of all, we should choose one of 13 different ranks for this. Now, we have 13 different choices. Now, out of the remaining 12 ranks, we have to choose the rank for the pair. So, that's number 12. So, we have 13 choices of rank here and 12 choices for rank here. Now, as soon as we have chosen the rank, nine and king, now we have to choose three cards out of four of that particular rank. So, number of choices is number of combinations from four by three. And for this guy, for the pair, we again need to choose two out of four kings. So, that's number of combinations from four by two. And that is the total number of different combinations, different elementary events, which comprise the event which we are interested in. So, this should be divided by 52 number of combinations from 52 to five. And that's the probability. I mean, obviously, you can go and calculate exactly what is this and this and this number, but I'm not going to waste my time on this. This is a simple arithmetic actually, right? And let's go to the next case. Next case is a straight. Now, what is straight? Now, straight is when you have five cards ranked sequentially one after another. For instance, you have two of something, three of something, four of something, five of something and six of something. Or you can have, let's say, ten, Jack, Queen, King and Ace. Or actually, there is one small detail. Ace can be both the most senior rank and the least senior rank, which means Ace, two, three, four and five is also a valid straight. I mean, maybe some casinos are not allowing this, but in my casino, this is also a valid combination as a straight. Now, the suits are not important here. However, there is one exception. If all of them are of the same suit, that's a different combination. It's not just a straight. It's called, how is it called, flush and royal flush or something like this. I don't remember. It's a much, much more rarely occurring combination. And that's why it's supposed to be excluded from the combinations which we are considering as a straight. So straight means that not all of them are of the same suit. So how many different elementary events comprise this one? Well, first of all, we have to establish what is our highest rank in this particular group. Or the lowest, if you wish, doesn't really matter. The question is how many of them? Because we have to really determine how many groups by rank we have. So let's say we top with Ace, that's one. Then we can top with a King, right? King, Queen, Queen, Jack, 10 and 9. Or we can start with a Queen, Jack, 10, 9, 8, etc. And the very last one would be this one. So the highest rank can be Ace, King, Queen, Jack, 10. It can be 9 and 8 and 7 and 6 and 5. We cannot have the highest 4 because there is not enough predecessors to form a group of 5 cards. I mean 3, 2, Ace as a 1, but that's it. There is no more. So starting from 5 and up to Ace, we can have as the top ranking member of the group. Now how many of them? 5, 6, 7, 8, 9, 10, Jack, Queen, King and Ace, 10. So we have 10 different choices for the top ranking card in a group. 10, that's important. So we have determined that we have 10 different choices for the top ranking card, which means top ranking determines rank of all of those down there, underneath. I mean if top ranking is 9, it determines the rank of the next one, which is 8 and then 7 and 6 and 5 and 5. Alright, so we have 10 different choices for the top ranking, which means 10 different choices for set of ranks, sequential ranks from 5 to 8. Alright, now, since we know the rank, we have 5 cards, right? We know their ranks, since we have already chosen the top one. How many combinations do we have of these 5 cards if I have already determined their rank? Well, we have 5 cards, we have 4 choices, 4 different suits for one card, 4 for another, 4 for the 3rd, 4 for the 4th and 5th. So we have 4 different choices for each of those, and I have to put it to 5th degree, because we have 5 different cards and each one has 4 different suits. So once the top card is chosen, I have determined basically the ranks of all other 5 cards and since each rank can be represented by one of 4 cards, one of 4 different suits, I'm multiplying by 4 to the 5th degree. However, I did not take into account the fact that all of them cannot be of the same suit. So let's just subtract from this number all the different combinations of sequential ranking cards, 5 cards, but of the same suit. Now, how many of them? Well, again, there are 10 different variations for the top card, but now for all of them we have to have exactly the same suit, right? So all 5 cards should be of the same suit, which one? It doesn't matter which one, as long as it's not the same, right? That's what we have to count as a real straight. So whenever they are all of the same suit and there are 4 suits, which means I have to multiply it by 4. That's the number which we have to exclude from this number to get the real number of straight. So these are the number of different elementary events, sets of 5 cards out of 52, which correspond to our condition of having exactly sequential ranks and being at least one of them should be different in suits from another. And obviously, if we're talking about probability, we have to divide it by number of combinations from 52 by 5. Alright, now my last problem is 3 of a kind. Now the first one, if you remember, was 4 of a kind. It was a little bit easier, because as soon as I chose the rank, I have basically established completely my 4 cards out of 5. And then all I had to do is multiply by the number of the remaining cards. This is just very much the same, but I have to really take into account a little bit more freedom. Well, 3 of a kind, that means I have to choose, let's say, 9 diamonds, 9, again, 9 hearts, 9 spades. And then I have to choose 2 others, but they're not supposed to be the same, and they're not supposed to be one of these. Because if they're one of these, if it's another 9, it will be 4 of a kind. If they are the same among themselves, let's say 2 kings, it will be full house, 3 plus 2, right? And that's not the same combination. So I need different ones. For instance, I can have king of spades and ace of diamonds, something like this. Now that would be 3 of a kind. Again, let's start with this group. We have, obviously, we have to choose which rank we are dealing with in this group of 3 cards. Well, there are 13 different ranks, so obviously I have to start with this. That's my freedom of choice. I can choose any rank for these 3. Now, how about their suits? I need only 3. There are 4 different suits. So there are a number of combinations from 4 by 3, and that's the number of different sets of 9s, for instance in this case, which I can get from the deck. It can be diamonds, hearts, spades, or it can be diamonds, clubs, spades, it can be whatever. I mean, there are different combinations. So 3 out of 4 I'm choosing. Now what's important is these are supposed to be of different rank than this and different among themselves. Now, how many ranks are left? Well, one is taken for this, so we have 12 different ranks, and I need 2 different ranks among these. So what's my freedom of choice? Well, I can choose number of combinations from 12 by 2 as different pairs of ranks, right? So now I know that these two cards have different ranks. Now, what's the choices of the suit for each one of them? Well, quite frankly I don't care, even if they are the same suit or different suit, as long as their ranks are different. So which means I have 4 choices for this and 4 choices for that. So I have to multiply it by 4 square. And that's the number of elementary events which comprise my three of a kind, and they have to divide it by number of combinations from 52 by 2 by 5, and that's the probability I'm looking for. Alright, now what I suggest you to do is all these numbers, obviously they're all convertible into factorials, etc., so you can basically calculate. Just as an example, what is this? This is 13, this is 4 factorial divided by 3 factorial and 1 factorial. This is 12 factorial divided by 2 factorial and 10 factorial. This is 16, this is 52 factorial divided by 5 factorial and 47 factorial. So do all the necessary cancellation like in this particular case, for instance 4 factorial and 3 factorial. Obviously 1 factorial is going out, 3 factorial is number of 1 times 2 times 3 and 4 is the same thing but 4. So I'm leaving this thing. Now 12 and 10, so I have 12 and 11 here. Now 2 factorial is 2, so I can just leave 6 here. So that's what I have in my numerator. In my denominator I have, now if I divide this, I will have 52, 51, 50, 49 and 48. Right, without any factorials and 5 factorial I'm putting here. 5 factorial is 1, 2, 3, 4, 5 which is what? 120. So that's the number. You can just take the calculator and calculate it precisely what it is and you will get the probability of full house. I think it's very educational to know in decimal actually how small or how big this probability is. Obviously it's between 0 and 1, question is how big is relative to your understanding what's the probable event. So that's it, that's all the problems which I wanted to present to you right now. I do suggest you to do this again. I do suggest you to go through combinatorics chapter if you forgot it, if you not feel uncomfortable in combinatorics. Go to the combinatorics chapter of this course, very educational. And well, that's it for today. Thank you very much and good luck.