 Hi, I'm Zor. Welcome to New Zor Education. Well, we continue talking about probabilities. And in this particular case, I'm basically talking about certain examples of, in this case, it's card games with certain aspects of probability related to these games. I'm not going to go into all the different miniscule details of the rules of every card game. But I have four examples and from each game I'm using, I'm just trying to extract certain important quantitative aspect of that particular game and apply the theory of probabilities to quantitatively evaluate that particular case. So the lecture is obviously the part of the Advanced Mathematics course presented on Unisor.com. Every lecture usually has notes including this one and I do suggest you to read the notes and try to solve these problems yourself first before listening to this lecture. And at the end, after the lecture is finished, I suggest you to do exactly the same. Just try to solve these problems by yourself. Alright, so four examples which I have from four different card games. Poker, Bridge, Blackjack and Bakara. And again, I'm not going to examine all these games and all the details. From each game I just took a particular detail which might be probabilistically evaluated. Alright, for the game of Poker, that's the game when the dealer basically gives five cards to a player. And what I am talking about is one particular player, he gets the first five cards from the deck and the question is what kind of combination in these five cards can be formed. And one of the combinations is called Full House which is actually three cards of one particular rank and two cards of another, let's say three kings and two sevens, something like this. So my question is what's the probability of having this particular combination if you have just the five cards from the deck. Now, the complications obviously is you're not the only player, so there are other players before you, after you, they also receive cards, so that actually changes the picture dramatically. I'm talking about very simple approach. You have a deck of 52 cards, a standard deck, and you just pick randomly five cards out of these 52. And the question is what's the probability of having Full House among these five cards. First of all, when you are dealing with Serif probability, you have to realize what is your sample space, what are the elementary events we are dealing with. Now, we have 52 cards and you pick five, which means that any combination of five cards chosen from 52 is your elementary event. And all these elementary events, all these combinations of five cards out of 52 are equally possible. They have the same chances to occur, which means that the probability of each of them is equal to probability of each elementary event is equal to 1 divided by the number of these events. The number of these elementary events is number of combinations from 52 by 5. So that's the probability of elementary event. Each elementary event, each combination of five cards I picked from 52 standard deck is equally probable. And since the total probability is 1, I mean, we're talking about frequency, right? So what's the frequency of having something? Well, obviously, it happens all the time, so frequency is always 1. So that means that every one of them has 1 over C, the combination of 52 by 5. Okay, now we have to count how many of these elementary events constitute the combination called full house. So what kind of freedom of choosing these elements in the full house combination we have? We have two different ranks. We have 13 different ranks in a deck, right? Four suits. Each suit has 13 different ranks, from two to ace. So for choosing the full house, we have to choose the rank of the triplet, and there are 13 different variations, right? So we have chosen the rank for the triplet. Great. Now, after we have chosen the rank, let's say it's eight, or king, or something else. But there are four different representatives of this rank, because there are four suits. We have diamonds, we have spades, we have hearts, and we have, what else, clubs, right? So we have four different suits. So we have four different representatives of this particular rank. But we need only three. So which three I choose basically determines the full house, and different suits represent different combinations. So after I picked a particular rank, I have to pick from that rank three representatives out of four. So that can be done in number of combinations from four by three, right? And they have to be multiplied, because with each rank I have four choices, and out of them I'm choosing only three. Okay, now that determines my triplet. Now we have to choose the rank for the other two. Now that rank can be any other, but the one we have already chosen. So there are 13 ranks, so we have 12 left. So we have to multiply it by different choices, which we have to choose the rank for the pair. And now, since we have chosen the rank, now we have four representatives from which we need only two. So we can choose in number of combinations from four to two different ways, right? So this is a number of elementary events, which together make up an event called full house. So if I know the number of these elementary events, and I know that the probability of each one is this, then the probability of my event is obviously this. So that's the answer. Whatever that answer is, and if you will go to the notes for this particular website, Unizor.com, I think I even put the concrete number with decimal places, etc., which gives you a little bit better feel of how often or rare the combination full house actually occurs. So that's all for the first game of poker. As you see, I'm not actually going into all the details of poker. I'm just choosing poker as an example of one particular aspect of theory of probability. Now, the next game I'm going to use, the card game is called bridge. Now, in bridge, the deck of 52 cards is dealt among four players equally, 52 divided by 4 is 13 cards each. What I'm interested in my event is one ace for each player. So out of 13 cards for one player, I have to have one ace, and 13 cards for another player should contain one ace, and the third one and the fourth one. So we have four different players, each one should have an ace. Here is something which I think would be very handy. It's a generalized problem, which I'm going to explain in a general way, and then I will apply it to this particular case. Here is a generalized problem. Let's consider that you have a certain number of objects, which we divide evenly among certain number of groups. So you have n times k objects, k groups by n objects each. Question is, in how many different ways, I mean considering all objects are different, considering all objects are different. In how many ways n times k objects can be divided evenly by n among k groups? Well, that's kind of a typical problem in theory of probability. It's definitely related to combinatorics. So let's just think about how can it be done. Well, the way to do it is the following. Let's say these are your objects, and you divide it evenly among certain number of groups. So it's one, two, three, four groups, two objects per group. How can it be divided? Well, the way to do it is the following. Well, first of all, you draw these vertical lines, which separate n times k objects, in this case it's eight, into k groups, four groups in this case, by n objects each. So this is n objects, this is n objects, this is n objects, and this is n objects, and number of groups is k. So you put the objects in some order with divisors between each group, and that's the way you distribute basically among these groups. You change the order of these n times k objects, and you have something different, right? So basically, the number of different permutations of n times k objects is n times k factorial, right? Now, what exactly is wrong with this particular distribution? What's wrong is that these n, we can actually change the sequence within this group in any way we want, and it will be exactly the same distribution. The same group will contain exactly the same elements. As long as we don't move elements around, so every permutation within each group, this one, this, this, and this, results in the same distribution of nk elements by k groups, right? So I have to divide this by n factorial and by n factorial again, and by n factorial again and again, and altogether I have to divide it by n factorial to the power of k, right? To get the real different distributions, because for each permutation of main group of n times k objects, the permutations within this group and within this group and within this group and within any other groups do not really change the distribution of the objects among groups. So this is the answer. How to divide how many different ways to divide n times k objects among k groups by n objects in each group? This is the answer how many different ways to do it. And we will use it in this particular case, because what do we have to actually do? Let's consider two different types of objects which constitute the deck of cards. One group of objects is aces. Another group of objects is everything else. And we are specifically distributing aces among four people and then all others among four people evenly, this and evenly that, four aces one each, and the rest which is 52 minus 448. So we have four aces plus 48 non-aces. So this we have to divide by four people and this we have to divide evenly by four people. So in this particular case, each group contains one, so n is equal to 1 and k is equal to 4. In this case n is equal to 12 and k is equal to 4. And we can use this formula basically. So what can we say? Well, to divide four aces among four people, so it's one each actually, if I will substitute it into this formula it will be four factorial which is 24 divided by one factorial which is one, so this is 24. Now this would be 48 factorial divided by 12 factorial to the power of 4. So I have 24 different distributions of four aces among four people and this many distributions of 48 non-aces among four people and with each of these I can probably do each of those, right? Which means the total number of different distributions of the whole deck among four people with each person having only one ace is the multiplication of these two. And this is the number of good elementary events which constitute the event we are looking for. Now to get the probability we have to divide it by the total number of elementary events, our sampling space, right? So what's the sample space in bridge? Well, we have to divide 52 cars among four players, 13 cars each, right? So how many different ways this can be done? Well, that's exactly the same type of a problem we saw before in the same formula. So it's 52 factorial divided by 13 factorial to the power of 4. So this is our denominator. The multiplication of these are numerator and this is our denominator. So the same formula we're using three times. And by the way, this is one of the things which mathematics is very good about. Instead of solving a certain number of particular problems they're trying to solve one general problem and apply it to all particular cases. So we have solved one general problem. How to divide a certain number of objects even among certain number of groups? And we were using this formula in three different places. So that's the result. I mean, obviously you can reduce the formula, et cetera. If you're interested, you can go to Unisor.com and notes actually for this lecture contain this simplification. But these are just arithmetic things. We are not interested in arithmetic. We are interested in theory of probabilities. Next. Blackjack. Blackjack. All right. Now, blackjack is when a dealer gives a certain number of cards to players, every card has a certain value. The numbers from 2 to 10 have values from 2 to 10. Corresponding with whatever is written Jack, Queen and King has number of points 10 allocated. And Ace has either 11 or 1 depending on the choice of a player. Right. So I'm not going into all the details, obviously. I'm interested in only one particular probability that the first two cards amount to 20 points. So you pick two cards randomly from 52 from a complete deck and you are interested in event of getting 20 points. All right. So how can we get 20 points? Well, first of all, if one of the cards is 8 or less, you can get 20. What you can get 20 with is 9 is one card and then you need an Ace for another card. So that's one combination when you can get 20. Or you can get two 10s where 10 is either the card with a rank of 10 or one of the pictures like Jack, Queen or King. So a pair of any two cards which have the point 10 is good enough for you. So number one question always is what's our sample space? Now our sample space is two cards randomly chosen from 52. So we have a number of combinations from 52 to 2 as the sampling space. That's number of elementary events. So every two cards randomly picked represents an elementary event. All have exactly the same probability and that's why the total number is this, so the probability is this of every pair. Now let's count how many pairs are constituting our event. Well, there are different kinds of pairs, right? So let's just calculate each one of them. Okay. If I'm looking for a combination of two 10-points cards. Now how many different 10-points cards I have? I have four different suits and for each suit I have 10, Jack, Queen and King, right? All of these have the value of 10 points each. So we have four choices times four suits gives me 16 different cards. 16 cards of 10 points each. From these 16 cards I have to choose any two basically. And how many combinations? Obviously 120. So I have 120 combinations of this type. 10 points and 10 points. How many combinations of this type I have? Well, I have four different nines, four different suits and four different aces. So I have to have one of these and one of those. Four choices of this and each of them has four choices of that. So it's four times 416. So I have 16. So together I have 136 different elementary events that basically are good for me. They constitute 20 points I'm getting from two cards. So if I will divide the number of these elementary events by the total number of elementary events or if you wish I multiply by the probability of each one so this is the number of elementary events and this is the probability of each one. So if I will multiply them or divide this by 1326 I will get the probability of having 20 on my first two cards which is something around one tenths, about 10%. So in 10% of the cases approximately again we're talking about not exact number but approximate count as the number of experiments is increasing then I might get something like 10% of the cases I will get 20 points on the first two cards. That's it. And the last game I wanted to discuss is Bakara. Again no claims of basically explaining how this game is played but there are certain rules which I can just quote. So in the game of Bakara ranks from two to nine half points equal to rank two to nine. Now Ace has one point while others zero points. So number 10 and Jack, Queen and King they're all zero points cards. What else is important is that when you add the points you add it by modular 10 which means only the last digit of the sum is actually the result of your summation. So if you have let's say 8 and 7 which is 15 so the result is actually 5 not 15. So my question is how to get 8 points on the first two cards. So you pick up two cards from the deck and the question is what's the probability of getting 8 points. Alright. We start with sampling space. Now the sample space contains exactly the same by the way as in the previous case, two cards out of 52. So we have number of combinations from 52 by 2 which is 13.26. That's my number of elementary events. Number two, which elementary events constitute my 8 points? Alright there are different cases. Let's just consider different cases. When you have 8 points as a result of summation of the values of first two cards it can be 0 and 8. It can be 1 and 7, 2 and 6, 3 and 5, 4 and 4. And what's important can be 9 and 9 because 9 and 9 is 18 and we are doing everything by module 10 so it's also 8. So all these combinations of point value are good for us. So let's count each one of them. Now how many different cards with a point of 0 I have? That's 10, Jack, Queen and King which is 4 different ranks times 4 different suits. It's 16 cards so I have 16 choices for the point 0 and 4 choices for the point 8 because there is only one rank which is 8 and we have 4 suits so it's 4 different choices and I have to multiply because I have 16 choices for 0 and 4 choices for 8 that gives me 64 different variations. Okay, now 1 and 7. Well, okay, for 1 I have 1 and 8 so it's 2 different ranks. So it's 8 different choices for number 1 and I have 4 different choices for 7, 4 different suits. So that gives me 32. Next, 2 and 6, 4 choices for 2 and 4 choices for 6 3 and 5, same thing, 4 and 4. Well, so both are 4 and I have only 4 4's like 4 different suits for the rank of 4 so I have a choice of different combinations of 2 cards out of 4. So that gives me C from 4 to 2 which is 4 times 3 which is 6 and 9 and 9, same thing. It's C 4 to 6. Now the sum of all these, whatever the sum is divided by 13.26 is the probability of having in Bakara 8 points on the first 2 cards. Well, that's it. I do suggest you to read the notes for this lecture and try to solve all these problems just yourself. It's very helpful and as usually if you are a registered user on Unizor.com and registration is just a name and email, basically. That's it. Then you can take exams and you can engage your supervisor in basically enrolling you in certain courses or the entire course of advanced mathematics. So you have a little bit more control or your parents or supervisor actually have a better control over your educational process. So I do recommend to get engaged your parent or supervisor and together basically get through entire educational process and all this free, obviously. That's it for today. Thank you very much and good luck.