 Hi, I'm Zor. Welcome to Unizor Education. This is the second lecture devoted to combinatorics of the poker game. Now the first lecture contains certain calculations of some combinations and this lecture is for the rest of the combinations which are part of the poker game. Now this lecture as any other is just part of the Unizor.com advanced course of mathematics for teenagers. Now I do recommend you to go to this website and look at the notes for this lecture first because it contains all the problems which I am going to present right now and answers so you can basically check your own answers as well as the logic behind all the calculations. So all right so let's just finish up the poker game by analyzing few combinations which are still remaining to be counted. Okay the probably the most frequent but not probably definitely most frequently occurring combination and I'm talking about combination versus something which has no combinations which is even more often. So the combination is one pair. Now one pair you have five cards out of 52 complete deck standard deck and out of these five cards you have to have one pair of cards of the same rank let's say two kings or two nines. Now the other three should not actually complement this pair for instance you cannot have the third card of the same rank because in which in this case you would have the combination three of a kind or four obviously cards of the same rank and so basically you have to avoid these other combinations and only the only thing you have the only combination you have is a pair of two cards of the same rank. Okay so how can we calculate the number of combinations with one pair in them well first of all we know that there are two cards of the same rank so let's just pick up the rank so it's one of the 13 different from two to eight so we have 13 different choices to pick up the rank of the cards right now after the rank is picked now the rank means basically four different cards of the same rank let's say clubs diamonds cards and and space so we need only two so we have to really pick two cards of that rank out of the available four well that obviously can be done in this number of choices right number of combinations of two cards out of four so we have chosen the rank of our pair and we have chosen two specific cards within that rank now now we have to take care of the rest we have three other cards and for other cards we should actually make sure that they will not combine with these cards to create something more than just one pair maybe like a trailer kind or something like this now how can that be done well obviously we have to choose the rest of the three cards out of other ranks right now we have how many different ranks do we have we have chosen one particular one particular rank for our pair now the rest should be all different right it cannot be like two of the same rank or even three of the same rank among them or one of the rank which is the same of these two so we basically have to pick up three different ranks from whatever the ranks are remaining now we have chosen one for the first pair right so we have out of 13 we have 12 different and out of these 12 ranks we have to pick three different ones so that's why I put number of combinations of three out of 12 now that gives me three different ranks from which I have to choose these three cards now each one of them can have any of the four suits right so out of these three cards out of these three ranks which I have chosen for the first card I have four choices of any of the suits for the second card I have four choices and for the third so I have to multiply it by four three times and this is the number of combinations called one pair or a pair by specifying number of combinations from 12 I'm excluding this one and that's how I'm making sure that there are no other cards which have the same rank as this pair and now I'm using the number of combinations of three out of 12 which means I'm choosing three different ranks they do not pair with each other among these three all right that's it that's the first problem problem number two we have two pairs okay the combination which is called two pairs obviously means we have well something like three and three and King and King and the fifth one can be something which is not one of these because if it would be if it was one of these it would be full house 3 plus 2 for instance so it should be something like 10 for instance which is different so this is a two pair with certain suits which these cards belong to I'm talking about ranks right now so these are five cards of different ranks which form two pair combination one pair and another pair and the fifth card is completely different all right so how many of these combinations exist well let's do it this way first of all let's choose which two ranks we would like to be represented as pairs well obviously we have 13 different ranks so this would give me two out of 13 so great we have chosen two different ranks let's say three and King in this particular case or any other pair of ranks now this pair is only two cards with this rank and there are four all together right there are spades diamonds hearts and clubs right but we need only two of these four and we are free basically to choose any two which means that we have to multiply by number of combinations of two out of four for this particular pair and two out of four for this particular pair so this gives me two ranks one for each pair this gives me the freedom of choice within the first pair of the first rank and the second pair freedom of choice for the second rank now I have to worry about this guy it cannot be one of these right so two ranks have already been chosen for two pairs out of 13 ranks we have 11 left so the freedom of choice for this guy is I can choose any one of the 11 remaining ranks and for that particular rank I have four different suits which means I have to multiply it by the freedom of number of freedom of choices number of choices for for the suit so that gives me the total number of two pairs combination now another thing which I just wanted to mention it's a very simple case and I did consider in the previous lecture I did consider the straight flush now straight flush is straight and flush which means you have a consecutive ranks and at the same time they are of the same suit let's say 10 9 8 7 and 6 or ace king queen jack and 10 all of the same suit right so these are straight flush now out of these straight flush flushes which by the way we have counted 40 of them that was the previous lecture why 40 because we have to choose the top card which can be either ace or king or queen there are 10 different combinations 10 different cards and different ranks rather which can be the top rank and then everything else actually is predetermined so you don't have any freedom of choice so you have only one freedom of choice the top card so there are 10 different ranks and four different suits that makes it 40 now one particular straight flush combination has its own name it's called royal flash and the royal flash is when the top card is ace so this is ace king queen that's cute jack and 10 so this is a royal flash if they all are of the same suit how many of these combinations are well basically only four because there are four different suits so the number of royal flash combinations is equal to four and these are usually included into straight flush they have just a special name okay now the last problem which I have is basically we have exhausted all the different combinations so the last problem is how many combinations which do not belong to any of the other category there are no pairs no straights no flushes nothing among these five cards now in the game of poker if you have one hand which contains basically no combination and another hand which has no combination the highest card actually wins if highest card are the same then the next highest etc so that's why this particular well pseudo combination is called high card but basically it's no combination all right so what's the number of no combinations what's the number of high card combinations if you wish well first of all they have to be of different ranks right because if there are two cards at least of the same rank that would be already a pair or or three of a kind or four of a kind so all ranks must be different which means we can just pick up the combinations which contain five different ranks how many of them are well there are 13 ranks so out of 13 ranks we need five different ranks now what's wrong with this picture well there is one combination and it's called straight when different ranks actually do form a combination if these ranks are consecutive right okay so how many different consecutive ranks exist well we already counted this it's 10 starting from ace from King from Queen from Jack from 10 from 9 from 8 from 7 from 6 down like 6 5 4 3 and 1 and 2 and even from 5 down because we can use ace as the lowest with the rank of 1 right so we have 10 different ranks which we should really exclude 10 different combinations of ranks I would say which we have to exclude from this number so minus 10 that gives me number of ranks which we can use for our combination so all different combinations of ranks of this number are good now let's talk about what kind of suit we want these cards to be well if any card can be of any suit then we would have this number of combination multiplied by 4 to the fifth degree right we have four choices for the first card four choices for the second card etc to the fifth card again there is one exception if they are of the same suit then that would make a combination and we're talking about no combination right so the the five cards of the same suit make a flush and we should exclude it so how many different flushes if if suits are already chosen then how many different flushes are in this case well four because there are four different suits so we have to subtract four from this so these four distributions of our ranks among suits are not supposed to be counted because these would be the combinations called flush so the product of these is actually the number of no combinations combinations or high card combinations all right so that actually ends up all these counting exercises which I wanted to present to you and what I did also I put on the board all the different combinations which I was talking about in two different lectures this one and the previous one and I have calculated all these number of combinations which I used like formulas I have counted the real numbers here and so this is the number of choices for this last one which I did the high card this is the greatest and I have sorted in decreasing a number of combinations order so the most frequently occurring combination is no combination at all then one pair combination and then two pairs etc so this is the order and in the game of poker obviously any combination which has a rare occurrence is considered to be stronger than the one which has more number of occurrences well for obvious reasons now what's interesting is to add these numbers together no combination and these combinations which are completely different from each other and I'm getting this number two million and a half and something right now this is supposed to be all the different combinations which can be constructed from from the five cards out of the standard deck so which means that this number should be equal to this number of combinations of five cards out of 52 cards in a deck well if you count what this number is you will get exactly this what does it mean well that's a great check of all our calculations if these are summed together give you the number which is equal to this one and they do I did check that actually proves that all our calculations are correct you see again let me just emphasize it once more in the combinatorics the the number which you obtain it's a pure logical conclusion which you make and you might make a mistake so like undercount something or overcount something so you have to be very very careful and existence of some checking procedure like in this particular case this is a greatest checking procedure it proves that you are correct in your calculations another way is just if you solve the problem in two different ways that would be the good so the good checking as well okay I do suggest you to go again through the notes for this lecture on the Unisor.com again try to solve all these problems just yourself have your own formulas it would be great if you can derive exactly the same formula it would be even better if you derive it in some other way and get the same number one of these and basically that's it for the game of poker I think I will use the playing cards more in different problems related to combinatorics and definitely I will return to these problems when I will discuss the probabilities because obviously the combinatorics is basically the introduction into the probabilities it's very very important topic all right that's it for today thank you very much and good luck