![]() ![]() ![]() Because the choice of these two determines which value will require $3$ cards and which $2$ in our full house we are effectively distinguishing between a full house of $(A,A,K,K,K)$ and $(A,A,A,K,K)$ with this permutation of $P(2,1)$. the $2$ values just chosen in $C(13,2)$, to choose the grouping of $3$ and $1$ option for the grouping of $2$. We care not that we chose $(A,K)$ versus $(K,A)$ After this choice, we can ask now how many ways can I pick my grouping of $3$ and my grouping of $2$? Well one would have $2$ options, i.e. $C(13,2)$ is the clearer way because, like the $5$-of-a-kind example we are merely choosing cards. The note above's arguement would have you say that not only are you choosing two of the $13$ values for your full house but you are simultaneously choosing which will be your group of $3$ and which your group of $2$, and though this is technically correct I believe it skips a step and is the reason the OP was confused. I would argue that the order of these two chosen values is not relevant and therefore calls for a $C$ to be calculated. And I agree that it gives the correct value I just argue that it is arrived at via misleading reasoning based on the order mattering definitions differentiating $P$ and $C$). So exaclty like in the $5$-of-a-kind example we can choose $2$ of the $13$ possible values in a standard deck of $52$ cards to determine the number of ways in which choosing these two values is possible (NOTE that some may disagree that clearly there are $13$ options for the first value in a full house and $12$ for the other and thus $P(13,2)$ is correct. We know that a full house consists of $2$ values of card, $2$ of the one and $3$ of a second. I actually quite dislike the answer given in the OP, that of $P(13,2)C(4,2)C(4,3)$ as I am of the impression that it is hiding the real permutation in the $P(13,2)$ term. This leads immediately to the thought of $C(4,1)$ for five cards yiedling the given answer of $C(13,5)*(C(4,1))^5$. Now however, we have to realize that each card can come in four different suit, and thus for each of the five cards we chose with $C(13,5)$ we have $4$ options to choose from. As such, to choose $5$ distinctly valued cards need employ $C(13,5)$. a hand of $A,2,3,4,5$ all fo hearts is identical to $5,4,3,2,A$ all of hearts. How to find the total number of full houses in a poker handįor the first, as was previous mentioned, the order of the hand of five distinctly valued cards does not change the hand we have i.e. ![]() How to find the total number of $5$-of-a-kinds in a poker hand.I however, wanted to shed some light on the implied problems of the OP which were: The other two answers correctly mentioned the difference between order mattering and not mattering for permutations and combinations respectively. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |