Given a deck of cards, what are the odds of consecutively drawing higher cards for a given number of draws?

First of all, as I can see, you dont know very much about probability, so my suggestion is that, if you are interested in these kind of questions about probability of games, try to read a book about it! By example books as Introduction to probability with Texas Hold'em!, or The mathematics of Games: An Introduction to Probability, or Probability and Games, or Probability and Statistics. A Didactic Introduction seems suitable for you.

For the case in hand: probability in many situations can be evaluated as number of valid cases divided by the number of all possible cases.

A simpler general case:

If you have a list of numbers from one to $n$ (for some $n\in {\mathbb N}$) and you want to know what is the probability to get a list of $j$ numbers, taken without replacement such that each number is greater than the precedent then you need to count two things:

  1. The number of different lists of length $j$ of numbers with the desired properties, taken from a bag with $n$ distinct numbers.

  2. The total number of different list of numbers of length $j$, taken from a bag with $n$ distinct numbers.

To count the lists of 1. we can notice that after draw any list of length $j$ it can be ordered from the lowest to the higher number, therefore for any distinct set of $j$ numbers there is a unique ordered list, so 1. is equivalent to count all possible distinct subsets of $\{1,\ldots,n\}$ of length $j$, this number is just $\binom{n}{j}$.

The second number, this said in 2., is just all possible ordered lists of length $j$ given $n$ possible distinct values for each number of the list, this is known in combinatorics as variations of length $j$ out of $n$ possible numbers, and is just $n^{\underline{j}}:= \prod_{k=0}^{n-j+1}(n-k)$.

Therefore the probability would be $$ \frac{\binom{n}{j}}{n^{\underline{j}}}= \frac1{j!} $$ where $j!:=1\cdot 2\cdots j$.

The probability of the question:

Your case with cards is a little more complicate to handle because a deck of 52 cards have each rank repeated four times (that is, there are four cards of the same "value"), then the number of lists in 1. cannot be calculated in the same way for this probability. However is not too much complicate: we only need to count for every list of 1. (with $n=13$) all possible variations where each card can be of one of four suits, that is for each valid list of length $j$ there are $4^j$ distinct possible variations of the same depending of the suits of the cards, therefore the probability of your question is $$ \frac{\binom{13}{j}\cdot 4^j}{52^{\underline{j}}} $$ This number doesn't seem that could be simplified significantly, so I leave it like this.

If you start from a given rank, namely the rank $m$ (of $13$ possible distinct ranks) and a list of ranks of length $j$, then the number of valid lists of length $j$ and ranks greater than $m$ are in one to one correspondence with the number of subsets of $\{m+1,\ldots,13\}$ of length $j$, each one multiplied by $4^j$ that represent the number of variations due to the four possible suits for each card, therefore $ \binom{13-m}{j}4^j$ are all possible distinct increasing lists of length $j$ of ranks greater than $m$, therefore the corresponding probability will be $$ p_{m,j}:=\frac{\binom{13-m}{j}\cdot 4^j}{52^{\underline{j}}} $$

Now suppose that you want to know what is the probability to succeed in $n$ tries, where a try is not a draw: it is just an attempt to get $j$ cards of the deck suck that they are increasing in it ranks and the minimum rank is greater than $m$. If you dont get a valid list then you put the cards again in the deck, shuffle, and try again; then the possibility to get a valid list of cards in $n$ tries would be $F_X(n)$, where $F_X$ is the cumulative distribution function of a random variable $X$ with geometric distribution of parameter $p_{m,j}$, that is, it would be $F_X(n)=1-(1-p_{m,j})^n$.

Now: to know the probability, after $n$ draws, to get some increasing sublist of length $j$ in this list of $n$ cards, of ranks greater than $m$, you need to count all these lists and divide by the total number of lists of length $n$. However this count is not easy to handle, it seems better suited for a computer.

Tags:

Probability

Related

If $A$ is a real or complex algebra and $a\in A$ is such that $ab=0$ for all $b\in A$, then $a=0$? Prove that for all $\alpha + \beta + \gamma = \pi$, $\sum_{cyc}\frac{\sin\beta}{\cos\beta + 1} = \frac{\sum_{cyc}\cos\beta + 3}{\sum_{cyc}\sin\beta}$. When commutative ring hom $A\to A/I$ flat? Show $\int_0^t (t-x)P_n(x)\,dx\leq \frac{t^2}{2}\int_0^1 P_n(x)\,\mathrm dx $ where $P_n(x)=(x(1-x))^{n}$ For any sequence of real numbers, one can always find a subsequence that is monotone How to Motivate Open-Cover Formulation of Compactness in a Metric Space? Is the Katětov extension of $\Bbb N$ zero-dimensional? A nice combinatorial identity: $\sum_{k=1}^{n-1}\frac{\binom{k-1}{n-k-1}+\binom{k}{n-k-1}}{\binom nk}=1$ Construct a circle tangent to sides $BC$ and $CD$ and s.t. its meetings with the diagonal $BD$ are tangent points from tangents draw from point $A$ Given $f$ holomorphic, which are the necessary conditions on $\phi$ in order to make $\phi \circ f \circ \phi^{-1}$ holomorphic? Doubt in finding the integral of $f(x)=x$ using the upper and lower sum definition of the integrals. What is the significance of the term "separable" in the context of countability properties?

Recent Posts