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

Question

Lets say you're trying to draw n cards, each higher than the last. How likely are you to succeed in consecutively drawing higher for a given starting number & number of iterations?

Somebody said to add my thoughts to the question! We've had a lot (my brother and i are trying to solve it!) so wait just a few minutes while I write it all out!

So first we tried by thinking about some easy cases. Lets say you only wanted 3 consecutive higher draws. If you drew a king or a queen you're done right off the bat, the odds of that are 8/52. If you drew a Jack, you have to get the sequence of JQK to succeed. That's $4/52* 4/51*4/50$. From this we can't yet establish too much, but we found a general way to answer this problem for three draws. If you start with a Jack, there is one way, a 10 in three ways, a 9 in six ways, an 8 in 10 ways... each drop in card number adds one more positive outcome than the last drop did (K-Q added zero, Q-J added one, J-10 added two, 10-9 added three etc). So the equation is $1(4/52 * 4/51 * 4/50) + 3(4/52 * 4/51 * 4/50) + 6(4/52 * 4/51 * 4/50)... + 66(4/52 * 4/51 * 4/50)$. (<- if we aren't mistaken)

Around here is where we have gotten stuck as brute force gets more difficult and we don't yet have an overarching formula to apply.

Any ideas of where to go from here to apply this to higher numbers of draws?

BIG EDIT (unless we're super wrong... not 100% sure of proper math notation) Okay... so our thought process from where we left off is as follows:

If we're drawing four cards instead of three, the sum is the same except the 66 term is lobbed off (because the 1,3,6,10... series starts a number lower [in the case of 4 cards, you can't draw a jack first] and ends at the same place, thereby eliminating one term) and the inside term gains 4/49 (signifying the probability of another drawn card). The equation would be: $1(4/52 * 4/51 * 4/50 * 4/49) + 3(4/52 * 4/51 * 4/50 * 4/49) + 6(4/52 * 4/51 * 4/50 * 4/49)... + 55(4/52 * 4/51 * 4/50 * 4/49)$. Notice the similarities between the equation for drawing three cards and drawing four.

From this, we derived the (hopefully correct) formula for any number of draws. let x = number of cards you want to draw. let y = the value of the first card drawn. Note: in the following, $n_{-1} = 0$

$$\sum_{n=0}^{14-x}((n + n_{n-1} +1)*(\prod_{n=0}^{x}(4/(52-n))))$$

Posting for now, any questions will be addressed after we eat dinner.

Answer 1

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.

Answer 2

The probability changes after the 1st draw, but it also changes after the 2nd draw. If you have three draws: i, j,k corresponding to the 1st, 2nd, and 3rd draw, then you have:

$$ P(j>i) = \sum_{j=i+1}^{13} \frac{4}{51} $$

$$ P(k>j) = \sum_{k=j+1}^{13} \frac{4}{50} $$

$$ P(k>j) \cap P(j>i) = \sum_{j=i+1}^{13} \sum_{k=j+1}^{13} \frac{4}{51} * \frac{4}{50} $$

Plugging in your 1st draw, I, should give you the probability of getting two higher consecutive draws.

For a test case, let us take the Jack, 11, for the 1st draw, i. You have a 4/51 chance of getting the Queen (if you get the King, you lose) then you have a 4/50 chance of getting the king.

Another test case, let us take the 10 as the 1st draw, i. You have a 8/51 chance of getting the Jack or Queen. If you get the Jack then you have a 8/50 chance of getting a Queen or King. If you get the Queen, you have a 4/50 chance of getting the king. So 8/51*8/50 + 8/51*4/50 = 0.018.

python code:

def proba_higher_consecutive_draws(i):
    result = 0
    for j in np.arange(i+1,14):
        for k in np.arange(j+1,14):
            result+=4/51*4/50
    return result

i=11
proba_higher_consecutive_draws(i)