This is based on an experiment I did with a standard pack of playing cards. From the pack of 52 cards, a set of six cards was selected at random (i.e. the first six cards at the top of the pack after thoroughly shuffling the pack before each draw) and the number of spades among the six randomly selected cards was noted. This was done 68 times i.e. 68 sets of six cards were drawn randomly and the number of spades in each set was noted. This is a summary of the data: Frequency and relative frequency table for the data $x$ is the number of spades in a set of six cards. $f_x$ is the frequency of each $x$ value i.e. the number of times a set of six cards with $x$ number of spades were drawn. $r_x$ is the relative frequency correct to three decimal places.

Q1: can this situation be modelled by a theoretical probability distribution as follows:$$P(x)={\binom{13}{x}\binom{39}{6-x}\over \binom{52}{6}}$$where $x=0,1,2,3,4,5,6$ and $x$ is the number of spades in a randomly selected set of six cards?

Q2: is it reasonable to consider the 68 randomly selected sets of six cards as a random sample from a population comprising all the possible sets of six cards that can be formed from the 52 playing cards (numbering $\binom{52}{6}$ or roughly 20.4 million)?

Q3: can the situation be modelled also by a theoretical binomial distribution $Bin(n,p)$ where the number of trials $n=6$ and the probability of success, $p$, is the probability of drawing a spade in a single draw i.e. $p=0.25$?


Q1 gives the correct probability of finding $x$ spades in a randomly-chosen set of six cards. Choosing the first six is fine, assuming your shuffling procedure generates every possible reordering of the deck with equally probability. (The question of how different card rearrangements are distributed after shuffling is itself an interesting problem. See https://www.nytimes.com/1990/01/09/science/in-shuffling-cards-7-is-winning-number.html.)

Q2 is correct if you are considering a random sample of $68$ six-card sets with replacement, since your experiment could (very rarely) yield the same six-card set more than once in the $68$ trials.

Q3 isn’t a precise model, however, because you are drawing the six cards in your set without replacement, and therefore the separate cards’ probabilities of being spades are not independent. Instead of the binomial distribution, use the hypergeometric distribution. https://en.wikipedia.org/wiki/Hypergeometric_distribution

  • $\begingroup$ Thanks for pointing out the fault with using a binomial distribution. The article on different card arrangements after shuffling looks very interesting $\endgroup$ – Indula May 16 at 2:24
  • $\begingroup$ This also looks like a lovely paper on shuffling: dartmouth.edu/~chance/teaching_aids/Mann.pdf $\endgroup$ – Steve Kass May 16 at 2:47
  • $\begingroup$ Thanks. And yes, each draw was after replacing the cards from the previous draw and reshuffling. $\endgroup$ – Indula May 16 at 2:53

Q1. Yes. This is the correct probabilistic modelling of this system.

Q2. According to what measure? There exists many theorems on necessary sample sizes to ensure a lower bound for error margin (good old hypothesis testing and confidence intervals) for statistical significance.

Your sample size is too small even compared to the sample space, therefore you cannot get significant experimental data. But, the data you obtain will be random (again, due to the fact that sample size being extremely small) in the sense that individual points will be extremely likely to be unique.

Q3. No, because the deck is not extremely big such as to diminish the effects of reducing items from it. It is obvious that the numbers are not the same, but the subtlety is to compare this error:

$$P_{real}(6) = {\binom{13}{6}\binom{39}{0}\over \binom{52}{6}}=8.43\times10^{-5}$$ $$P_{binomial}(6) = (0.25)^{6}\cdot (0.75)^0\cdot \binom{6}{0} = 2.44\times10^{-4}$$

The problem is, as the spades get taken out of the deck, their probability of getting drawn together drop because of diminishing individual step probabilities. This phenomenon is seen the most pressingly on the example above where the error is 189%.

  • $\begingroup$ Thanks! I edited my post to include a snapshot of the data of the experiment. I suppose increases the number of draws the resulting sample would be a more representative of the 20.4 million odd population? and would that remedy the error in using a small sample that @Steve Kass refers to? $\endgroup$ – Indula May 16 at 2:31
  • $\begingroup$ @Indula I cannot give you the exact numbers right now, but using a bigger sample set always improves your model in statistics (given you are following the protocols as I and Steve Kass have mentioned). The total scope of this exchange is statistics and hypothesis testing specifically, which are big subjects by themselves. But the main idea is: Bigger the sample, more revealing the data. $\endgroup$ – acarturk May 16 at 2:46
  • $\begingroup$ No need! I'm just getting started on statistics. But I would love to know what those those theorems are for determining the size of samples. Just the names. And sorry about all the typos in my earlier comment. $\endgroup$ – Indula May 16 at 2:55
  • $\begingroup$ I wouldn’t say there was an “error” in experimenting with $N=68$. Your experiment of $68$ trials yields a size-$68$ random sample with replacement of all six-card subsets of a deck. How such a sample would be expected to represent the true distribution of number of spades in randomly chosen six-card sets (more precisely, how the distribution of numbers of spades in such samples vary from the true distribution over all six-card sets), and what inferences with what confidence could be drawn from the results of one experiment is, as acarturk notes is a question about inferential statistics. $\endgroup$ – Steve Kass May 16 at 2:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.