Theoretical Two-Category Discrete Probability Distribution Without Replacement, Statistical Note 31

Calculate the theoretical discrete probability distribution of black cards in 20 cards drawn without replacement from a deck of 52 cards.

Theoretical probability distribution gives an idea of an ideal probability distribution, what a distribution should be given the parameters. Distribution of black cards in 20 cards drawn without replacement from a deck of cards is an example of the two-category discrete probability distribution of sampling without replacement. Refer to my earlier Statistical Notes for clarity on calculating the two-category discrete probability of samples drawn without replacement using tree diagram, formula and Excel software function.

Why 20 cards were chosen in this example?

One may pose why 20 cards were chosen, why not other numbers. Drawing 20 cards without replacement from a deck of 52 cards is a case of Hypergeometric Distribution, which is symmetric when the sample size is even number.

Theoretical Discrete Probability Distribution Without Replacement

This example has three characteristic features. First, the example has a finite population of 52 cards, denoted by ‘N’. Second, each card can be characterized as a success or a failure. Since the question asks the probability of black cards, the selection of a black card is considered as a success, and there are 26 black cards in the population. Third, a sample of 20 cards, denoted by ‘n’, is drawn without replacement in a way that each sample of 20 cards is equally likely to be selected.

Let X be a random variable of interest that takes one of 0 to  20 values as the number of black cards in the sample of 20 cards drawn without replacement, denoted by ‘x’. The probability distribution of X depends on the parameters, ‘n’, ‘M’ and ‘N’, and is given by the expression

P(X=x) = h(x;n,M,N) = Number of outcomes having X=x divided by total number of outcomes

P(X=x) = h(x;n,M,N) = [C(M,x) X C(N-M,n-x)]/C(N,n)

This distribution is referred to as Hypergeometric distribution.

In this example, n=20, M=26, N=52 and ‘x’ takes the value 0 to 2. Putting these values in the above formula, one gets

P(X=10) = [C(26,10) X C(26,10)/C(52,20)] = (26! X 26! X 32! X 20!) / (16! X 10! X 16! X 10! X 52!) = 0.223934379

This value is equal to the one presented in table 1. In the same way, the probability for other number of black cards in 20 cards drawn without replacement can be calculated.

Table 1: Number of success, failure and Probability of successes in 20 cards drawn without replacement from a deck of 52 cards

Graphical Presentation

Chart 1 shows the theoretical two category discrete probability distribution of black cards in 20 cards drawn without replacement from a deck of cards.  This clearly shows the bell-shaped curve, the symmetric line chart of theoretical probability distribution.

Conclusion

Looking at the table and the chart one can see that the occurrence of 10 black cards in 20 cards drawn without replacement is highly likely. Two extreme number of heads, 0 and 20, have the least chance of occurrence.