Same card probability in a deck

65 Views Asked by At

Working on my first board game design. To simplify, having 4 types of cards:

5 blue ones, 4 red ones, 3 yellow ones, 2 green ones, so 14 cards in total.

How can I calulate the possibility that when I draw a given number of them I will find at least two of a given color?

For example, when given 5 cards at least two of them to be blue.

1

There are 1 best solutions below

0
On

Note that in a deck of $n$ cards, there are $\binom{n}{k}$ number of different $k$-card hands with each hand being equally likely.

In an equiprobable setting like this, for an event $A$ we define $Pr(A) := \frac{|A|}{|S|}$ where $|A|$ is the number of ways that $A$ could occur and $|S|$ is the total number of possibilities regardless of whether or not $A$ occurs. (in our case here, $|S| = \binom{n}{k}$)

Note also by the rule of sum that: Pr(at least 2 blue) = Pr(exactly 2 blue) + Pr(exactly 3 blue) + ... + Pr(exactly 5 blue)

Let us look at your specific example setting $n=14, k=5$ and the number of blue cards to be $5$.

To calculate Pr(exactly 2 blue), count the number of different hands with exactly two blue cards. Which two blue cards did you use? $\binom{5}{2}$ number of choices. Which three notblue cards did you use? $\binom{9}{3}$ number of choices. By multiplication principle there are then $\binom{5}{2}\binom{9}{3}$ number of different hands with exactly two blue cards. Calculate the other cases similarly and then divide by the total number of possible hands to obtain:

$$Pr(\text{at least two blue}) = \frac{\binom{5}{2}\binom{9}{3}+\binom{5}{3}\binom{9}{2}+\binom{5}{4}\binom{9}{1}+\binom{5}{5}\binom{9}{0}}{\binom{14}{5}}$$

Note that the method of calculating "at least two blue and at least one red" is similar but more involved as it requires a heavier case analysis. Again, in general, to calculate probability you take the number of different outcomes which satisfy your condition divided by the number of outcomes total regardless of whether or not the condition is satisfied. To count number of outcomes apply multiplication principle and addition principle using counting techniques often involving combinations.