Chapter 10.1, Problem 26E

Solution

To calculate: The probability of the event that a hand includes two jacks of the same color (called the right and left bower).

The probability of the event that a hand includes two jacks of the same color is $\frac{5}{92}$ .

Given information:

A version of the card game “bid Euchre” that uses a pack of 24 cards, consisting of ace, king, queen, jack, 10, and 9 in each of the four suits (spades, hearts, diamonds, and clubs). In bid Euchre, a hand consists of 6 cards.

Calculation:

The number of events that can happen is $C{}_6$ .

There is only $2$ red jacks and we get six cards in one hand so that leaves $4$ more cards of possible $22$ .That leaves $C$ combinations.

The possibility of getting two red jacks is:

  $\begin{array}{l}\frac{C{}_4}{C{}_6}=\frac{\frac{20!}{2!\left(20-4\right)!}}{\frac{24!}{6!\left(24-6\right)!}}\\ =\frac{\frac{20\times 19\times 18\times 17\times 16!}{2\times 1\times 16!}}{\frac{24\times 23\times 22\times 21\times 20\times 19\times 18!}{6\times 5\times 4\times 3\times 2\times 1\times 18!}}\\ =\frac{\frac{20\times 19\times 18\times 17}{2}}{\frac{24\times 23\times 22\times 21\times 20\times 19}{6\times 5\times 4\times 3\times 2}}\\ =\frac{7315}{134592}\\ =\frac{5}{92}\end{array}$

The probability of the event that a hand includes two jacks of the same color is $\frac{5}{92}$ .