Question: Miscounting (9 points)

Consider the question: what is the probability of getting a 7-card poker hand (order doesn't matter) that contains at least two 3-of-a-kind (3-of-a-kind means three cards of the same rank). For example, this would be a valid hand: ace of hearts, ace of diamonds, ace of spaces, 7 of clubs, 7 of spades, 7 of hearts and queen of clubs. (Note that a hand consisting of all 4 aces and three of the 7s is also valid.)

Here is how we might compute this:

Each of the \(\left( \begin{matrix} 52 \\ 7 \\ \end{matrix} \right)\) hands is equally likely. Let E be the event that the hand selected contains at least two 3-of-a-kinds. Then

To compute |E| apply the product rule. First pick two ranks that have a 3-of-a-kind (e.g. ace and

7 in the example above). For the lower rank of these, pick the suits of the three cards. Then for

the higher rank of these, pick the suits of the three cards. Then out of the remaining 52 - 6 = 46

cards, pick one. Therefore

\(|E|=\left( \begin{matrix} 13 \\ 2 \\ \end{matrix} \right)\left( \begin{matrix} 4 \\ 3 \\ \end{matrix} \right)\left( \begin{matrix} 4 \\ 3 \\ \end{matrix} \right)\left( \begin{matrix} 46 \\ 1 \\ \end{matrix} \right)\)

and hence

Explain what is wrong with this solution. If there is over-counting in |E|, characterize all hands that are counted more than once, and how many times each such hand is counted. If there is under-counting in |E|, explain which hands are not counted.

Also, give the correct answer for Pr(E).

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