A set of 9 cards are numbered 1, 1, 2, 2, 3, 3, 4, 5 and 6. A round of matching game is played where two cards are drawn, one at a time without replacement. A match occurs when the two cards drawn have the same number.

1. Calculate the probability that the sum of the two cards drawn is less than 5.

Let $S$ be the event that the sum of the two cards is less than $5$, and $M$ be the event of a match.
The possible sequence of outcomes for a sum $< 5$ are $(1, 1)$, $(1, 2)$, $(2, 1)$, $(1, 3)$, $(3, 1)$, and $(2, 2)$.
Calculating the probability of each outcome:
$\text{P}(S) = \left(\frac{2}{9}\right)\left(\frac{1}{8}\right) + \left(\frac{2}{9}\right)\left(\frac{2}{8}\right) + \left(\frac{2}{9}\right)\left(\frac{2}{8}\right) + \left(\frac{2}{9}\right)\left(\frac{2}{8}\right) + \left(\frac{2}{9}\right)\left(\frac{2}{8}\right) + \left(\frac{2}{9}\right)\left(\frac{1}{8}\right)$
$\text{P}(S) = \frac{2 + 4 + 4 + 4 + 4 + 2}{72} = \frac{20}{72} = \frac{5}{18}$

2. Find the conditional probability of a match given the sum is less than 5.

The outcomes where the sum is $< 5$ and a match occurs are $(1, 1)$ and $(2, 2)$.
$\text{P}(M \cap S) = \text{P}(1, 1) + \text{P}(2, 2) = \frac{2}{72} + \frac{2}{72} = \frac{4}{72} = \frac{1}{18}$
The conditional probability of a match given the sum is less than $5$ is:
$\text{P}(M \mid S) = \frac{\text{P}(M \cap S)}{\text{P}(S)} = \frac{1/18}{5/18} = \frac{1}{5}$

3. Calculate the overall probability of a match and compare to test for independence.

For the events to be independent, we must have $\text{P}(M \mid S) = \text{P}(M)$.
The matching outcomes in the game are $(1, 1)$, $(2, 2)$, and $(3, 3)$. Note that cards 4, 5, and 6 only appear once.
$\text{P}(M) = \text{P}(1, 1) + \text{P}(2, 2) + \text{P}(3, 3) = \frac{2}{72} + \frac{2}{72} + \frac{2}{72} = \frac{6}{72} = \frac{1}{12}$
Since $\text{P}(M \mid S) = \frac{1}{5} \neq \frac{1}{12} = \text{P}(M)$, having a match is not independent of having a sum $< 5$.

1. Find the probability of obtaining a match in a single round

In a round, two cards are drawn from $9$. The pairs that give a match are $(1,1)$, $(2,2)$ and $(3,3)$.
$P(\text{Match}) = P(1,1) + P(2,2) + P(3,3)$
$P(\text{Match}) = 3 \times \left(\frac{2}{9} \times \frac{1}{8}\right) = \frac{6}{72} = \frac{1}{12}$

2. Formulate the inequality for the cumulative probability of obtaining a match

Let $n$ be the number of rounds.
$P(\text{No match in 1 round}) = 1 - \frac{1}{12} = \frac{11}{12}$
The probability of obtaining at least one match in $n$ independent rounds is $1 - P(\text{No match in } n \text{ rounds})$.
$1 - \left(\frac{11}{12}\right)^n \ge 0.75$

3. Solve the exponential inequality using logarithms

$\left(\frac{11}{12}\right)^n \le 0.25$
Taking natural logarithms on both sides:
$n \ln\left(\frac{11}{12}\right) \le \ln 0.25$
Since $\ln\left(\frac{11}{12}\right)$ is negative, the inequality sign flips upon division:
$n \ge \frac{\ln 0.25}{\ln(11/12)}$
$n \ge 15.931$ (to 5 s.f.)

4. Determine the least integer number of rounds

Since $n$ must be an integer, the least number of rounds Dave must attempt is $16$.