The $52$ cards in a deck are numbered $1, 2, \cdots, 52$. Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked, The two persons with lower numbered cards from a team, and the two persons with higher numbered cards form another team. Let $p(a)$ be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards $a$ and $a+9$, and Dylan picks the other of these two cards. The minimum value of $p(a)$ for which $p(a)\ge\frac{1}{2}$ can be written as $\frac{m}{n}$. where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Once the two cards are drawn, there are $\dbinom{50}{2} = 1225$ ways for the other two people to draw. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below $a$, which occurs in $\dbinom{a-1}{2}$ ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above $a+9$, which occurs in $\dbinom{43-a}{2}$ ways. Thus,\[p(a)=\frac{\dbinom{43-a}{2}+\dbinom{a-1}{2}}{1225}.\]Simplifying, we get $p(a)=\frac{(43-a)(42-a)+(a-1)(a-2)}{2\cdot1225}$, so we need $(43-a)(42-a)+(a-1)(a-2)\ge (1225)$. If $a=22+b$, then\begin{align*}(43-a)(42-a)+(a-1)(a-2)&=(21-b)(20-b)+(21+b)(20+b)=2b^2+2(21)(20)\ge (1225) \\ b^2\ge \frac{385}{2} &= 192.5 >13^2 \end{align*}So $b> 13$ or $b< -13$, and $a=22+b<9$ or $a>35$, so $a=8$ or $a=36$. Thus, $p(8) = \frac{616}{1225} = \frac{88}{175}$, and the answer is $88+175 = \boxed{263}$.
263
Level 5
Counting & Probability
Problem: The $52$ cards in a deck are numbered $1, 2, \cdots, 52$. Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked, The two persons with lower numbered cards from a team, and the two persons with higher numbered cards form another team. Let $p(a)$ be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards $a$ and $a+9$, and Dylan picks the other of these two cards. The minimum value of $p(a)$ for which $p(a)\ge\frac{1}{2}$ can be written as $\frac{m}{n}$. where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the minimum value of \( p(a) \) such that \( p(a) \ge \frac{1}{2} \), we need to analyze the conditions under which Alex and Dylan are on the same team based on their card selections. 

1. Alex picks card \( a \) or \( a + 9 \). Dylan picks the other card. The possible values for \( a \) range from \( 1 \) to \( 43 \) because \( a + 9 \) must not exceed \( 52 \).

2. The teams are formed based on the two lowest and two highest cards among the four players. We need to calculate the probability that Alex and Dylan end up on the same team.

3. For Alex to be on the same team as Dylan, either both must have lower numbered cards than the other two players or both must have higher numbered cards. 

4. We denote the cards picked by Blair and Corey as \( b \) and \( c \), respectively. The conditions for Alex and Dylan to be on the same team can be expressed as:
   - \( \min(a, a + 9) < \min(b, c) \) and \( \max(a, a + 9) < \max(b, c) \) (both are lower)
   - or
   - \( \min(b, c) < \min(a, a + 9) \) and \( \max(b, c) < \max(a, a + 9) \) (both are higher)

5. The total number of ways to choose \( b \) and \( c \) from the remaining \( 50 \) cards is \( \binom{50}{2} \).

6. We need to count the favorable outcomes for each case:
   - For the first case, if \( a \) is the lower card, then \( b \) and \( c \) must be chosen from the cards numbered less than \( a \). If \( a + 9 \) is the lower card, then \( b \) and \( c \) must be chosen from cards numbered less than \( a + 9 \).
   - For the second case, if \( a \) is the higher card, then \( b \) and \( c \) must be chosen from cards numbered greater than \( a + 9 \). If \( a + 9 \) is the higher card, then \( b \) and \( c \) must be chosen from cards numbered greater than \( a \).

7. The probability \( p(a) \) can be expressed as the ratio of the number of favorable outcomes to the total outcomes. 

8. We derive \( p(a) \) and set the inequality \( p(a) \ge \frac{1}{2} \) to find the minimum \( a \).

9. After performing the calculations and simplifying, we find the minimum value of \( a \) that satisfies the condition.

10. The final answer is expressed in the form \( \frac{m}{n} \) where \( m \) and \( n \) are coprime integers. 

11. The sum \( m + n \) is calculated to yield the final result of \( 263 \).<|END_OF_LATENT|> The final answer is $263$.